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Question 1: If
Solution:
Answer:(d)
Apply log on both sides
log (y1/3x1/2) = log(x+y)5/6
(1/3) log y+(1/2) log x = (5/6) log(x+y)
(1/3y)(dy/dx)+(1/2x) = (5/6(x+y))(1+dy/dx)
(1/3y)(dy/dx)- (5/6(x+y))dy/dx =(5/6(x+y)) -(1/2x)
(dy/dx)[(1/3y)-(5/6(x+y)] = 5x-3(x+y)/6x(x+y)
(dy/dx) = [(2x-3y)/6x(x+y)][6y(x+y)/(2x-3y)]
dy/dx = y/x
Question 2: Rolle's theorem is not applicable in which one of the following cases?
Solution:
Answer:(c)
Check option:
Option (a):
(1) f(x) is continuous function in [1, 3]
(2) f(x) is differentiable in (1, 3)
Option (b):
(1) f(x) is continuous function in [0, 1]
(2) f(x) is differentiable in (0, 1)
Option (c):
(1) f(x) is continuous function in [–2, 2]
(2) f(x) is not differentiable at point x = 0 in (–2, 2)
Option (d):
(1) f(x) is continuous function in [2.5, 2.7]
(2) f(x) is differentiable in (2.5, 2.7)
Question 3: The interval in which the function f(x) = x3-6x2+9x+10 is increasing in
Solution:
Answer:(b)
f(x) = x3-6x2+9x+10
Differentiate with respect to x
f’(x) = 3x2 -12x+9
f’(x) ≥ 0
⇒ 3x2-12x+9 ≥0
⇒ x2-4x+3 ≥ 0
⇒ x2-3x -x+3 ≥ 0
⇒ (x-3) (x-1) ≥ 0
x ∈ (-∞, 1] ⋃[3, ∞ )
Question 4: The side of an equilateral triangle are increasing at the rate of 4cm/sec. The rate at which its area is increasing, when the side is 14 cm
Solution:
Answer:(G)
Bonus
Question 5: The value of √24.99 is
Solution:
Answer:(a)
f(x) = √x
⇒ dy/dx = 1/2√x
∆y = (dy/dx) ∆x …(i)
Here x = 25 and ∆x = 0.1
∆y = √x-√(x-∆x)
(dy/dx) ∆x = √25-√(25-0.1)
√24.99 = 5-(1/2×5)×0.1
= 5-0.01
= 4.99
Question 6: If 3x-5≤2 then
Solution:
Answer:(b)
|3x-5| ≤ 2
⇒ -2 ≤3x–5 ≤ 2
⇒ 3 ≤ 3x ≤ 7
⇒ 1 ≤ x ≤7/3
Question 7: A random variable 'X' has the following probability distribution:
X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X) | k-1 | 3k | k | 3k | 3k2 | k2 | k2+k |
Solution:
Answer:(a)
⇒ K-1+3K+K+3K+3K2+K2+K2+K = 1
⇒ 5K2+9K-2 = 0
⇒5K2+10K-K-2 = 0
⇒ (5K-1)(K+2) = 0
5K-1 = 0 or K+2 = 0
K = 1/5 or K = -2 (not possible)
Question 8: If A and B are two events of a sample space S such that P(A) =0.2, P(B) =0.6 and P(AB)=0.5 then P(A’|B)=
Solution:
Answer:(c)
Given that
P(A) = 0.2
P(B) = 0.6
P(A|B) = 0.5
Now
P(A’|B) = 1-P(A|B)
= 1-0.5
= 0.5
= 1/2
Question 9: If ‘X’ has a binomial distribution with parameters n = 6, p and P(X = 2) = 12, P(X = 3) = 5 then P=
Solution:
Answer:(G)
Bonus
*G indicates one grace mark
Question 10: A man speaks truth 2 out of 3 times. He picks one of the natural numbers in the set S = {1, 2, 3, 4, 5, 6, 7} and reports that it is even. The probability that it is actually even is
Solution:
Answer:(d)
S = {1,2,3,4,5,6,7}
Total probability = 7
P(S1) = Probability of even number = 3/7
P(S2) = Probability of odd number = 4/7
P(ES1) = 2/3
P(ES2) = 1-2/3 = 1/3
Question 11: If U is the universal set with 100 elements; A and B are two sets such that n(A) = 50, n(B) = 60, n(A⋂B) = 20 then n(A’⋂B’) = ?
