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KCET 2019 - Maths
Question 1: If \(\sqrt[3]{y}\sqrt{x}=\sqrt[6]{(x+y)^{5}}\), then dy/dx equal to:
- a. x/y
- b. x+y
- c. x-y
- d. y/x
Solution:
Answer:(d)
\(\sqrt[3]{y}\sqrt{x}=\sqrt[6]{(x+y)^{5}}\)
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?
- a. f(x) = x2-4x+5 in [1, 3]
- b. f(x) = x2-x in [0, 1]
- c. f(x) = x in [-2, 2]
- d. f(x) = [x] in [2.5, 2.7]
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
- a. (-∞,1) ⋃ (3, ∞)
- b. (-∞,-1) ⋃ (-3, ∞)
- c. [1, 3]
- d. (-∞,-1) ⋃ [3, ∞)
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
- a. 10√3 cm2/sec
- b. 14√3 cm2/sec
- c. 42 cm2/sec
- d. 14 cm2/sec
Solution:
Answer:(G)
Bonus
Question 5: The value of √24.99 is
- a. 4.999
- b. 4.899
- c. 5.001
- d. 4.897
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
- a. -1≤x≤7/3
- b. 1≤x≤7/3
- c. 1≤x≤9/3
- d. -1≤x≤9/3
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 |
- a. 1/5
- b. -2
- c. 2/7
- d. 1/10
Solution:
Answer:(a)
\(\sum P(K)=1\)
⇒ 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)=
- a. 3/10
- b. 2/3
- c. 1/2
- d. 1/3
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=
- a. 5/12
- b. 16/21
- c. ½
- d. 5/16
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
- a. 2/5
- b. 1/5
- c. 1/10
- d. 3/5
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’) = ?
- a. 40
- b. 20
- c. 90
- d. 10
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)
- a. (-∞,3] ⋃[4, ∞)
- b. (-∞,3] ⋃(4, ∞)
- c. (-∞,3] ⋂ [4, ∞)
- d. (3,4)
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
- a. x = nπ ± π/4, n∈Z
- b. x = 2nπ ± π/4, n∈Z
- c. x = nπ+(-1)n π/4, n∈Z
- d. x = (2n+1)π ± π/4, n∈Z
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 =
- a. 2
- b. 3
- c. 4
- d. 1
Solution:
Answer:(c)
Question 15: If P(n): 2n <n! then the smallest positive integer for which P(n) is true, is
- a. 2
- b. 3
- c. 4
- d. 5
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
- a. (-1,4,3)
- b. (0,-4,-7)
- c. (1,2,-3)
- d. (-3,5,2)
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:
- a. cos-1(9/√364)
- b. sin-1(9/√364)
- c. cos-1(5/2√13)
- d. sin-1(5/2√13)
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
- a. 2√3/5
- b. 2√5/3
- c. √5/3
- d. 20/3
Solution:
Answer:(b)
Let L: (x-1)/2 = (y-2)/1 = (z-3)/2 = λ
\(\overrightarrow{PA}\) = Position vector of A - position of vector of P
Vector PA perpendicular to L.
So \(\overrightarrow{PA}.L=0\)
Question 19: XY-plane divides the line joining the points A(2,3,-5) and B(-1,-2,-3) in the ratio
- a. 2:1 internally
- b. 3:2 externally
- c. 5:3 internally
- d. 5:3 externally
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
- a. 4x+5y≥20, 3x+10y≤30, x≤6, x, y≥0.
- b. 4x+5y≥20, 3x+10y≤30, x≥6, x, y≥0.
- c. 4x+5y≤20, 3x+10y≤30, x≤6, x, y≥0.
- d. 4x+5y≤20, 3x+10y≤30, x≥6, x, y≥0.
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 \(y= C_{1}e^{C_{2}+x}+C_{3}e^{C_{4}+x}\)
- a. 1
- b. 2
- c. 3
- d. 4
Solution:
Answer:(a)
\(y= C_{1}e^{C_{2}+x}+C_{3}e^{C_{4}+x}\)
Differentiate with respect to x
dy/dx = \(C_{1}e^{C_{2}+x}+C_{3}e^{C_{4}+x}\)
dy/dx = y
(dy/dx)-y = 0
Hence the order is 1.
