Cls Jeead-15-16 Xii Mat Target-6 Set-2 Chapter-6
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Chapter
6
Application of Derivatives Solutions SECTION - A Objective Type Questions (One option is correct) 1.
Suppose x1 and x2 are the point of maximum and the point of minimum respectively of the function f(x) = 2x3 – 9ax2 + 12a2x + 1 respectively, (a > 0) then for the equality x12 = x2 to be true the value of ‘a’ must be (1) 0
(2)
2
(3)
1
(4)
1 4
Sol. Answer (2) 6(x2 – 3ax + 2a2) = 6(x – a)(x – 2a);
a>0
x = a is point of maxima x = 2a is point of minima a2 = 2a a = 0 or a = 2 but a > 0 a = 2 2.
2
2
Point ‘A’ lies on the curve y e x and has the coordinate ( x, e x ) where x > 0. Point B has the coordinates (x, 0). If ‘O’ is the origin, then the maximum area of the triangle AOB is (1)
1
(2)
2e
1 4e
(3)
1 e
(4)
1 8e
Sol. Answer (4) A=
2 1 x e x 2
dA 1 x2 2 x2 = [e 2 x e ] = 0 dx 2
x2 =
1 2 1
1 1 e 2 = A= 2 2
1 8e
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84 3.
Application of Derivatives
Solution of Assignment (Set-2)
The radius of a right circular cylinder increases at the rate of 0.1 cm/minute, and the height decreases at the rate of 0.2 cm/minute. The rate of change of the volume of the cylinder, in cm3/minute, when the radius is 2 cm and the height is 3 cm is (1) –2
(2)
8 5
(3)
3 5
(4)
2 5
(4)
3
(4)
Sol. Answer (4) V = r2h dV = 2rh dr + r2 dh dV = 2 × 2 × 3 × (0.1) + × 22 × (–0.2) = 1.2 – 0.8 = 0.4 = 4.
2 5
The number of points of maxima/minima of f(x) = x(x + 1) (x + 2) (x + 3) is (1) 0
(2)
1
(3)
2
Sol. Answer (4) y = x (x + 1) (x + 2) (x + 3)
Clearly, option (4). 5.
The difference between the greatest and the least values of the function,
⎡ ⎤ f(x) = sin 2x – x on ⎢ , ⎥ is ⎣ 2 2⎦ (1)
(2)
⎞ ⎛ ⎜ 3 3⎟ ⎝ ⎠
(3)
3 2 3
3 2 2 3
Sol. Answer (1) y = sin 2x – x y' = 2 cos 2x – 1 = 0 cos 2x = 2x =
1 2
3
x=
6
⎛⎞ y⎜ ⎟ = ⎝6⎠
3 2 6
3 ⎛ ⎞ y ⎜ ⎟ = ⎝ 6⎠ 2 6 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
85
3 = 2 2 6
⎛ ⎞ y ⎜ ⎟ = ⎝ 2⎠ 2 ymax =
2
2 Difference =
ymin =
6.
If a variable tangent to the curve x2y = c3 makes intercepts a, b on x and y axis respectively, then the value of a2b is (1) 27 c3
(2)
4 3 c 27
(3)
27 3 c 4
(4)
4 3 c 9
(4)
1
Sol. Answer (3) x2y = c3 y=
c3 x2
dy 2c 3 = 3 dx x
Variable tangent : (y – y1) = a=
2c 3 ( x x1 ) x13
y1x13 x1 2c 3
=
x1 x1 2
=
3 x1 2
b = y1
2c 3 3c 3 = 2 x1 x12
9 2 3c 3 27 3 c a b = x1 2 = 4 x1 4 2
7.
Difference between the greatest and the least values of the function f(x) = x (ln x – 2) on [1, e2] is (1) 2
(2)
e
(3)
e2
Sol. Answer (2) f(x) = x (ln x – 2) f'(x) = (ln x – 2) + 1 = ln x – 1 =0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
86
Application of Derivatives
Solution of Assignment (Set-2)
x=e f''(x) = f''(e) =
1 x 1 0 e
x = e is minima Minimum value = e (ln e – 2) = –e f(1) = 1 (0 – 2) = –2 f(e2) = e2 (2 – 2) = 0 Maximum = 0 Difference = 0 – (– e) = e 8.
The sum of lengths of the hypotenuse and another side of a right angle triangle is given. The area of the triangle will be maximum if the angle between them is (1)
6
(2)
4
(3)
3
(4)
5 12
Sol. Answer (3) A
1 bc 2
A
1 b a2 b2 2
A
1 b (b k )2 b 2 where a + b = k 2
dA k 2k 0 ⇒ b and a db 3 3
Hence, cos
9.
1 , 2 3
Let C be the curve y = x3 (where x takes all real values). The tangent at A except (0, 0) meets the curve again at B. If the gradient at B is k times the gradient at A, then k is equal to (1) 4
(2)
2
(3)
–2
(4)
1 4
Sol. Answer (1) Tangent at A
A
(y – y1) = 3x12 (x – x1)
x1
y – x13 = 3x12 x – 3x13 y = 3x12 x – 2x13
–2 x
Solving tangent with curve 3x12 x – 2x13 = x3 Or, x3 – 3x12 x + 2x13 = 0
B
(x – x1) (x2 + x x1 – 2x12) = 0 (x – x1)2 (x + 2x1) = 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
87
x = (x1, – 2x1) Gradient at A = 3x12 Gradient at B = 3 (4 x12) k = 4 10. The interval on which f(x) = 2x3 + 9x2 + 12x – 1 is decreasing in (1) ( 1, )
(2)
(–2, –1)
(3)
( , 2)
(4)
(–1, 1)
(3)
( , )
(4)
(0, )
Sol. Answer (2) f(x) = 2x3 + 9x2 + 12x – 1 f'(x) = 6x2 + 18x + 12 For decreasing f'(x) < 0 6x| + 18x + 12 < 0 x2 + 3x + 2 < 0 (x + 1) (x + 2) < 0 x (–2, –1) 11. The function f(x) = cot–1 x + x increases in the interval (1) (1, )
(2)
( 1, )
Sol. Answer (3) f(x) = cot–1 x + x f'(x) =
1 x2 1 = 1 x2 1 x2
f'(x) > 0 xR 12. Divide 64 into two parts such that sum of the cubes of two parts is minimum. The two parts are (1) 44, 20
(2)
16, 48
(3)
32, 32
(4)
50, 14
Sol. Answer (3) x + y = 64 S = x3 + y3 S = x\ + (64 – x)3 S' = 3x2 – 3 (64 – x)2 For maxima and minima, S' = 0 x2 = (64 – x)2 x = 64 – x x = 32 13. If f(x) = x5 – 5x4 + 5x3 – 10 has local maximum and minimum at x = a and x = b respectively then (a, b) is (1) (0, 1)
(2)
(1, 3)
(3)
(1, 0)
(4)
(2, 4)
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88
Application of Derivatives
Solution of Assignment (Set-2)
Sol. Answer (2) f(x) = x5 – 5x4 + 5x3 – 10 f'(x) = 5x4 – 20x3 + 15x2 For maxima and minima, f'(x) = 0 5x2 (x2 – 4x + 3) = 0 x2 (x – 3) (x – 1) = 0 x=0
neither maxima nor minima
x = 3, 1 f''(x) = 20x3 – 60x2 + 30x = 10x (2x2 – 6x + 3) f''(3) > 0
Point of minima
f''(1) < 0
Point of maxima
14. A point on the parabola y2 = 18x at which the ordinate increases at twice the rate of abscissa is (1) (2, 4)
(2)
(2, –4)
(3)
⎛ 9 9⎞ ⎜ 8,2⎟ ⎝ ⎠
(4)
⎛9 9⎞ ⎜ 8,2 ⎟ ⎝ ⎠
Sol. Answer (4) y2 = 18x 2y
dy dx 18 dt dt
2y
2dx dx 18 dt dt
4y = 18
y=
9 2
81 = 18x 4
x=
9 8
15. The real number x when added to it’s reciprocal gives the minimum value of sum at x equal to (x > 0) (1) 1
(2)
–1
(3)
–2
(4)
2
Sol. Answer (1) y = x
1 ⎛ ⎞ 1 , x > 0 ⎜ x 2 if x 0 and minima occurs at x 1⎟ x x ⎝ ⎠
16. Which of the following statement is true for the function? ⎧ x , x 1 ⎪ 3 ⎪x , 0 x 1 f (x) ⎨ 3 ⎪x ⎪⎩ 3 4 x , x 0
(1) It is monotonic increasing x R (2) f(x) fails to exist for 3 distinct real values of x (3) f(x) changes its sign twice as x varies from (–, ) (4) Function attains its extreme values at x1 and x2, such that x1x2 > 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
89
Sol. Answer (3)
⎧ x x 1 ⎪ 3 ⎪x 0 x 1 f(x) = ⎨ 3 ⎪x ⎪⎩ 3 4 x x 0 ⎧ 1 x 1 ⎪ ⎪2 x f'(x) = ⎨ 2 0 x 1 ⎪3 x ⎪⎩ x 2 4 x 0 f'(x) changes its sign two times at x = 0, –2 17. The function f ( x ) x 3 3 x is (1) Increasing in (– , –1) (1, ) and decreasing in (–1, 1) (2) Decreasing in (– , –1) (1, ) and increasing in (–1, 1) (3) Increasing in (0, ) and decreasing in (–, 0) (4) Decreasing in (0, ) and increasing in (–, 0) Sol. Answer (1) f(x) = x3 – 3x f'(x) = 3x2 – 3 = 3 (x2 – 1) f'(x) = 3 (x + 1) (x – 1) For increasing function f'(x) > 0 x (–, –1) (1, ) 18. Coffee is draining from a conical filter of height and diameter both 15 cm into a cylindrical coffee pot of diameter 15 cm. The rate at which coffee drains from the filter into the pot is 100 cm3/minute. The rate in cm/minute at which the level in the pot is rising at the instant when the coffee in the pot is 10 cm, is (1)
9 16
(2)
25 9
(3)
5 3
(4)
16 9
Sol. Answer (4) Rate of change of volume in the cylinder = Rate of change of volume in conical filter. r2h= V Differentiate w.r.t. t dV 2 dh = r dt dt 2
⎛ 15 ⎞ dh 100 = ⎜ ⎟ ⎝ 2 ⎠ dt
dh 16 cm/s = dt 9
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Application of Derivatives
Solution of Assignment (Set-2)
19. A curve is represented by the equations, x = sec2 t and y = cot t where t is a parameter. If the tangent at the point P on the curve where t
meets the curve again at the 4
point Q then, | PQ | is equal to : (1)
5 3 2
(2)
5 5 2
(3)
2 5 3
(4)
3 5 2
Sol. Answer (4) x = sec2 t y = cot t
dx 2 sec t sec t tan t dt dy cosec 2 t dt
dy cosec 2 t = sin t dx 2 cos3 t ⎛ dy ⎞ 1 3 ⎜ ⎟ = (cot t ) ⎝ dx ⎠ 2
1 dy ⎞ ⎛⎜ ⎟ = 2 ⎝ dx ⎠t 4
Equation of tangent 1 y – 1 = ( x 2) 2
2y – 2 = – x + 2 x + 2y = 4 Equation of curve
x
1 =1 y2
1 =1–x y2
y2 =
1 1 x
(2y – 3) y2 = –1 2y3 – 3y2 + 1 = 0 For point Q,
y
1 , x 5 2
Distance |PQ| =
⎛ 1 ⎞ (5 2)2 ⎜ 1⎟ ⎝ 2 ⎠
2
=
9
9 = 4
45 3 5 = 4 2
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Solution of Assignment (Set-2)
Application of Derivatives
91
20. For all a, b R, the function f(x) = 3x4 – 4x3 + 6x2 + ax + b has (1) No extremum
(2)
Exactly one extremum
(3) Exactly two extremum
(4)
Three extremum
Sol. Answer (2) f'(x) = 12x3 – 12x2+ 12x + a = 12x (x2 – x + 1) + a f''(x) = 12 (3x2 – 2x + 1) f''(x) > 0 f'(x) is always increasing So, there is one point at which f'(x) = 0. 21. Maximum area of a rectangle which can be inscribed in a circle of a given radius R is (1)
R 2 3
(2)
3R 2
(3)
2
2R2
(4)
3R2
Sol. Answer (3) Let rectangle has width b and height h A = bh
R
also, b2 + h2 = (2R)2 = 4R2 2
b 4R h
R
2
h
b