Solution:
Answer:(d)
Given that
n(A) = 50
n(B) = 60
n(A⋂B) = 20
Now
n(A’⋂ B’) = P(A⋂B)’
= 100-P(A⋂B)
= 100-[P(A)+P(B)- P(A⋂B)]
= 100-[50+60-20]
= 100-90
= 10
Question 12: The domain of the function f:R→R defined by f(x) = √(x2-x+12)
Solution:
Answer:(a)
f(x) = √(x2-7x+12)
Now
x2-7x+12≥0
x2-4x-3x+12≥0
(x-4)(x-3) ≥0
x∈(-∞,3] ⋃[4, ∞)
Question 13: If cos x = sin x then, the general solution is
Solution:
Answer:(b)
cos x = |sinx|
Case-1:
cos x = sin x for x∈ [0, π ]
tan x = 1 for x = π/4
Case-2:
s x = -sin x for x ∈ [π, 2π]
tan x = -1 for x∈ 7π/4
Combining both the cases, we get the general is solution 2nπ±π/4
Question 14: √3 cosec 200- sec 200 =
Solution:
Answer:(c)
Question 15: If P(n): 2n <n! then the smallest positive integer for which P(n) is true, is
Solution:
Answer:(c)
P(n): 2n< n!
Since
P(1): 2 < 1 is false.
P(2): 22< 2! is false.
P(3): 23< 3! is false.
P(4): 24< 4! is true.
P(5): 25< 5! is true.
So, the smallest positive integer is 4.
Question 16: Foot of the perpendicular drawn from the point (1,3,4) to the plane 2x-y+z+3=0 is
Solution:
Answer:(a)
The Dr’s of PA are
x1-1, y1-3, z1-4
The Dr’s n are 2, -1, 1
The Dr’s of PA and n are parallel
Question 17: Acute angle between the line (x-5)/2 = (y+1)/-1 = (z+4)/1 and the plane 3x-4y-z+5 = 0 is:
Solution:
Answer:(c)
L: (x-5)/2 = (y+1)/-1 = (z+4)/1
Dr’s of line: 2, -1, 1
P: 3x-4y-z + 5 = 0
Dr’s of normal to plane: -3, 4, 1
Question 18: The distance of the point (1,2,1) form the line (x-1)/2 = (y-2)/1 = (z-3)/2 is
Solution:
Answer:(b)
Let L: (x-1)/2 = (y-2)/1 = (z-3)/2 = λ
Vector PA perpendicular to L.
So
Question 19: XY-plane divides the line joining the points A(2,3,-5) and B(-1,-2,-3) in the ratio
Solution:
Answer:(d)
A(2,3,-5), B(-1,-2,-3)
The ratio that xy plane divide line joining the points (x1,y1,z1) and (x2,y2,z2) = -z1:z2
If result is positive, it divides internally otherwise externally.
The ratio that xy plane divides the points (2,3,-5) and (-1,-2,-3)
= -(-5) :(-3)
= -5 : 3
So, 5 : 3 externally in the ratio.
Question 20: The shaded region in the figure is the solution set of the inequations
Solution:
Answer:(a)
(1) Since the region lies above the x-axis
So y ≥0
(2) Shaded region is on right side of the y-axis
So x ≥0
(3) Equation of line passing through (5,0) and (0, 4) is
(y-0) = (4/-5) (x-5)
-5y = 4x- 20
4x+5y = 20
(4) Equation of line passing through (10,0) and (0, 3) is
(y-0) = (-3/10) (x-10)
-10y = 3x-30
3x+10y = 30
(5) Shaded region is on left of x = 6.