Question 22: If \(\left | \vec{a} \right |=16\), \(\left | \vec{b} \right |=4\), then \(\sqrt{\left | \vec{a\times \vec{b}} \right |^{2}+\left | \vec{a}.\vec{b} \right |^{2}}\)
- a. 4
- b. 8
- c. 16
- d. 64
Solution:
Answer:(d)
\(\left | \vec{a} \right |=16\)
\(\left | \vec{b} \right |=4\)
Now,
\(\sqrt{\left | \vec{a\times \vec{b}} \right |^{2}+\left | \vec{a}.\vec{b} \right |^{2}}= \sqrt{\left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}\sin ^{2}\theta +\left | \vec{a} \right |^{2}\left | \vec{b} \right |^{2}\cos ^{2}\theta}\)
= \(\left | \vec{a} \right |\left | \vec{b} \right |\sqrt{\sin ^{2}\theta +\cos ^{2}\theta }\)
= 16×4×1
= 64
Question 23: If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{2\pi }{3}\) and the projection of \(\vec{a}\) in the direction of \(\vec{b}\) is -2, then \(\left | \vec{a} \right |\) =
- a. 4
- b. 3
- c. 2
- d. 1
Solution:
Answer:(a)
Given angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{2\pi }{3}\)
And \(\frac{\vec{a}.\vec{b}}{\left | \vec{b} \right |}=-2\)
⇒\(\frac{\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta }{\left | \vec{b} \right |}=-2\)
⇒\(\left | \vec{a} \right |\cos \frac{2\pi }{3}=-2\)
⇒\(\left | \vec{a} \right |\times \frac{-1}{2}=-2\)
⇒\(\left | \vec{a} \right |=4\)
Question 24: A unit vector perpendicular to the plane containing the vectors \(\hat{i}+2\hat{j}+\hat{k}\) and \(-2\hat{i}+\hat{j}+3\hat{k}\) is:
- a. \(\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\)
- b. \(\frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\)
- c. \(\frac{-\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\)
- d. \(\frac{-\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}}\)
Solution:
Answer:(c)
The vector which is perpendicular to two vectors is given by the cross product of the two vectors.
\(\vec{P}\times \vec{Q}=(-2\hat{i}+\hat{j}+3\hat{k})\times (\hat{i}+2\hat{j}+\hat{k})\)
= \(-5\hat{i}+5\hat{j}-5\hat{k}\)
\(\left | \vec{P} \times \vec{Q}\right |=\sqrt{5^{2}+5^{2}+5^{2}}=5\sqrt{3}\)
Unit vector perpendicular to P and Q is given by
\(C=\frac{\vec{P}\times \vec{Q}}{\left | \vec{P}\times \vec{Q} \right |}\)
= \(\frac{-5\hat{i}+5\hat{j}-5\hat{k}}{5\sqrt{3}}\)
= \(\frac{-\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\)
Question 25: \(\left [ \vec{a}+2\vec{b} -\vec{c}, \vec{a}-\vec{b},\vec{a}-\vec{b}-\vec{c}\right ]\)
- a. 0
- b. \(\left [ \vec{a},\vec{b} ,\vec{c}\right ]\)
- c. \(2\left [ \vec{a},\vec{b} ,\vec{c}\right ]\)
- d. \(3\left [ \vec{a},\vec{b} ,\vec{c}\right ]\)
Solution:
Answer:(d)
\((\vec{a}+2 \vec{b}-\vec{c}) \cdot\{\vec{a}-\vec{b} \times \vec{a}-\vec{b}-\vec{c}\}\)
= \((\vec{a}+2 \vec{b}-\vec{c}) \cdot\{-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{b} \times \vec{a}+\vec{b} \times \vec{c}\}\)
= \([\vec{a} \: \vec{b}\; \vec{c}]-2[\vec{b}\: \vec{a}\: \vec{c}]\)
= \([\vec{a} \: \vec{b}\: \vec{c}]+2[\vec{a} \: \vec{b}\: \vec{c}]\)
= \(3[\vec{a}\: \vec{b}\: \vec{c}]\)
Question 26: \(\int_{-3}^{3}\cot ^{-1}x\: dx=\)
- a. 3π
- b. 0
- c. 6π
- d. 3
Solution:
Answer:(a)
Let I = \(\int_{-3}^{3}\cot ^{-1}x\: dx\)
Question 27: \(\int \frac{1}{\sqrt{x}+x\sqrt{x}}dx=\)
- a. 2 log (√x+1)+C
