# Jee 2014 Booklet1 Hwt Solutions Quadratic Equations & Inequations

August 28, 2017 | Author: varunkohliin | Category: Quadratic Equation, Equations, Elementary Mathematics, Mathematical Analysis, Mathematical Concepts

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Jee 2014 Booklet1 Hwt Solutions Quadratic Equations & Inequations...

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Vidyamandir Classes

Aggarwal Corporate Heights, 3rd Floor, Plot No. A - 7, Netaji Subhash Place, Pitam Pura, Delhi - 110034 Phone: 011-45221190-93.

Solutions to Home Work Test/Mathematics Quadratic Equations & Inequations

HWT - 1

3

2.

 2  2  3   3          

27  a

3

 2 a

3

 3   2   3   3     3      a   a  a    2 a  

18 a 2  18a  27 2a 2

18a  27

As a < 0, 18a – 27 < 0 and 2a2 > 0

2a 2

0

So, answer is option (D). 3.

a  b  a

and

ab = b

a = 1 and b  2

2

So, the equation is x  x  2  0 Least value 

8.

  

10.

 D 9  4a 4

b a

b q  2h  a p

and

  h      h       2 h 

or

h

q p

1b q     2 a p

Let  be the common root of x 2  px  qr  0 and x 2  qx  rp  0 . Then and

 2  p  qr  0

. . . .(i)

2

  q  rp  0

. . . .(ii)

Subtracting (ii) from (i), we get :

 p  q  r q  p  0

 p  q   r   0

or

 r

(Since p  q).

Similarly, p and q are other two common roots.  Product of common roots = pqr

Let

f  x    x  a1    x  a2   . . . .   x  an  2

2

2

 nx 2  2  a1  a2  . . . .  an  x  a12  a22  . . . .  an 2

f (x) assumes least value when x 

HWT - 2

  2  a1  a2  . . . .  an  

2n

a  a  . . . .  an  1 2 n

1

HWT-Solutions/Mathematics

Vidyamandir Classes 3.

ax 2  2cx  b  0

ax 2  2bx  c  0 have a common root if

and

 a  2b    a  2c   .  2c  c    2b  b     a  c    a  b 

8.

or

 2ab  2ac  .  2c 2  2b2    ac  ab 2

or

4a  c  b   a 2

(Since c  b)

or

4c  4b  a

or

2

4a  c  b 

or

2

 c  b   a 2  c  b 2

a  4b  4c  0

Let  be a root of x 2  x  m  0 . Then 2 is a root of x 2  3 x  2m  0

 2   m  0

and

4 2  6  2m  0

2

or 2  2  2m  0 Subtracting (ii) from (i), we get :   m  0 or  = –m

2 2  3  m  0

or

. . . .(ii)

. . . .(i)

Putting value of  in (i), we get :

2m 2  4m  0 10.

m  0,  2

or

,  are roots of the equation  x  a  x  b   c  0 or

x 2   a  b  x  ab  c  0

   ab

. . . .(i)

and

  ab  c

. . . .(ii)

Let p, q be roots of the equation  x  c    x  c     c or

x 2      2c  x  c 2  c     1    0

p  q      2c

. . . .(iii)

and

pq  c 2  c     1  

. . . .(iv)

Putting value of  +  from (i) into (iii), (iv) and value of  from (ii) into (iv), we get : p  q  a  b  2c   a  c    b  c 

pq  c 2  c  a  b  1  ab  c

and

  a  c  . b  c 

p  a  c and q  b  c

Quadratic Equations & Inequations 5. or

7.

2e 2 log k  1  31

or

2e 2 log k  32

k 2  16

or

k =  4.

HWT - 3 e 2 log k  16

or

x2  2 x  a

gives  y  1 x 2   4 y  2  x   3ay  a   0 x 2  4 x  3a Since x is real D  0 ,

y

 4 y  2 2  4  y  1 3ay  a   0

4 y

and

 4a  4 2  4  4  3a 1  a   0

2

 4 y  1  a  y  1 3 y  1  0

or

 4  3a  y 2   4a  4  y  1  a   0

Since

yR

or

a

4 3

and

16 a 2  2a  1  4 3a 2  7 a  4  0

a

4 3

and

4a 2  4a  0

4  3a  0

 

or

2

HWT-Solutions/Mathematics

Vidyamandir Classes

8.

a

4 3

and

a  a  1  0

a

4 3

and

a 0 , 1

a 0 , 1

Let roots be   p  qi and   p  qi (Roots will exist in conjugate-pair) 

10.

   

p2  q2

Now,

c    p  qi   p  qi   p 2  q 2  a

Since

a  c,

c 1 a

or

p2  q2  1

or

   

p2  q2  1

Determinant of f  x   D1  b 2  4ac Determinant of g  x   D2  b 2  4ac

D1  D2  2b 2  0  

Atleast one of D1 and D2 ≥ 0

f  x  . g  x   0 has at least two real roots.