so, area A( h ) h 4R 2 h 2 Area is maximum when A2 is maxima. A2 = h2(4R2 – h2) f(h) = h2(4R2 – h2) For maxima f'(h) = 0 f'(h) = h2 (–2h) + (4R2 – h2)2h = 0 –h2 + 4R2 – h2 = 0 h2 = 2R2 h 2R
From physical nature of problem it is clear that this should be maximum area since minimum area will tend towards zero. Amax = =
2R 4R 2 2R 2 2R 2R = 2R2
22. If atmosphere pressure at a height of h units is given by the function P(h) = he–h then pressure is maximum at the height of (1) 1 unit
(2)
1 e
units
(3)
e units
(4)
2 units
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92
Application of Derivatives
Solution of Assignment (Set-2)
Sol. Answer (1) P(h) = he–h P'(h) = –he–h + e–h = e–h(1 – h) P'(h) = 0 h = 1 at h = 1, P'(h) changes sign from positive to negative so it is point of maxima. 23. The radius of cylinder of maximum volume which can be inscribed in a right circular cone of radius R and height H (axis of cylinder and cone are same) H given by (1)
R 2
(2)
R 3
(3)
2R 3
(4)
2R 5
(4)
4
Sol. Answer (3) Let cylinder has radius r and height h. AO'B and AOC are similar so,
AO O B AO OC
H h r = H R
1
h r H R
h r 1 H R
r ⎞ ⎛ h H ⎜1 ⎟ ⎝ R⎠ V = r2h
r ⎞ ⎛ V (r ) r 2 H ⎜ 1 ⎟ R ⎝ ⎠ ⎛ r3 ⎞ V (r ) H ⎜ r 2 ⎟ R⎠ ⎝
For maximum volume ⎛ 3r 2 ⎞ V (r ) H ⎜ 2r ⎟0 R ⎠ ⎝
2r
3r 2 0 R
2
3r 0 R
r
2R 3
(r 0)
; x3 ⎧ | x 2 | 24. Let f ( x ) ⎨ 2 ⎩ x 2x 4 ; x 3
Then the number of critical points on the graph of the function is (1) 1
(2)
2
(3)
3
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Solution of Assignment (Set-2)
Application of Derivatives
93
Sol. Answer (2) Critical point is the point where f'(x) = 0 or f'(x) does not exist. Function changes its definition at a point x = 3 so we will check at x = 3 Also, |x – 2| = (x – 2) if x > 2 = –(x – 2) if x < 2
; x2 ⎧x 2 ⎪ so, f ( x ) ⎨( x 2) ; 2 x 3 ⎪ x 2 2x 4 ; x 3 ⎩ if ⎧1 ⎪ f ( x ) ⎨1 if ⎪2 x 2 if ⎩
x2 2x3 x3
If f'(x) = 0 then x > 3 and
2 x 2 0 ⇒ x 1 (3, ) so f'(x) cannot be zero anywhere. Now, f'(x) does not exist at x = 2 since, f'(2 – 0) = 1 and f'(2 + 0) = –1 Also, f'(3 – 0) = –1 f'(3 + 0) = 2(3) – 2 = 4 so, f'(x) does not exist at x = 3 Hence, there are two critical points x = 2 and x = 3 25. Consider the function f(x) = x cos x – sin x, then identify the statement which is correct (1) f is neither odd nor even
(2)
⎞ ⎛ f is monotonic increasing in ⎜⎝ 0, ⎟⎠ 2
(3) f has a maxima at x = –
(4)
f has a minima at x = –
(3)
8 3
Sol. Answer (3) f(x) is odd function f'(x) = – x sin x + cos x – cos x f'(x) = – x sin x f''(x) = –x cos x – sin x at x = – f''(x) < 0 point of maxima 26. If x = 2t – 3t2 and y = 6t3 then (1)
1 3
(2)
dy at point (–1, 6) is dx
9 2
(4)
0
Sol. Answer (2) dx 2 6t dt Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
94
Application of Derivatives
Solution of Assignment (Set-2)
dy 18t 2 dt
dy dy 18t 2 dt dx dx 2 6t dt If y = 6 = 6t3
t=1 18 12 18 9 ⎛ dy ⎞ ⎜ dx ⎟ 2 6 1 4 2 ⎝ ⎠t 1
27. For the cubic, f(x) = 2x3 + 9x2 + 12x + 1, which one of the following statement, does not hold good? (1) f(x) is non-monotonic (2) Increasing in (– , –2) (–1, ) and decreasing in (–2, –1) (3) f : R R is bijective (4) Inflection point occurs at x
3 2
Sol. Answer (3) f'(x) = 6x2 + 18x + 12 = 6 (x2 + 3x + 2) = 6 (x + 1) (x + 2), ‘f’ is increasing in (–, –2) (–1, ) and decreasing in (–2, –1). f’’(x) = 6(2x + 3) f’’(x) changes sing at x = −
3 2
Point of inflection occur at x = −
3 2
28. The function ‘f’ is defined by f(x) = xp (1 – x)q for all x R where p, q are positive integer, has a maximum value, for x equal to (1)
pq pq
(2)
–1
(3)
1 2
(4)
p pq
Sol. Answer (4) f'(x) = xp × q (1 – x)q – 1 (–1) + p(1 – x)q xp
–1
q ⎤ p q ⎡p = x (1 x ) ⎢ ⎥ ⎣ x (1 x ) ⎦ f'(x) = 0
Then, x = 1, (As x 0) p q =0 x 1 x p(1 – x) – qx = 0
and
p – x(p + q) = 0 x=
p pq
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Solution of Assignment (Set-2)
Application of Derivatives
95
29. Let h(x) be a twice continuously differentiable positive function on an open interval J. Let g(x) = ln(h(x)) for each x J. Suppose (h(x))2 > (h(x))h(x) for each x J then (1) g is increasing on J
(2)
g is decreasing on J
(3) g is concave up on J
(4)
g(x) is decreasing function
Sol. Answer (4) g(x) = ln (h(x)) g'(x) =
1 h( x ) h( x )
g''(x) =
h( x ) h( x ) ( h( x ))2 h2 ( x )
g''(x) < 0 So, g'(x) is decreasing function and g is concave down on J. 30. The least area of a circle circumscribing any right triangle of area S is (1) S
(2)
2S
(3)
2S
(4)
4S
Sol. Answer (1)
1 2r r sin 2
Area of ABC = S = r2 sin r2 =
S sin
Area of circle =
S sin
Least area = S. 31. If f(x) = x(1 – x)3 then which of following is true? (1) f(x) has local maxima at x = 1
(2)
f(x) has local minima at x = 1
1 4
(4)
f(x) has local maxima at x
(3) f(x) has local minima at x
1 4
Sol. Answer (4)
f ( x ) x 3(1 x )2 ( 1) (1 x )3 1 f ( x ) (1 x )2 [ 3 x 1 x ] f ( x ) (1 x )2 [1 4 x ] x = 1, f'(x) does not change sign so x = 1 is not maxima/minima At x
1 it has maxima since f¢(x) changes sign from positive to negative. 4
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Application of Derivatives
Solution of Assignment (Set-2)
32. A function y = f(x) is given by x
1 1 y for all t > 0, then f is 2 and t (1 t 2 ) 1 t
⎛ 3⎞ 3 (1) Increasing for t ⎜⎝ 0, ⎟⎠ and decreasing for t ⎛⎜ , ⎞⎟ 2 ⎝2 ⎠ 1⎞ ⎛ (2) Decreasing for t ⎜ 0, ⎟ ⎝ 2⎠
(3) Increasing for t (0, ) (4) Decreasing for t (0, 1) Sol. Answer (3) x=
1 1 y 2 , t (1 t2) 1 t
dx 2t dy (1 3t 2 ) dt (1 t 2 )2 , dt t 2 (1 t 2 )2 dy (1 3t 2 ) = 2 dx t 2t dy 1 3t 2 = dx 2t 3 dy > 0, if t > 0 dx
33. Find the equation of tangent to the curve y
x at the point where tangent drawn makes an angle
with 4
‘+’ x-axis : (1) x y
1 4
(2)
xy
1 4
(3)
xy
1 2
(4)
xy
1 2
Sol. Answer (1)
1 dy = dx 2 x 1 2 x
=1 1 x = 2
x=
1 4
y=
1 2
1⎞ ⎛ 1 ⎜x ⎟ = y ⎝ 4⎠ 2
x–y=
1 4
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Solution of Assignment (Set-2)
Application of Derivatives
97
34. The radius of a right circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as radius. When the radius is 1 cm the altitude is 6 cm. When the radius is 6 cm, the volume is increasing at the rate of 1 cm/s. When the radius is 36 cm, the volume is increasing at a rate of n cm/s. The value of ‘n’ is equal to (1) 12
(2)
22
(3)
30
(4)
33
Sol. Answer (4) dr = k dt
dh = 3k dt
h = r + c dh dr = dt dt
=3 h = 3r + c h = 6, r = 1 c=3 Volume = r2 (3r + 3) dr dV 2 = 3 [3r 2r ] dt dt
1 = 3 [108 12]
dr dt
dr 1 = dt 360 dV dr = 3r (3r 2) dt dt 1 40 9 = 33 cubic cm/second.
= 3 36(110)
35. Two side of a triangle are to have lengths ‘a’ cm and ‘b’ cm. If the triangle is to have the maximum area, then the length of the median from the vertex containing the side ‘a’ and ‘b’ is (1)
1 2 a b2 2
(2)
2a b 3
(3)
a2 b2 2
(4)
a 2b 3
Sol. Answer (1)
1 ab sin 2 For maximum area = 90º Area of triangle =
Length of median OD =
a2 b2 2
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Application of Derivatives
Solution of Assignment (Set-2)
36. The cost function at American Gadget is C(x) = x3 – 6x2 + 15x (x is thousands of units and x > 0). The production level at which average cost is minimum is (1) 2
(2)
3
(3)
5
(4)
4
Sol. Answer (2) Average cost = A( x ) =
(cx ) = x 2 − 6 x + 15 x
A(x) = 2x – 6 A(x)= 0 x=3 3
37. A particle moves along the curve y x 2 in the first quadrant in such a way that its distance from the origin increases at the rate of 11 units per second. The value of (1) 4
(2)
9 2
(3)
dx where x = 3 is, t is time in seconds dt
3 3 2
(4)
12
Sol. Answer (1) 3
y = x2 y2 = x3 Distance from origin
D x2 x3 Squaring both side and differentiating 2 2DD' = (2 x 3 x )
2D
dx dt
( x,
3 x2 )
2
2 x 3 x dx dD = dt 2x (1 x ) dt
11 =
2 3 x dx 2 1 x dt
11 2 2 ⎛ dx ⎞ = =4 ⎜ ⎟ 11 ⎝ dt ⎠ x 3 38. Let f(x) = ax2 – b|x|, where a and b are constant. Then at x = 0, f(x) has (1) A maxima whenever a > 0, b > 0 (2) A maxima whenever a > 0, b < 0 (3) Minima whenever a > 0, b > 0 (4) Neither maxima nor minima whenever a > 0, b < 0 Sol. Answer (1) f(x) = ax2 – b|x| f'(x) = 2ax – b, x > 0 = 2ax + b, x < 0 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
99
39. The co-ordinates of the point on the curve 9y2 = x3 where the normal of positive slope to the curve makes equal intercepts with the axis is
⎛ 1⎞ (1) ⎜ 1, ⎟ ⎝ 3⎠
(2)
(3, 3)
(3)
⎛ 8⎞ ⎜ 4, 3 ⎟ ⎝ ⎠
(4)
⎛6 2 6 ⎞ ⎜⎜ , ⎟⎟ ⎝5 5 5 ⎠
Sol. Answer (3) 9y2 = x3 18 y
dy = 3x2 dx
x3 1 6 xy
3y = 2x 9y2 =
27 3 y 8
y2 (8 – 3y) = 0
x2 dy = =1 6y dx
9y 2 =1 6 xy
⎛3 ⎞ 9y = ⎜ y ⎟ ⎝2 ⎠
3
2
8y2 – 3y3 = 0 y=
8 3
40. The tangent of angle made by the tangent of the curve x = a (t + sin t cos t), y = a (1 + sin t)2 with the xaxis at any point on it is (1)
1 ( 2t ) 4
