 # KCET 2020 Mathematics Paper with Solutions

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### KCET 2020 - Maths

1. If $$y = 2x^{n+1} + \frac{3}{x^n}$$, then$$x^2 \frac{d^2y}{dx^2}$$is

1. a. y
2. b. 6n(n + 1)y
3. c. n(n +1 )y
4. d. $$x \frac{dy}{dx} + y$$

Solution:

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

1. a. 8
2. b. 4
3. c. 2√2
4. d. 2

Solution:

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

1. a. $$\frac{e^x}{x(y-1)}$$
2. b. $$\frac{log \ x}{(1 + log \ x)^2}$$
3. c. $$\frac{1}{(1 - log \ x)^2}$$
4. d. $$\frac{log \ x}{1+log \ x}$$

Solution:

(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

1. a. 20%
2. b. 10%
3. c. 60%
4. d. 6%

Solution:

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 $$\int \frac{1+x^4}{1+x^6}dx$$is

1. a. tan-1x +(1/3)tan-1x2 + C
2. b. tan-1 x+tan-1x3+C
3. c. tan-1x+(1/3)tan-1 x3+C
4. d. tan-1x-(1/3)tan-1 x3+C

Solution: 6. The maximum value of $$\frac{log_e \ x}{x}$$, if x>0 is

1. a. -1/e
2. b. e
3. c. 1
4. d. 1/e

Solution:

$$y = \frac{log_e \ x}{x}$$

⇒ $$\frac{dy}{dx}=\frac{1 - log_e x}{x^2}$$

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 $$\int e^{sin x} sin2x dx$$ is

1. a. 2esin x(cosx-1)+C
2. b. 2esin x(sinx-1)+C
3. c. 2esin x(sinx+1)+C
4. d. 2esin x(cosx+1)+C

Solution:

$$I = \int e^{sinx} \ sin 2x \ dx = 2 \int e^{sinx} \ sin x \ cos x \ dx$$

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 $$\int^{1/2}_{-1/2} cos^{-1}x \ dx$$ is

1. a. π2/2
2. b. π
3. c. π/2
4. d. 1

Solution: 9. If $$\int \frac{3x+1}{(x-1)(x-2)(x-3)}dx = A \ log|x-1|+B \ log|x-2| + C \ log|x-3| + C$$, then the values of A, B and C are respectively,

1. a. 2, -7, 5
2. b. 5, -7, -5
3. c. 2,-7, -5
4. d. 5, -7, 5

Solution: 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$$\int_0^1 \frac{log(1+x)}{1+x^2}dx$$is

1. a. (π/8)log2
2. b. (π/2)log 2
3. c. (π/4)log 2
4. d. 1/2

Solution: 11. The area of the region bounded by the curve y2 = 8x and the line y = 2x is

1. a. (8/3) sq. units
2. b. (16/3) sq. units
3. c. (4/3) sq. units
4. d. (3/4) sq. units

Solution:

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 $$\int^{\pi/2}_{- \pi/2} \frac{cos \ x}{1 + e^x}dx$$is

1. a. -2
2. b. 2
3. c. 0
4. d. 1

Solution: 13. The order of the differential equation obtained by eliminating arbitrary constants in the family of curves c1y = (c2+c3)e x+c4 is

1. a. 4
2. b. 1
3. c. 2
4. d. 3

Solution:

$$y = [(\frac{C_2 + C_3}{C_1})e^{C_4}]e^x = A e^x,$$

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

1. a. y = cos x+cx2
2. b. y = x2 sin x+cx2
3. c. y = x2sin x+cx
4. d. y = sin x+cx2

Solution:

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

1. a. 5
2. b. 9/4
3. c. 2
4. d. 5/2

Solution: 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$$\hat i + \hat j + \hat k$$ and$$\hat i + 3 \hat j + 5\hat k$$represent the two sides$$\vec {AB}$$ and $$\vec {AC}$$ respectively of a ∆ABC. The length of the median through A is

1. a. 14
2. b. 14/2
3. c. 14
4. d. 7

Solution: 17. If $$\vec {a}$$and$$\vec {b}$$ are unit vectors and θ is the angle between$$\vec {a}$$ and$$\vec {b}$$, then sin(θ/2)is Solution: 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