So x ≤6
Consider the figure and observe the shaded region the inequalities can be written as
x ≥ 0,
y ≥ 0,
x ≤ 6
4x+5y ≥20,
3x+10y ≤ 30
Question 21: The order of the differential equation is
Solution:
Answer:(a)
Differentiate with respect to x
dy/dx =
dy/dx = y
(dy/dx)-y = 0
Hence the order is 1.
Question 22: If
Solution:
Answer:(d)
Now,
=
= 16×4×1
= 64
Question 23: If the angle between
Solution:
Answer:(a)
Given angle between
And
⇒
⇒
⇒
⇒
Question 24: A unit vector perpendicular to the plane containing the vectors
Solution:
Answer:(c)
The vector which is perpendicular to two vectors is given by the cross product of the two vectors.
=
Unit vector perpendicular to P and Q is given by
=
=
Question 25:
Solution:
Answer:(d)
=
=
=
=
Question 26:
Solution:
Answer:(a)
Let I =
Question 27:
Solution:
Answer:(d)
Let I =
Let √x = t
x = t2
dx = 2t dt
So I =
I =
= 2 tan-1t +C
= 2 tan-1√x+C
Question 28:
Solution:
Answer:(d)
Let I =
Now,
(2x-1)/(x-1)(x+2)(x-3) = (A/(x-1) )+(B/(x+2))+(C/(x-3))
2x-1 = A(x+2)(x-3)+B(x-1)(x-3)+C(x -1)(x+2)
Putting x = 3, we get
6 -1 = C(3-1)(3+2)
C = ½
Putting x = 1, we get
2-1 = A(1+2)(1-3)
A = -1/6
Putting x = -2, we get
-4-1 = B(-2-1)(-2-3)
B = -1/3
A.B,C are -1/6, -1/3, 1/2.
Question 29:
Solution:
Answer:(a)
Let I =
I = (√2-1)+2(√3-√2)+3(2-√3)
= √2-1+2√3-2√2+6-3√3
= 5-√2-√3
Question 30:
Solution:
Answer:(b)
Let I =
= (π/2)-0-(0-1)
= (π/2)+1
Question 31: If α and β are the roots of the equation x2+x+1 = 0 then α2+β2 is:>
Solution:
Answer:(d)
Given equation is x2+x+1 = 0
α+β = -1
αβ = 1
Now, α2+β2 = (α+β)2-2αβ
= (-1)2-2(1)
= 1-2
= -1
Question 32: The number of 4-digit numbers without repetition that can be formed using the digits 1,2,3,4,5,6,7 in which each number has two odd digits and two even digits is
Solution:
Answer:(a)
Given digits are 1, 2, 3, 4, 5, 6, 7
Two even digits can be selected in 3C2
Two odd digits can be selected in 4C2 ways
These selected 4 digits can be arranged in 4! ways
The total number of ways = 4C2 ×3C2 ×4!
= 6×3×24
= 432
Question 33: The number of terms in the expansion of (x2+y2)25-(x2-y2)25 after simplification is
Solution:
Answer:(b)
(x2+y2)25-(x2-y2)25
(a+b)n-(a-b)n if n is odd
On simplification we get (n+1)/2 terms
i.e. (25+1)/2 = 26/2
= 13
Question 34: The third term of a G.P. is 9. The product of its first five terms i
Solution:
Answer:(c)
The third term of G.P. is 9
a.2 = 9
Now, the product of five terms = T1T2T3T4T5
= (a)(ar)(ar2)(ar3)(ar4)
= a5r10
= (ar2)5
= (9)5
= 310
Question 35: A line cuts off equal intercepts on the co-ordinate axes. The angle made by this line with the positive direction of X-axis is
Solution:
Answer:(d)
We know that equation of a line in intercept form is
(x/a)+(y/b) = 1
Here, both intercept are equal
So (x/a)+(y/a) = 1
y = –x+a
Now comparing it with
y = mx + c
We get m = -1 = slope
We know that
Slope = tan θ
tan θ = -1
θ = 1350
Question 36:.