(2)
1 sin t cos t
(3)
1 (2t ) 4
(4)
1 sin t cos 2t
Sol. Answer (2) x = a (t + sin t . cos t) y = a (1 + sin t)2 dx = a (1 + cos 2t) dt dy = 2a (1 + sin t) cos t dt
2a (1 sin t ) cos t dy 2(1 sin t ) cos t = = a (1 cos 2t ) dx 2cos2 t ⎛ 1 sin t ⎞ tan = ⎜ ⎟ ⎝ cos t ⎠
41. A cube of ice melts without changing shape at the uniform rate of 4 cm3/minute. The rate of change of the surface area of the cube, in cm2/minute, when the volume of the cube is 125 cm3, is (1) –4
(2)
16 5
(3)
16 6
(4)
8 15
Sol. Answer (2) V = a3
dV 2 da = 3a dt dt
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100
Application of Derivatives
4 cm3/minute = 3 25 S = 6a2
Solution of Assignment (Set-2)
da dt
da dS = 12a dt dt
dS 4 16 = 12 5 = dt 75 5
42. Consider the curve represented parametrically by the equation x = t3 – 4t2 – 3t and y = 2t2 + 3t – 5 where t R If H denotes the number of point on the curve where the tangent is horizontal and V the number of point where the tangent is vertical then (1) H = 2 and V = 1
(2)
H = 1 and V = 2
(3) H = 2 and V = 2
(4)
H = 1 and V = 1
Sol. Answer (2) x = t3 – 4t2 – 3t y = 2t2 + 3t – 5
dx = 3t2 – 8t – 3 dt
= 3t2 – 9t + t – 3 = (3t + 1) (t – 3)
dy = 4t + 3 dt
(4t 3) dy = (3t 1)(t 3) dx
For vertical tangents
dy 1 t 3, dx 3
For Horizontal tangents
3 dy 0 t 4 dx
So, H = 1, V = 2 43. The point on the curve y = 6x – x2 where the tangent is parallel to x-axis is (1) (0, 0)
(2)
(2, 8)
(3)
(6, 0)
(4)
(3, 9)
Sol. Answer (4) y = 6x – x2
dy = 6 – 2x = 0 dx
x=3 y = 18 – 9 = 9 i.e., (3, 9) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
101
44. For the curve y = 3sin . cos, x = e sin , 0 , the tangent is parallel to x-axis when is (1) 0
(2)
2
(3)
4
(4)
6
Sol. Answer (3) y = 3 sin . cos =
3 sin2 2 dy = 3 cos 2 d
x = e sin
dx = e cos e sin d dx = e (sin cos ) d dy 3cos 2 0 , when = dx 4 e (sin cos )
2 , then point is 3 (4) (a, a)
45. The slope of normal to the curve x3 = 8a2y, a > 0 at a point in the first quadrant is (1) (2a, –a)
(2)
(2a, a)
(3)
(a, 2a)
⎛ a ⎞ (1) It passes through ⎜ , a ⎟ ⎝ 2 ⎠
(2)
Is at a constant distance from origin
(3) It passes through origin
(4)
⎛ ⎞ It makes an angle ⎜⎝ ⎟⎠ with the ‘+’ x-axis 2
Sol. Answer (2) x3 = 8a2 y 3x2 = 8a 2
dy ,a>0 dx
dy 3 x 2 3 dx 8a2 2
x2 = 4a2
(when x = 2a, y = a)
x = ± 2a 46. The normal to the curve x = a(cos + sin), y = a (sin – cos) at any point ‘’ such that
Sol. Answer (2) x = a (cos + sin )
dx = a ( sin cos sin ) d
= a cos Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
102
Application of Derivatives
Solution of Assignment (Set-2)
y = a (sin – cos )
dy = a (cos sin cos ) = a sin d dy = tan dx
[y – a(sin – cos )] =
cos x a (cos sin ) sin
y sin + x cos = a[(cos + sin ) . cos + sin (sin – cos )] = a[1] x cos + y sin = a 47. The equation of tangent to the curve x = a cos3 t, y = a sin3 t at ‘t’ is (1) x sec t y cosec t a
(2)
x sec t y cosec t a
(3) x cosec t y cos t a
(4)
x sec t y cos t a
Sol. Answer (2) x = a cos3 t y = a sin3 t dx = –3a cos2 t . sin t dt dy dt
= 3a sin2 t . cos t
dy dx
=
sin2 t cos t = – tan t cos2 t sin t
Equation of tangent (y – a sin3 t) = – tan t (x – a cos3 t) y cos t – a sin3 t . cos t = – x sin t + a cos3 t . sin t x sin t + y cos t = a sin t . cos t x sec t + y cosec t = a 48. For the function f(x) = x2 – 6x + 8, 2 x 4, the value of x for which f(x) vanishes is (1) 3
(2)
5 2
(3)
9 4
(4)
7 2
Sol. Answer (1) f'(x) = 2x – 6 f'(x) = 0 For x = 3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives x
49. The slope of the tangent to the curve y
dx
∫ 1 x 0
1 2 Sol. Answer (1)
(2)
(1)
1
3
103
at the point where x = 1 is
(3)
1 4
(4)
1 5
The equation of the curve is y
x
dx
∫ 1 x3
0
dy 1 dx 1 x 3
1 1 ⎛ dy ⎞ ⎜ at x 1 ⎟ 1 1 2 ⎝ dx ⎠
Slope of the tangent to the curve at x = 1 is
1 2
50. If at each point of the curve y = x3 – ax2 + x + 1 the tangent is inclined at an acute angle with the positive direction of the x-axis then (1) a > 0
(2)
a 3
(3)
–
3 a 3
(4)
2a3
(4)
|m| 1
Sol. Answer (3) 3 2 We have, y x ax x 1
dy 3 x 2 2ax 1 dx
tangent make an acute angle with positive direction of x-axis Hence
dy 0 dx
3x2 – 2ax + 1 0, x R 4a2 – 12 0 3 a 3 a [ 3, 3] 51. If m be the slope of a tangent to the curve ey = 1 + x2 then (1) |m| > 1
(2)
m 0 and 2 sinx 2, x R x
f'(x) > 0 for x > 0 Hence f(x) is increasing in (0, ) When x < 0, x
1 – 2 but 2 2sinx 2, x R x
f'(0) < 0hence f(x) is decreasing in (–, 0) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
107
60. Let f(x) = xex(1 – x), then f(x) is (1) Increasing on R
(2)
⎡ 1 Increasing on ⎢ , ⎣ 2
(3) Decreasing on R
(4)
⎡ 1 Decreasing on ⎢ , ⎣ 2
⎤ 1⎥ ⎦ ⎤ 1⎥ ⎦
Sol. Answer (2) f ( x ) xe x (1 x ) f ( x ) e x (1 x ) (1 x 2 x 2 ) f '( x ) ( x 1)(2 x 1)e x (1 x )
⎡ 1 ⎤ Clearly f(x) is increasing on ⎢ , 1⎥ ⎣ 2 ⎦ 61. Which of the following is correct? (1) ln(1 + x) < x
–1 < x 0
(2) ln(1 + x) < x
0 x
x>0
(4) ln(1 + x) < x
x > –1
Sol. Answer (2) We know that If x > 0
ex 1 x
x2 x3 ... 2! 3!
ex 1 x
x ln(1 x )
Second Method Let f(x) = x - ln(1 + x) f'(x) = 1
1 x 1 x 1 x
f'(x) > 0 for x (–, -1) (0, ) and f'(x) < 0 for x (–1, 0) f(x) > f(0) for x (0, ) x – ln(1 + x) > 0 x > ln(1 + x), x > 0 ⎡ ⎤ 62. Let the equation x – sin x = a has a unique root in ⎢ , ⎥ , then ⎣ 2 2⎦ ⎡ ⎞ (1) a ⎢ 1, ⎟ 2 ⎣ ⎠
(2)
⎤ ⎛ a ⎜ , 1⎥ 2 ⎝ ⎦
⎡ ⎤ (3) a ⎢1 , 1⎥ ⎣ 2 2 ⎦
(4)
⎞ ⎛ a R ⎜1 , 1⎟ 2 2 ⎠ ⎝
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108
Application of Derivatives
Solution of Assignment (Set-2)
Sol. Answer (3) sin x = x – a Case–1
y=x–a
If a > 0 If y = x – a touch sin x at x a
2
1 2
A
1
2 B
–1
Case–2
2
If a < 0 If y = x – a touch y = sin x at x a 1
2
2
2
1
B
⎡ ⎤ Hence a ⎢1 , 1⎥ . ⎣ 2 2 ⎦
A
–1
2
y=x–a 63. Number of real roots of the equation ex – 1 – x = 0 is (1) 1
(2)
2
(3)
3
(4)
0
Sol. Answer (1) The number of real roots of the equation ex-1 = x is the number of points of intersection of the graph of the two curves y = ex-1 and y = x.
y = ex – 1
ex – 1 = x
1 1
y=
x
From graph it is clear that the number of real root = 1.
64.
f (x)
x x and g ( x ) , where 0 < x 1, then in the interval sin x tan x
(1) Both f(x) and g(x) are increasing functions (2) Both f(x) and g(x) are decreasing functions (3) f(x) is an increasing function (4) g(x) is an increasing function Sol. Answer (3) f (x)
x sin x
f '( x )
Let sin x x cos x h( x )
sin x x cos x sin2 x
h '( x ) cos x [cos x x sin x ] = x sin x
h'(x) > 0 x (0, 1)
f'(x) > 0 i.e., f(x) is increasing function in (0, 1] Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
109
x tan x x sec 2 x g '( x ) tan x tan2 x Let tan x – x sec2 x = (x)
Now g ( x )
'( x ) sec 2 x [sec 2 x 2 x sec 2 x tan x ] = – 2x sec x tan x '( x ) 0, x (0, 1]
g(x) is decreasing in (0, 1] 65. If f ( x )
P2 –1 2
x 3 – 3 x log 2 is a decreasing function of x in R then the set of possible values of P
P 1 (independent of x) is
(1) [–1, 1]
(2)
(1, )
(3)
(–, 1]
(4)
(2, )
Sol. Answer (1) f (x)
p2 1 p2 1
x 3 3 x log2
⎛ p2 1 ⎞ 2 f '( x ) ⎜ ⎟ 3x 3 ⎜ p2 1 ⎟ ⎝ ⎠ For decreasing function f'(x) < 0
⎛ p2 1 ⎞ ⎛ p2 1⎞ 2 ⎜ 2 ⎟x 1 0 ⎜ ⎟0 ⎜ p 1⎟ ⎜ p2 1 ⎟ ⎝ ⎠ ⎝ ⎠ –1 p 1
p [–1, 1]
66. Let f(x) = (x – p)2 + (x – q)2 + (x – r)2. Then f(x) has a minimum at x = , where is equal to (1)
pqr 3
(2)
3
pqr
Sol. Answer (1)
(3)
3 1 1 1 p q r
(4)
3 pqr
f(x) = (x – p)2 + (x – q)2 + (x – r)2 = 3x2 – 2x(p + q + r) + (p2 + q2 + r2) We know that f(x) = ax2 + bx + c(if a > 0) is minimum at x =
b 2a
pqr 3 Second method :
Hence
For minimum value of f(x), We observe that x=
pqr 3
d 2f ( x ) dx 2
df ( x ) d 2f ( x ) 0 and 0 dx dx 2
df ( x ) 6 x 2( p q r ) 0 dx
6 = +ve
Hence f(x) is minimum at x =
pqr 3
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110
Application of Derivatives
Solution of Assignment (Set-2)
⎛ 1⎞ 67. The maximum value of ⎜ ⎟ ⎝x⎠
(1) e
2x 2
is
(2)
⎛ 1⎞ ⎜ ⎟
e⎝ e ⎠
(3)
1
(4)
ee
(4)
a=
1 1 ,b=– 2 2
(4)
ac
b2 4
Sol. Answer (2) ⎛ 1⎞ y ⎜ ⎟ ⎝x⎠
2x2
2 y x 2x
2 dy x 2x [–2x – 4x log x] dx
For maxima or minima of y,
x
1
dy 0 dx
as x 0
e
⎛ d 2y ⎞ 0 and ⎜ 2 ⎟ ⎜ dx ⎟ 1 ⎝ ⎠x e
Hence y attains maximum value at x =
1 e
1
and ymax = e e e e
68. If y = a log |x| + bx2 + x has its extremum values at x = –1 and x = 2 then (1) a = 2, b = –1
(2)
a = 2, b =
1 2
(3)
a = –2, b =
1 2
Sol. Answer (2) y a log | x | bx 2 x
dy a 2bx 1 dx x
f(x) has extremum at x = –1 and x = 2
Hence f'(–1) = 0 and f'(2) = 0 –a – 2b + 1 = 0 and
69. If ax
1 a + 4b + 1 = 0 a = 2 and b = 2 2
b c , x > 0 and a, b, c are positive constants then x
(1) ab
c2 4
(2)
ab
c2 4
(3)
bc
a2 4
Sol. Answer (1) Given ax
b cx0 x
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Solution of Assignment (Set-2)
Application of Derivatives
111
A.M. G.M.
