Most-important-100-maths-solved-questions-for-jee-main-2019.pdf

Question 1: 100

 1  x 

50

The coefficient of x in the expansion of

r

is

r 0

 a  100C50  b  101C52  c  100C51  d  101C51 Solution: (d) 100

Consider,

 1  x 

r

 1  1  x   1  x   ...  1  x  2

100

r 0

 1  x 101  1     1  x  1  

1 101 1  x   1 x 50

Now, coefficient of x in

100

 1  x 

r

= coefficient of x51 in 1  x 

101

r 0

=

101

C51

Question 2: The greatest value of the term independent of x in the expansion of  x sin   x 1 cos   ,   R 10

is

 a  25 b c

10!

 5!

2

10! 2   5! 5

2

 d  None of these

Solution: (c) Let (r + 1)th term be independent of x. Now, Tr 1  10Cr  x sin  

10  r

 10Cr x102 r  sin  

10 r

 cos      x 

 cos  

r

r

This is independent of x when the power of x will be 0. 10  2r  0 r 5

Substituting r = 5 in Tr 1  10Cr x102 r  sin  

10 r

T6  C5  x sin  

10 5

10

 cos      x 

 10C5  sin    cos   5

5

5

 sin 2   C5    2   10C5 25  Maximum value of sin 2 is 1 5

10



10! 2  5! 5

2

Question 3: The smaller of 99100 + 100100 and 101100, is (a) 99100 + 100100 (b) Both are equal (c) 101100 (d) None of these Solution: (a)

 cos  

r

, we get

99100  100100  101100  100  1

100

 100  1

100

 100100

100 100   100  1  100  1   100100   3 99  2  100C1100  100C3 100   ...  100C99 100    100100  0   100 100 100  99  100  101  100

 99100  100100  101100

Question 4: Let a1, a2, a3,.... be in harmonic progression with a1 = 10 and a20 = 50. The least positive integer n for which an < 0 is (a) 25 (b) 23 (c) 24 (d) 21 Solution: (a) If a1, a2, a3,.... are in harmonic progression, then First term of AP,

1 1  a1 10

20th term of AP,

1 1  a20 50

1 1 1 , , .... are in AP. a1 a2 a3

1 1  19d  10 50 4 d  19  50 We have to find the least positive integer n for which an < 0 1    n  1 d  0 10 1  4     n  1  0 10  19  50  95   n  1  4  n  24.75 

Question 5: The sum of the first 10 term of the sequence 0.6, 0.66, 0.666,…, is

2  1 (a)  89    27   10 

  

10

1  1  89    27   10 

10

  

2  1 (c)  89    27   10 

10

  

2 1 (d)  89     3  10 

  

(b)

10

Solution: (c) We have, 0.6  0.66  0.666  ...upto10 terms

6  0.1  0.11  0.111  ...upto10 terms   1 11 111   6   2  3  ...upto10 terms   10 10 10  6  9 99 999     2  3  ...upto10 terms  9  10 10 10  2  1  1   1     1    1    1    ...upto10 terms  3   10   100   1000   2 1 1 1   10      ...upto10 terms   3  10 100 1000 

    1 10       1      2 1 10  10          10  1  3 1       10        10 2 1   1    10  1      3  9   10   

2  1   89     27   10 

10

  

Question 6: The sum to infinity of the series

1 2 6 10 14      ... is 2 6 18 54 162

(a) 2/3 (b) 3/2 (c) 6 (d) 3 Solution: (b) Consider,

1 2 6 10 14 1  2 6 10 14       ...  1   2  3  4  ...  2 6 18 54 162 2 3 3 3 3 

2 6 10 14 Let us suppose S  1   2  3  4  ... 3 3 3 3 2 6 10 14     ... 3 32 33 34 S  1 2 6 10 14   2  3  4  5  ... 3 3 3 3 3  S 1 

... 1

 2

Subtracting (2) from (1), we get 4 2  S  1 2 32    3 3 1 1 3 2  S  1 2 2    3 3 3 S 3

Therefore,

1  2 6 10 14  3 1   2  3  4  ...   2 3 3 3 3  2

Question 7: If

1 1 1 1 1   0 and x, y, z are in G.P, then  log x 2  ,  log xz  ,  log z 2  are in y 1 z 1

1  x 1

(a) A.P. (b) G.P. (c) H.P. (d) None of these Solution: (c) We are given that,

1  x 1

1 1  0 y 1 z 1

 x  1, y  1, z  1 Also, we are given that x, y, z are in G.P.

 log x, log y, log z are in A.P.  2 log x, 2 log y, 2 log z are in A.P.  log x 2 , log y 2 , log z 2 are in A.P.   log x 2  ,  log y 2  ,  log z 2  are in H.P. 1

1

1

  log x 2  ,  log xz  ,  log z 2  are in H.P 1

1

1

 y 2  xz 

Question 8: If three successive terms of a G.P with common ratio r (r > 1) form the sides of a triangle ABC and [r] denotes greatest integer function, then [r] + [−r] is (a) 0 (b) 1 (c) −1 (d) None of these

Solution: (c) Let us suppose the sides of the triangle be a, ar and ar2 (being the largest side as r > 1) Now, a  ar  ar 2 (Sum of two sides is greater than the third side)  ar 2  ar  a  0  r 2  r 1  0 1 5 1 5 r 2 2 1 5 1 r   r  1 2  r   1 

1 5  r  1 2   r   2 Also, 

  r     r   1  2  1 Question 9: If a, x, b are in A.P., a, y, b are in G.P., a, z, b are in H.P. such that x = 9z and a > 0, b > 0, then (a) |y| = 3z and x = 3|y| (b) y = 2|z| and |x| = 3y (c) 2y = x + z (d) None of these Solution: (a) We are given that a, x, b are in A.P., a, y, b are in G.P., a, z, b are in H.P. Therefore, x, y, z are A.M., G.M. and H.M. of a and b which are in G.P.  y 2  xz  y2  z 9z   y  3z

 x  9 z   y  0

Substitute z 

x in y  3z 9

 x y  3  9  x3 y Question 10: If (0, 6) and (0, 3) are respectively the vertex and focus of a parabola, the its equation is (a) x2 + 12y = 36 (b) y2 + 12x = 36 (c) x2 − 12y = 36 (d) y2 − 12x = 36 Solution: (a) It is given that the vertex and focus of a parabola are (0, 6) and (0, 3) respectively. Therefore, the axis of parabola will be y-axis and the concavity of the curve is downward. Also, the distance between the vertex and focus is 0   6  3  3 . 2

Now, the length of the latusrectum is four times the distance. Hence, latusrectum = 4 × 3 = 12 Therefore, the equation of the parabola is given by

 x  0

2

 12  y  3

 x 2  12 y  36 Question 11: The common tangents to the circle x 2  y 2  2 and the parabola y 2  8 x touch the circle at the points P, Q and the parabola at the points R, S. Then, the area of the quadrilateral PQRS is (a) 3 (b) 6 (c) 9

(d) 15 Solution: (d) Let the slope of common tangent be m. Then, the equation of tangent to parabola is y  mx  It is also tangent to the circle x 2  y 2  2 Then,

2 m 1  m2

 2

 m4  m2  2  m4  m2  2  0   m 2  1 m 2  2   0  m 2  1, m 2  2  m 2  2 rejected  So, the equation of tangents are y  x  2, y   x  2

Equation of chord PQ is, 2 x  2  x  1 Equation of chord PQ is, 4  x  2  0 x2

Coordinates of P, Q, R, S are P(–1, 1) Q(–1, –1) R(2, – 4) & S(2, 4)

2 m

Area of PQRS =

 2  8  3  15 sq unit 2

Question 12: The slope of the line touching both the parabolas y 2  2 x and x 2  16 y is 1 2 3 (b) 2 1 (c) 3 4 1 (d) 4

(a)

Solution: (c) Let the equation of the line touching the both parabolas be 1 …(i) y  mx  m We have,

x 2  16 y From (i) and (ii) 1 x2  mx    m 16  x 2  16mx 

16 0 m

D  0

 16   4   0 m 1  m3  4 1 m 3 4

16m 

2

Question 13:

…(ii)

A class has n students, we have to form a team of the students including at least two students and also excluding at least two students. The number of ways of forming the team is Options: (a) 2n–2n (b) 2n–2n–2 (c) 2n–2n–4 (d) 2n–2n–6 Solution: (b) Required number of ways

 n C2  n C3  n C4  ....  n Cn  3  n Cn  2  2n  (n C0  n C1  n Cn  1  n Cn )  2n  2(1  n)  2n  2n  2 Question 14: Eighteen guests have to be seated, half on each side a long table. Four particular guest desires to sit on one particular side and three others on the other side. The number of ways in which the seating arrangement can be made, is (a) 9!  9! (b) 11C5  9!  9! 11!  9!  9! 5! (d) 11C5

(c)

Solution: (b) Given that 4 particular guests will sit on a particular side A (say) and three other on the other side B (say). Therefore, we are to arrange 11 guests so that 5 guests will sit on side A and remaining 6 will sit on side B. This can be done in the following ways-C5  6C6  11C5

11

Now 9 guests on each side can arrange among themselves in 9! ways.

