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KCET 2020 - Maths
1. If
- a. y
- b. 6n(n + 1)y
- c. n(n +1 )y
- d. \(x \frac{dy}{dx} + y\)
Solution:
Answer: (c)
y= (2x)(n+1) +(3x)(-n)
⇒ dy/dx = 2(n+1)xn-(3nx)(-n-1)
⇒ (d2 y)/(dx2) = 2n(n+1)x(n-1)+3n(n+1)x(-n-2)
⇒ x2 (d2 y)/(dx2) = n(n+1) [(2x)(n+1)+3/xn ]
⇒ x2 (d2 y)/(dx2) = n(n+1)y
2. If the curves are 2x = y2 and 2xy = K intersect perpendicularly, then the value of K2 is
- a. 8
- b. 4
- c. 2√2
- d. 2
Solution:
Answer: (a)
2x = y2 . . . (1)
2xy = K . . . (2)
Solving (1) and (2), we get
(x,y) = (K(2/3)/2, K(1/3))
Differentiating (1) and (2) w.r.t. x
m1 = dy/dx = 1/y ...(3)
m2 = dy/dx = -y/x ...(4)
∵ Both curves intersect each other perpendicularly
∴m1 m2 = -1
⇒ -1/x = -1
⇒ x = 1
⇒ K(2/3) = 2
⇒ K2 = 8
3. If (xe)y = ex, then dy/dx is
- a. \(\frac{e^x}{x(y-1)}\)
- b. \(\frac{log \ x}{(1 + log \ x)^2}\)
- c. \(\frac{1}{(1 - log \ x)^2}\)
- d. \(\frac{log \ x}{1+log \ x}\)
Solution:
Answer: (b)
(xe)y = ex
⇒y (log x + 1) = x
⇒ y = x/(logx+1)
∴
\(\frac{dy}{dx} = \frac{log \ x}{(log \ x + 1)^2}\)
4. If the side of a cube is increased by 5%, then the surface area of a cube is increased by
- a. 20%
- b. 10%
- c. 60%
- d. 6%
Solution:
Answer: (b)
Let one side of the cube be x and surface area be A
So, dx = 5% = 5x/100
Then, A = (6x)2
⇒dA/dx = 12x
⇒dA = (12x)dx
⇒dA = (12x) (5x/100)
⇒dA = 10A/100
⇒dA = 10%
5. The value of
- a. tan-1x +(1/3)tan-1x2 + C
- b. tan-1 x+tan-1x3+C
- c. tan-1x+(1/3)tan-1 x3+C
- d. tan-1x-(1/3)tan-1 x3+C
Solution:
Answer: (c)
6. The maximum value of
- a. -1/e
- b. e
- c. 1
- d. 1/e
Solution:
Answer: (d)
\(y = \frac{log_e \ x}{x}\)⇒
\(\frac{dy}{dx}=\frac{1 - log_e x}{x^2}\)For maxima, dy/dx = 0
⇒ 1 - loge x = 0
⇒ x = e
dy/dx changes sign from positive to negative at x = e
∴ ymax = 1/e
7. The value of
- a. 2esin x(cosx-1)+C
- b. 2esin x(sinx-1)+C
- c. 2esin x(sinx+1)+C
- d. 2esin x(cosx+1)+C
Solution:
Answer: (b)
\(I = \int e^{sinx} \ sin 2x \ dx = 2 \int e^{sinx} \ sin x \ cos x \ dx\)Let t =sinx
⇒ dt=cosx dx
Therefore, I = 2∫tet dt
= 2(tet - et) + C
= 2(sinx - 1)esin x + C
8. The value of
- a. π2/2
- b. π
- c. π/2
- d. 1
Solution:
Answer: (c)
9. If
- a. 2, -7, 5
- b. 5, -7, -5
- c. 2,-7, -5
- d. 5, -7, 5
Solution:
Answer: (a)
Now, 3x + 1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2) [From eqn. (1)]
Putting x = 1, x = 2, x = 3 in the above equation one at a time, we get
A = 2, B = -7, C = 5.