ax
b 2 ab x
ab
c2 4
Second Method : We have, ax +
b c, x > 0 x
ax2 – cx + b 0
c2 – 4ab 0 and a > 0
x>0
c2 4
ab
70. Let x be a number which exceeds its square by the greatest possible quantity. Then x is equal to (1)
1 2
1 4
(2)
(3)
3 4
(4)
1
(4)
(2, –2)
Sol. Answer (1)
Let y = x – x2
dy 1 2x dx
For maximum or minimum value of y,
We observe that
d 2y dx
2
dy 0 dx
2 = –negative
1 – 2x = 0
x
y is maximum for x =
1 2
1 2
71. The point (0, 3) is nearest to the curve x2 = 2y at (1)
2
2, 0
(2)
(0, 0)
(3)
(2, 2)
Sol. Answer (3)
⎛ 1 2⎞ Let point on the curve x2 = 2y be ⎜ t , t ⎟ ⎝ 2 ⎠ ⎛ 1 2⎞ The distance of the point (0, 3) from a point P ⎜ t , t ⎟ on ⎝ 2 ⎠ the given curve x2 = 2y is ⎛ t2 ⎞ d t ⎜ 3⎟ ⎜2 ⎟ ⎝ ⎠
2
⎛ t2 ⎞ d2 t2 ⎜ 3⎟ ⎜2 ⎟ ⎝ ⎠
2
(0,3)
1 t, 2 t2
2
Let d2 = z d is maximum or minimum according as z is maximum or minimum
⎛ t2 ⎞ z t2 ⎜ 3⎟ ⎜2 ⎟ ⎝ ⎠
2
⎛ t2 ⎞ dz 2t 2t ⎜ 3 ⎟ ⎜2 ⎟ dt ⎝ ⎠
For maximum or minimum value of z, t = 0, 2, –2
d 2z dt 2
dz 0 dt
3t 2 4
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112
Application of Derivatives
d 2z dt 2
Solution of Assignment (Set-2)
0 at t = ± 2
Hence point is (2, 2) or (–2, 2) Thus the point on the given curve which is nearest from (0, 3) is either (2, 2) or (–2, 2). 72. Let f(x) be a function defined as below : f(x) = sin(x2 – 3x), x 0 = 6x + 5x2, x > 0 then at x = 0, f(x) (1) Has a local maximum
(2)
Has a local minimum
(3) Is discontinuous
(4)
Point of inflexion
Sol. Answer (2) ⎧sin( x 2 3 x ), x 0 We have, f ( x ) ⎪⎨ x0 ⎪⎩ 6 x 5 x 2 ,
We observe that f(0) = 0 f'(0 + h) = 6 + 10h > 0, for h > 0 and f'(0 – h) = – (2h + 3) cos (h2 + 3h) < 0 Here we observe that f'(x) changes its sign from -ve to +ve in the neighbourhood of x = 0, hence f(x) has a local minimum at x = 0 Hence x = 0 is a point of local minima. 73. If , be real numbers such that x3 – x2 + x – 6 = 0 has its roots real and positive then minimum value of is (1) 3 3 36
(2)
11
(3)
0
(4)
1
Sol. Answer (1) Let roots of x3 – x2 + x – 6 = 0 be a, ,
++=
+ + = = 6 Applying A.M. – G.M. inequality 1 1 1 1/3 ⎛ 1 1 1⎞ ⎜ . . ⎟ 3 ⎝ ⎠
2/3 3
2/3 3(6)
3(36)1/3 Minimum 3.(36)1/3 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
113
74. If a, b, c, d are real numbers such that 3a 2b 3 0 c d 2
then the equation ax3 + bx2 + cx + d = 0 has (1) At least one root in [–2, 0]
(2)
At least one root in [0, 2]
(3) At least two roots in [–2, 2]
(4)
No root in [–2, 2]
Sol. Answer (2) 3a 2b 3 0 c d 2
Let f ( x )
6a + 4b + 3c + 4d = 0
ax 4 bx 3 cx 2 dx + e 4 3 2
f(0) = e f (2)
2 [6a 4b 3c 3d ] e e 3
Since f(x) is continuous and differentiable in (0, 2) and f(0) = f(2) = e Hence according to Rolle’s theorem equation f'(x) = ax3 + bx2 + cx + d = 0 has at least one root in (0, 2). 75. If the least area of triangle formed by tangent, normal at any point P on the curve y = f(x) and x-axis is 4 sq. unit. Then the ordinate of the point P (P lies in first quadrant) is (1) 1
(2)
1 2
(3)
1 4
(4)
2
Sol. Answer (4) From figure PQ = y1
T
MQ y1 cot
QN = y1 tan Area of PMN = 2 1
Thus 4 = y
P(x1, y1)
MN = MQ + QN = y1(cot + tan )
1 1 1 MN PQ = y12 (tan cot ) y12 2 y12 2 2 2
M
Q
N
y1 = 2
76. Let f (x) > 0 x R and let g(x) = f(x) + f(2 – x) then interval of x for which g(x) is increasing is (1) (1, )
(2)
R
(3)
[–1, 1]
(4)
[–2, )
Sol. Answer (1) g(x) = f(x) + f(2 – x) g'(x) = f'(x) – f'(2 – x) for g to be increase x>2–x x>1 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
114
Application of Derivatives
Solution of Assignment (Set-2)
77. The number of tangents to the curve x3/2 + y3/2 = a3/2, where the tangents are equally inclined to axes, is (1) 2
(2)
1
(3)
0
(4)
4
(3)
2
(4)
6
Sol. Answer (2) The equation of the curve is x 3/2 y 3/2 a3/2
…(1)
dy x dx y
Since the tangent is equally inclined to the axes hence
dy 1 dx
x 1 y
x 1 y
y=x
Putting x = y in (1), we get, 2x 3/2 a3/2
⎛ 1⎞ x ⎜ ⎟ ⎝2⎠
2/3
a
⎛ 1⎞ y ⎜ ⎟ ⎝2⎠
2/3
a
Hence number of tangent = 1 78. The number of tangents to the curve y2 – 2x3 – 4y + 8 = 0 that pass through (1, 2) is (1) 3
(2)
1
Sol. Answer (3) y2 – 2x3 – 4y + 8 = 0 Let point be (x1, y1) 3 x12 ⎛ dy ⎞ ⎜ dx ⎟ ⎝ ⎠( x1,y1) y1 2 Equation of tangent at (x1, y1) is given by y y1
3 x12 ( x x1) y1 2
Since the tangent passes through (1, 2) Hence, 2 y1
3 x12 (1 x1) y1 2
4 y12 4 y1 3 x13 3 x12
3 x13 3 x12 y12 4 y1 4 0
…(1)
Also (x1, y1) lies on y2 – 2x3 – 4y + 8 = 0 Hence y12 2 x13 4 y1 8 0
…(2)
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Solution of Assignment (Set-2)
Application of Derivatives
115
Solving (1) and (2) x13 3 x12 4 0 ( x1 1)( x12 4 x1 4) 0 ( x1 1)( x1 2)2 0 x1 = –1, 2 From (2) at x = –1. y12 – 4y1 + 10 = 0 Imaginary roots at x = 2 y12 4 y1 8 0
4 16 32 = 2 3 2
y1
Points are (2, 2 3) hence number of tangent = 2.
79. The curve
xn a
n
yn b
n
2 touches the line
(1) (b, a)
(2)
x y 2 at the point a b
(a, b)
(3)
(1, 1)
(4)
⎛ 1 1⎞ ⎜ , ⎟ ⎝b a⎠
Sol. Answer (2)
xn an
yn
2
bn
Differentiating it w.r.t. x we get
nx n 1
dy b n x n 1 ny n 1 dy 0 dx a n y n 1 bn dx
an Let the point be (x1, y1)
The equation of tangent
xx1n 1 an
yy1n 1 bn
x1n an
y1n
…(1)
bn
According to question tangent is
x y 2 a b
…(2)
Equations (1) & (2) represent same line Hence
x1n 1
y n 1 1 ⎡ x1n y1n ⎤ 1 ⎢ ⎥ a n 1 b n 1 2 ⎢⎣ a n b n ⎥⎦
x1 = a, y1 = b Hence point is (a, b) Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Application of Derivatives
Solution of Assignment (Set-2)
80. Let the equation of a curve be x = a( + sin ), y = a(1 – cos ). If changes at a constant rate k then the rate of change of the slope of the tangent to the curve at is 3 2k k (1) (2) (3) k 3 3
(4)
2k 3
Sol. Answer (4) We have x = a ( + sin), y = a (1 – cos)
dy tan dx 2
dm 1 d sec 2 dt 2 2 dt
m tan
= Slope of the tangent to the given curve 2
1 4 ⎛ dm ⎞ ⎜⎝ dt ⎟⎠ 2 3 k 3
The rate of change of the slope of the tangent =
2k 3
81. Two cyclists start from the junction of two perpendicular roads, their velocities being 3v metres/minute and 4v metres/minute. The rate at which the two cyclists are separating is (1)
7 v m/min 2
(2)
5v m/min
(3)
v m/min
(4)
7v m/min
(4)
3y = 9x + 2
Sol. Answer (2) 2 = x2 + y2 dx dy 3vm / minute, 4vm / minute dt dt
y
d dx dy d 2x 2y x 3v y 4v . 2 dt dt dt dt
x
3 4 d x y 3v 4v = 3v 4v 5v 5 5 dt
82. The equation of the common tangent to the curves y2 = 8x and xy = –1 is (1) y = x + 2
(2)
y = 2x + 1
(3)
2y = x + 8
Sol. Answer (1) Equation of tangent to the curve y2 = 8x is
2 m It is also the tangent to xy = –1 y mx
…(1)
2⎞ ⎛ Hence x ⎜ mx ⎟ = –1 m⎠ ⎝ i.e., mx 2 2 x 1 0 have equal real roots. m Consequently b2 – 4ac = 0
4
4m 0 m = 1. m2 Hence equation of the tangent common to the given curves is y = x + 2. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
117
83. If is a positive acute angle then (1) tan < < sin
(2)
< sin < tan
(3) sin < tan <
(4)
sin < < tan
Sol. Answer (4) Let f() = – sin f'() = 1 – cos
⎛ ⎞ ⎛ ⎞ f ' ( ) 0 ⎜ 0, ⎟ which shows that f() is increasing in ⎜ 0, ⎟ ⎝ 2⎠ ⎝ 2⎠ Hence f() > f(0) sin > 0 > sin
…(1)
Again let g() = tan – g'() = sec2 – 1
⎛ ⎞ ⎛ ⎞ g '() 0, ⎜ 0, ⎟ which shows that g(q) is increasing in ⎜ 0, ⎟ ⎝ 2⎠ ⎝ 2⎠ g() > g(0) tan – > 0 tan >
…(2)
Combining (1) and (2) tan > > sin 84. If f(x) = x2 + 2bx + 2c2 and g(x) = –x2 – 2cx + b2 are such that, min f(x) > max g(x), then (1) 0 c
b 2
(2)
|c||b| 2
(3)
|c||b| 2
(4)
|b||c| 2
3
(4)
–
Sol. Answer (3) 2 2 We have, f ( x ) x 2 2bx 2c 2 , and g ( x ) x 2cx b 2 2 2 f ( x ) ( x b )2 2c 2 b2 and g ( x ) b c ( x c ) 2 2 min f ( x ) 2c 2 b2 and max g ( x ) c b
Since min f(x) > max g(x) Hence 2c2 – b2 > c2 + b2 c2 > 2b2 |c| > |b|
2
85. If x [–1, 1] then the minimum value of f(x) = x2 + x + 1 is (1)
3 4
(2)
1
(3)
3 4
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118
Application of Derivatives
Solution of Assignment (Set-2)
Sol. Answer (1) 2
1⎞ 3 3 ⎛ ,xR f(x) = x2 + x + 1 = ⎜ x ⎟ 4 2 4 ⎝ ⎠
Minimum f(x) =
3 4
86. The maximum value of cos[cos (sin x)] is (1) cos (cos 1)
(2)
cos 1
(3)
1
(4)
0
Sol. Answer (1) f(x) = cos [cos(sin x)] f'(x) = sin[cos(sin x)] sin(sin x) cos x for maxima–minima of f(x)
f'(x) = 0
x = (2n 1) , m 2 i.e, f(x) is maximum at x = (2n + 1) 2 2
f"(x) < 0 at x (2n 1)
maximum value of f(x) = cos(cos 1) 87. In a ABC, B = 90º and a + b = 4. The area of the triangle is the maximum when C is
4 Sol. Answer (3) A
1 1 ac = a b 2 a 2 2 2
A2
A2
6
(2)
(1)
(3)
3
(4)
2
A
1 2 2 a (b a2 ) 4
b
c
1 2 a [(4 a )2 a 2 ) 4
B
1 A a 2 [16 8a] 4 2 A 4a 2 2a3 2
90º a
C
Let A2 y
y 4a2 2a3 Area is maximum when y is maximum dy 8a 6a 2 da
For maximum a
dy 0 da
4 8 ,b 3 3
From figure, we have cos C
a b
cos C
1 2
C
3
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Solution of Assignment (Set-2)
Application of Derivatives
119
88. The number of critical points of the function x
∫
f ( x ) e t (t 1)(t 2)(t 3)dt is 0
(1) 3
(2)
4
(3)
12
(4)
2
Sol. Answer (1) x
f ( x ) ∫ et (t 1)(t 2)(t 3)dt 0
f '( x ) e x ( x 1)( x 2)( x 3) For critical points of f(x), we must have f'(x) = 0 = ex(x – 1) (x – 2) (x – 3) x = 1, 2, 3 Number of critical point = 3. 89. If the equation x3 + px2 + qx + 1 = 0 (p < q) has only one real root , then belongs to (1) (–2, –1)
(2)
(–1, 0)
(3)
(0, 1)
(4)
(1, 2)
Sol. Answer (2)
x 3 px 2 qx 1 0 ( p q ) Let f ( x ) x 3 px 2 qx 1 f (0) 1 ⇒ f (0) 0 f ( 1) p q ⇒ f ( 1) 0
f(0) . f(–1) < 0
f(x) = 0 has one real root in (-1, 0)
(–1, 0)
90. The value of C (if exists) in Lagrange's theorem for the function |x| in the interval [–1, 1] is (1) 0 (3) –
1 2
(2)
1 2
(4)
Non-existent in the interval
Sol. Answer (4) | x | is not differentiable at x = 0 hence conditions of Lagrange’s theorem is not satisfied consequently c does not exist. 91. The equation ex – 1 + x – 2 = 0 has (1) One real root
(2)
Two real roots
(3) Three real roots
(4)
Four real roots
Sol. Answer (1) The given equation can be put in the form ex–1 = 2 – x Let us draw the graphs of y = ex–1 and y = 2 – x and the point of intersection of the two curves gives the number of real roots of the equation.