Therefore, total number of arrangements

C5  9!  9!

11

Question 15: The letters of the word "RANDOM" are written in possible orders and these words are written out as in a dictionary, then the rank of the word "RANDOM" is (a) 614 (b) 615 (c) 613 (d) 616 Solution: (a) In a dictionary the words at each stage are arranged in alphabetical order. So here we consider the words beginning with A, D, M, N, O and R in order. Now keeping A at the first place, the remaining 5 letters can be arranged in 5! ways. Similarly, D, M, N, 0 will occur in the first place the same number of times i.e., 5! Number of words starting with A = 5! = 120 Number of words starting with D = 5! = 120 Number of words starting with M = 5! = 120 Number of words starting with N = 5! = 120 Number of words starting with O = 5! = 120 Number of words starting with R = 5! = 120, but one of these word is the word RANDOM. So, number of words starting with RAD = 3! Number of words starting with RAM = 3! Now, the words beginning with RAN must follow. The first word beginning with RAN is the word RANDMO and the next word is RANDOM. Therefore, the rank of RANDOM is = 5 × 5! + 3! + 3! + 2 = 614

Question 16: The sum of all four digit numbers that can be med using the digits 1, 2, 3, 4, when repetition of digits is not allowed, is (a) 36600 (b) 66000 (c) 36000 (d) 66660 Solution: (d) The sum of all n-digits numbers formed by using n-digits from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, is

 10n  1  Sum of digits n  1 ! given by       9 

 104  1  Therefore, required sum = 10  3!    9 

 66660 Question 17: Tangents drawn from point P (1, 8) to the circle x2  y 2  6 x  4 y  11  0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (a) x2  y 2  4 x  6 y  19  0 (b) x2  y 2  4 x  10 y  19  0 (c) x2  y 2  2 x  6 y  29  0 (d) x2  y 2  6 x  4 y  19  0 Solution: (b) We have,

x2  y 2  6 x  4 y  11  0 x2  6 x  9  y 2  4 y  4  11  13  0

 x  3   y  2  2

2

 24

Let C be the centre of the circle. So, the coordinates of the centre of the circle is C (3, 2). Since CA and CB are perpendicular to PA and PB, CP is the diameter of the circumcircle of triangle PAB. So, the equation of the circle with the end points of the diameter as (1, 8) and (3, 2) is  x 1 x  3   y  8 y  2  0

 x2  y 2  4 x  10 y  19  0 Question 18: The locus of the centre of a circle, which touches externally the circle x2 + y2  6x  6y + 14 = 0 and also touches the y-axis, is given by the equation (a) x2  6x  10y + 14 = 0 (b) x2 10x  6y + 14 = 0 (c) y2  6 x  10y + 14 = 0 (d) y2  10x  6y + 14 = 0 Solution: (d) We have x2 + y2  6x  6y + 14 = 0 x2  6 x  9  y 2  6 y  9  4  0 x2  6 x  9  y 2  6 y  9  4

 x  32   y  32  22 So, the centre and the radius of the given circle are: Centre: (3, 3) Radius: 2 units Let the centre of the circle which touches externally the given circle be  g , f  According to distance formula, The distance between point  g , f  and (3, 3) is

 g  3   f  3 2

2

The distance between centres of two circles is sum of radii of two circles i.e. g + 2

g 2

 g  3   f  3 2

2

On squaring both side,

 g  3   f  3 2

  g  2

2

2

 g 2  f 2  6 g  6 f  18   2  g 

2

 f2  10g – 6f + 14 = 0  Locus of centre (g, f) is y2  10x  6y + 14 = 0 Question 19: The circle passing through the point  1, 0  and touching the y-axis at  0, 2  also passes through the point  3  (a)   , 0   2   5  (b)   , 2   2 

 3 5 (c)   ,   2 2 (d)  4, 0  Solution: (d) Equation of a circle passing through a point  x1 , y1  and touching line L is:

 x  x1    y  y1  2

2

 L  0

Equation of a circle passing through a point (0, 2) and touching line x = 0 is: 2 2  x  0   y  2   x  0

…(i)

The circle also passes through (−1, 0) 1 4    0  5 Putting the value of  in equation (i), we get

 x  0   y  2 2

2

 5x  0

 x2  y 2  5x  4 y  4  0

Put y = 0 for x-intercept

 x2  5x  4  0  x  1, 4 So, circle passes through (−4, 0) Question 20: Let C be the circle with centre at (1, 1) and radius 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to

(a)

3 2

3 2 1 (c) 2 1 (d) 4

(b)

Solution: (a)

Let the coordinate of the centre of circle be (0, y). Distance between their centre

k  1  1   k  1

2

 k  1  1  k 2  1  2k On squaring both sides, we get  k  1  k 2  2  2k

 k 2  1  2k  k 2  2  2k 1 k  4

Question 21: 3

x  9  2  e x  28  Find lim x 0

x

Options: (a) 0 (b) DNE (c) 4 (d) 11/2 Solution: (d) 3

We have f  x    x  9  2  e x  28 Now, we can write the limit as lim x 0

Now, 1 3  x  9 2  ex 2 1 3  f '  0    0  9  2  e0 2 9  1 2 11  2

f ' x 

Question 22:

r3  8 Solve lim  3 n  r 3 r  8 n

(a)

2 3

f  x   f  0 which is the definition of f '  0  . x0

(b)

2 5

(c)

2 7

(d)

1 7

Solution: (c)

r3  8 3 n  r 3 r  8 n

Consider, lim 

 32  8  43  8   n2  8   lim  2 ......................  3   2  n  3  8   4  8   n 8  3  2 32  4  2  3  4  2 42  4  2  4    lim  . .   n  3  2 32  4  2  3   4  2 4 2  4  2  4       n  2 n2  4  2  n   ...........................................  . 2  n  2 n  4  2  n    n 2 3  2 4  2 5  2  lim  . . ...............  n  3  2 4  2 5  2 n  2  2 2 n 2  4  2  n    3  4  2  3 4  4  2  4  . ...............  2  2 n 2  4  2  n    3  4  2  3 4  4  2  4 

19.28.39.52.63......... 1.2.3.4.5.6.......     lim    n   5.6.7.8.......   7.12.19.28.39.52.63..............   1  2  lim 1.2.3.4.   n   7.12  7

Question 23:

n n n   lim sin 2 2  sin 2 2  ...  sin 2 2   n  n 1 n 1   n 1

 4  b 2 c

a

 d  2 Solution: (a)

x3 For x  0, x   sin x  x 3! n

So,

n r 1

n n 1 n  n  n n n   sin       2 2 2 2 2 2 2 2  r 3! r 1  n  r  r 1 n r r 1 n  r

n dx    2 2 2 n  1 x 4 r 1 n  r 0 1

n

Sandwich rule and lim  So, given limit   / 4. Question 24:

1k  3k  ....   2n  1  For k  0, lim n  nk 1 k

2k a k 2k b k 1 2k 1 c   k 2k 1 d  k 1 Solution: (b) Rewrite the given sum is

2k n

k  1  k  3  k  2n  1         ...      2n    2n   2n 

1

then the given limit is 2k . x k dx  0

2k k 1

Question 25: 2 1 n 1 n   2 lim    log k     log k    n  n  n k 1    k 1

(a) 1 (b) 2 (c) ½ (d) e

1 n 1 n  2 Solution: (a) an    log k     log k  n k 1  n k 1  2

1 n 1   k    1 n 1  k   .  log       log    n k 1   n    n k 1  n 

2

2

2

1  So, lim an    log x  dx    log xdx   2  1  1 n  0 0  1

2

Question 26: Let y(x) be the solution of the differential equation  x ln x  equals to: (a) e (b) 0 (c) 2 (d) 2e Solution: (c) We have,

dy  y  2 x ln x,  x  1 . Then y(e) is dx

dy  y  2 x ln x,  x  1 dx On putting x = 1 we get y = 0 On dividing the above equation by x logx it can be written as: dy y  2 dx  x ln x 

 x ln x 

1

IF = e x ln x  ln x Solution of the differential equation can be given as: y ln x   2 ln xdx dx

 y ln x  2  x ln x  x   c

At x = 1, y = 0 0 = c + 2(−1) c=2  y ln x  2  x ln x  x  2 Put x = e y  2  e ln e  e

 y2 Question 27: The solution of the differential equation 1 

dy  e x  y is dx

(a) (x + C) ex − y + 1 = 0 (b) (x − C) ex − y + 1 = 0 (c) (x − C) ex − y − 1 = 0 (d) (x − C) ex − y + 1 = 0 Solution: (c) Given differential equation is 1 

dy  e x y dx

On substituting x − y = t, we get

 1

dy dt  dx dx

Eq. (1) becomes

dt  et dx

…(1)

 et dt  dx  e  t  x  C 

1  xC e x y

 1 = (x + C) ex − y  (x + C) ex − y + 1 = 0

Question 28: Let the population of rabbits surviving at time t be governed by the differential equation dp  t  1  p  t   200 . If p(0) = 100, then p(t) is equals to dt 2 t

(a) 400  300e 2 (b) 300  200e



t 2

t

(c) 600  500e 2 (d) 400  300e



t 2

Solution: (a) Given, dp  t  1  p  t   200 dt 2 dp  t  1   p  t   200 dt 2 This is a linear differential equation. 1

  2 dx



t

IF = e e 2 The solution of this differential equation is given by p t   e



t 2

p t   e



t 2



t

  200  e 2 dt  400e



t 2

K t

 p  t   400  ke 2

If p(0) = 100, then k = −300

 p  t   400  300e

t 2

Question 29:



Solve



0

x 1  sin x

(a) 0 (b)

 3

(c)  (d)  Solution: (d)

I 



and I  



Let

0

0

x 1  sin x

  x  x dx   dx 0 1  sin x 1  sin(  x)

On adding Eqs. (1) and (2),we get

2I   



0









0

0

…(1)

1 dx 1  sin x

(1  sin x) dx (1  sin x) (1  sin x)

(1  sin x) dx cos 2 x



   (sec2 x  tan x  sec x)dx 0



   tan x  sec x  0

   tan   sec   tan 0  sec0

…(2)

  0  1  0  1  2I  2  I  

Questions 30: Solve

(a)

(b)

(c)

(d)



 /2

0

dx (a cos x  b2 sin 2 x)2 2

2

  a 2  b2    2  a3b3 

  a 2  b2    4  a3b3 

  a 2  b2    3  a3b3 

  a 2  b2    4  a3b3 

Solution: (b) Let

I 

 /2

0

dx (a cos x  b2 sin 2 x)2 2

2

Divide numerator and denominator by cos4x, we get

I 

 /2

0



 /2

0

Put tan x = t

sec4 x dx (a 2  b2 tan 2 x)2

(1  tan 2 x)sec2 x dx (a 2  b 2 tan 2 x)2

 sec2 x dx  dt As x  0, then t  0 and x 

Now,

 2

I 

then t  



0

1 t2 (a 2  b 2 t 2 ) 2

(1  t 2 ) (a 2  b 2 t 2 ) 2

[Let t 2  u ]

1 u A B  2  2 2 2 2 (a  b u ) ( a  b u ) ( a  b 2u ) 2 2

 1  u  A(a 2  b2u) B On comparing the coefficient of x and constant term on both sides, we get a2 A  B  1 b2 A  1

and

1 b2

A

a2  B 1 b2

Now, a 2 b2  a 2  B  1 2  b b2

I  



0

1  2 b 

1 b2









0

0

(1  t 2 ) (a 2  b 2 t 2 )2

dt b2  a 2  a 2  b 2t 2 b2





0

dt b2  a 2  2 b2 2a 2 b  2 t  b 

dt ( a  b 2t 2 ) 2 2





0

dt ( a  b 2t 2 ) 2 2

(1) (2)



1  1  tb   b2  a 2   1   3  tan        ab  b2  4 a3b   a  0 



1  b2  a 2 1 1   tan   tan 0   4  a3b3 ab3   

 2ab3



 b2  a 2 4



a b  3 3

 2a 2  b 2  a 2    a 2  b 2      3 3  4a3b3   4 a b  Questions 31: 50

If f(x) = [|x|], then

 f  x  dx is equal to (where [.] denotes the greatest integer function) 0

(a) 1225 (b) 2450 (c) 5000 (d) 2500 Solution: (a) 50

1

2

3

50

0

0

1

2

49

 f  x  dx   0dx   1dx   2dx  ....   49dx

49  50 2  1225 

Questions 32:



1/ 2 1/ 2

2 2    x  1  x  1    2 dx is         x  1 x  1    

3 (a) 4 ln   4  

3 (b) ln    4 3 (c) 4 ln   4  

81  (d) ln   256 



Solution: (d) 12 x  1 x  1  Let I  1 2   dx 2

 x 1



1 1 2

x  1

1/ 2 4x 4x dx  2 dx 0 x 1 x2  1



2



1/ 2 3  4  ln 2  1 0  4ln   4

4  4 ln   3 256    ln    81   81    ln    256 

Question 33:  /2

The value of

(a)

 2

(b) 4 (c)

(d)

 4

 8

Solution: (c)

sin 2 x dx is : x  1  2  /2

 /2

Let us suppose, I  b

b

a

a

 /2

I

... 1

f  x  dx   f  a  b  x  dx in



Use

sin 2 x  1  2x dx  /2

sin 2 x dx x  1  2  /2

 /2

sin 2 x  1  2x dx  /2

....  2 

Adding (1) and (2)  /2

2I 

 

sin 2 xdx

 /2

 /2

 2 I  2.  sin 2 xdx 0

 2I  2  I



 4

4

Question 34: sin x cos x sin 2 x  cos 2 x Let   x   0 then 1 1 1

0

 '  x  vanishes at least once in

 a  0,  / 2   b  / 2,    c  0,  / 4   d   / 2, 0  Solution: (a)

1

The function   x  is continuous on  0,  / 2 and differentiable on  0,  / 2 . Also   0   0 and       0 . Thus, by the Rolle’s theorem there exists at least one c   0,  / 2  such that 2  'c  0 .

