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:

  1. 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

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

Solution:

  1. 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

  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:

  1. 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

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

Solution:

  1. 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 \(\int \frac{1+x^4}{1+x^6}dx\)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:

  1. Answer: (c)

    KCET 2020 Solved Paper Maths


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:

  1. 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 \(\int e^{sin x} sin2x dx \) 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:

  1. 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 \(\int^{1/2}_{-1/2} cos^{-1}x \ dx\) is

  1. a. π2/2
  2. b. π
  3. c. π/2
  4. d. 1

Solution:

  1. Answer: (c)

    KCET  Solved Paper Maths 2020


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:

  1. Answer: (a)

    KCET  Solved Paper 2020 Maths

    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:

  1. Answer: (a)

    KCET Maths Solved Paper 2020


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:

  1. Answer: (c)

    Solving y2 = 8x and y = 2x, we get

    (x,y) = (0,0), (2,4)

    KCET 2020 Sample Paper Answer

    So, area bounded by the curve is

    2020 KCET Sample Paper Answers


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:

  1. Answer: (d)

    KCET Solutions Paper 2020 Maths


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:

  1. 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

  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:

  1. 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

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

Solution:

  1. Answer: (d)

    KCET 2020 Solutions Paper Maths

    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:

  1. Answer: (a)

    KCET 2020 Maths Solutions Paper


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

KCET 2020 Maths Paper Solutions

    Solution:

    1. Answer: (d)

      2020 KCET Solved Paper Maths


    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:

    1. 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 \(\left | \vec{a} \times \vec{b} \right |^{2} + \left | \vec{a} \cdot \vec{b} \right |^{2}\) 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:

    1. Answer: (d)

      2020 Solved Paper KCET Maths


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

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

    Solution:

    1. 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 \(2 \hat i-3 \hat j + 4 \hat k\), \(2 \hat i+ \hat j - \hat k\) 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:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. Answer: (a)

      Given line is\(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)

      and plane is 2x-2y+z = 5

      2020 Solved Paper Maths KCET


    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:

    1. 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

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

    Solution:

    1. 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

    2020 Maths KCET Solved Paper

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. Answer: (b)

      Let P(A|E1) = x

      By Bayes’ theorem,

      2020 KCET Maths Solved Paper


    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:

    1. 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.

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

    Solution:

    1. 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

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

    Solution:

    1. 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=

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. Answer: (a)

      2020 Maths Solved Paper KCET


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

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. 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

    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:

    1. 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

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

    Solution:

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

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

    Solution:

    1. Answer: (b)

      2020 KCET Solved Papers Maths


    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:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

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

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

    Solution:

    1. 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 \(\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 &0 \\ 1&0 &0 \end{pmatrix}\) , then A4 is equal to

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

    Solution:

    1. Answer: (d)

      2020 Solved Papers KCET Maths


    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:

    1. 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 \(\begin{pmatrix} 2 &1 \\ 3& 2 \end{pmatrix}A = \begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix}\), 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:

    1. Answer: (c)

      2020 Solved Papers Maths KCET


    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:

    1. 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

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

    Solution:

    1. 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

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

    Solution:

    1. Answer: (c)

      |A adj A| = |A| |adj A|

      = |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:

    1. Answer: (a)

      Given, f is continuous at x = 0.

      2020 Maths KCET Solved Papers


    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:

    1. Answer: (d)

      2020 KCET Maths Solved Papers


    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:

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

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

    Solution:

    1. Answer: (c)

      2020 Maths Solved Papers KCET


    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:

    1. Answer: (c)

      KCET 2020 Solutions Paper Maths


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