10. The value of
- a. (π/8)log2
- b. (π/2)log 2
- c. (π/4)log 2
- d. 1/2
Solution:
Answer: (a)
11. The area of the region bounded by the curve y2 = 8x and the line y = 2x is
- a. (8/3) sq. units
- b. (16/3) sq. units
- c. (4/3) sq. units
- d. (3/4) sq. units
Solution:
Answer: (c)
Solving y2 = 8x and y = 2x, we get
(x,y) = (0,0), (2,4)
So, area bounded by the curve is
12. The value of
- a. -2
- b. 2
- c. 0
- d. 1
Solution:
Answer: (d)
13. The order of the differential equation obtained by eliminating arbitrary constants in the family of curves c1y = (c2+c3)e x+c4 is
- a. 4
- b. 1
- c. 2
- d. 3
Solution:
Answer: (b)
\(y = [(\frac{C_2 + C_3}{C_1})e^{C_4}]e^x = A e^x,\)where A =
\((\frac{C_2 + C_3}{C_1})e^{C_4}\)Order = Number of independent arbitrary constants = 1
14. The general solution of the differential equation x2dy-2xydx = x4 cos x dx is
- a. y = cos x+cx2
- b. y = x2 sin x+cx2
- c. y = x2sin x+cx
- d. y = sin x+cx2
Solution:
Answer: (b)
x2dy - 2xydx = x4 cosx dx
\(\frac{dy}{dx} = \frac{x^4 \ cos x+2xy}{x^2}\)⇒dy/dx – 2y/x =x2 cos x
I.F. = e∫-2/x dx = e-2 logx = 1/x2
Therefore, the general solution is
\(y(\frac{1}{x^2}) = \int \frac{1}{x^2}(x^2 \ cos x)dx = sinx + c\)∴y = x2 (sinx+c)
= x2 sinx+cx2
15. The area of the region bounded by the line y = 2x+1, x- axis and the ordinates x = -1 and x = 1 is
- a. 5
- b. 9/4
- c. 2
- d. 5/2
Solution:
Answer: (d)
Area bounded by y = 2x+1 with x- axis
= (1/2) (1/2)(1) +(1/2) (3/2)(3) = 5/2 sq. units.
16. The two vectors
- a. 14
- b. 14/2
- c. 14
- d. 7
Solution:
Answer: (a)
17. If
Solution:
Answer: (d)
18. The curve passing through the point (1, 2) given that the slope of the tangent at any point (x,y) is 2x/y represents
- a. Hyperbola
- b. Circle
- c. Parabola
- d. Ellipse
Solution:
Answer: (a)
Given,
\(\frac{dy}{dx}= \frac{2x}{y}\)⇒ ydy = 2xdx
⇒ ∫ydy = ∫2x dx
⇒ y2/2 = x2+A, where A is a constant.
The above equation represents a hyperbola.
19. If
- a. 4
- b. 6
- c. 3
- d. 2
Solution:
Answer: (d)
20. The point(1,-3,4) lies in the octant
- a. Eighth
- b. Second
- c. Third
- d. Fourth
Solution:
Answer: (d)
Signs of x-coordinate, y-coordinate and z-coordinate are +,-,+ respectively.
∴(1,-3,4) lies in the fourth octant.
21. If the vector
- a. 5
- b. 6
- c. -5
- d. -6
Solution:
Answer: (b)
Given vectors are coplanar.
