KCET 2022 Mathematics Questions with Answers Key Solutions

Q1. If all permutations of the letters of the word MASK are arranged in the order as in dictionary with or without meaning, which one of the following is $19^{\text {th }}$ word?

A

KAMS

B

AKMS

C

SAKM

D

AMSK

Solution:

$A\, K\, M \,S$
$A \rightarrow 3 !$
$K \rightarrow 3 !$
$M \rightarrow 3 !$
$SAKM \rightarrow \frac{1}{19}$

Q2. If $a_{1}, a_{2}, a_{3}, \ldots, a_{10}$ is a geometric progression and $\frac{a_{3}}{a_{1}}=25$ , then $\frac{a_{9}}{a_{5}}$ equals

A

$3\left(5^{2}\right)$

B

$5^{3}$

C

$5^{4}$

D

$2\left(5^{2}\right)$

Solution:

$\frac{ a _{3}}{ a _{1}}=25$
$\frac{ ar ^{2}}{ a }=25$
$r ^{2}=5^{2}$
$\frac{ a _{4}}{ a _{5}}=\frac{ ar ^{8}}{ ar ^{4}}= r ^{4}=5^{4}$

Q3. If the straight line $2 x-3 y+17=0$ is perpendicular to the line passing through the points $(7,17)$ and $(15, \beta)$ , then $\beta$ equals

A

$-5$

B

$29$

C

$5$

D

$-29$

Solution:

Given $2 x-3 y+17=0$
is a line perpendicular to line passing through point $P (7,17) \& Q (15, \beta)$ then equation of line passing throught $P \& Q$ is
$y-17=\left(\frac{\beta-17}{15-7}\right)(x-7)$
$8 y-136=\beta x-7 \beta-17 x+119$
$(7-\beta) x+8 y-255=0 \ldots \ldots \ldots . .(2)$
Line (1) \& (2) are perpendicular to each other then
$m _{1} m _{2}=1$
Product of slopes of line (1) \& (2) is $-1$
Now,
$m _{1}=\frac{2}{3}$ and $m _{2}=\frac{\beta-7}{8}$
$m _{1} m _{2}=-1 \Rightarrow \frac{2}{3}\left(\frac{\beta-7}{48}\right)=-1$
$\beta-7=-12$
$\beta=-12+7=-5$
Hence $[\beta=-5]$ .

Q4. The octant in which the point $(2,-4,-7)$ lies is

A

Eighth

B

Fourth

C

Third

D

Fift

Q5. If $f(x)= \begin{cases}x^{2}-1, & 0 < x < 2 \\ 2 x+3, & 2 \leq x < 3\end{cases}$
the quadratic equation whose roots are $\displaystyle\lim _{x \rightarrow 2^{-}} f (x)$ and $\displaystyle\lim _{x \rightarrow 2^{+}} f (x)$ is

A

$x^{2}-14 x+49=0$

B

$x^{2}-6 x+9=0$

C

$x^{2}-10 x+21=0$

D

$x^{2}-7 x+8=0$

Solution:

$\alpha=\displaystyle\lim _{x \rightarrow 2^{-}} f(x)=\displaystyle\lim _{x \rightarrow 2} x^{2}-1=3$
$\beta=\displaystyle\lim _{x \rightarrow 2^{+}} f(x)=\displaystyle\lim _{x \rightarrow 2} 2 x+3=7$
$x^{2}-(\alpha+\beta) x+\alpha \beta=0$
$x^{2}-10 x+21=0$

Q6. If $3 x+ i (4 x-y)=6- i$ where $x$ and $y$ are real numbers, then the values of $x$ and $y$ are respectively,

A

3,9

B

2,9

C

2,4

D

3,4

Solution:

$3 x=6$
$x=2$
$4 x-y=-1$
$8-y=-1$
$9= y$

Q7. If the standard deviation of the numbers $-1,0,1, k$ is $\sqrt{5}$ where $k > 0$ , then $k$ is equal to

A

$4 \sqrt{\frac{5}{3}}$

B

$2 \sqrt{\frac{10}{3}}$

C

$\sqrt{6}$

D

$2 \sqrt{6}$

Solution:

$\sigma^{2}=5, \overline{ x }=\frac{ k }{4}$
$\frac{1}{4}\left(1+0+1+ k ^{2}\right)-\frac{ k ^{2}}{16}=5$
$\frac{ k ^{2}+2}{4} \frac{- k ^{2}}{16}=5$
$\frac{4 k ^{2}+8- k ^{2}}{16}=5$
$\Rightarrow 3 k ^{2}+8=80$
$3 k ^{2}=72$
$k ^{2}=24$
$k =\pm \sqrt{24}=2 \sqrt{6}$

Q8. If the set $x$ contains $7$ elements and set $y$ contains 8 elements, then the number of bijections from $x$ to $y$ is

A

$0$

B

$7!$

C

$8 P_{7}$

D

$8 !$

Solution:

$n ( A ) \neq n ( B )$
Number of bijections is zero

Q9. If $f: R \rightarrow R$ be defined by
$f(x)=\begin{cases}2 x & : & x>3 \\x^{2} & : & 1 < x \leq 3 \\3 x & : & x \leq 1\end{cases}$
then $f(-1)+f(2)+f(4)$ is

A

5

B

9

C

10

D

14

Solution:

$f(-1)=3(-1)=-3$
$f(2)=2^{2}=4$
$f(4)=2(4)=8$
$f(-1)+f(2)+f(4)$
$=-3+4+8=9$

Q10. Let the relation $R$ is defined in $N$ by $a R b$ , if $3 a+2 b=27$ then $R$ is

A

$\{(1,12)(3,9)(5,6)(7,3)\}$

B

$\{(1,12)(3,9)(5,6)(7,3)(9,0)\}$

C

$\left\{\left(0, \frac{27}{2}\right)(1,12)(3,9)(5,6)(7,3)\right\}$

D

$\{(2,1)(9,3)(6,5)(3,7)\}$

Solution:

$2 b =27-3 a$
$b =\frac{27-3 a }{2}$
$R =\{(1,2),(3,9),(5,6),(7,3)\}$

Q11. $\displaystyle\lim _{y \rightarrow 0} \frac{\sqrt{3+y^{3}}-\sqrt{3}}{y^{3}}=$

A

$\frac{1}{2 \sqrt{3}}$

B

$2 \sqrt{3}$

C

$\frac{1}{3 \sqrt{2}}$

D

$3 \sqrt{2}$

Solution:

$\displaystyle\lim _{y \rightarrow 0} \frac{3+y^{3}-3}{y\left(\sqrt{3+y^{3}}+\sqrt{3}\right)}=\frac{1}{2 \sqrt{3}}$

Q12. If $A$ is a matrix of order $3 \times 3$ , then $\left(A^{2}\right)^{-1}$ is equal to

A

$\left(-A^{2}\right)^{2}$

B

$A ^{2}$

C

$\left(A^{-1}\right)^{2}$

D

$(-A)^{-2}$

Solution:

$\left( A ^{2}\right)^{-1}=\left( A ^{-1}\right)^{2}$

Q13. If $A=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}$ , then the inverse of the matrix $A^{3}$ is

A

$A$

B

$1$

C

$-1$

D

$- A$

Solution:

$A=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}$
$A^{-1}=\frac{1}{-1}\begin{bmatrix}-2 & 1 \\ -3 & 2\end{bmatrix}=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}=A$
$A^{2}=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}$
$=\begin{bmatrix}4-3 & -2+2 \\ 6-6 & -3+4\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}=I$
$\therefore A^{3}=A$

Q14. If $A$ is a skew symmetric matrix, then $A^{2021}$ is

A

Row matrix

B

Symmetric matrix

C

Column matrix

D

Skew symmetric matrix

Solution:

$A ^{ T }=- A$ or $A ^{ n }$ is slow symmetric if $n$ is odd
$P = A ^{2021}$
$P ^{ T }=\left[ A ^{2021}\right]^{ T }=\left[ A ^{ T }\right]^{2021}=(- A )^{2021}=- P$