Question 35: Let A be a 3 × 3 matrix such that det  A  2 . Then det  2 A1   2det  A is equal to (a) -2 (b) -4 (c) 7 (d) None of these Solution: (d) As A1 is 3  3 matrix, det  2 A1    2  det  A1    2   det  A  3

3

1

 1   8     4  2 Hence det  2 A1   2det A  4  2   2   0 Question 36:

 4 3 Let A   then A482 , A700 , A345 are respectively   7 5

 a  A  I , A,  A  I  b  A,  A, I  c  A  I ,  A, _ I  d  A  I ,  A  I , I Solution: (c) Using characteristic equation 4  

3

7

5

 0  20     2  21  0

 2    I  0  A2  A  I  0  A3   I So, A482   A3 

160

A700   A3 

A2    I 

233

.A   A

115

 I

A345   A3 

160

A I  A I

Question 37: 1 2 0  Let A  1 1 1  by observing orthogonally among the column vectors of A one may obtain   1 1 1 the inverse of A as 1 1 1 (a) A   2 1 1     0 1 1

 1  3  2 (b)    6   0 

1 3 1 6 1 2

1  3   1  6   1   2 

 2 2 2 (c)  2 1 1     0 3 3   2 2 0  1 (d) 2 1 3   6  2 1 3

Solution: (b) If it is column wise orthogonal then take  1 1 1  1 2 0   3 0 0  A A   2 1 1  1 1 1  0 6 0   0 1 1 1 1 1 0 0 2 T

Prefactor matrix will become inverse of A if 1st row is divided by 3, 2nd by 6, 3rd by 2. Question 38: xp  y The determinant xp  z

x y

y z

xp  y

xp  z

0 0

x y

y z

  p 2 x  py    py  z 

px  y

py  z

0

(a) x, y, z are in A.P. (b) x, y, z are in G.P (c) x, y, z are in H. P. (d) xy, yz, zx are in A. P. Solution: (b) Using C1  C1  pC2  C3

 0, if

  y 2  xz  p 2 x  py  py  z   0

x, y, z are in G.P.

Question 39: 3

1  f 1

If  ,   0 and f  n    n   n and 1  f 1 1  f  2  1  f  2  1  f  3

then K is equals to: (a)  (b)

1



(c) 1 (d) −1 Solution: (c) We have, f n   n   n  f 1      f  2   2   2  f  3   3   3  f  4   4   4

Now,

1  f  2 1  f  3  K 1    1        2

1  f  4

2

2

111   1   

1   

1  2   2

1  2   2

1  3   3

1  3   3

1  4   4

1  2   2 1 1 1

2

1   1 2 2

On expanding, we get

  1    1        2

2

2

 1    1         K 1    1        2

2

2

2

2

2

 K 1

Question 39: Question 40:

x 1 1 The least value of the product xyz for which the determinant 1 y 1 is non negative is 1 1 z (a) −8 (b) −1 (c) 2 2 (d) 16 2 Solution: (a) Given,

x 1 1 1 y 1 0 1 1 z Expanding the determinant along the first row, we get

x  yz  1  1 z  1  11  y   0  xyz  x  z  1  1  y  0  xyz  x  z  2  y  0 1

 xyz  2  x  y  z  3  xyz  3 1

 xyz  3  t  t 3  3t  2  0

  t  1 t  2   0

 t  2

 t 3  8  xyz  8

Question 41: The area bounded by the curve x2  2 x  y  3  0, the x-axis and the tangent at the point where it meets the y-axis is (a)

7 sq. units 12

(b)

12 sq. units 7

(c)

7 sq. units 6

(d)

6 sq. units 7

Solution: (a) The equation of the curve is y – 4 = −(x + 1)2. The curve meets y-axis at (0, 3) and x - axis at (−3, 0) and (1, 0). The tangent at (0, 3) is y − 3 = −2x

1





The required area = Area of OAB   4   x  1 dx 2

0

1

3 x  1   9  8  1 7   9    4 x      4      4  3  3  3  12  0 4 

Question 42: The area between the curves y  xe x and y  xe x and the line x = 1 is (a) 2e (b) e (c) 2/e (d) 1/e Solution: (c)  1 The line x = 1 meets the curves in A(1, e) and B 1,  . Both the curves pass through origin.  e

1

2 The required area    xe x  xe x dx  sq.units e 0

Question 43: The area bounded by the curve, y  f  x   x 4  2 x3  x 2  3 , the x – axis and the ordinates corresponding to the minimum of function f(x) is (a) 1 (b) 91/30 (c) 30/9 (d) 4 Solution: (b) We have, f  x   x 4  2 x3  x 2  3

 f '  x   4 x3  6 x 2  2 x Now, f '  x   0  4 x3  6 x 2  2 x  0 1  x  0, 1, 2 f ''  x   12 x 2  12 x  2  2  6 x 2  6 x  1 f "  x  x 0  0  Minimum exists at x = 0 and x = 1 f "  x  x 0  0  Minimum exists at x 

1 2

Therefore, the curve is bounded by the ordinates x = 0 and x = 1 1

 x5 x 4 x3   Required area    x  2 x  x  3dx      3x   5 2 3 0 0 1

4



91 sq. units 31

3

2

Question 44: A letter is known to have come either from TATANAGAR or CALCUTTA. On the envelope just two consecutive letters are visible. The probability that the letter came from CALCUTTA is (a) 4/11 (b) 7/11 (c) 5/11 (d) none of these Solution: (a) Let El and E2 denote the event that letters came from TATANAGAR and CALCUTTA respectively. Let A denotes the event that two consecutive visible letters on the envelop are 'TA’ We have

1 2 1 P  E1   P  E2   , P  A | E1   and P  A | E2   . 2 8 7 Using the Bayes' theorem, we get P  E2 | A  

P  E2  P  A | E2  P  E1  P  A | E1   P  E2  P  A | E2 

1 1 . 4 2 7   1 2 1 1 11 .  . 2 8 2 7

Question 45: A coffee connoisseur claims that he can distinguish between a cup of instant coffee and a cup of percolator coffee 75% of time. It is agreed that his claim will be accepted if he correctly identifies at least 5 of the 6 cups. His chances of having the claim accepted is (a) 0.534 (b) 0.466

(c) 0.763 (d) none of these Solution: (a) Let p denotes the probability of a correct distinction between a cup of instant coffee and a cup of percolator coffee. Then, p 

75 3 1   q  1  p  and n = 6 100 4 4

Let random variable X denotes the number of correct distinctions. x

3 1 Then, P  X  x   p  x  6 Cx     4 4

6 x

, x  0,1,......, 6

The probability of the claim being accepted is 5

3 1 P  X  5  p  5  p  6  6 C5     4 4

6 5

 3  6 C6   4

6

= 0.534 (approx.) Question 46:

X  ; X  1, 2,3, 4,5 . Then The probability distribution of a random variable is P  X   15  0, otherwise 5 1  P   X  X 1  2 2 

(a) 1/5 (b) 1/7 (c) 2/15 (d) 1/15 Solution: (b)

X  ; X  1, 2,3, 4,5 . Consider, P  X   15  0, otherwise 1 2 1   15 15 5  1 5  P   X    X  1 2 5 1  2  Now, P   X  | x  1   2 P  X  1 2  2 P  X  1 or 2    X  1 P  X  2 1    15  P  X  1 1  P  X  1 1  1 7 15

 P  X  1 or 2   P  X  1  P  X  2  

Question 47: A perfect die is thrown twice. The expected value of the product of the number of points obtained in two throws is (a) 7/2 (b) 7 (c) 49/4 (d) none of these Solution: (c) Let X and Y denote respectively the number of points obtained in the first and the second throws. Then both of them take the values 1, 2, 3, 4, 5 and 6 with each probability 1/6.

1 7  E  X   1  2  3  4  5  6    . 6 2

7 Similarly, E Y   . 2 7 7 49 Thus, E  XY   E  X  E Y   .  2 2 4 Question 48: There are 4 white and 3 black balls in a box. In another box there are 3 white and 4 black balls. An unbiased dice is rolled. If it shows a number less than or equal to 3 then a ball is drawn from the first box but if it shows a number more than 3 then a ball is drawn from the second box. If the ball drawn is black then the probability that the ball was drawn from the first box is (a) 1/2 (b) 6/7 (c) 4/7 (d) 3/7 Solution: (d) Let E1 : The ball drawn from the first box E2: The ball drawn from the second box A: The drawn ball is black.

 P  E1   P  E2  

3 1  6 2

3 4 Now, P  A | E1   andP  A | E2   . 7 7  P  E1 | A  

P  E1  .P  A | E1  P  E1  .P  A | E1   P  E2  .P  A | E2 

1 3 . 3 2 7   1 3 1 4 7 .  . 2 7 2 7

Question 49: If f(x) and g(x) are differentiable functions in [0, 1] such that f (0) = 2, f (1) = 6, g(0) = 0, g(1) = 2, then there exists 0 < c < 1 such that (a) f '(c) = g' (c) (b) f '(c) = -g’ (c) (c) f '(c) = 2g’(c) (d) 2f '(c) = g'(c) Solution: (c) Let F  x   f  x   2 g  x 

F  0  f  0  2g  0  2 F 1  f 1  2 g 1  2 F(x) satisfies the condition for Rolle’s theorem. So, there exist c, such that F’(c) = 0 or f’(c) – 2g’(c) = 0 for 0 < c < 1.