⇒
\(\begin{vmatrix} 2 &-3 &4 \\ 2& 1 &-1 \\ \lambda & -1 &2 \end{vmatrix} = 0\)⇒ 2(1) + 3(4 + λ) + 4(-2 - λ) = 0
⇒ 2 + 12 + 3λ - 8 - 4λ = 0
⇒ λ = 6
22. The distance of the point (1,2,-4) from the line (x-3)/2 = (y-3)/3 = (z+5)/6 is
- a. √293/49
- b. 293/7
- c. √293/7
- d. 293/49
Solution:
Answer: (c)
Let
\(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}=t\)⇒ (x, y, z) = (2t + 3, 3t + 3, 6t - 5)
∴ d.r.’s of the line perpendicular to
\(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}\)and joining (2t+3,3t+3,6t-5)and (1,2,-4) is (2t+2,3t+1,6t-1)
∴2(2t+2)+3(3t+1)+6(6t-1)=0
⇒t = -1/49
∴ Distance =
\(\sqrt{(2t+2)^2+(3t+1)^2+(6t-1)^2} = \sqrt{49t^2+2t+6}\)=
\(\sqrt{\frac{1}{49}-\frac{2}{49}+6}=\frac{\sqrt{293}}{7}\)
23. The sine of the angle between the straight line (x-2)/3 = (3-y)/(-4) = (z-4)/5 and the plane
- a. √2/10
- b. 3/√50
- c. 3/50
- d. 4/5√2
Solution:
Answer: (a)
Given line is
\(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)and plane is 2x-2y+z = 5
24. If a line makes an angle of π/3 with each of x and y-axis, then the acute angle made by z-axis is
- a. π/2
- b. π/4
- c. π/6
- d. π/3
Solution:
Answer: (b)
Given, α = β = π/3
Let acute angle made by z- axis be γ.
Then,
\(cos^2 \alpha + cos^2 \beta + cos^2 \gamma\)⇒
\((\frac{1}{2})^2+(\frac{1}{2})^2 + cos^2 \gamma = 1\)⇒
\(\frac{1}{4}+\frac{1}{4} + cos^2 \gamma = 1 ⇒ cos^2 \gamma = \frac{1}{2}\)⇒
\(cos \gamma = \pm \frac{1}{\sqrt 2}⇒ \gamma = \frac{\pi}{4}\)[∵γ is acute]
25. Corner points of the feasible region determined by the system of linear constraints are (0,3),(1,1) and (3,0). Let z = px+qy, where p,q>0. Condition on p and q so that the minimum of z occurs at (3,0) and (1,1) is
- a. p = q
- b. p = 2q
- c. p = q/2
- d. p = 3q
Solution:
Answer: (c)
Given corner points are (0,3),(1,1),(3,0)
z = px + qy
At (3,0), z = 3p
At (1,1), z = p + q
It is given that the minimum of z occurs at (3, 0) and (1, 1)
⇒ 3p = p + q
⇒ 2p = q
⇒ p = q/2
26. The feasible region of an LPP is shown in the figure. If Z = 11x+7y, then the maximum value of Z occurs at
- a. (3,2)
- b. (0,5)
- c. (3,3)
- d. (5,0)
Solution:
Answer: (a)
y-intercept of x+y = 5 is (0,5)
y-intercept of x+3y = 9 is (0,3)
The intersection point of x+y = 5 and x+3y = 9 is (3,2)
Therefore, the corner points are (0,5),(0,3),(3,2)
At (0,5), Z = 35
At (0,3), Z = 21
At (3,2), Z = 47
So, Zmax= 47 at (3,2).