x–1
y=e
y=2–x
From the graph of the two curves it follows that the given equation has one real root. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
120
Application of Derivatives
92. If 0 < < < <
x
Solution of Assignment (Set-2)
and 2
sin sin sin then cos cos cos
(1) x [tan , tan ]
(2)
x (tan , tan )
(3) x (cot , cot )
(4)
x [cot , cot ]
Sol. Answer (2)
0
2
sin < sin < sin 3 sin < sin + sin + sin < 3 sin Also cos > cos cos
…(1)
3 cos > cos + cos + cos > 3 sin
1 1 1 3cos cos cos cos 3cos
…(2)
Multiplying (1) and (2), we get
tan
sin sin sin tan cos cos cos
93. Let A be the area formed by the positive x-axis and the normal and tangent to the circle x2 + y2 = 4 at (1, 3 ) then A2 equals (1) 9
(2)
12
⎛ dy ⎞ ⎜ ⎟ ⎝ dx ⎠(1,
(3)
15
(4)
18
Sol. Answer (2) x2 + y2 = 4
dy x dx y
The equation of tangent at (1,
y 3
1 3
P(1, 3)
1
3)
3
3 ) is
O
( x 1)
Q(4, 0)
and the point of intersection with x axis is (4, 0) The equation of normal at (1, 3) is
y 3 3( x 1) and point of intersection with x-axis is (0, 0) Hence A =
1 4 3 = 2 3 2
A2 = 12 ⎛ ⎞ 94. The minimum value of 64 sec + 27 cosec where ⎜ 0, ⎟ is ⎝ 2⎠
(1) 91
(2)
12
(3)
36
(4)
125
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Solution of Assignment (Set-2)
Application of Derivatives
121
Sol. Answer (4) Let f () 64 sec 27 cosec f '() 64 sec tan 27cosec cot =
64 sin3 27 cos3 sin cos
For maxima–minima of f() tan
f '() 0
3 4
Since f"() > 0, f() has minimum value for tan =
3 4
and Min f() = 125 95. If the tangent to the curve 2y3 = ax2 + x3 at the point (a, a) cuts off intercepts and on the co-ordinate axes, (where 2 + 2 = 61) then a2 equals (1) 400
(2)
600
(3)
900
(4)
800
Sol. Answer (3) We have, 2y 3 ax 2 x 3
5 ⎛ dy ⎞ ⎜ ⎟ dx ⎝ ⎠(a,a ) 6
dy 2ax 3 x 2 dx 6y 2
The equation of tangent to the given curve at (a, a) is y a x intercept of the tangent =
1 a 5
and y intercept of the tangent =
1 a 6
2 2 61 =
96. If the curves
x2 a2
5 ( x a) 6
a 2 a 2 61a 2 a2 = 900 25 36 900 y2 1 and y3 = 16x intersects at right angles, then 3a2 is equal to 4
(1) 1
(2)
2
(3)
3
(4)
4
Sol. Answer (4) Let point of intersection be ( x1, y 1 ) x2 a2
y2 1 4
…(1)
dy 4 x 2 dx a y 4 x ⎛ dy ⎞ m1 ⎜ 2. 1 ⎟ dx a y1 ⎝ ⎠( x1, y1 ) y3 = 16x
…(2)
16 ⎛ dy ⎞ m2 ⎜ ⎟ ⎝ dx ⎠( x1, y1 ) 3 y12 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
122
Application of Derivatives
Solution of Assignment (Set-2)
The two curves intersect at right angles then m1m2 = –1
⎛ 4 x1 ⎞ ⎛ 16 ⎞ ⎜ 2 ⎟ ⎜⎜ 2 ⎟⎟ 1 ⎝ a y1 ⎠ ⎝ 3 y1 ⎠ x1
3a 2 y13 64 2 3
But y13 16 x1 16 3a y1 64
y13 16 x1 32 = 4 97. The area of the triangle formed by the tangent to the curve y (1) 1
(2)
2
(3)
8 4 x2
at x = 2 and the co-ordinate axes is
3
(4)
4
Sol. Answer (4) Given curve is y
8 4 x2
dy 8 2x dx (4 x 2 )2
When x = 2, y = 1
point is (2, 1)
m
Y
32 1 64 2
B (2, 1)
Equation of tangent at P(2, 1) is
1 y 1 ( x 2) 2
X'
X'
O
xy – 2 = – x + 2
A
X
x + xy = 4 A = (4, 0), B = (0, 2)
ar ( OAB )
1 4 2 4 sq. units 2
Y'
SECTION - B Objective Type Questions (More than one options are correct) 1.
If the line, ax + by + c = 0 is normal to the curve xy = 2, then (1) a < 0, b > 0
(2)
a > 0, b < 0
(3) a > 0, b > 0
(4)
a < 0, b < 0
Sol. Answer (1, 2) Slope of the normal (ax by c 0)
a b
The given curve is xy = 2
dy y b b 0 dx x a a
a < 0, b > 0 or a > 0, b < 0
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Solution of Assignment (Set-2)
2.
Application of Derivatives
123
The equation of the tangents to the curve y = x4 from the point (2, 0), are given by (1) y = 0 (3) y
4096 2048 ⎛ 8⎞ x ⎟ ⎜ 81 27 ⎝ 3⎠
(2)
y – 1 = 5(x – 1)
(4)
y
32 80 ⎛ 2⎞ ⎜x ⎟ 243 81 ⎝ 3⎠
Sol. Answer (1, 3) Let P(x1, y1) be a point on the curve y = x4
dy 4x3 dx
⎛ dy ⎞ 4 x13 m⎜ ⎟ ⎝ dx ⎠( x1,y1 ) The equation of tangent to the given curve at (x1, y1) is given by
y y1 4 x13 ( x x1 ) Which passes through (2, 0) hence
y1 4 x13 (2 x1 ) 8 x13 4 x14 y1 0
4 x14 8 x13 y1 0
…(1)
(x1, y1) lie on y x 4 ⇒ y1 x14
x1 0,
8 3
4 x14 8 x13 x14 0
y1 0,
4096 2048 m = 0 and 81 27
Thus the equations of tangents are y = 0 and y
3.
4096 2048 ⎛ 8⎞ x ⎟ 81 27 ⎜⎝ 3⎠
Suppose f '(x) exists for each x and h(x) = f(x) – (f(x))2 + (f(x))3 x R, then (1) h is increasing whenever f is increasing
(2)
h is increasing whenever f is decreasing
(3) h is decreasing whenever f is decreasing
(4)
Nothing can be said in general
Sol. Answer (1, 3) We have, h( x ) f ( x ) (f ( x ))2 f (( x ))3 h '( x ) f '( x ) 2f ( x )f '( x ) 3(f ( x ))2 f '( x ) = f '( x )[3(f ( x ))2 2f ( x ) 1] Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
124
Application of Derivatives
Solution of Assignment (Set-2)
Since 3(f ( x ))2 2f ( x ) 1 0 for real values of f(x) as 4 – 12 < 0 h'(x) > 0 if f'(x) > 0 and h'(x) < 0 if f'(x) < 0 Hence h(x) is increasing or decreasing function according as f(x) increases or decreases. 4.
Let f '(x) = g(x) (x – a)2, where g(a) 0 and g is continuous at x = a then (1) f is increasing near a if g(a) < 0
(2)
f is decreasing near a if g(a) > 0
(3) f is decreasing near a if g(a) < 0
(4)
f is increasing near a if g(a) > 0
Sol. Answer (3, 4)
f '( x ) g ( x )( x a)2 In the neighbourhood of x = a, we have x = a ± h as g(x) is continuous at x = a f '(a ) g (a)h2 if g (a ) 0 ⇒ f '(a ) 0 if g (a ) 0 ⇒ f '(a ) 0 . f(x) is increasing in the neighbourhood of a if g(a) > 0 and is decreasing if g(a) < 0 5.
Let f(x) = (x2 – 1)n (x2 + x – 1) then f(x) has local minimum at x = 1, when (1) n = 2
(2)
n=3
(3)
n=4
(4)
n=5
Sol. Answer (1, 3)
f ( x ) ( x 2 1)n ( x 2 x 1) f(x) has local extremum at x = 1 then either f (1) f (1 h ) and f (1) f (1 h ) or f (1) f (1 h ) and f (1) f (1 h )
where f(1) = 0. Thus for local minimum at x = 1 f (1 h ) 0 and f (1 h ) 0 or f (1 h ) 0 and f (1 h ) 0
If f (1 h ) 0 and f (1 h ) 0 h 0
( h )n (2 h )n (h2 3h 1) and (h )n (2 h )n (h2 3h 1) 0 i.e., ( h )n 0 n even number Hence n = 2, 4. 6.
Which of the following is correct? (1) e2 > 2e
(2)
e3 > 3e
(3)
e > e
(4)
⎛⎞ e ⎜ ⎟ ⎝2⎠
e
Sol. Answer (1, 2, 3, 4) Consider the function 1 x f ( x ) x1/ x = e
ln x
1/ x 1 ⎡ 1 ⎤ f '( x ) x1/ x ⎢ 2 2 n x ⎥ = x [1 n x ] x ⎣x ⎦ x2
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Solution of Assignment (Set-2)
Application of Derivatives
125
if x e ⇒ f '( x ) 0 i.e., f(x) is increasing in (0, e) if x e ⇒ f '( x ) 0 i.e., f(x) is decreasing in (e, ) Now 2 21/2
e2 > 2e option (1) is correct Also e < 3 and f(x) decreases in (e, ) f(e) > f(3)
e3 > 3e option (2) is correct
e f() < f(e)
e1/ > e1/e
e > e
option (3) is correct
Moreover 1
⎛⎞ ⎛ ⎞ /2 f ⎜ ⎟ f (e ) ⎜ ⎟ e1/ e ⎝2⎠ 2 ⎝ ⎠
e 2
⎛⎞ e /2 ⎜ ⎟ ⎝2⎠
e
⎛⎞ e ⎜ ⎟ ⎝2⎠
7.
e
option (4) is correct.
An extremum value of the function f(x) = (sin–1 x)3 + (cos–1 x)3 (–1 x 1) is (1)
7 3 8
(2)
3 8
(3)
3 32
(4)
3 16
Sol. Answer (1, 3) Let sin1 x y cos1 x
y 2
⎛ ⎞ f (y ) y 3 ⎜ y ⎟ ⎝2 ⎠
3
2
⎛ ⎞ f ( y ) 3 y 2 3 ⎜ y ⎟ ⇒ y 4 ⎝2 ⎠ is point of minima 4
min
max
3 32
73 at y 2 8
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126 8.
Application of Derivatives
Solution of Assignment (Set-2)
Let the function x
f (x)
∫ t (e
t
1) (t 1) (t 2) (t 3)dt
2
has a point of minima at x equal to (1) 0
(2)
–1
(3)
2
(4)
3
Sol. Answer (2, 4) f'(x) = x(ex – 1) (x + 1) (x – 2) (x – 3)
– 9.
