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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    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions
    KCET 2020 Maths Question Paper with Solutions