Question 50: If f  x  

x3

dt

 log t , x  0, then

x2

(a) f(x) is maximum at x = 1 (b) f (x) is an increasing function x  R  only (c) f(x) is minimum at x = 1 (d) f (x) is neither maximum nor minimum at x = 1 Solution: (d)

Here f  x  

x3

1

 log t dt

x2

3x 2 2x  f ' x   3log x 2 log x x2  x  log x 

x  x  1 log x

For extremum, we put f’(x) = 0 Now, f’(x) = 0 gives x (x – 1) = 0

 Critical points are x = 0 and x = 1 Let us consider x = 1 as x > 0 Now for x  1, f '  x    ve  ve      ve  0 and for x  1, f '  x    ve  ve      ve  0

 f’(x) does not change its sign in the immediate neighbourhood of x = 1. So x = 1 is neither the point of maxima nor minima. Question 51: x

If y  f  x    2  t 2 dt. Then 1

(a) f(x) increases if x  2 (b) f(x) decreases if x  2 (c) f(x) increases if x  2 (d) none of these Solution: (a)

x

Consider, f  x    2  t 2 dt 1

f '  x   2  x2

d  x   2  x2 dx

For an increasing function f ' x  0  2  x2  0  2  x2  0

 2  x  2  x   0   x  2  x  2   0 

 x 2

Question 52: The curve y  ax3  bx2  cx  5 touches the x-axis at P(−2,0) and cut the y-axis at a point Q where its gradient is 3. Then, a + b + c = (a) 9/4 (b) 7/4 (c) 5/4 (d) 3/4 Solution: (b) Consider, y  f  x   f  2  f '  2   0, f '  0   3

Now, f  0   5

5 2 f '  x    x  2   ax   4  5 2  f '  x   2  x  2   ax     x  2  a 4  f '  0  3

 5  4a  3 a

1 2

x 5 2 f  x    x  2      2 4 27  a  b  c  5  f 1  4 7  abc  4 Question 53: A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is (a)

1 cm/min. 18

(b)

1 cm/min. 36

(c)

5 cm/min. 6

(d)

1 cm/min. 54

Solution: (a)

4 3 Consider, v    y  10  where y is thickness of ice 3 dv 2 dy  4 10  y  dt dt 50  dy     2  dt t 5 4 15 





 dv  3  dt  50 cm / mn 

1 cm/min. 18

Question 54: In a class of 60 students, 34 have passed in Mathematics and 38 in physics. If each of the students pass at least in one of the subjects, then the number of students who have passed in physics only is (a) 12 (b) 26 (c) 38 (d) 34 Solution: (b) Let A be the set of students who passed in Mathematics and B be the set of students who passed in physics.  n  A  B   60

Now, n  A  B   n  A  n  B   n  B   n  A  B   n  A  34 and n  B   38

 60  34  38  n  A  B   n  A  B   72  60  12

 The number of students who have passed in physics only  n  B   n  A  B   38  12  26

Question 55: Let Y  1, 2,3, 4,5,6,7 , A  3, 4 , B  1, 2,5 , then Y  A  Y  B  is (a) Y (b) A (c) B (d)  Solution: (d) Consider, A  3, 4 , B  1, 2,5 , A  B  

Y  A  Y  B   Y   A  B   Y     Question 56: If A is the set of even natural numbers less than 10 and B is the set of prime numbers less than 5, then the number of relations from A to B is (a) 28 (b) 82 (c) 34 (d) 29 − 1 Solution: (a) Here, A = {2, 4, 6, 8} and B = {2, 3} ∴n(A) = 4 and n(B) = 2 Now, number of relations  242  28 Question 57:   The area enclosed by the curves y = sinx + cosx and y  cos x  sin x over the interval 0,  is  2

(a) 4



(b) 2 2 (c) 2



2 1







2 1



2 1

(d) 2 2





2 1

Solution: (c) Let,

  y1  sin x  cos x  2 sin  x   4    y2  sin x  cos x  2 sin   x  4 





4 0

Area =

= 42 2 =2 2





 sin x  cos x    cos x  sin x dx  2  sin x  cos x    sin x  cos x dx 4



2 1

Question 58: The area of the region bounded by y = 1 +|x + l|, x =  3, x = 3, y = 0 is (a) 14 sq units (b) 12 sq units (c) 16 sq units (d) 17 sq units Solution: (c) We have, y = 1 + |x + 1|, x =  3, x = 3 and y = 0 if x < 1  x or y    x  2, if x  1

1

3

3

1

Now, Area of shaded region, A    x dx   1

 x  2 dx

3

 x2   x2         2x 2 2  1 1  1 9  9        6   2 2  2 2 2

= [4] + [8 + 4] = 12 + 4 = 16 sq units Question 59:   The area enclosed by the curves y = sinx + cosx and y  cos x  sin x over the interval 0,  is  2

(a) 4



(b) 2 2 (c) 2





2 1





2 1

(d) 2 2





2 1

Solution: (b) Let,



2 1

  y1  sin x  cos x  2 sin  x   4    y2  sin x  cos x  2 sin   x  4  Area =





4 0

2

  sin x  cos x    cos x  sin x dx    sin x  cos x    sin x  cos x dx 4

= 42 2 =2 2





2 1

Question 60: PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 450, 300 and 300, then the height of the tower (in m) is(a) 50 (b) 100 3 (c) 50 2 (d) 100 Solution: (d)

Let us suppose the height of tower MN is ‘h’ In QMN MN 0 We have QM  tan 30

QM  3h  MR.......  i  Now in MNP

MN  PM ....  ii  In PMQ MP 

 200 

2





3h



2

From equation (ii), we get

 200 

2





3h



2

h

 h  100 m Question 61: The equation 3sin 2 x  10cos x  6  0 is satisfied, if (a) x  n  cos1 1/ 3 (b) x  2n  cos1 1/ 3 (c) x  n  cos1 1/ 6  (d) x  2n  cos1 1/ 6  Solution: (b) We have, 3sin 2 x  10cos x  6  0

 3 1  cos 2 x   10 cos x  6  0  3  3cos 2 x  10 cos x  6  0  3cos 2 x  10 cos x  3  0   cos x  3 3cos x  1  0

Either cos x  3 (not possible)

or cos x 

1 3

 x  2n  cos1 1/ 3 , n  Z

Question 62: In a triangle, the sum of two sides is x and the product of the same two sides is y. If x 2  c 2  y ,where c is the third side of the triangle, then the ratio of the inradius to the circum radius of the triangle is

(a)

3y 2x  x  c 

(b)

3y 2c  x  c 

(c)

3y 4x  x  c 

(d)

3y 4c  x  c 

Solution: (a) Let the sides of the triangle be a, b and c respectively. We have, …(i) x  ab And, y  ab Also, x 2  c 2  y From (i) and (ii)

 a  b

2

…(ii)

 c2  y

 a 2  b 2  2ab  c 2  ab  a 2  b 2  ab  c 2  0  a 2  b 2  c 2  ab a 2  b 2  c 2 ab  2ab 2ab 2  C  3



Let R be the radius of circumcircle and r be the radius of the incircle. Then, abc  R ,r  4 s abc R 4 abc s abc  s       r 4  4 2 s r 4 2  R abc  s 

1  2   ab sin    r 2  3   xc R  y c 2 r 3y  R 2c  x  c 

2

Question 63: If 12cot 2   31cos ec  32  0 , then the value of sin  is (a)

3 or 1 5

(b)

2 2 or 3 3

(c)

4 3 or 5 4

(d) 