27. A die is thrown 10 times, the probability that an odd number will come up atleast one time is
- a. 1013/1024
- b. 1/1024
- c. 1023/1024
- d. 11/1024
Solution:
Answer: (c)
Given n = 10
Probability of odd number, p = ½
∴q=1/2
Required probability = P(X≥1)
=1-P(X=0)
=
\(1-^{10}C_0 (1/2)^{10-0}(1/2)^0\)= 1 – 1/210
= 1 - 1/1024
= 1023/1024
28. If A and B are two events such that P(A) = 1/3, P(B) = 1/2 and P(A∩B) = 1/6, then P(A'|B) is
- a. 1/12
- b. 2/3
- c. 1/3
- d. 1/2
Solution:
Answer: (b)
Given P(A) = 1/3, P(B) = 1/2, P(A ∩ B) = 1/6
So, P(A’|B) = 1 – P(A|B)
=
\(1 - \frac{P(A \cap B)}{P(B)}\)= 1 - 1/3
= 2/3
29. Events E1 and E2 form a partition of the sample space S.A is any event such that P(E1) = P(E2) = 1/2, P(E2│A) = 1/2 and (A│E2) = 2/3 . Then P(E1│A) is
- a. 1/4
- b. 1/2
- c. 2/3
- d. 1
Solution:
Answer: (b)
Let P(A|E1) = x
By Bayes’ theorem,
30. The probability of solving a problem by three persons A,B and C independently is 1/2,1/4 and 1/3 respectively. Then the probability that the problem is solved by any two of them is
- a. 1/8
- b. 1/12
- c. 1/4
- d. 1/24
Solution:
Answer: (c)
Required probability = P(A'BC) + P(AB'C) + P(ABC')
= 1/2 × 1/4 × 1/3 + 1/2 × 3/4 × 1/3 + 1/2 × 1/4 × 2/3
= 1/24 + 1/8 + 1/12
=(1+3+2)/24
= 1/4
31. If n(A) = 2 and total number of possible relations from set A to set B is 1024, then n(B) is.
- a. 5
- b. 512
- c. 20
- d. 10
Solution:
Answer: (a)
n(A) = 2
Given, 2(n(A)⋅n(B)) = 1024
⇒(2)(2⋅n(B)) = (2)10
⇒2⋅n(B) = 10
⇒n(B) = 5
32. The value of sin2 510 + sin2 390 is
- a. cos 120
- b. 1
- c. 0
- d. sin 120
Solution:
Answer: (b)
sin2 51°+sin2 39°
= cos2 39° + sin2 39°
= 1
33. If tan A+cot A=2, then the value of tan4 A+cot4 A=
- a. 5
- b. 2
- c. 1
- d. 4
Solution:
Answer: (b)
tan A + cot A = 2
⇒ (tan A + cotA )2 = 4
⇒tan2 A + cot2 A + 2tan A cotA = 4
⇒tan2 A+cot2 A=2
⇒(tan2 A+cot2 A) 2=4
⇒tan4 A+cot4 A+2 tan2 A cot2 A=4
⇒tan4 A+cot4 A=2
34. If A={1,2,3,4,5,6}, then the number of subsets of A which contain atleast two elements is
- a. 58
- b. 64
- c. 63
- d. 57
Solution:
Answer: (d)
Total number of subsets of A is 2n(A) = 26 = 64
Number of subsets of A which contain at least two elements is
64-( 6C0 + 6C1 )
= 64-(1+6)
= 57
35. If z = x + iy, then the equation |z+1| = |z-1| represents
- a. y-axis
- b. a circle
- c. a parabola
- d. x-axis
Solution:
Answer: (a)
36. The value of 16 C9+16 C10- 16C6- 16C7 is
- a. 17C2
- b. 0
- c. 1
- d. 17C10
Solution:
Answer: (b)
16C9+ 16C10- 16C6- 16C7
= 16C9+ 16C10- 16C10- 16C9=0 (∵ nCr = nC(n-r))
37. The number of terms in the expansion of (x+y+z)10 is
- a. 110
- b. 66
- c. 142
- d. 11
Solution:
Answer: (b)
Number of terms in the expansion of (x+y+z)10 is (10+3-1)C10
= 12C10=12!/(2! 10!)=66
38. If P(n) ∶2n< n!,then the smallest positive integer for which P(n)is true if
- a. 5
- b. 2
- c. 3
- d. 4
Solution:
Answer: (d)
For n = 1, 2, 3, 2n> n!
P(4) ∶ 24< 4!