1. a. Hyperbola
2. b. Circle
3. c. Parabola
4. d. Ellipse

Solution:

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 $$\left | \vec{a} \times \vec{b} \right |^{2} + \left | \vec{a} \cdot \vec{b} \right |^{2}$$ and$$|\vec a|$$ = 6, then$$\left | \vec{b} \right |$$ is equal to

1. a. 4
2. b. 6
3. c. 3
4. d. 2

Solution: 20. The point(1,-3,4) lies in the octant

1. a. Eighth
2. b. Second
3. c. Third
4. d. Fourth

Solution:

Signs of x-coordinate, y-coordinate and z-coordinate are +,-,+ respectively.

∴(1,-3,4) lies in the fourth octant.

21. If the vector $$2 \hat i-3 \hat j + 4 \hat k$$, $$2 \hat i+ \hat j - \hat k$$ and $$\lambda \hat i- \hat j + 2 \hat k$$ are coplanar, then the value of λ is

1. a. 5
2. b. 6
3. c. -5
4. d. -6

Solution:

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

1. a. √293/49
2. b. 293/7
3. c. √293/7
4. d. 293/49

Solution:

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

1. a. √2/10
2. b. 3/√50
3. c. 3/50
4. d. 4/5√2

Solution:

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

1. a. π/2
2. b. π/4
3. c. π/6
4. d. π/3

Solution:

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

1. a. p = q
2. b. p = 2q
3. c. p = q/2
4. d. p = 3q

Solution:

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 1. a. (3,2)
2. b. (0,5)
3. c. (3,3)
4. d. (5,0)

Solution:

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

1. a. 1013/1024
2. b. 1/1024
3. c. 1023/1024
4. d. 11/1024

Solution:

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

1. a. 1/12
2. b. 2/3
3. c. 1/3
4. d. 1/2

Solution:

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

1. a. 1/4
2. b. 1/2
3. c. 2/3
4. d. 1

Solution:

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

1. a. 1/8
2. b. 1/12
3. c. 1/4
4. d. 1/24

Solution:

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.

1. a. 5
2. b. 512
3. c. 20
4. d. 10

Solution:

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

1. a. cos 120
2. b. 1
3. c. 0
4. d. sin 120

Solution:

sin2 51°+sin2 39°

= cos2 39° + sin2 39°

= 1

33. If tan A+cot A=2, then the value of tan4 A+cot4 A=

1. a. 5
2. b. 2
3. c. 1
4. d. 4

Solution:

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

1. a. 58
2. b. 64
3. c. 63
4. d. 57

Solution:

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

1. a. y-axis
2. b. a circle
3. c. a parabola
4. d. x-axis

Solution: 36. The value of 16 C9+16 C10- 16C6- 16C7 is

1. a. 17C2
2. b. 0
3. c. 1
4. d. 17C10

Solution:

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

1. a. 110
2. b. 66
3. c. 142
4. d. 11

Solution:

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

1. a. 5
2. b. 2
3. c. 3
4. d. 4

Solution:

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

1. a. lm' + ml'= 0
2. b. ll'+mm'=0
3. c. lm'=ml'
4. d. lm+l'm'=0

Solution:

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

1. a. 8
2. b. 1
3. c. 4
4. d. 2

Solution:

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

1. a. 6
2. b. 4
3. c. 1
4. d. 2

Solution:

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

1. a. for some real numbers x and y, x - y = y - x
2. b. for all real numbers x and y, x+y ≠ y+x
3. c. for some real numbers x and y, x+y = y+x
4. d. for some real numbers x and y,x+y ≠ y+x

Solution:

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

1. a. 10
2. b. 2
3. c. 10
4. d. 2

Solution:

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. $$lim_{x \rightarrow 0}(\frac{tan \ x}{\sqrt{2x+4}-2})$$is equal to

1. a. 6
2. b. 2
3. c. 3
4. d. 4

Solution: 45. If a relation R on the set {1,2,3} be defined by R = {(1,1)}, then R is