1 2

Solution: (c) Consider, 12cot 2   31cos ec  32  0

 12  cosec 2  1  31cosec  32  0  12cosec 2  31cosec  20  0  12cosec 2  16cosec  15cosec  20  0   4cosec  5  3cosec  4   0 5 4  cosec  , 4 3 4 3  sin   , 5 4 Question 64: The number of roots of the equation x + 2 tan x = (a) 1 (b) 2 (c) 3 (d) infinite Solution: (c) We have,

π 2 π  2 tan x   x 2 π x  tan x   4 2 Let, y = tan x π x y  4 2 x  2 tan x 

 in the interval [0, 2] is 2

The curves y = tan x and y 

π x  in interval [0, 2], intersect at three points. 4 2

So, there are three roots of the equation, x + 2 tan x =

Question 65:

 sin

2

dx x  tan 2 x

1 1  tan x  (a)  cot x  tan 1  c 2 2 2  2  (b)

1 1  x  cot x  tan 1  c 2 2 2  2

(c)

1 1 cot x  tanx  c 2 2

 tan x  (d)  cot x  tan 1  c  2  Solution: (a) Let I  

dx sec2 xdx  sin 2 x  tan 2 x tan 2 x  2  tan 2 x 

Put tan x = t

 2

1 1

dt

 t  2  t   2   t 2

2

2



1  dt 2  t2 

1 1 t  tan 1 c 2t 2 2 2 1 1  1    cot x  tan 1  tan x   c 2 2 2  2  

Question 66: If

 tan x  1  2 tan x   tan x  1  tan xdx  a tan 1    c, then a + b =  bIn  tan x  1  2 tan x  2 tan x   



(a)

(b)

(c)

(d)

1 2 1 2 1 2 2

1 4

Solution: (c) Let I   tan xdx Put tan x  t 2

t 2 dt t 2 1 t 2 1  dt t4 1  t4 1 1 1 1 2 1 2 t dt  t dt   1 1 t2  2 t2  2 t t  1  1 d t   d t   t t   2    2   1  1 t    2 t    2  t  t  1  1  t  t  2   1 1  tan 1  t   ln  t c 1 2 2 2 2   t  2     t  I  2

 tan x  1  2 tan x  1 1  tan x  1  tan 1  ln  c  2  2 tan x  2 2  tan x  1  2 tan x  1 1 1 a  b    2 2 2 2 2 

Question 67: The integral



dx x  3 x2

represents the function

(a) 6  3 x 2  3 x  log 1  3 x   c   (b) 3 3 x  6 6 x  6log 1  6 x  c (c) 3 3 x  6 6 x  6log 1  6 x  c (d) 6 3 x 2  3 3 x  6log 1  3 x  c Solution: (b)

Put x  t 6  dx  6t 5 dt

dx





6t 5 dt t3  t4

x  3 x2 t 2 dt  6 t 1 1    6  t  1  dt t 1   3t 2  6t  6 log t  1  c  3 3 x  6 6 x  6 log

6

x 1  c

Question 68:

 x  1

 x 1  xe



(a)

1 x  ex

(b)

1 1  xe x

(c)

(d)



x 2

dx  log 1  f  x   f  x   c , then f(x) is

1

1  xe 

x 2

1

x e 

x 2

Solution: (b) Let I  

 x  1

x 1  xe x 

Put xe x  t  e x 1  x  dx  dt

2

dx

I  

dt t 1  t 

Now, let

2

1 t 1  t 

2



A B C   t 1  t 1  t 2

 1  A 1  t   Bt 1  t   Ct 2

Put t  0 , then, 1  A Put t  1, then C  1 Equating the coefficients of t2, we get

0  A  B  B   A  1 1 dt dt  I   dt    2 t 1 t 1  t   log t  log 1  t   log

1 c 1 t

t 1 1  1    C  log 1  c  1 t 1 t  1 t  1 t

Question 69: Evaluate:

sin  x  a   sin  x  a dx

 x - a  cos 2a - sin 2a log sin  x  a   c

 x  a  cos 2a - sin 2a log sin  x - a   c  x  a  cos 2a - sin 2a log sin  x  a   c  x - a  cos 2a - sin 2a log sin  x - a   c

Solution: (c)

Consider, 

sin  x  a  dx sin  x  a 

Let x  a  t dx  dt



sin  t  2a 

dt sin t sin t cos 2a  cos t sin 2a  dt sin t sin t cos 2a cos t sin 2a  dt   dt sin t sin t  cos 2a  dt  sin 2a  cot t.dt  t cos 2a  sin 2a log sin t  c   x  a  cos 2a  sin 2a log sin  x  a   c Question 70: If a and b are two vectors such that a  b  29 and a   2i  3 j  4k    2i  3 j  4k   b ,then a possible value of  a  b    7i  2 j  3k  is (a) 0 (b) 3 (c) 4 (d) 8 Solution: (c) We have,

a   2i  3 j  4k    2i  3 j  4k   b  a   2i  3 j  4k    2i  3 j  4k   b  0  a   2i  3 j  4k   b   2i  3 j  4k   0   a  b    2i  3 j  4k   0 So,  a  b  and  2i  3 j  4k  are parallel vectors.

  a  b     2i  3 j  4k 

(Since  a  b     b  a  )

  22  32  42  29

   1  a  b    2i  3 j  4k  Now,

 a  b  . 7i  2 j  3k     2i  3 j  4k    7i  2 j  3k     14  6  12   4 Question 71:

     Let a  i  j  k , b  i  j  k and c  i  j  k be three vectors. A vector v in the plane of a and   1 , is given by b , whose projection on c is 3 (a) i  3 j  3k (b) 3i  3 j  k (c) 3i  j  3k (d) i  3 j  3k Solution: (c)    A vector v in the plane of a and b is    v  a  b   v  i  j  k   i  j  k



 



  1 Projection of v on c = 3   vc 1    3 c



1     1     1     3

 1     1     1     1

 2

1 3

 So, v  3i  j  3k Question 72:

 Two adjacent sides of a parallelogram ABCD are given AB  2i  10 j  11k and  AD  i  2 j  2k . The side AD is rotated by an acute angle  in the plane of the parallelogram so that AD becomes AD ' . If AD ' makes a right angle with the side AB, then the cosine of angle  is given by 8 (a) 9

17 9

(b) (c)

1 9

4 5 9 Solution: (a) (d)

We have,  AB  2i  10 j  11k  and AD  i  2 j  2k

  Let the angle between AB and AD be  .   AB  AD  cos     AB  AD

  cos  

8 9

2  20  22 8  9 15 3

Question 73: If a, b and c are unit vectors satisfying a  b  b  c  c  a  9 , then 2a  5b  5c , is 2

2

2

(a) 2 (b) 3 (c) 1 (d) 0 Solution: (b) We have, a b  b c  c a  9 2

2

2

 a  b  2a  b  b  c  2b  c  c  a  2c  a  9 2

2

2

2

2

2

Since, a  b  c  1

 6  2a b  b  c  c  a  9  2  a  b  b  c  c  a   3  a b  b  c  c  a  

3 2

…(i)

We know that,





1 2 2 2 a b c 2 3 …(ii)  a b  bc  c  a   2 a b  bc  c  a  

From (i) and (ii) a  b  c  0 Also, 2a  5b  5c  2a  5  b  c 

 2a  5a  3

Question 74: Consider a branch of the hyperbola x2  2 y 2  2 2 x  4 2 y  6  0 with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is

(a) 1 

2 3

3 1 2

(b)

(c) 1 

2 3

3 1 2

(d)

Solution: (b) We have, x2  2 y 2  2 2 x  4 2 y  6  0

This can be written as:

x  2   y  2 2

4 Now,

2

2

1

b2 2 6 3 e  1 2  1   a 4 4 2

1 b2 1  3  2 3 a  e  1    2   1   1 2 a 2  2  2 2 Question 75: Area =

If the line 2 x  6 y  2 touches the hyperbola x 2  2 y 2  4 then the point of contact is

  (b)  5, 2 6  (a) 2, 6

1 1  (c)  ,  2 6



(d) 4,  6



Solution: (b)

We have equation of hyperbola

x2  2 y 2  4 Equation of tangent to above hyperbola at point

xx1  2 yy1  4

 x1 , y1  is

…(i)