Q15. If $A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ then $(a I+b A)^{n}$ is (where $I$ is the identity matrix of order $2$ )

A

$a ^{2} I+a^{n-1} b \cdot A$

B

$a^{n} I+n a^{n} b A$

C

$a^{n} I+n \cdot a^{n-1} b \cdot A$

D

$a^{n} I+b^{n} A$

Solution:

$A =\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$
${[ aI + bA ]^{1}=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}+\begin{bmatrix}0 & b \\ 0 & 0\end{bmatrix}=\begin{bmatrix} a & b \\ 0 & a \end{bmatrix} }$
${[ aI + IA ]^{2}=\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}=\begin{bmatrix} a ^{2} & 2 ab \\ 0 & a ^{2}\end{bmatrix} }$
${[ aI + bA ]^{3}=\begin{bmatrix} a ^{2} & 2 ab \\ 0 & a ^{2}\end{bmatrix}\begin{bmatrix}a & b \\ 0 & a \end{bmatrix}=\begin{bmatrix} a ^{3} & 3^{2} b \\ 0 & a ^{3}\end{bmatrix} }$
$\therefore[ aI + bA ]^{ n }=\begin{bmatrix} a ^{ n } & na a ^{ n -1} b \\ 0 & a ^{ n }\end{bmatrix}$
$= a ^{ n } I + n \cdot a ^{ n -1} bA$

Q16. If $A$ is a $3 \times 3$ matrix such that $\mid 5 \cdot$ adj $A \mid=5$ then $|A|$ is equal to

A

$\pm 1$

B

$\pm 1 / 5$

C

$\pm 1 / 25$

D

$\pm 5$

Solution:

$A_{3 \times 3}$ matrix $|5 . \operatorname{AdjA}|=5$
$\Rightarrow 5^{3}| A |^{2}=5$
$\Rightarrow\left| A ^{2}\right|=\frac{1}{5^{2}}$
$| A |=\pm \frac{1}{5}$

Q17. If there are two values of ' $a$ ' which makes determinant
$\Delta=\begin{vmatrix}1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a\end{vmatrix}=86$
Then the sum of these numbers is

A

$-4$

B

4

C

9

D

5

Solution:

$\Delta=1\left(2 a^{2}+4\right)+2(4 a-0)+5(8)=86$
$2 a^{2}+8 a+44-86=0$
$2 a^{2}+8 a-42=0$
$a^{2}+4 a-21=0$
Sum of numbers $=-4\left(\therefore-\frac{ b }{ a }=\alpha+\beta\right)$

Q18. If the vertices of a triangle are $(-2,6)(3,-6)$ and $(1,5)$ , then the area of the triangle is

A

40 sq. units

B

30 sq. units

C

15.5 sq. units

D

35 sq. units

Solution:

$\Delta=\frac{1}{2}\begin{vmatrix}-2-3 & -2-1 \\ 6+6 & 6-5\end{vmatrix}=\frac{1}{5}\begin{vmatrix}-5 & -3 \\ 12 & 1\end{vmatrix}$
$=\frac{1}{2}|-5+36|=\frac{31}{2}=15.5$

Q19. Domain of $\cos ^{-1}[x]$ is, where $[ . ]$ denotes a greatest integer function

A

$(-1,2]$

B

$[-1,2]$

C

$(-1,2)$

D

$[-1,2)$

Solution:

$\cos ^{-1}[x]$
$-1 \leq[x] \leq 1$
$\Rightarrow[x]=\{-1,0,1\}$
$x \in[-1,2)$

Q20. If $y=\left(1+x^{2}\right) \tan ^{-1} x-x$ then $\frac{d y}{d x}$ is

A

$2 x \tan ^{-1} x$

B

$x^{2} \tan ^{-1} x$

C

$\frac{\tan ^{-1} x}{x}$

D

$x \tan ^{-1} x$

Solution:

$y=\left(1+x^{2}\right) \tan ^{-1} x-x$
$\frac{d y}{d x}=\frac{\left(1+x^{2}\right)}{1+x^{2}}+\tan ^{-1} x \cdot(2 x)-1$
$=2 x \tan ^{-1} x$