–1
+
0
+
2
–
3
+
Let f(x) = |sinx + cosx| then for 'f' (1) Point of relative maxima in [0, 4] is 4
(2)
Point of relative maxima in [0, 4] is 7
(3) Point of relative minima in (0, 4) is 4
(4)
Point of relative minima in (0, 4) is 8
Sol. Answer (1, 3)
3 4
7 4
11 4
15 4
10. For every pair of continuous functions f, g : [0, 1] » such that max {f(x) : x [0, 1]} = max {g(x) : x [0, 1]}, the correct statement(s) is(are) [JEE(Advanced)-2014] (1) (f(c))2 + 3f(c) = (g(c))2 + 3g(c) for some c [0, 1] (2) (f(c))2 + f(c) = (g(c))2 + 3g(c) for some c [0, 1] (3) (f(c))2 + 3f(c) = (g(c))2 + g(c) for some c [0, 1] (4) (f(c))2 = (g(c))2 for some c [0, 1] Sol. Answer (1, 4) Let f and g be maximum at c1 and c2 respectively, c1, c2 (0, 1). Let h(x) = f(x) – g(x) Now h(c1) = f(c1) – g(c1) = +ve and h(c2) = f(c2) – g(c2) = –ve h(x) = 0 has at least one root in (c1, c2) f(x) = g(x) for some x = c (c1, c2) f(c) = g(c) for some c (0, 1) Clearly (1, 4) are correct 11. Let a » and let f : » » be given by f(x) = x5 – 5x + a. Then
[JEE(Advanced)-2014]
(1) f(x) has three real roots if a > 4
(2)
f(x) has only one real root if a > 4
(3) f(x) has three real roots if a < –4
(4)
f(x) has three real roots if –4 < a < 4
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Solution of Assignment (Set-2)
Application of Derivatives
127
Sol. Answer (2, 4) f(x) = x5 – 5x + a if f(x) = 0 a = 5x – x5 = g(x) g(x) = 0 x = 0, 5
1
4,
–5
1 4
4
g’(x) = 0 5 – 5x4 = 0 x = 1, –1
5
g(–1) = –4
1 4
O
–1 –4
1
5
1 4
g(1) = 4 If a (–4, 4)
f(x) = 0 has 3 real roots
if a > 4 or a < –4 f(x) = 0 has only 1 real root.
SECTION - C Linked Comprehension Type Questions Comprehension I Let the solution set of the inequation ⎡ ⎤ ⎛ 1 ⎞ ⎟⎟ 0 in ⎢ , ⎥ be A and let solution set of equation sin–1 (3x – 4x3) = 3 sin–1 x be B. Now ⎜⎜ sin x 2 ⎣ ⎦ 2⎠ ⎝ define a function f : A B.
1⎞ ⎛ ⎜ sin x ⎟ 2⎠ ⎝
1.
There are exactly two linear onto functions that can be drawn from A to B, one of them is (3)
12 x 19 2
(1) cos = f(x), for some x A, R
(2)
sin + cos = f(x), for some x A, R
(3) tan = f(x) for some x A, R
(4)
sec = f(x) for some x A and R
(1) 2.
3.
12 x 19 2
(2)
12 x 19 2
(4)
12
19 x 2
Among the following statements which is true
on to If f : A B is a strictly decreasing function, then ordered pair (, ) satisfying the equation
2 6
37 f ( ) is 4
⎡ 3 1 ⎤ (1) ⎢ , ⎥ ⎣ 4 2⎦
(2)
⎡ 5 ⎤ ⎢ 6 , 3⎥ ⎣ ⎦
(3)
⎡ 3 ⎤ ⎢ 4 , 3⎥ ⎣ ⎦
(4)
(3, )
Solutions of Comprehension I 1 ⎞⎛ 1 ⎞ ⎛ ⎟0 ⎜ sin x 2 ⎟ ⎜ sin x 2⎠ ⎝ ⎠⎝
⎡ 3 5 ⎤ x⎢ . ⎥ = A ⎣4 6⎦
and sin1(3 x 4 x 3 ) 3 sin1( x )
⎡ 1 1⎤ x ⎢ , ⎥ = B ⎣ 2 2⎦ f:AB
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128 1.
Application of Derivatives
Solution of Assignment (Set-2)
Answer (1) Let f ( x ) ax b Case -1
3 5 ⎛ ⎞ ⎛ 5 ⎞ f⎜ ⎟a b and f ⎜ ⎟ a b 4 4 6 6 ⎝ ⎠ ⎝ ⎠
1 5a b 2 6
1 3 ab 2 4
On solving a
i.e., f ( x )
12 , b 19 2
12 19 x 2
Case - 2
⎛ 3 ⎞ 3 ⎛ 5 ⎞ 5 f⎜ ⎟ a b and f ⎜ ⎟ ab ⎝ 4 ⎠ 4 ⎝ 6 ⎠ 6
1 3 ab 2 4
1 5 ab 2 6
On solving a
12 19 , b 2
f (x)
2.
Answer (4)
3.
Answer (3)
12 19 x 2
onto As f : A B is strictly decreasing function
then
⎡ 3π 5π ⎤ ⎡ 1 1 ⎤ f : ⎢ , ⎥ → ⎢− , ⎥ ⎣ 4 6 ⎦ ⎣ 2 2⎦ then
3π 5π f ( x ) > f ⎛⎜ ⎞⎟ ⎝ 4 ⎠ ⎝ 6 ⎠
Now 2 6
37 f ( ) 4
If = 3 f ( )
( 3)2
1 f ( ) 4
3 1 i.e., 4 2
⎛ 3 ⎞ i.e., ordered pair (, ) is ⎜ , 3 ⎟ . ⎝ 4 ⎠ Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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129
Comprehension II When the product of two positive numbers is constant,then the minimum value of their sum is 2 times the square root of their product. Also when the sum of two positive numbers is constant, the maximum value of their-product is
1 time the square of their sum. We get these minimum and maximum values only when the two numbers are 4
equal. 1.
The minimum value of f ( x ) 2
(1)
1 { x } , (where {x} is fractional part of x) is 2{ x } 2 1
(2)
(3)
2 2
(4)
2 1
(3)
2 ⎞ ⎛ ⎜ 0, ⎟ ⎜ 6 ⎟⎠ ⎝
(4)
0, 2
(3)
(–, 0]
(4)
(0, )
Sol. Answer (1)
1 { x } { x } 2 2 2{ x } 2{ x } 2.
The range of g(x) = (cot–1x) (cot–1(–x)) is
⎛ 2 2 ⎞ ⎟ (1) ⎜⎜ , 4 ⎟⎠ ⎝ 6
(2)
2 ⎞ ⎛ ⎜ 0, ⎟ ⎜ 4 ⎟⎠ ⎝
Sol. Answer (2)
g ( x ) cot 1 x( cot 1 x ) 1 1 2 = cot x (cot x ) 2
⎞ 2 ⎛ 1 = ⎜ cot x ⎟ 4⎠ 4 ⎝ Completing the squares
⎛ 2 ⎞ ⎜⎜ 0, ⎟ 4 ⎟⎠ ⎝
3.
e⎤ ⎡ x The range of f ( x ) ln ⎢ 3 ⎥ is x⎦ ⎣ 4e (1) [0, )
(2)
[–1, )
Sol. Answer (2)
e⎤ ⎡ x f ( x ) ln ⎢ 3 ⎥ x⎦ ⎣ 4e ⎛ ⎛ 1⎞ ⎞ ⎜ ln ⎜ ⎟ , ln ⎟ e ⎝ ⎠ ⎝ ⎠
x 4e
3
[–1, ]
e 1 1 2 2 x e 4e
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Application of Derivatives
Solution of Assignment (Set-2)
Comprehension III Among several applications of maxima and minima is finding the largest term of a sequence. Let < an > be a n x consider f ( x ) on [1, ) sequence. Consider f(x) obtained by replacing x by n e.g., let an n 1 x 1
f '(x)
1 (1 x )2
0 x R.
Hence max. f ( x ) lim f ( x ) 1 x
1.
The largest term of the sequence an
(1)
29 453
n2 n 3 200
is
49 543
(2)
(3)
43 543
(4)
41 451
(3)
1
(4)
1 7
Sol. Answer (2) f (x)
x2 x 3 200
f '( x )
x(400 x 3 ) ( x 3 200)2
f '( x ) 0 if 0 x (400)1/3 f '( x ) 0 if x (400)1/3
i.e., f(x) is maximum at x (400)1/3 Since 7 (400)1/3 8 Hence max term of the given sequence an =
2.
n2 3
n 200
is a7
49 543
The largest term of the sequence an (1)
3 19
n 2
n 10
2 13
(2)
is
Sol. Answer (1) Let an
n 2
n 10
Let f ( x ) f '( x )
x 2
x 10 x 2 10 2 x 2 2
2
( x 10)
=
10 x 2 2
2
( x 10)
=
( x 10)( x 10) ( x 2 10)2
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Solution of Assignment (Set-2)
Application of Derivatives
131
f '( x ) 0 if x 10 and f '( x ) 0 if x 10 i.e., f(x) is max at x 10 since 3 10 4 and a3
9 4 a4 19 26
Hence largest term of the sequence
an 3.
n 2
n 10
is a3
3 19
If f(x) is the function required to find largest term in question 1 then (1) f(x) increase x
(2)
f(x) decrease x
(3) f(x) has a maxima at x = (400)1/3
(4)
f(x) increases on [0, 9]
Sol. Answer (3) From (1), f(x) has maxima at x (400)1/3
SECTION - D Assertion-Reason Type Questions 1.
STATEMENT-1 : Let f(x) = |x2 – 1|, x [–2, 2] f(–2) = f(2) and hence there must be at least one c (–2, 2) so that f (c) = 0. and STATEMENT-2 : f (0) = 0, where f(x) is the function of S1.
Sol. Answer (4) Statement–1 : f (x ) is not differentiable at x = ± 1 hence Rolle’s theorem is not applicable consequently Statement–1 is false.
–2 –1
1
2
From graph f'(0), a = 0 i.e., Statement–2 is True. 2.
STATEMENT-1 : Let f (x) and g (x) be two decreasing function then f (g(x)) must be an increasing function. and STATEMENT-2 : f (g(2)) > f (g(1)); where f and g are two decreasing function.
Sol. Answer (2) Statement–1 : We have f'(x) < 0 and g'(x) < 0 Let h(x) = f[g(x)] h '( x ) f '(g ( x )).g '( x )
h '( x ) 0 hence f (g ( x )) is increasing
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Application of Derivatives
Solution of Assignment (Set-2)
Since g (2) g (1) , as g(x) is a decreasing function f [g (2)] f [g (1)] as f(x) is a decreasing function Hence Statement–2 is true but it is not correct explanation of statement-1. 3.
STATEMENT-1 : f(x) = x3 is a one-one function. and STATEMENT-2 : Any monotonic function is a one-one function.
Sol. Answer (1) Clearly from graph x3 is one-one. Statement–1 is true. By monotonic function we mean either it is monotonic increasing or monotonic decreasing.
y = x3
Any monotonic function is one-one hence Statement–2 is true and correct explanation of Statement–1. 4.
STATEMENT-1 : f(x) = |x – 1| + |x – 2| +|x – 3| has a local minima. and STATEMENT-2 : Any differentiable function f(x) may have a local maxima or minima if f '(x) = 0 at some points.
Sol. Answer (2) Statement–1 : f ( x ) | x 1| | x 2 | | x 3 |
⎧ ⎪ 3 x 6, x 1 ⎪⎪ f ( x ) ⎨ x 4, 1 x 2 ⎪ x, 2x3 ⎪ ⎪⎩ 3 x 6, x 3 f(x) has minima at x = 2.
6 5 4 3
(1, 3)
2
(3, 3)
(2, 2)
1 1
2
3
4
5
If f'(x) = 0 then it is either point of maxima or minima hence Statement–2 is true but not correct explanation.
5.
STATEMENT-1 : Let a function f(x) be satisfying f ' ( x ) f (b ) f (a ) , x (a, b) f(x) is a differentiable ba function on [a, b]. and STATEMENT-2 : According to LMVT if f(x) is differentiable on [a, b] then f ' ( x )
f ( b ) f (a ) ,a x b . (b a)
Sol. Answer (4) Statement–1 : If f(x) is continuous on [a, b] and differentiable on (a, b) then f ( b ) f (a ) , x (a, b ) ba Hence Statement–1 is false. f '( x )
Statement–2 is True. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
6.
Application of Derivatives
133
STATEMENT-1 : Let f(x) = ex and g(x) = e2x f(x) . g(x) be an increasing function. and STATEMENT-2 : Multiplication of two increasing function is an increasing function.
Sol. Answer (3) Statement–1 : f ( x ) e x , g ( x ) e2 x
h( x ) f ( x ).g ( x ) e3 x h '( x ) 3e3 x . h '( x ) 0 x R hence h(x ) is increasing i.e., Statement–1 is True. It is not necessary that multiplication of two increasing function is increasing hence Statement–2 is false. 7.
Consider the function f(x) = 2x3 + 3x2 + 6x. STATEMENT-1 : f (x) > 0. and STATEMENT-2 : f is an increasing function.