1. a. Only symmetric
2. b. Reflexive and symmetric
3. c. Reflexive and transitive
4. d. Symmetric and transitive

Solution:

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

1. a. [5,∞)
2. b. (-∞,∞)
3. c. [1, ∞)
4. d. (1,∞)

Solution:

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

1. a. 4/11
2. b. 1/11
3. c. 2/11
4. d. 3/11

Solution:

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

1. a. [0,1]
2. b. [1,2]
3. c. [0,2]
4. d. [-1,1]

Solution:

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 $$cos(sin^{-1} \frac{\pi}{3} + cos^{-1} \frac{\pi}{3})$$ is

1. a. Does not exist
2. b. 0
3. c. 1
4. d. -1

Solution:

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 $$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 &0 \\ 1&0 &0 \end{pmatrix}$$ , then A4 is equal to

1. a. 4A
2. b. A
3. c. 2A
4. d. I

Solution: 51. If A = {a,b,c}, then the number of binary operations on A is

1. a. 39
2. b. 3
3. c. 36
4. d. 33

Solution:

A = {a,b,c}

n(A) = 3

Number of binary operations is n(A)(n(A))^2 = 3(3^2) = 39

52. If $$\begin{pmatrix} 2 &1 \\ 3& 2 \end{pmatrix}A = \begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}$$, then the matrix A is

1. a. $$\begin{pmatrix} 2 &-1 \\ 3& 2 \end{pmatrix}$$
2. b. $$\begin{pmatrix} 2 &1 \\ 3& 2 \end{pmatrix}$$
3. c. $$\begin{pmatrix} 2 &-1 \\ -3& 2 \end{pmatrix}$$
4. d. $$\begin{pmatrix} -2 &1 \\ 3&- 2 \end{pmatrix}$$

Solution: 53. If f(x) = $$\begin{vmatrix} x^3-x &a+x&b+x\\ x-a & x^2-x &c+x \\ x-b &x-c& 0 \end{vmatrix}$$, then

1. a. f(-1) = 0
2. b. f(1) = 0
3. c. f(2) = 0
4. d. f(0) = 0

Solution:

$$f(0) = \begin{vmatrix} 0& a& b\\ -a& 0&c \\-b & -c& 0\end{vmatrix}$$

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

1. a. Skew symmetric matrix
2. b. Symmetric matrix
3. c. Null matrix
4. d. Diagonal matrix

Solution:

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

1. a. 625
2. b. 5
3. c. 125
4. d. 25

Solution:

= |A||A|3-1

= 5 52

= 125

56. If $$f(x) = \left\{\begin{matrix} \frac{1-cos \ kx}{x \ sin x} & if x \neq 0 \\ \frac{1}{2} & if x = 0 \end{matrix}\right.$$ is continuous at x = 0, then the value of K is

1. a. ±1
2. b. ±1/2
3. c. 0
4. d. ±2

Solution:

Given, f is continuous at x = 0. 57. If a1,a2,a3,……,a9are in A.P. then the value of $$\begin{vmatrix} a_1 &a_2 &a_3 \\ a_4&a_5 &a_6 \\ a_7& a_8 &a_9 \end{vmatrix}$$ is

1. a. 1
2. b. (9/2)(a1+a9)
3. c. a1+a9
4. d. loge(logee)

Solution: 58. If 2x + 2y = 2(x+y), then dy/dx is

1. a. $$\frac{2^y-1}{2^x-1}$$
2. b. 2y-x
3. c. -2y-x
4. d. 2x-y

Solution:

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 $$f(x) = sin^{-1}(\frac{2x}{1+x^2})$$, then f’(√3)is

1. a. -1/√3
2. b. -1/2
3. c. 1/2
4. d. 1/√3

Solution: 60. The right hand and left limit of the function$$f(x) = \left\{\begin{matrix} \frac{e^{1/x}-1}{e^{1/x}+1} & if \ x \neq 0\\ 0 & if \ x = 0 \end{matrix}\right.$$ are respectively

1. a. -1 and 1
2. b. 1 and 1
3. c. 1 and -1
4. d. -1 and -1

Solution: ### KCET 2020 Maths Question Paper with Solutions                      