- b. (½) tan-1√x+C
- c. tan-1√x+C
- d. 2 tan-1√x+C
Solution:
Answer:(d)
Let I = \(\int \frac{1}{\sqrt{x}+x\sqrt{x}}dx\)
\(\int \frac{1}{\sqrt{x}(x+1)}dx\)
Let √x = t
x = t2
dx = 2t dt
So I = \(\int \frac{1}{t(1+t^{2})}2t\: dt\)
I = \(2\int \frac{1}{(1+t^{2})}\: dt\)
= 2 tan-1t +C
= 2 tan-1√x+C
Question 28: \(\int \frac{2x-1}{(x-1)(x+2)(x-3)}dx=A\: log\left | x-1 \right |+B\: log\left | x+2 \right |+C\: log \left | x-3 \right |+K\), then A, B, C are respectively:
- a. -1/6, 1/3, -1/2
- b. 1/6, 1/3, -1/5
- c. 1/6, -1/3, 1/3
- d. -1/6, -1/3, 1/2
Solution:
Answer:(d)
Let I = \(\int \frac{2x-1}{(x-1)(x+2)(x-3)}dx\)
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: \(\int_{0}^{2}\left [ x^{2} \right ]dx=\)
- a. 5-√2-√3
- b. 5+√2-√3
- c. 5-√2+√3
- d. -5-√2-√3
Solution:
Answer:(a)
Let I = \(\int_{0}^{2}\left [ x^{2} \right ]dx\)
I = (√2-1)+2(√3-√2)+3(2-√3)
= √2-1+2√3-2√2+6-3√3
= 5-√2-√3
Question 30: \(\int_{0}^{1}\sqrt{\frac{1+x}{1-x}}dx=\)
- a. (π/2)-1
- b. (π/2)+1
- c. π/2
- d. 1/2
Solution:
Answer:(b)
Let I = \(\int_{0}^{1}\sqrt{\frac{1+x}{1-x}}dx\)
= (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:>
- a. 1
- b. (-1+i√3)/2
- c. (-1-i√3)/2
- d. -1
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
- a. 432
- b. 436
- c. 450
- d. 454
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
- a. 0
- b. 13
- c. 26
- d. 50
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
- a. 35
- b. 39
- c. 310
- d. 312
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
- a. 450
- b. 900
- c. 1200
- d. 1350
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:. \(\int x^{3}\sin 3xdx=\)
Solution:
Answer:(c)
Let I = \(\int x^{3}\sin 3xdx\)
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
- a. (π/4)-(3/2)
- b. (3/2)-( π/4)
- c. (2/3)-( π/4)
- d. (π/4)-(2/3)
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
- a. √2-1 Sq. units
- b. √2 Sq. units
- c. √2+1 sq. units
- d. 2-√2 Sq. units
Solution:
Answer:(a)
Required area = \(\int_{0}^{\frac{\pi }{4}}(\cos x-\sin x)dx\)
= \([\cos x-\sin x]_{0}^{\frac{\pi }{4}}\)
= (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
- a. e1/y
- b. -1/y2
- c. 1/x
- d. 1/y2
Solution:
Answer:(d)
(2x+3y2)dy = y dx
(2x+3y2)/y = dx/dy
(dx/dy)-(2/y)x = 3y
I.F = \(e^{\int \frac{-2}{y}dy}\)
= 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
- a. 2 log x = y2-1
- b. 2 log y = x2+1
- c. 2 log y = x2-1
- d. 2 log x = y2+1
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
- a. 4/5
- b. 3/5
- c. ¾
- d. 9/16
Solution:
Answer:(a)
9x2+25y2 = 225
(x2/52)+(y2/32) = 1
Eccentricity (e) = √(a2-b2)/a
= √(25-9)/5
= 4/5
Question 42: \(\sum_{r=1}^{n}(2r-1)=x\) then \(\lim_{n \to \infty }\left [ \frac{1^{3}}{x^{2}} +\frac{2^{3}}{x^{2}}+\frac{3^{3}}{x^{2}}+...+\frac{n^{3}}{x^{2}}\right ] =\)
- a. ½
- b. ¼
- c. 1
- d. 4
Solution:
Answer:(b)
\(\sum_{r=1}^{n}(2r-1)=x\)
= (1/4)×1
= 1/4
Question 43: The negation of the statement ''All continuous functions are differentiable.''
- a. All continuous functions are not differentiable
- b. Some continuous functions are differentiable.
- c. Some continuous functions are not differentiable.
- d. All differentiable functions are continuous.
