Solution

CUMULATIVE TEST-2 (CT-2) (JEE ADVANCE PATTERN)

TARGET : JEE (MAIN+ADVANCED) 2015

COURSE : VIJETA (JP)

DATE : 06-07-2014

HINTS & SOLUTIONS ¼la d sr ,oagy ½ g (f (x)) . f(x) = 1 g (f (x)) =

Paper-1 Part-I Mathematics 1.

g (f (x)) =

Which of the following ……………….



1 x .(1  n x)

fuEufy f[kr Q y uksad s;qXeksa……………….

1

 g(x)

and f(x) = g(x)

g (x) =

fod Yi (B) esa, f rFkk g d k izkUr l eku gSA



x

g  (f(g(x))) =

Sol. In (B) option, Domain of f and g is same

1 f  (x)

g(x)

.(1  n(g(x)))

1 x (1  ng(x))

vkSj f(x) = g(x) 2.

5.

The value of tan 100º + 4 sin ………………. tan 100º + 4 sin 100º

d k ……………….

Sol. tan 100º + 4 sin 100º

 a11 a12  ;fn vkO;wg a21 a22 a31 a32

sin100  2 sin 200 = cos100 =

sin100  sin200  sin 200 cos100

3.

a23 =

cos 50  cos 70 2 sin 60.sin10 = =–  sin10  sin10

J s.kh 1 +

3 5 7 + + 3 + . .. .... .. ... .... .. . . 2 5 5 5

Sol . S n = 1 +

3 5 7 (2n – 1) + + 3 + .. .. .. + 5 52 5 5n–1

cofactor of (3, 2)th entry of given matrix |A|

– 



4 2 2 2 (2n – 1) Sn = 1 + + 2 +…. .+ n–1 – 5 5 5 5 5n

3 (4n  3) – 2 2.5n

Let f(x) = xx ; x (1, ) and g(x) be inverse ……………….

 a23 =

1 1 0

1

10

Alternate solution

10 a23 = –1

fod Yi gy %

 1 2 1 1 0 0    0 2 1 0 1 0  0 0 5 0 0 1

1 1 3 5 (2n – 3) (2n – 1) Sn = + + 3 +.. .. . + + 5 5 52 5 5n–1 5n Subtrac ting ?kVkus ij

4.

a13   a23  ………………. a33 

(3, 2)okavo; o d k lg[k.M |A|

3

3 5 The s um of t he s eries 1 + + + ………………. 5 52

=

a13   a23  is the inverse ………………. a33 

Sol. |A| = 10

2 sin150 cos 50  sin 20 = cos100 =

 a11 a12  If matrix a21 a22  a31 a32

~

0   1 2 1 1 0   0 1 1/ 2 0 1/ 2 0  0 0 1 0 0 1/ 5 

1/ 5  1 2 0 1 0   ~ 0 1 0 0 1/ 2 1/10   0 0 1 0 0 1/ 5 

ekuk f(x) = xx ; x (1, ) rFkk……………………………….... Sol. we have fn;k x;k gS f(g(x)) = g(x) g(x) = x



10a23 = – 1

u% g (f (x) ) = x also iq Corporate Office (New Campus): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555

JPCT2060714C0-1



6.

If (tan–1x)2 + (cot –1x)2 ……………….

;fn (tan–1x)2 + (cot –1x)2 =

52 , rc ………………. 8

5 2  –1  Sol. (tan x + cot x) – 2tan x  – tan x  = 8 2  –1

2

–1

10.

If the product (sin1º) (sin3º) (sin5º) (sin7º)...........(sin89º) ……………….

;fn xq.kuQ y (sin1º) (sin3º) (sin5º) ……………….

–1

Sol.

(sin1º) (sin3º) (sin5º) (sin7º)...........(sin89º) =

3 2   tan–1(x) – =0  8 2

[sin = cos(90 – )]

2(tan–1x)2 – 2 

tan–1x = –

1 2n

1

(cos1º) (cos3º) (cos5º) ...........(cos89º) =

 4

cos1º cos 2º cos 3º........cos 88º cos89º cos 2º cos 4º cos 6º........cos88º



x=–1

2n

 7.

If the function f : [1, )  [1, ) is defined ……………….

(2cos1º sin1º )(2cos 2º sin2º )........(2cos 44º sin 44º )

;fn d ksbZQ y u f : [1, )  [1, ) bl izd kj ……………….

244 cos 2º cos 4º cos 6º........cos88º

y = f(x) = 2x(x – 1)

Sol.

 x(x – 1) = log2y

 1  sin2º sin 4º........sin88º ×  sin 45º    44 2  2 cos 2º cos 4º........cos88º 

 x2 – x – log2 y = 0  x=

1  1  4log2 y

1

2

2

x1 x=

1  1  4log2 y

 f–1(x) =

1 

1

1/ 2

2



89 / 2

2

n =

89 2

The curve y  exy + x = 0 has a vertical ……………….

11.

2

2

1 ×

44

fuEu esalsft l fcUnqij oØ y  exy + x = 0 d s……….

1 (1  1  4log2 x ) 2

Sol. y  exy + x = 0

{k vod y u y d slkis

Differentiating w.r.t. to y

8.