Equation of tangent Given, 2x  6 y  2

 4x  2 6 y  4

…(ii)

Comparing eq(i) and (ii) we get x1  4 And, 2 y1  2 6  y1   6





So, the point of contact is 4,  6 . Now the focus of new parabola is same. a1e1  1

 e1 sin   1  e1  cos ec

And, b12  a12 1  e12 



 b12   sin   1   cos ec  2

2



 b12  cos 2   b1  cos  So, the equation of the required hyperbola is: x2 y2  1 sin 2  cos 2  x 2 cos ec 2  y 2 sec2   1 Question 76: The equation of the hyperbola having foci (0,  10) and passing through the point (2,3) is (a) y 2  x 2  25 (b) y 2  x 2  5 (c) y 2  x2  15

(d) y 2  x 2  35 Solution: (b) Since, foci lies on y-axis so the equation of the hyperbola is of the form It is given that c  10 . Also, the hyperbola passes through the point (2,3) .  c2  a 2  b2 

 

2

10

 a 2  b2

 10  a 2  b 2  b 2  10  a 2

32 22  1 a 2 b2 9 4  2  2 1 a b 9 4  2 1 a 10  a 2 90  9a 2  4a 2  2 1 a 10  a 2 



 90  13a 2  10a 2  a 4  a 4  23a 2  90  0  a 4  18a 2  5a 2  90  0  a 2  a 2  18   5  a 2  18    a 2  18  a 2  5   a 2  18,5 For a 2  18 b2  10  a 2  10  18  8 which is not possible

For a 2  5 b2  10  a 2  10  5  5

y 2 x2   1. a 2 b2

Thus, the required equation is

y 2 x2   1 or y 2  x 2  5 . 5 5

Question 77: x The function f  x   e is

(a)

continuous everywhere but not differentiable at x = 0

(b)

continuous and differentiable everywhere

(c)

not continuous at x = 0

(d)

None of the above

Solution: (a) Let u  x   x and v (x) = ex 

f(x) = vou(x) = v[u (x)]

 v x  ex Since, u(x) and v(x) are both continuous functions. So, f(x) is also continuous function but u  x   x is not differentiable at x = 0, whereas v(x) = ex is differentiable at everywhere. Hence, f(x) is continuous everywhere but not differentiable at x = 0. Question 78: Which of the following functions is differentiable at x = 0? (a) cos| x | | x | (b) cos| x | | x | (c) sin | x | | x |

(d) sin | x | | x | Solution: (d) f(x) = sin|x|  |x| f x  x 0 = sinx + x  f (x) = cosx + 1 f x  x 0 = sinx  x

 f (x) = cos x  1

 Lim f x  = Lim f x  = 0 and Lf (0) = Rf (0) x 0

x 0

Question 79:

 k x  1 0  x  3 If the function g  x    is differentiable, then the value of k + m is   mx  2 3  x  5 (a) 2 16 (b) 5 10 (c) 3 (d) 4 Solution: (a) We have,

k x  1 0  x  3  g  x     mx  2 3  x  5 R.H.D g  3  h   g  3 lim h 0 h m  3  h   2  2k  lim h 0 h  3m  2k   mh  2  m lim h 0 h  3m  2k  2  0 L.H.D

lim

k

 3  h   1  2k h

h 0

k lim



4h 2 h

h 0

k lim



4h 2 h

h 0

lim h 0







4h 2

4h 2

k  4  h  4  h



4h 2









k 4

From above, k  m and 3m  2k  2  0 4 2 8  m  ,k  5 5 k m 2 Question 80: If g is the inverse of a function f and f '  x  

1 , then g '  x  is equal to 1  x5

(a) 1  x5 (b) 5x 4 (c)

1 1   g  x 

5

(d) 1   g  x 

5

Solution: (d) Since g is inverse of f …(i) f  g  x  x On differentiating (i) with respect to x using chain rule, we get

f ' g  x   g '  x   1  g ' x 

1 f ' g  x 

 g ' x 

1 1

1   g  x 

5

 g '  x   1   g  x 

5

Question 81: x 2 y 3 z 4 x 1 y  4 z  5 and are coplanar, then k can have     If the lines 1 1 k k 2 1

(a) any value (b) exactly one value (c) exactly two values (d) exactly three values Solution: (c) Given lines are: x 2 y 3 z 4   1 1 k

…(i)

x 1 y  4 z  5 …(ii)   k 2 1 Two lines are coplanar if x1  x2 y1  y2 z1  z2

l1

m1

n1  0

l2

m2

n2

 x1 , y1 , z1  and  x2 , y2 , z2  are the points lying on the line (i) and (ii).  l1 , m1 , n1  and  l2 , m2 , n2  are the directional cosines of line (i) and (ii).

2 1  1 k

3 4 1 2

45 k 0 1

 11  2k   1  k 2    2  k   0  k 2  2k  k  0  k 2  3k  0  k  0, 3

Question 82: The angle between the lines whose direction cosines satisfy the equations l  m  n  0 and

l 2  m2  n2 is  (a) 3  (b) 4  (c) 6  (d) 2 Solution: (a) Angle between two lines is given as a1a2  b1b2  c1c2 cos   2 a1  b12  c12 a2 2  b2 2  c2 2 We have, l 2  m2  n 2 l mn  0

…(i)

 l   m  n  l 2  m  n

2

From (i), 2mn = 0 When, m  0  l  n Hence,  l , m, n  is 1, 0, 1 .

When, n  0  l  m Hence,  l , m, n  is 1, 1, 0  .

1 0  0

 cos    

12  02   1

2

12   1  02 2



1 2

 3

Question 83: xad ya z ad       

The lines

and

x b c y b z b c are    r   

coplanar and then equation to

the plane in which they lie, is (a) x + y + z = 0 (b) x  y + z = 0 (c) x  2y + z = 0 (d) x + y  2z = 0

Solution: (a) a d bc a b a  d bc      0 The lines will be coplanar, if  r   r

Add 3rd column to first and it becomes twice the second and hence the determinant is zero, as the two columns are identical. Again, the equation of the plane in which they lie is xad    r

ya

 

zad    0  r

On adding 1st and 3rd columns and subtracting twice the 2nd, we get xz 0 0

ya

 

zad    0  r

 [( + r) ( +)] (x + z 2y) = 0  x + z 2y = 0

Question 84: Let a = 2i + j + k, b = i + 2j  k, and a unit vector c be coplanar. If c is perpendicular to a, then c is (a)

1  j  k  2

(b)

1  i  j  k  3

(c)

1 i  2 j  5

(d)

1 i  j  k  3

Solution: (a) c is coplanar with a and b. c=xa+yb  c = x (2i + j + k) + y(i + 2j k)  c = (2x + y)i + (x + 2y).j + (x y)k ac=0 2 (2x + y) + x + 2y + x y = 0 y = 2x c = 3xj + 3xk = 3x(-j + k) |c| = i  9x2 + 9x2 = 1 x c 

1 3 2

1 j  k  2

Question 85: The distance of the point (1, 0, 2) from the point of intersection of the line and the plane x  y  z  16 , is

(a) 2 14

x  2 y 1 z  2   3 4 12

(b) 8 (c) 3 21 (d) 13 Solution: (d) Let,

 5  1   3  0  14  2  2

2

2

 169  13

The above coordinate will also satisfy the equation of plane   3k  2    4k  1  12k  2   16  k 1 So, the coordinate of the point of intersection of line and plane are: (5, 3, 14)

Required distance =

 5  1   3  0  14  2  2

2

2

 169  13

Question 86: A ray of light along x  3 y  3 get reflected upon reaching x-axis, the equation of the reflected ray is: (a) y  x  3 (b)

3y  x  3

(c) y  3x  3 (d)

3 y  x 1

Solution: (b)

Take any point B(0, 1) on given line AB. Then, the coordinates of B’(0, −1).