Let

HWT - 4

f  x    x  a  x  b    x  b  x  c    x  c  x  a  f  a    a  b  a  c  , f  b    b  c  b  a  , f  c    c  a  c  b 

Since f (x) is symmetric with respect to a, b and c. Let us assume that a < b < c. Then f  a   0 , f  b   0 and f  c   0 .  

f  a  . f b   0

and

f c  . f b   0

f (x) has one root in x   a, b  and other root in x   b, c 

f (x) has real roots.

HWT - 5

a b   1 gives a  x  b   b  x  a    x  a  x  b  xa xb

or

x 2  2  a  b  x  3ab  0

Since roots are equal in magnitude but opposite in sign, sum of roots = 0 = 2(a + b) 2.

a+b=0

         2        Since

 ,  are roots of x 2  px  r  0 ,      p and   r

         2  p  r

Since  is a root of x 2  px  q  0,  2  p  q  0 or  2  p  q 

3.

          q  r

Let  ,  be roots of x 2  px  q  0 . Then k , k  can be taken as roots of x 2  lx  m  0 

    p

and

  q and k      

and

k 2  m

From above equations we get : 2

k2 

m    2    2 q  p  p

p2m  l 2q

3

HWT-Solutions/Mathematics

Vidyamandir Classes Quadratic Equations & Inequations 1.

Discriminant of x 2  b1 x  c1  D1  b12  4c1

HWT - 6

Discriminant of x 2  b2 x  c2  D2  b22  4c2

 

Consider  b1  b2   b12  b22  2b1b2  b12  b22  2  2  c1  c2    b12  4c1  b22  4c2  D1  D2 2

Since  b1  b2   0 , D1  D2  0 2

 2.

Atleast one of D1 and D2 is ≥ 0.

Atleast one of the equations has real roots.

Consider x 2  4 x  3   x  1 x  3

x 2  4 x  3  0 for x    ,  3  1,   and x 2  4 x  3  0 for x   3,  1

Let us take x    ,  3  1,   .

Then x 2  4 x  3   2 x  5   0 gives x 2  4 x  3   2 x  5   0

x2  6 x  8  0

or

6  2  2 ,  4 2 x = –4

Reject x = –2 Now take x   3,  1

x

So,

Then x 2  4 x  3   2 x  5   0 gives  x 2  4 x  3  2 x  5  0

5.

or

 x2  2 x  2  0

or

x2  2 x  2  0

Reject

x  1  3

So,

x  1  3

y or

x

or

2  2 3  1  3 2

 x  b  x  c  Gives y  x  a    x 2   b  c  x  bc    x  a x 2   b  c  y  x  bc  ay  0

Since

x is real, D  0

 b  c  y 2  4 1 bc  ay   0 Since

y 2  2  b  c  2a  y   b  c   0 y  R D≤ 0

or

 2  b  c  2a   4  b  c   0  2c  2a  2b  2a   0

b 2  c 2  2bc  y 2  2by  2cy  4ay   0  

or

2

or

2

2

 b  c  2a 2   b  c 2  0  c  a  b  a   0

or or

So one of the expressions  c  a  and  b  a  is less than equal to zero and other one is greater than equal to zero. It is possible if c  a  b or b  a  c 9.

2x 1 2 x3  3 x 2  x

 0 gives

2x 1 0 x  x  1 2 x  1

+

–1

+

–1/2

0

+

1/2

 1  1  Solution : x    ,  1    , 0    ,    2  2 

3

x

HWT - 7

. 2  x 1

Consider x  2 The equation becomes 3 x .  2  x   1 or or

 x  2  . 3 x  1

 x  2  3x

From the graph we get a solution of the equation.

4

HWT-Solutions/Mathematics

Vidyamandir Classes Now, consider 2  x  0 The equation becomes 3 x .  x  2   1

or

x  2  3x

We get another solution of the equation. Similarly, there exists a solution when 0  x  2 and there exists a solution when x > 2. So, there are 4 solutions. 4.

See that x = 1 is a solution of the given quadratic equation. Hence, (D) is the correct option.

5.

x 4  4 x  1  0 gives x 4  4 x  1 From the figure it is clear that the given equation has 2 positive real roots.

6.

x 4  4 x  1  0 gives x 4  4 x  1 From the graph it is clear that the given equation has one negative real root.

7.

From the above graph we see that there are 2 real roots for the equation x 4  4 x  1  0 . The equation has 4 roots. Hence remaining 2 roots are complex roots.

8.

 a  b 2 x 2  2  a  b  2c  x  1  0 . Let us check D. D  4  a  b  2c   4  a  b   4  2b  2c  2a  2c    16  b  c  a  c   2

As

2

a  c  b,  b  c   0 and  a  c   0 

D 0, then f (0) > 0 6.

The equation x 2  ax  b  0 has equal roots which are x 

x 2  cx  d  0 

a 2 ac   d  0 or a 2  2ac  4d  0 4 2

3c  0

or c < 0.

a a . According to the question x  2 2

is a root of equation

. . . .(i)

We also have D = 0 for the equation x 2  ax  b  0  a 2  4b  0 or a 2  4b Using (ii) in (i) we get : 4  b  d   2ac  0

. . . .(ii)

or ac  2  b  d 