So, smallest positive integer, n = 4.
39. The two lines lx+my = n and l'x+m'y= n’ are perpendicular if
- a. lm' + ml'= 0
- b. ll'+mm'=0
- c. lm'=ml'
- d. lm+l'm'=0
Solution:
Answer: (b)
Product of slopes = -1
⇒ll’ + mm’ = 0
40. If the parabola x2 = 4ay passes through the point (2, 1), then the length of the latus rectum is
- a. 8
- b. 1
- c. 4
- d. 2
Solution:
Answer: (c)
x2=4ay
Given parabola passes through (2,1).
⇒ 22= 4a
⇒ a = 1
Length of latus rectum= 4a = 4.
41. If the sum of n terms of an A.P. is given by Sn = n2 + n, then the common difference of the A.P. is
- a. 6
- b. 4
- c. 1
- d. 2
Solution:
Answer: (d)
Sn= n2+n
S1 = 1+1 = 2 = T1
S2 = 22+2 = 6 = T1+T2
∴T2 = S2-S1= 4
Common difference, d= T2-T1
= 4-2
= 2
42. The negation of the statement “For all real numbers x and y, x + y = y + x” is
- a. for some real numbers x and y, x - y = y - x
- b. for all real numbers x and y, x+y ≠ y+x
- c. for some real numbers x and y, x+y = y+x
- d. for some real numbers x and y,x+y ≠ y+x
Solution:
Answer: (d)
Negation: For some real numbers x and y, x + y ≠ y + x.
43. The standard deviation of the data 6,7,8,9,10 is
- a. 10
- b. 2
- c. 10
- d. 2
Solution:
Answer: (b)
Mean,
\(\bar{x} = \frac{6+7+8+9+10}{5} = \frac{40}{5} = 8\)Standard deviation,
\(\sigma = \sqrt{\frac{1}{5}(4+1+0+1+4)}\)Because,
\(\sigma = \sqrt{\frac{1}{n} \sum(x_i - \bar{x})^2}\)⇒
\(\sigma = \sqrt{\frac{10}{5}} = \sqrt{2}\)
44.
- a. 6
- b. 2
- c. 3
- d. 4
Solution:
Answer: (b)
45. If a relation R on the set {1,2,3} be defined by R = {(1,1)}, then R is
- a. Only symmetric
- b. Reflexive and symmetric
- c. Reflexive and transitive
- d. Symmetric and transitive
Solution:
Answer: (d)
R={(1,1)} on set {1,2,3}
Clearly, R is symmetric and transitive.
46. Let f ∶ [2, ∞] → R be the function defined by f(x) = x2-4x+5, then the range of f is
- a. [5,∞)
- b. (-∞,∞)
- c. [1, ∞)
- d. (1,∞)
Solution:
Answer: (c)
f(x) = (x-2)2 + 1 ≥ 1, ∀ x ∈ [2,∞)
fmin = 1 at x = 2
∴ Range of f is [1, ∞)
47. If A,B,C are three mutually exclusive and exhaustive events of an experiment such that P(A) = 2P(B) = 3P(C), then P(B) is equal to
- a. 4/11
- b. 1/11
- c. 2/11
- d. 3/11
Solution:
Answer: (d)
Given, P(A) = 2P(B) = 3P(C)
⇒ P(C) = 2/3 P(B)
Since A,B,C are three mutually exclusive and exhaustive events
∴P(A)+P(B)+P(C) = 1
⇒ P(B) = 3/11
48. The domain of the function defined by f(x) = cos-1 √(x-1)is
- a. [0,1]
- b. [1,2]
- c. [0,2]
- d. [-1,1]
Solution:
Answer: (b)
For f to be defined,
x -1 ≥ 0 and -1 ≤ √(x-1) ≤ 1
⇒ x ≥ 1 and 0 ≤ x-1 ≤ 1
⇒ x ≥ 1 and 1 ≤ x ≤ 2
⇒ x ∈ [1,2]
Hence, domain of f is [1, 2].