Q21. If $x= e ^{\theta} \sin \theta, y= e ^{\theta} \cos \theta$ where $\theta$ is a parameter, then $\frac{ dy }{ d x}$ at $(1,1)$ is equal to

A

$0$

B

$-\frac{1}{2}$

C

$\frac{1}{2}$

D

$-\frac{1}{4}$

Solution:

$x = e ^{\theta} \sin \theta=1$
$y = e ^{\theta} \cos \theta=1, \frac{ x }{ y }=\tan \theta=1$
$\Rightarrow \theta=\frac{\pi}{4}$
$\frac{ dy }{ dx }=\left|\frac{ dy / d \theta}{ dx / d \theta}\right|=\frac{- e ^{\theta} \sin \theta+\cos \theta \cdot e ^{\theta}}{ e ^{\theta} \cos \theta+\sin \theta e ^{\theta}}=\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}$
$=\tan \left(\frac{\pi}{4}-\theta\right)=0$

Q22. If $y=e^{\sqrt{x \sqrt{x \sqrt{x}}}} x > 1$ then $\frac{d^{2} y}{d x^{2}}$ at $x=\log _{e}^{3}$ is

A

3

B

0

C

5

D

1

Solution:

$Y = e^{\sqrt{x\sqrt{x\sqrt{x}}} ........} = e^{x^{\frac{1}{2}} x^{\frac{1}{4}},x^{\frac{1}{8}} ..... = e^{x^{\frac{1}{2}} x^{\frac{1}{4}},x^{\frac{1}{8}} .....}}$
$e ^{ x ^{\frac{1}{2}\left[1+\frac{1}{2}+\frac{1}{4}+\ldots\right]}}= e ^{ x ^{\frac{1}{2}}}= e ^{ x ^{1}}= e ^{ x }$
$\frac{d y}{d x}=e^{x}$
$\frac{ d ^{2} y }{ dx x ^{2}}= e ^{ x } x \log _{ e }^{3}= e ^{\log _{ e }^{3}}=$

Q23. If $f (1)=1, f' (1)=3$ then the derivative of $f ( f ( f (x)))+(f(x))^{2}$ at $x=1$ is

A

10

B

35

C

33

D

12

Solution:

$f (1)=1, f ^{\prime}(1)=3$
$\frac{ d }{ dx }\left[ f ( f ( f ( x )))+( f ( x ))^{2}\right]$
${\left[ f ^{\prime}( f ( f ( x ))) \cdot f ^{\prime} f ( x ) \cdot f ^{\prime}( x )+2 f ( x ) \cdot f ^{\prime}( x )\right] }$
$= f ^{\prime}( f ( f (1))) f ^{\prime}( f (1)) \cdot f ^{\prime}(1)+2 f (1) \cdot f ^{\prime}(1)$
$f ^{\prime}( f (1)) f ^{\prime}(1) \cdot 3+2 \cdot(1) 3$
$= f ^{\prime}(1) \cdot 3 \cdot 3+6$
$=27+6=33$

Q24. If $y=x^{\sin x}+(\sin x)^{x}$ then $\frac{d y}{d x}$ at $x=\frac{\pi}{2}$ is

A

$\frac{4}{\pi}$

B

$1$

C

$\pi \log \frac{\pi}{2}$

D

$\frac{\pi^{2}}{2}$

Solution:

$y=x^{\sin x}+(\sin x)^{x}$
$\frac{d y}{d x}=\left[x^{\sin x}\right]\left[\frac{\sin x}{x}+\cos x \cdot \log x\right]+$
$(\sin x)^{x}[x \cos x+\log \sin x]$
$x=\frac{\pi}{2}$
$=\frac{\pi}{2}\left[\frac{2}{\pi}\right]+1[0+0]=1$

Q25. If $A_{n}=\begin{bmatrix}1-n & n \\ n & 1-n\end{bmatrix}$ then $\left|A_{1}\right|+\left|A_{2}\right|+\ldots+\left|A_{2021}\right|=$