The value of k for which roots ………………. k d seku ft ud sfy , l ehd j.k ………………. Sol. Let ekuk f(x) = 4x2 – 4kx + k2 + 3k – 66 (i) f(2) > 0  k2 – 5k – 50 > 0 (k + 5) (k – 10) > 0 k  (– , –5)  (10, )

1 – exy

d jusij

 dx  dx 0  . y  x   dy  dy

dx =0 dy 1 – xexy = 0 xexy = 1

x=1,y=0

Point is fcUnq (1, 0) (ii)

–

(–4k) >2  k>4 2(4) 2

12.

If

3 2 3 2 3 2 lim a(2x – x )  b(x  5x – 1) – c(3x  x ) ….

x 

2

(iii) D 0  16k – 4.4 (k + 3k – 66)  0 – 16.3 (k – 22)  0  k  22 Hence vr% k  (10, 22]

a(5x 4 – x) – bx 4  c (4x 4  1)  2x2  5x

;fn lim ………………………….. x 

Sol. For limit to exist and equal to 1 coefficient of x4 in denominator = 0

Number of values of x ………………. lehd j.k |2x – 3| + ………………. Sol. Let ekuk f(x) = |2x – 3| + |2x + 3| g(x) = – |x| + 6 9.

r

r

(as degree of x in D > N ) Now degree of x in Dr is 2 and degree of x in Nr is 3.  coefficient of x3 in Nr = 0 otherwise L  1 and will be infinity and

coefficient of x 2 in Nr coefficient of x 2 in Dr 4

=1

r

Now coefficient of x in D = 5a – b + 4c = 0 .........(1) 3

r

 coefficient of x in N = 2a + b – 3c = 0  Only one solution at x = 0



d soy ,d gy x = 0 ij

.........(2)

–a  5b – c = 1 ;  a - 5b + c + 2 = 0 ..........(3) 2

On solving (1), (2) and (3)

Corporate Office (New Campus): CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555

JPCT2060714C0-2

–2 46 14 ,b= and c = 109 109 109

a=

 2 tan 

=

58 p  a+b+c= =  p + q = 167 109 q

.

 2n =  tan tan

2n tan

lhek d sfo|eku gksusrFkk bld k eku 1 gksusd sfy ,

Hindi.

f(n,) =

gj esax4 d k xq.kkad = 0 (pawfd va'k esax d h ?kkr > Nr) vc gj es ax d h ?kkr 2 gSrFkk va'k esax d h ?kkr 3 gSA. 

va'k esax 2 d k xq.kkad gj esax 2 d k xq.kkad



=1 The locus of point of intersection ……………….

15.

y2 = 4x

3

va'k esax d k xq.kkad = 2a + b – 3c = 0 .........(2)

Sol.

–a  5b – c = 1 ;  a - 5b + c + 2 = 0 ..........(3) 2 (1), (2) rFkk (3) gy d jusij

d h mu Li'kZjs [kkvksad sizfrPNsn ……………….

m2 2 m1



a=

tan >  as  0

 0

vc gj es ax4 d k xq.kkad = 5a – b + 4c = 0 .........(1) 

as n 

   lim   = 0 asD;ksafd  tan 

va'k esax3 d k xq.kkad = 0 vU;Fkk L  1 rFkk vuUr gksxkA

rFkk

   2 tan 3 2 tan n 2 ..... 2 22 .    tan 2 tan n–1 tan 2 2 2

2 tan

2 tan

–2 46 14 ,b= rFkk c = 109 109 109

y = mx +

1 m

k = mh +

1 m

km = m2h + 1

58 p  a+b+c= =  p + q = 167 109 q

m1 2

m h – km + 1 = 0

m2 13. The coordinates of the feet of ……………….

fcUnq(14, 7) lsy2 – 16x – 8y = 0 ………………. 2

Sol. Given curve is y – 16x – 8y = 0

...(1)

Let P  (14, 7)

m1m2 =

Equation (1) is (y – 4)2 = 16(x + 1) Equation of normal at (–1 + 4t2, 4 + 8t) is tx + y – 4t3 – 7t – 4 = 0 passes through, (14, 7) 

3 1 , foot are (0, 0) (3, – 4) (8, 16) 2 2 fn;k x;k oØ y2 – 16x – 8y = 0 ...(1)

2m12  2  k2

ekukP  (14, 7)

9h

lehd j.k(1) gS(y – 4)2 = 16(x + 1) (–1 + 4t2, 4 + 8t) ij

vfHky Ec d h lehd j.ktx + y – 4t3 – 7t – 4

3 1 , ikn gksxs(0, 0) (3, – 4) (8, 16) 2 2

  t = –1,

n

14.

If f(n,) =



 1– tan r 1

2

   then the value ……………. 2r 

16.

Sol.



r 1

2

1 h

If pth, qth, rth terms of an A.P. ……………….

Sol. Givenfn;k x;k gS, k =

   2r 

a  (q  1)d a  (r  1)d = a  (p  1)d a  (q  1)d

=

a  (q  1) d  a  (r  1)d a  (p  1)d  a  (q  1)d

=

(q  r)d (q  r) = (p  q)d p  q



 1– tan



;fn fd lh l ekUrj Js