Solution:
Answer:(c)
Let I =
Question 37: The area of the region above X-axis included between the parabola y2= x and the circle x2+y2 = 2x in square units is
Solution:
Answer:(d)
y2 = x …(1)
x2+ y2 = 2x…(2)
(x-1)2 +(y-0)2 = 1
Equation (2) is a circle with centre (1, 0) and radius 1.
Solving eq.(1) and eq.(2)
(x-1)2 + x = 1
x2-x = 0
x(x-1) = 0
x = 0, 1
We get the points of intersection (0, 0) and (1, 1)
= [0+ π/4]-(2/3)
= (-2/3)+( π/4)
Question 38: The area of the region bounded by Y-axis, y = cos x and y = sin x; 0 ≤x≤π/2 is
Solution:
Answer:(a)
Required area =
=
= (1/√2)+(1/√2)-1)
= (√2-1) Sq. units
Question 39: The integrating factor of the differential equation (2x+3y2)dy = y dx (y > 0) is
Solution:
Answer:(d)
(2x+3y2)dy = y dx
(2x+3y2)/y = dx/dy
(dx/dy)-(2/y)x = 3y
I.F =
= e-2 log y
= y-2
= 1/y2
Question 40: The equation of the curve passing through the point (1,1) such that the slope of the tangent at any point (x,y) is equal to the product of its coordinates is
Solution:
Answer:(c)
(dy/dx) = xy
(1/y)dy = x dx
Integrate
log y = (x2/2)+C ..(i)
Equation (i) passing through (1,2)
log(1) = (12/2)+C
C = -1/2
Put in (i)
log y = (x2/2)-(1/2)
2 log y = x2-1
Question 41: The eccentricity of the ellipse 9x2+25y2 = 225 is
Solution:
Answer:(a)
9x2+25y2 = 225
(x2/52)+(y2/32) = 1
Eccentricity (e) = √(a2-b2)/a
= √(25-9)/5
= 4/5
Question 42:
Solution:
Answer:(b)
= (1/4)×1
= 1/4
Question 43: The negation of the statement ''All continuous functions are differentiable.''
Solution:
Answer:(c)
Always all differentiable functions are continuous. But all continuous functions are not differentiable. So, negation of the given statement is some continuous functions are not differentiable. Ex: |x|
Question 44: Mean and standard deviation of 100 items are 50 and 4 respectively. The sum of all squares of the items is
Solution:
Answer:(a)
Here n = 100
A = 50
σ = 4
Now A = Σx/n
Σx = An
Σx = 100×50
= 5000
Again from the formula,
σ2+A2 = Σx2/n
Σx2 = n(σ2+A2)
= 100(16+2500)
= 251600
Question 45: Two letters are chosen from the letters of the word 'EQUATIONS'. The probability that one is vowel and the other is consonant is
Solution:
Answer:(d)
Vowel: E, A, U, I, O ⇒ P(v) = 5C1
Consonant: Q, T, N, S ⇒ P(c) = 4C1
Probability = 5C1 × 4C1/9C2
= 5×4/36
= 20/36
= 5/9
Question 46: The constant term in the expansion of
Solution:
Answer:(d)
Put x = 0
= -3+3+6
= 6
Question 47: If [x] represents the greatest integer function and f(x) = x-[x]-cos x then f’(π/2) =
Solution:
Answer:(c)
f(x) = x-[x]-cos x
since x = [x]+{x}
f(x) = {x}-cos x
f’(x) = 1+sin x
f’(π/2) = 1+ sin (π/2)
= 1+1
= 2
Question 48: If
Solution:
Answer:(G)
Bonus
*G indicates one grace mark
Question 49: If f(x) = sin-1[(2x+1 )/(1+4x)], then f’(0) =
Solution:
Answer:(b)
f(x) = sin-1[(2x+1 )/(1+4x)]
= sin-1[(2×2x)/(1+(2x)2)]
= 2 tan-1(2x)
Differentiating w.r.t.x
f’(x) = (2/(1+4x))(2x log 2)
f’(0) = (2/(1+1)) log 2
f’(0) = log 2
Question 50: If x = a sec2 θ, y = a tan2 θ then d2y/dx2 =
Solution:
Answer:(c)
x = a sec2 θ
y = a tan2 θ
y = a(sec2 θ-1)
y = a(x-1)
dy/dx = a
d2y/dx2 = 0
Question 51: The inverse of the matrix
Solution:
Answer:(c)
Let A =
= 1
A-1= adj A/|A|
Question 52: If P and Q are symmetric matrices of the same order then PQ-QP is
Solution:
Answer:(d)
Given P = P’ and Q = Q’
Now,
(PQ-QP)’ = (PQ)’-(QP)’
= Q’P’-P’Q’
= QP-PQ
= -[PQ-QP]
So (PQ-QP) is skew-symmetric.
Question 53: If 3A+4B’ =
Solution:
Answer:(a)
3A+4B’ =
2B-3A’ =
Question 54: If A =
Solution:
Answer:(b)
A =
B =
B’ =
Now, |ABB’| = |A| |B| |B’|
= (2-12)(4+1)(4+1)
= -10×5×5
= -250
Question 55: If the value of a third-order determinant is 16, then the value of the determinant formed by replacing each of its elements by its cofactor is
Solution:
Answer:(c)
Given that |A| = 16
Order of determinant A = 3
Now, |adj A| = |A|2
= 162
= 256
Question 56:
Solution:
Answer:(b)
f(x) = x2
g(x) = √x
Now,
fog(x) = f[g(x)]
= f[√x ]
= x
fog(2) = 2
fog(-4) = -4
And
gof(x) = g[f(x)]
gof(x) = g[x2]
gof(-2) = g(4)
= √4
= 2
Question 57: A = {x|x∈N, x ≤5}, B = {x|x∈Z, x2-5x+6 = 0}, then the number of onto functions from A to B is
Solution:
Answer:(c)
A = {x|x∈N, x ≤5}
A = {1,2,3,4,5}
B = {x|x∈Z, x2-5x+6 = 0}
B = {2,3}
= (5+10)×2
= 30
Question 58: On the set of positive rationals, a binary operation * is defined by a*b = 2ab/5 . If 2*x = 3-1, then x =
Solution:
Answer:(b)
Given condition
a*b = 2ab/5
Now,
2*x = 3-1
2(2x)/5= 1/3
4x = 5/3
x = 5/12
Question 59: cos[2 sin-1 (3/4)+cos-1(3/4)] =
Solution:
Answer:(a)
cos[2 sin-1 (3/4)+cos-1(3/4)] = cos [2 sin-1(3/4)+cos-1(3/4)]
= cos [sin-1(3/4)+ sin-1(3/4)+cos-1(3/4)]
= cos [(π/2)+sin-1(3/4)]
= -sin(sin-1(3/4)
= -3/4
Question 60: If a+ π/2 <2 tan-1x+3 cot-1x<b then ‘a’ and ‘b’ are respectively:
Solution:
Answer:(b)
We know that 0 < cot-1x < π
⇒ π+0<2 (π/2)+ cot-1x < π+ π
⇒ π<2(tan-1x+cot-1x)+cot-1x<2π
⇒ π<2tan-1x+3cot-1x<2π ..(1)
compare equation (1) to a+ π/2 <2 tan-1x+3 cot-1x<b
a = π/2
b = 2π
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