49. The value of
- a. Does not exist
- b. 0
- c. 1
- d. -1
Solution:
Answer: (a)
π/3 ∉ [-1,1] which is the domain of sin-1 x, cos-1 x
So, cos(sin-1 π/3 + cos-1 π/3) does not exist.
50. If
- a. 4A
- b. A
- c. 2A
- d. I
Solution:
Answer: (d)
51. If A = {a,b,c}, then the number of binary operations on A is
- a. 39
- b. 3
- c. 36
- d. 33
Solution:
Answer: (a)
A = {a,b,c}
n(A) = 3
Number of binary operations is n(A)(n(A))^2 = 3(3^2) = 39
52. If
- a. \(\begin{pmatrix} 2 &-1 \\ 3& 2 \end{pmatrix}\)
- b. \(\begin{pmatrix} 2 &1 \\ 3& 2 \end{pmatrix}\)
- c. \(\begin{pmatrix} 2 &-1 \\ -3& 2 \end{pmatrix}\)
- d. \(\begin{pmatrix} -2 &1 \\ 3&- 2 \end{pmatrix}\)
Solution:
Answer: (c)
53. If f(x) =
- a. f(-1) = 0
- b. f(1) = 0
- c. f(2) = 0
- d. f(0) = 0
Solution:
Answer: (d)
\(f(0) = \begin{vmatrix} 0& a& b\\ -a& 0&c \\-b & -c& 0\end{vmatrix}\)f(0) is the determinant of skew-symmetric matrix of order 3 (odd).
∴f(0) = 0.
54. If A and B are square matrices of same order and B is a skew symmetric matrix, then A’BA is
- a. Skew symmetric matrix
- b. Symmetric matrix
- c. Null matrix
- d. Diagonal matrix
Solution:
Answer: (a)
B is a skew symmetric matrix.
⇒ B' = -B
Now, (A'BA)' = A'B'(A')'
= A'(-B)A
= -(A' BA)
Hence, A'BA is a skew symmetric matrix.
55. If A is a square matrix of order 3 and |A| = 5, then |A adj A| is
- a. 625
- b. 5
- c. 125
- d. 25
Solution:
Answer: (c)
|A adj A| = |A| |adj A|
= |A||A|3-1
= 5 52
= 125
56. If
- a. ±1
- b. ±1/2
- c. 0
- d. ±2
Solution:
Answer: (a)
Given, f is continuous at x = 0.
57. If a1,a2,a3,……,a9are in A.P. then the value of
- a. 1
- b. (9/2)(a1+a9)
- c. a1+a9
- d. loge(logee)
Solution:
Answer: (d)
58. If 2x + 2y = 2(x+y), then dy/dx is
- a. \(\frac{2^y-1}{2^x-1}\)
- b. 2y-x
- c. -2y-x
- d. 2x-y
Solution:
Answer: (c)
2x + 2y = 2(x+y).....(1)
Differentiating both sides w.r.t. x, we get
2x ln 2 + 2y y' ln2 = 2(x+y) ln2 (1+y')
⇒2x+2y y'=2(x+y)(1+y')
⇒2x+2y⋅y'= 2(x+y)+2(x+y)⋅y'
⇒2x-2(x+y)=y'(2(x+y)-2y)
⇒2x-2x-2y=y'(2x+2y-2y) [From eqn.(1)]
⇒-2y=y'2x
⇒y'= dy/dx=-2y-x
59. If
- a. -1/√3
- b. -1/2
- c. 1/2
- d. 1/√3
Solution:
Answer: (c)
60. The right hand and left limit of the function
- a. -1 and 1
- b. 1 and 1
- c. 1 and -1
- d. -1 and -1
Solution:
Answer: (c)
Video Lessons - KCET 2020 - Maths
KCET 2020 Maths Question Paper with Solutions