Equation of AB’ 1 y x 3 3







 3y  x  3



Question 87: The x-coordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is (a) 2  2 (b) 2  2 (c) 1  2 (d) 1  2 Solution: (b) The coordinates of mid points of the sides of triangle are as (0, 1) (1, 1) and (1, 0). So, the required triangle is as shown below:

x-coordinate of the incentre = =

2 0  2 2  0  2 2 222 2

2 2 2

 2 2 Question 88: The image of the line

x 1 y  3 z  4 in the plane 2 x  y  z  3  0 is the line   3 1 5

x 3 y 5 z 2   3 1 5 x 3 y 5 z  2 (b)   3 1 5 x 3 y 5 z 2 (c)   3 1 5 x 3 y 5 z 2 (d)   3 1 5 Solution: (a)

(a)

We have, Equation of line passing through A x 1 y  3 z  4   3 1 5 Equation of plane 2x  y  z  3  0 So, the directional cosines of the normal to the plane is (2, −1, 1) So, the equation of line AB is x 1 y  3 z  4   k 2 1 1

So, the coordinates of point B are  2k  1,3  k , k  4  The coordinates of point B will also satisfy the equation of the plane 2  2k  1   3  k    k  4   3  0  k  1 Coordinates of point B are: (−1, 4, 3) By applying mid point formula: Coordinates of C are: (−3, 5, 2) So, equation of line passing through point C and parallel to the line passing through point B is x 3 y 5 z 2   3 1 5

Question 89: The value of λ, for which the equation 6x2 + 11 xy  10y2 + x + 3 + y + λ = 0 represents a pair of straight lines, is (a) 15 (b) 0 (c) 2 (d) None of these Solution: (d) Given, 6x2 + 11 xy  10y2 + x + 3 + y + λ = 0 The given equation will represent a pair of straight lines, if  = abc + 2fgh  af2  bg2 - ch2 = 0 Here, a  6, h 

11 , b  10, 2

1 1 g  , f  ,c  3  2 2

2   1 2    1 2   11   1  1  11   6       10        λ     2       60  λ    2    2   2  2  2  2       121  11  1  λ     60 λ  4  4  361λ =15

λ

15 361

Question 90: A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is: (a) 2x + 3y = xy (b) 3x + 2y = xy (c) 3x + 2y = 6xy (d) 3x + 2y = 6 Solution: (d)

The equation of PQ is given as

x y  1 h k

Since, the line passes through (2, 3) Hence,

2 3  1 h k

Therefore, the locus would be

 3x  2 y  xy

2 3  1 x y

Question 91: A line L passing through the focus of the parabola y2 = 4(x  1), intersects the parabola in two distinct points. If ’m’ be the slope of the line ‘L’ then Options: (a) 1 < m < 1 (b) m< 1or m> 1 (c) m  R (d) None of the above Solution: (d) Let y = Y, x 1 = X Then, the equation becomes Y2 = 4X. So, the focus = (2, 0) Any line through the focus is y = m(x  2). y2 = 4 (x  1)

On solving this with

m2(x  2)2 = 4(x  1) 

m2x2 4 (m2 + 1) x + 4 (m2 +1) = 0

If m  0, D = 16 (m2 + 1)2 16m2(m2 + 1) = 16(m2 + 1) > 0, for all m But, if m = 0, then x does not have two real distinct values. So, m  R except m = 0  m  R  {0} Question 92: The tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x+12y + c = 0 at (a) (6, 7) (b) (– 6, 7)

(c)(6, –7) (d)( –6, –7) Solution: (d) The tangent at (1,7) to the parabola x2 = y − 6 is x

1  y  7  6 2



2x = y+7 − 12



y = 2x + 5

which is also a tangent to the circle x2 + y2 + 16x + 12y + c =0 

x2 + (2x + 5)2 + 16x + 12(2x + 5) + c =0



5x2 + 60x +85 + c =0

must have equal roots. Let  and β be the roots of the equation Then,

 + β = − 12   = − 6

 x = − 6 and

(  = β)

y = 2x + 5 = −7

 Point of contact is (− 6, − 7), Question 93: If z is a complex number such that z  2 , then the minimum value of z 

(a) is equal to

5 2

(b) lies in the interval (1, 2) (c) is strictly greater than

5 2

1 2

(d) is strictly greater than

3 5 but less than 2 2

Solution: (b)

z  2 is the region on or outside circle whose centre is (0, 0) and radius is 2. Minimum z 

1 is distance of z, which lie on circle z  2 from (−1/2, 0). 2

2

1 3    2    0  2 2  Question 94: If z1, z2, z3 ,. z4 are the four complex numbers represented by the vertices of a quadrilateral taken z z   in order such that z1 – z4 = z2 − z3 and amp  4 1   , then the quadrilateral is a  z2  z1  2 (a) rhombus (b) square (c) rectangle (d) cyclic quadrilateral Solution: (d) Given,

z z   amp  4 1    z2  z1  2

Also,

z1 - z4 = z2 – z3

 z3 - z4 = z2 – z1

   z2  z3     amp    z3  z4  2 z z    amp  2 3    z4  z3  2 Since two angle of the quadrilateral is 90 and the sum of opposite angles of a quadrilateral is 180 so, the quadrilateral ABCD is cyclic quadrilateral. Question 95:

If (  1) is a cube root of unity and i =

1 1 i  2 2 1  2  1 is equal to 1 , then 1  i i i    1 1

(a) 0 (b) 1 (c) i (d)  Solution: (a) Applying R2  R2 – (R1 + R3), then

1 1 i  2 2 1 1 i  2 2 1 i 1  2 1  0 0 0 i i    1 1 i i    1 1

Since all the elements of the second row of the determinant is 0, so the value of determinant is zero.

1 1 i  2 2  0 0 0 0 i i    1 1 Question 96:

1 1 1 1 3 2 Let     i where i  1 , then the value of the determinant 1 1    2 is 2 2 1 2 4 (a) 3 (b) 3 ( − 1) (c) 32 (d) 3 (1 − ) Solution: (b)

1 1 1 3 1 1 2 2 Applying C1  C1 + C2 + C3 , then 1 1     0   2 1 2 4 0 2  Expanding the resultant determinant along first column of the determinant

3 0

1

1

  2  3 ( 2   4 )  3 ( 2   3   )  3 ( 2   )  3   1 0 2 

Question 97:

 1 3 If i  1 , then 4  5    i  2   2 (a) 1  i 3 (b) 1  i 3 (c) i 3

334

 1 i 3 3     2   2

365

is equal to



3

 1

(d) i 3 Solution: (c) We know that, 1 i 3    2 2

Also,  3  1, 1     2  0

 1 i 3 4  5     2   2

334

 1 i 3 3     2   2

365

= 4 + 5334 + 3365

= 4 + 5 + 32 = 1 + 2 + 3 (1 +  + 2)

 1  i 3   1  2    0 2   =1−1+ i 3

i 3 Question 98: For the equation 10z2  4iz  m = 0 where z is a complex variable and i  1 , which of the following is true? (a) Roots are purely imaginary for all negative realvalues of m (b) Roots are not real for all complex numbers m (c) Roots are purely irnaginary for all positive real values of m (d) None of the above Solution: (a) Given: 10z2 4iz m = 0 By using quadratic formula, we get

z

  4i  

 4i 

2

 4 10  m 

20

4i  16i 2  40m 20

z

We know i 2  1  z

4i  16  40m 20



4i  2 4  10m 20



2i  4  10m 10

D = 4 + 10 m If m is negative, then D will also be negative. So, the roots of z will be purely imaginary. Question 99: The Boolean expression   p  q     p  q  is equivalent to : (a) p (b) q (c) ~q (d) ~p Solution: (d) Consider,   p  q     p  q 

   p  q     p  q   p    q  q   p  1  p

Question 100: If

9

9

i 1

i 1

  xi  5  9 and   xi  5

2

 45 , then the standard deviation of the 9 items x1, x2 …..x9 is

(a) 4 (b) 2 (c) 3 (d) 9 Solution: (b) 9

We have

  x  5  9   x i

i 1

9

Also,

  x  5 i 1

i

i

2

 54

 45

  xi2  10. xi  9  25   45

Substitute  xi  54

  xi2  540  225  45   xi2  360

x The variance is given by 9

360  54     9  9   40  36 4

2 i

  xi    9

  

2

2

Therefore, the standard deviation is 2 which is equal to the square root of variation.