Sol. Answer (2)
f ( x ) 2x 3 3 x 2 6 x f '( x ) 6 x 2 6 x 6 = 6( x 2 x 1) 2 ⎡⎛ 1⎞ 3⎤ = 6 ⎢⎜ x ⎟ ⎥ 2⎠ 4 ⎥⎦ ⎢⎣⎝
f'(x) > 0 Statement–1 is true. f(x) is increasing function (but it is not given continuous) hence f(x) may not be differentiable. i.e., Statement–2 is true but not correct explanation.
SECTION - E Matrix-Match Type Questions 1.
Match the greatest value of the function in column I to column II. Column-I 2 (A) y 100 x on [–6, 8]
(B) y = 2 tanx – tan2x on ⎡ 0, ⎞⎟ ⎢ 2⎠ ⎣ 1 x 1 on [0, 1] (C) y tan 1 x (D) y
a2 b2 on (0, 1), a > 0, b > 0 x 1 x
Column-II (p) No greatest value (q) 10 (r)
4
(s) 1
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Application of Derivatives
Solution of Assignment (Set-2)
Sol. Answer A(q), B(s), C(r), D(p) (A) y =
100 x 2 , domain x [ 10, 10]
dy x dx 100 x 2 critical points of y = f(x) are x = 0, ±10 We find that f (0) 10 f (10) 0 f ( 10) 0
ymax = 10 2 (B) y 2 tan x tan x
dy 2 sec 2 x 2 tan x.sec 2 x dx
for maxima–minima of y dy 0 ⇒x dx 4
⎛ d 2y ⎞ 0 ⎜⎜ 2 ⎟⎟ ⎝ dx ⎠ x 4
ymax = 1 1 (C) y tan
0
1 x ,0x1 1 x
1 x 1 1 x
⎛ 1 x ⎞ 0 tan1 ⎜ ⎟ ⎝ 1 x ⎠ 4 y max
(D) y
4
dy a2 b2 a2 b2 2 dx x (1 x )2 x 1 x
for maxima–minima of y dy a 0 ⇒x dx ab
⎛ d 2y ⎞ ⎜⎜ 2 ⎟⎟ ⎝ dx ⎠ x
a ab
0 i.e., it is minima.
hence no maximum value. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
2.
Application of Derivatives
135
⎞ ⎛ Let the function defined in Colomn-I have domain ⎜ 0, ⎟ then match the followings. 2 ⎝ ⎠
Column-I
Column-II
(A) x2 + 2cos x + 2 has
⎛2⎞ (p) Local maximum at cos–1 ⎜ ⎟ ⎝3⎠
(B) 9x – 4 tan x has
(q) Maximum at x
x2 x ⎛1 ⎞ x x x cos sin ⎜ ⎟ (C) has 4 ⎝2 ⎠
(r) No local extremum
1 ⎛1 ⎞ (D) ⎜ x ⎟ cos ( x 3) sin ( x 3) has ⎝2 ⎠
(s) Minima at x = 1
1 2
Sol. Answer A(r), B(p), C(q), D(q, s) (A) f ( x ) x 2 2cos x 2 f '( x ) 2 x 2sin x = 2( x sin x ) For maxima–minima of f(x)
⎛ ⎞ f '( x ) 0 x = sin x no value of x lies in ⎜ 0, ⎟ ⎝ 2⎠ i.e., No extremum. (B) f ( x ) 9 x 4 tan x
f '( x ) 9 4 sec 2 x For maxima–minima of f(x)
f '( x ) 0 sec 2 x
9 4
sec x
3 2
cos x
2 3
⎛2⎞ x cos1 ⎜ ⎟ ⎝3⎠
f ''( x ) 8 sec 2 x tan x 2⎞ ⎛ f '' ⎜ cos1 ⎟ 0 3 ⎝ ⎠ 1 ⎛ 2 ⎞ Hence f(x) attains maxima for x = cos ⎜ ⎟ ⎝3⎠
1 x2 x (C) f ( x ) ⎛⎜ x ⎞⎟ cos x sin x 4 ⎝2 ⎠
f '( x ) sin x x sin x cos x cos x 2 x 1 2 4 =
1⎞ ⎛ x 1⎞ ⎛ sin x 2x 1 x sin x = sin x ⎜ x ⎟ ⎜ ⎟ 2 2 4 ⎝ ⎠ ⎝2 4⎠
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Application of Derivatives
Solution of Assignment (Set-2)
For maxima–minima of f(x)
f '( x ) 0 ⇒ x
1 2
1⎞ 1 ⎛ Now f ''( x ) ⎜ x ⎟ cos x sin x 2⎠ 2 ⎝ 1 1 ⎛ 1⎞ f '' ⎜ ⎟ sin 0 2 2 2 ⎝ ⎠ Hence f(x) attains maximum value at x =
1 2
1 ⎛1 ⎞ (D) f ( x ) ⎜ x ⎟ cos ( x 3) sin ( x 3) 2 ⎝ ⎠ 1⎞ ⎛ f '( x ) ⎜ x ⎟ sin ( x 3) 2⎠ ⎝ For maxima–minima of f(x)
f '( x ) 0 ⇒ x
1 ,1 2
1⎞ ⎛ f ''( x ) 2 ⎜ x ⎟ cos ( x 3) sin ( x 3) 2⎠ ⎝ ⎛ 1⎞ f '' ⎜ ⎟ 0 ⎝2⎠
f '' 1 3.
2 0 2
f(x) attains maximum value at x =
1 2
f(x) attains minima in the neighbourhood of x = 1
Match the followings Column I
Column II 2 –x
(A) The length of interval in which f(x) = x e is monotonic increasing is
(p) 1
(B) If y = kx3 + 3x2 + (2k + 1) x + 1000 is strictly increasing for all values of x, then the least of positive integral value which k can attain, is
(q) 2
(C) If the sum of the squares of intercepts on axes made by a tangent at any point on the curve x2/3 + y2/3 = a2/3 is ak then k is
(r) Irrational number
(D) If
log x attains maximum value at x = k then k is x
(s) Greater than 1 (t) An integer
Sol. Answer A(q, s, t), B(p, t), C(q, s, t), D(r, s) (A) f(x) = x2e–x f’(x) = –xe–x(x – 2) > 0 for x (0, 2) hence the length of interval is 2. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
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Application of Derivatives
137
(B) y = kx3 + 3x2 + (2k + 1)x + 1000 y' = f'(x)= 3kx2 + 6x + (2k + 1) > 0, x R k > 0 and 62 – 12k(2k + 1) < 0 k > 1 we observe that for k = 1, f(x) = (x + 1)3 + 999, which is clearly monotonic increasing. Hence the least of positive integer which k attain is 1. (C) Tangent at point (, ) is
x y 1/3 a 2/3 1/3 4
2
2
Now, a 3 3 3 a k 4
2
k a 3 .a 3 a 2 k a a ⇒k 2
(D) y
log x 1 log x ⇒ y' x x2
The point of maximum is e Which is an irrational number greater than 1. 4.
Match the followings according to the given conditions. Column I
Column II
(A) Let k be the number which can be the slope of a tangent
(p) Less than 1
to the curve x2 + y2 – 2x + 2y + 2xy = 1, then k is (B) f(x) = 2sec x –
3 tan x, then minimum value of f(x) is
(C) If the function f x
ax b
x 1 x 4
has a local minima at
(q) 1 (r) Whole number
(2, –1), then a is (D) Let f x max sin x , cos x x R then minimum value of f(x) is
(s) Irrational number
(t)
1 2
Sol. Answer A(q, r), B(q, r), C(q, r), D(p, s, t) (A) x2 + y2 – 2x – 2y – 2xy = 1
…(1)
(x – y)2 = 2x + 2y + 1 This is a parabolic curve with slope of axis = 1, therefore there can be no tangent with slope = 1 Differentiating (1) w.r.t. x
dy x y 1 dx x y 1 As
dy 1, we have x – y + 1 = x – y – 1 dx
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Application of Derivatives
Solution of Assignment (Set-2)
(B) The minimum value of f x a sec x b tan x is a 2 b 2 Hence minimum of 2sec x 3 tan x is 22 3 1 2
(C) f(2) = –1 f’(2) = 0
2a + b = 2
…(1)
…(2)
b=0
From (1) and (2) 2a = 2
a=1
(D) y = maximum {|sin x|, |cos x|}
gives minimum value
1 ⎛⎞ fmin f ⎜ ⎟ ⎝4⎠ 2 5.
Match the interval of monotonicity of functions in column I to column II. Column-I
Column-II
(A) y = x – ex
(p) (–, )
(B) y log ( x x 2 1)
(q) (0, 3)
(C) y x 4 x x 2
(r) (0, )
(D) y
10
(s) (1, )
3
4x 9x 2 6x
Sol. Answer A(q, r, s), B(p, q, r, s), C(q), D(s) (A) y x e x dy 1 ex dx
⎧ dy ⎪⎪ dx 0, if x 0 ⎨ ⎪ dy 0, if x 0 ⎪⎩ dx
i.e., y is monotonic in (0, 3), (0, ) and (1, )
2 (B) y log x x 1
⎞⎛ ⎞ dy ⎛ 1 2x ⎜ ⎟⎜ 1 ⎟ ⎜ ⎟⎜ dx ⎝ x x 2 1 ⎠⎝ 2 x 2 1 ⎟⎠
1 x2 1
dy 0 x R dx
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Solution of Assignment (Set-2)
Application of Derivatives
139
(C) y x 4 x x 2
4 x x 2 2 x 2 x 2 2(3 x x 2 ) dy 1 2 4x x x (4 2 x ) = dx 4x x2 4x x2 2 4x x 2
dy 0, x (0, 3) dx
(D) y
6.
10 3
4x 9x 2 6x dy 10(12 x 2 18 x 6) dx (4 x 3 9 x 2 6 x )2
dy 60(2 x 2 3 x 1) dx (4 x 3 9 x 2 6 x )2
dy ( x 1)(2 x 1) 60 dx (4 x 3 9 x 2 6 x )2
⎧ 1⎞ ⎛ ⎪ 0, ⎜ , 2 ⎟ (1, ) dy ⎪ ⎝ ⎠ ⎨ dx ⎪ ⎛1 ⎞ 0, ⎜ , 1⎟ ⎪⎩ ⎝2 ⎠
Match the extrema of the functions in column I to column II. Column-I
Column-II
(A) y x 1 x 2
(p) ymin = 0
(B) y = x2e–x
(q) ymin = 1
(C) y x 2 x 2
(r) ymax =
e ax e ax 2 Sol. Answer A(r, s), B(p), C(s), D(q) (D) y
2
(s) ymin = –1
(A) f ( x ) x 1 x 2 x = cos
f ( x ) cos | sin |, sin 0 [0, ]
f ( x )max 2 Now,
f ( x ) cos sin
x 2
f ( x )min 1
(B) y x 2e x
dy x e x (2 x ) dx
at x = 0 sign of
dy changes from negative to positive dx
Hence y = f(x) attains minima at x = 0 and min f(x) = 0 at x = 2 sign of
dy changes from positive to negative dx
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Application of Derivatives
Solution of Assignment (Set-2)
Hence y = f(x) attains maxima at x = 2 and ymax = ymin = 0, ymax =
4 e2
4 e2
(C) y x 2 x 2 , domain x [ 2, 2] x2
dy 2(1 x 2 ) dx 2 x2
dy dx
dy ⎪⎧ 0, 1 x 1 ⎨ dx ⎩⎪ 0, x ( 2, 1) (1, 2)
2 x2
2 x2
In the neighbourhood of x = –1,
In the neighbourhood of x = 1,
dy changes from negative to positive hence y attains minima at x = –1. dx dy changes from positive to negative hence y attains maxima at x = 1. dx
ymin = –1, ymax = 1 (D) y
eax e ax 2
From A.M. – G. M. inequally, we have
eax e ax 1 2 ymin = 1. 7.
Match the maxima of the function in column I to column II. Column-I (A)
x2e–x
(B)
2
Column-II (p) –1
4x
(q) 0
x 4
(C) –x2 (x – 2)2/5 (D)
(r) f (2)
14
(s) 1
4
x 8x 2 2
Sol. Answer A(r), B(r, s), C(q), D(p) (A) y x 2e x
– –
–
+ 0
2
dy x 2 x 2 e x 2 xe x = x(2 - x)e-x e ( x 2x ) dx
We observe that f'(x) changes its sign from +ve to –ve in the neighbourhood of x = 2 and from –ve to +ve in the neighbourhood of x = 0 f(2) is maxima. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
(B) y
4x
y
2
x 4
Application of Derivatives
141
4 x
4 x
A.M. – G.M. inequally x
4 4, x 0 x
4 x
4 x
1
ymax = 1 (C) y x 2 ( x 2)2/5 ⎡ x 5 x 10 ⎤ 2 x (6 x 10) 2 dy x2 = 2 x( x 2)2/5 = 2 x ⎢ 3/5 ⎥ 3/5 dx 5 ( x 2) 5( x 2)3/5 ⎣ 5( x 2) ⎦
dy 4 x (3 x 5) dx 5( x 2)3/5
Critical points of the given function are 0,
5 ,2 3
f (0) 0 f (2) 0 1/5
25 ⎛ 1 ⎞ ⎛5⎞ f⎜ ⎟ ⎜ ⎟ 3 ⎝9⎠ ⎝3⎠ y max 0 .