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
- a. 251600
- b. 256100
- c. 266000
- d. 261600
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
- a. 8/9
- b. 4/9
- c. 3/9
- d. 5/9
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 \(\begin{vmatrix} 3x+1 & 2x-1 &x+2 \\ 5x-1 & 3x+2 & x+1\\ 7x-2 & 3x+1 & 4x-1 \end{vmatrix}\) is:
- a. 0
- b. 2
- c. -10
- d. 6
Solution:
Answer:(d)
\(\Delta =\begin{vmatrix} 3x+1 & 2x-1 &x+2 \\ 5x-1 & 3x+2 & x+1\\ 7x-2 & 3x+1 & 4x-1 \end{vmatrix}\)
Put x = 0
\(\Delta =\begin{vmatrix} 1 & -1 &2 \\ -1 & 2 & 1\\ -2& 1 & -1 \end{vmatrix}\)
\(\Delta\) = 1(-3)+1(3)+2(-1+4)
= -3+3+6
= 6
Question 47: If [x] represents the greatest integer function and f(x) = x-[x]-cos x then f’(π/2) =
- a. 0
- b. 1
- c. 2
- d. does not exist
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
- a. 3/2
- b. 9/5
- c. 1/2
- d. 2/3
Solution:
Answer:(G)
Bonus
*G indicates one grace mark
Question 49: If f(x) = sin-1[(2x+1 )/(1+4x)], then f’(0) =
- a. 2 log 2
- b. log 2
- c. (2 log 2)/5
- d. (4 log 2)/5
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 =
- a. 2a
- b. 1
- c. 0
- d. 4
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 \(\begin{bmatrix} 2 &5 &0 \\ 0 & 1 &1 \\ -1 & 0 & 3 \end{bmatrix}\) is:
- a. \(\begin{bmatrix} 3 & -1 & 1\\ -15& 6 &-5 \\ 5 & -2 & 2 \end{bmatrix}\)
- b. \(\begin{bmatrix} 3 & -5 & 5\\ -1& -6 &-2 \\ 1 & -5 & 2 \end{bmatrix}\)
- c. \(\begin{bmatrix} 3 & -15 & 5\\ -1 &6 & -2\\ 1& -5 & 2 \end{bmatrix}\)
- d. \(\begin{bmatrix} 3 & -15 & 5\\ -1 &6 & -2\\ 1& -5 & -2 \end{bmatrix}\)
Solution:
Answer:(c)
Let A = \(\begin{bmatrix} 2 &5 &0 \\ 0 & 1 &1 \\ -1 & 0 & 3 \end{bmatrix}\)
\(\left | A \right |=2(3-0)-5(0+1)+0\)
= 1
A-1= adj A/|A|
Question 52: If P and Q are symmetric matrices of the same order then PQ-QP is
- a. Identity matrix
- b. Symmetric matrix
- c. zero matrix
- d. Skew symmetric matrix
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’ = \(\begin{bmatrix} 7 & -10 & 17\\ 0 &6 & 31 \end{bmatrix}\) and 2B-3A’ = \(\begin{bmatrix} -1 & 18\\ 4 & 0\\ -5& -7 \end{bmatrix}\) then B =
Solution:
Answer:(a)
3A+4B’ = \(\begin{bmatrix} 7 & -10 & 17\\ 0 &6 & 31 \end{bmatrix}\) ..(1)
2B-3A’ = \(\begin{bmatrix} -1 & 18\\ 4 & 0\\ -5& -7 \end{bmatrix}\) ..(2)
Question 54: If A = \(\begin{bmatrix} 1 & 3\\ 4 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} 2 & -1\\ 1 & 2 \end{bmatrix}\), then |ABB’|=
- a. 50
- b. -250
- c. 100
- d. 250
Solution:
Answer:(b)
A = \(\begin{bmatrix} 1 & 3\\ 4 & 2 \end{bmatrix}\)
B = \(\begin{bmatrix} 2 & -1\\ 1 & 2 \end{bmatrix}\)
B’ = \(\begin{bmatrix} 2 & 1\\ -1 & 2 \end{bmatrix}\)
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
- a. 96
- b. 48
- c. 256
- d. 16
Solution:
Answer:(c)
Given that |A| = 16
Order of determinant A = 3
Now, |adj A| = |A|2
= 162
= 256
Question 56: \(f:R\rightarrow R\) and \(g:[0,\infty )\rightarrow R\) is defined by f(x) = x2 and g(x) = √x. Which one of the following is not true ?
- a. gof (4) = 4
- b. fog (-4) = 4
- c. fog (2) = 2
- d. gof (-2) = 2
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
- a. 2
- b. 23
- c. 30
- d. 32
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 =
- a. 1/6
- b. 5/12
- c. 2/5
- d. 125/48
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)] =
- a. -3/4
- b. 3/4
- c. 3/5
- d. does not exist
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:
- a. 0 and π
- b. π/2 and 2π
- c. 0 and 2π
- d. -π/2 and π/2
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
Video Lessons - KCET 2019 - Maths
KCET 2019 Maths Question Paper with Solutions
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