(D) y
14 4
x 8x 2 2
y is maximum when x4 – 8x2 + 2 is minimum Let z = x4 – 8x2 + 2
dz 4 x 3 16 x dx
For maxima–minima of z dz 0 ⇒ x 0, 2 dx
⎛ d 2z ⎞ 2 ⎜⎜ 2 ⎟⎟ 12 x 16 dx ⎝ ⎠ ⎛ d 2z ⎞ 0 ⎜⎜ 2 ⎟⎟ ⎝ dx ⎠ x 2 Hence ymax =
z is minimum for x = ± 2
14 = –1 16 32 2
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Application of Derivatives
Solution of Assignment (Set-2)
SECTION - F Integer Answer Type Questions 1.
The ratio of absolute maxima and minima of f x
x2 x 1 , x R is___________ . x2 x 1
Sol. Answer (9) The absolute maxima = 3 and the absolute minima = The required ratio = [we know
2.
1 3
3 9 1/ 3
1 x2 x 1 3] 3 x2 x 1
Consider the function f x x of interval of increase, then
1 ⎛ 1 9⎞ , x ⎜ , ⎟ . If is the length of interval of decrease and be the length ⎝ 2 2⎠ x
is ________ .
Sol. Answer (7) f x x
1
3.
1 ⎛1 9⎞ ⎛1 ⎞ ⎛ 9⎞ , x ⎜ , ⎟ is decreasing in ⎜ , 1⎟ and increasing in ⎜ 1, ⎟ , x ⎝2 2⎠ ⎝2 ⎠ ⎝ 2⎠
1 1 , 2 2
9 7 1 2 2
7
If the function f x
1 is downward concave in (, ) then [ – ] is ______ ( [.] is greatest integer function.) 1 x2
Sol. Answer (1)
1⎞ ⎛ 6 ⎜ x2 ⎟ 3 ⎝ ⎠ 0 in ⎛ 1 , 1 ⎞ . Hence 1 , 1 f "x ⎜ ⎟ 3 3⎠ ⎝ 3 3 1 x 2 3
⎛ 1 ⎞ ⎤ ⎡ 2 ⎤ ⎜ ⎟⎥ ⎢ ⎥ 1 ⎣ 3 ⎝ 3 ⎠⎦ ⎣ 3 ⎦ ⎡ 1
⎢ 4.
The slope of the tangent to the curve (y – x5)2 = x(1 + x2)2 at the point (1, 3) is
[JEE(Advanced)-2014]
Sol. Answer (8) (y – x5)2 = x(1 + x2)2
⎛ dy ⎞ 2( y x 5 ) ⎜ 5 x 4 ⎟ = (1 + x2)2 + 2x(1 + x2).2x ⎝ dx ⎠ x = 1, y = 3
⎛ dy ⎞ 2(3 1) ⎜ 5 ⎟ (1 1)2 4(1 1) dx ⎝ ⎠
⎛ dy ⎞ 4⎜ 5 ⎟ 12 ⎝ dx ⎠
dy dy 5 3 8 dx dx
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Solution of Assignment (Set-2)
Application of Derivatives
143
SECTION - G Multiple True-False Type Questions 1.
STATEMENT-1 : If two sides of a triangle are given, then its area will be maximum if the angle between the given sides be
2
STATEMENT-2 : The function f x
1 is increases in the interval (0, ). 1 x2
STATEMENT-3 : The length of the subnormal of the curve y3 = 12ax at y = 1 is 4a. (1) T F T
(2)
TTT
(3)
FFF
(4)
FFT
(4)
FFT
Sol. Answer (1) STATEMENT-1
True,
Area =
1 1 ab sin C ab sin C sin C 1 2 2
Area maximum = STATEMENT-2
False,
y f x
1 ⎡ ⎤ ab sin C 1 at C ⎥ 2 ⎢⎣ 2⎦
1 1 x2
f(x) decreases in (0, ) STATEMENT-3 2.
True,
⎛ 4a ⎞ subnormal = yy ' y . ⎜ 2 ⎟ 4a, at y = 1 ⎝y ⎠
STATEMENT-1 : The maximum value of 3sinx + 4cos x is 7. STATEMENT-2 : In the interval [–1, 1] the greatest value of f x
1 x is . 2 5 4 x x
STATEMENT-3 : The greatest value of 4sin2x + 3cos2x is 4. (1) T F T
(2)
TTT
(3)
FFF
Sol. Answer (4) STATEMENT-1
False, Maximum value of 3 sin x 4cos x 32 42 5
STATEMENT-2
False, f x
x x2 4 , f 'x 0 for x(–2, 2) 2 4xx x 2 x 4 2
f(x) is increasing in [–1, 1] fmax f 1
STATEMENT-3
True,
1 6
f(x) = 4sin2x + 3cos2x = 3 + sin2x 4.
Hence f(x)max = 4 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
144
Application of Derivatives
Solution of Assignment (Set-2)
SECTION - H Aakash Challengers Questions 1.
The least natural number a for which x
Sol. x
a x2
a x2
2 x (0, ) is ______.
2
x 3 2x 2 a 0
Let f ( x ) x 3 2x 2 a f '( x ) 3 x 2 4 x
f '( x ) x (3 x 4)
For maximum or minimum value of f ( x ) 0 ⇒ x
4 ,0 3
f ''( x ) 6 x 4
⎛4⎞ f '' ⎜ ⎟ = 4 > 0 ⎝3⎠
f(x) attains minimum value at x =
4 3
32 ⎛4⎞ and (f(x))min = f ⎜ ⎟ a 3 27 ⎝ ⎠ for f ( x ) 0
⎛4⎞ f⎜ ⎟0 ⎝3⎠
2.
a
32 , i.e., least natural number value of a = 2. 27
x2 y 2 1 meets the co-ordinate axes in the points A and B such 8 18 that the area of the OAB is least, then the point P is of the form (m, n) where m + n + 10 is
Any tangent at a point p(x, y) to the ellipse
Tangent at P is x.2 2 cos y .3 2 sin 1 8 18
Sol.
x 2 2 sec
y 3 2 sec
1
A : (2 2 sec ,0), B : (0, 3 2 cosec) ar ( OAB )
12 1 2 2 sec 3 2cosec = sin 2 2
ar(OAB) is least on sin2 = 1 2
2
Y B(0, 3 2 cos) P(2 2 cos, 3 2 sin) X'
X A(2 2 sin, 0)
O
Y'
4
⎛ 1 1 ⎞ P ⎜2 2 ,3 2 ⎟ (2, 3) ( m, n ) 2 2⎠ ⎝
m + n + 10 = 2 + 3 + 10 = 15 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
3.
Application of Derivatives
145
If the tangent at P(1, 1) on the curve y2 = x(2 – x)2 meets the curve again at Q, then the points is of the form ⎛ 3a a ⎞ 2 2 ⎜ b , 2b ⎟ , where a + b is ⎝ ⎠
Sol. Given equation is y2 = x(2 –x)2 dy (2 x )2 2 x (2 x ) 2y dx = 4 – 4x + x2 – 4x + 2x2 2
Y
…(i)
P(1, 1) X'
Q O
⎛ 3a a ⎞ ⎜ , ⎟ ⎝ b 2b ⎠ X
= 3x – 8x + 4
dy 3 x 2 8 x 4 dx 2y
dy 384 1 dx m (1,1) 2.1 2
Tangent at P is y –1 = –
Y'
1 (x – 1) 2
2y – 2 = – x + 1 x + 2y = 3 From (1) & (2), we get 2
⎛3x⎞ 2 ⎜ 2 ⎟ x( x – 2) ⎝ ⎠
(3 – x)2 = 4x(x – 2)2 4x3 – 17x2 + 22x – 9 = 0 (x – 1) (4x2 – 13x + 9) = 0 (x – 1) (x – 1) (4x – 9) = 0 x = 1,
9 4
3x y 2
9 4 3 2 8
3
⎛ 9 3 ⎞ ⎛ 3a a ⎞ Q⎜ , ⎟⎜ , ⎟ ⎝ 4 8 ⎠ ⎝ b 2b ⎠
3a = 9, b = 4 a = 3, b = 4 a2 + b2 = 32 + 42 = 25 4.
The curve y = ax3 + bx2 + cx is inclined at 45° to x-axis at (0, 0) but it touches x-axis at (1, 0), then a + b + c + 10 is
Sol. Given curve is y = ax3 + bx2 + cx
dy 3ax 2 2bx c dx
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
146
Application of Derivatives
Solution of Assignment (Set-2)
Given tan 45° = 1 3ax2 + 2bx + c = 1 ⎛ dy ⎞ ⇒ c 1 Now, ⎜ ⎟ ⎝ dx ⎠(0,0) ⎛ dy ⎞ ⇒ 3a 2b c 0 3a + 2b + 1 = 0 ⎜ dx ⎟ ⎝ ⎠(1, 0)
Also, a + b + c = 0 3a + 2b + 1 = 0 a+b+1=0 2a + 2b + 2 = 0 –
– –
a–1=0 a=1b=–2 a + b + c + 10 = 1 – 2 + 1 + 10 = 10 5.
Let f(x) = 2x3 + ax2 + bx – 3cos2x is an increasing function for all x R such that ma2 + nb + 18 < 0 then the value of m + n + 7 is
Sol. Given f(x) = 2x3 + ax2 + bx – 3cos2x f'(x) = 6x2 + 2ax + b + 3sin2x > 0 6x2 + 2ax + b – 3 > 0 (∵ sin2x –1) 6x2 + 2ax + (b – 3) > 0 D 0
…(A)
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
Solution of Assignment (Set-2)
Application of Derivatives
151
Given f"(x) > 0 f'(x) is increase f'{(cotx + 1)2 + 1} > f'(1). (∵ f'(1) = 0) g'(x) < 0 if (cotx + 1) > 0
from (A)
⎛ 3 ⎞ x ⎜ 0, ⎟ ⎝ 4 ⎠
⎛ 3 ⎞ (a, b ) ⎜ 0, ⎟ ⎝ 4 ⎠
ab
3 0 4 4 4
15. If the equation x3 – 3x + k = 0 has two distinct roots is (0, 1), then the value of k is Sol. Let and such that be the roots of f(x) = x3 – 3x + k = 0 Then 0 < < < 1 and f() = 0 = f() Also f(x) is differentiable in (0, 1) and continuous in [0, 1] and f() = 0 = f() Thus, all the conditions of Rolle’s theorem are satisfied. Now, we have to show that a point. c (, ) such that f'(c) = 0 3c2 – 3c = 0 c = 0, 1. (0, 1) Thus k = 16. Let 'f' be a real function whose derivative upto third order exist and for same pair a, b R such that a < b, ⎛ f (a ) f (a ) f (a ) ⎞ f (c ) is log ⎜ ⎟ a b then there exist c (a, b) for which f (c ) ⎝ f (b ) f (b ) f (b ) ⎠
Sol. Let (x) = (f(x) + f'(x) + f"(x))e–x (x) is differential on (a, b) and continuous on [a, b]. Also,
(a) f (a) f (a) f (a) e b a (b) f (b) f (b) f (b) = ea–b × eb–a = ea–b + b – a = e0 = 1
(a) = (b) Thus, Rolle’s theorem is satisfied Now, there exists a point c (a, b) such that 'c) = 0 (f'(c)+ f"(c) + f'''(c))e–c – (f(c) + f'(c) + f''(c))e–C = 0 f'(c) + f"(c) + f'''(c) – f(c) – f'(c) – f"(c) = 0 f'''(c) = f(c)
f (c ) 1 f (c )
Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456
152
Application of Derivatives
Solution of Assignment (Set-2)
17. A function f : R R is defined as f(x) = |x|m |x – 1|n x R, m, n N. Then the maximum value of the functions is ______
Sol.
⎧( 1)m n x m ( x 1)n ⎪⎪ f ( x ) ⎨ ( 1)n x m ( x 1)n ⎪ x m ( x 1)n ⎪⎩
,
x0
, 0 x 1 ,
x 1
Let g(x) = xm(x – 1)n g'(x) = mxm–1(x – 1)n + nxm(x – 1)n – 1 = xm–1(x – 1)n–1 {mx – m + nx} = 0 Now, f'(x) = 0 g'(x) = 0
m mn Now, f(0) = 0, f(1) = 1
x 0,
mn n m mn n m ( 1)n ⎛ m ⎞ n 1 And f ⎜ = ( 1) mn ⎟ mn m n ( ) m n m n ( ) ⎝ ⎠ Hence, the maximum value =
mn n m (m n )m n
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