A

$-2021$

B

$(2021)^{2}$

C

$-(2021)^{2}$

D

$4042$

Solution:

$A_{n}=\begin{bmatrix}1-n & n \\ n & 1-n\end{bmatrix}$
$\left|A_{n}\right|=(1-n)^{2}-n^{2}$
$=1+ n ^{2}-2 n - n ^{2}$

Q26. The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

A

$(-\infty, \infty)$

B

$(-1, \infty)$

C

$(\infty,-1)$

D

$(-\infty, 0)$

Solution:

$f^{\prime}(x)=\frac{x^{2}}{(x+1)(2+x)^{2}}>0$
$x+1>0 \Rightarrow x>-1$

Q27. The co-ordinates of the point on the $\sqrt{x}+\sqrt{y}=6$ at which the tangent is equally inclined the axes is

A

$(4,4)$

B

$(9,9)$

C

$(1,1)$

D

$(6,6)$

Solution:

$\frac{d y}{d x}=-\sqrt{\frac{y}{x}}=-1$
$y=x$
$\sqrt{x}+\sqrt{x}=6$
$x=9, y=9$

Q28. The function $f (x)=4 \sin ^{3} x-6 \sin ^{2} x+12 \sin x+100$ is strictly

A

decreasing in $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$

B

increasing in $\left(\pi, \frac{3 \pi}{2}\right)$

C

decreasing in $\left[0, \frac{\pi}{2}\right]$

D

decreasing in $\left(\frac{\pi}{2}, \pi\right)$

Solution:

$f^{\prime}(x)=\left(12 \sin ^{2} x-12 \sin x+12\right) \cos x$
$f^{\prime}(x)=12\left(\sin ^{2} x-\sin x+1\right) \cos x$
$\sin ^{2} x-\sin x+1>0$
$x \in\left(\frac{\pi}{2}, \pi\right) \cos x < 0$

Q29. If $[x]$ is the greatest integer function not greater than $x$ then $\int\limits_{0}^{8}[x] d x$ is equal to

A

28

B

29

C

30

D

20

Solution:

$\int\limits_{0}^{8}[x] d x=1+2+3+\ldots .+7$
$=\frac{7(7+1)}{2}=28$

Q30. $\int\limits_{0}^{\pi / 2} \sqrt{\sin \theta} \cos ^{3} \theta d \theta$ is equal to

A

$\frac{8}{23}$

B

$\frac{8}{21}$

C

$\frac{7}{23}$

D

$\frac{7}{21}$

Solution:

Put $\sin \theta= t$
$\int\limits_{0}^{1} t^{1 / 2}\left(1-t^{2}\right) d t=\frac{8}{21}$

Q31. If $e^{y}+x y=e$ the ordered pair $\left(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}\right)$ at $x=0$ is equal to

A

$\left(\frac{1}{e}, \frac{1}{e^{2}}\right)$

B

$\left(\frac{1}{e}, \frac{-1}{e^{2}}\right)$

C

$\left(\frac{-1}{e}, \frac{-1}{e^{2}}\right)$

D

$\left(\frac{-1}{e}, \frac{1}{e^{2}}\right)$

Solution:

$x=0 \Rightarrow y=1$
$\frac{d y}{d x}=\frac{-y}{e^{y}+x}$
$\left(\frac{d y}{d x}\right)_{(0,1)}=-\frac{1}{e}$
$\left(\frac{d^{2} y}{d x^{2}}\right)_{(0,1)}=\frac{1}{e^{2}}$

Q32. $\int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x$ is equal to

A

$2(\sin x-x \cos \alpha)+c$

B

$2(\sin x-2 x \cos \alpha)+c$

C

$2(\sin x+x \cos \alpha)+ c$

D

$2(\sin x+2 x \cos \alpha)+c$

Solution:

$2 \int \frac{\cos ^{2} x-\cos ^{2} \alpha}{\cos x-\cos \alpha} d x=2 \int(\cos x+\cos \alpha) d x$
$=2[\sin x+x \cos \alpha]$

Q33. $\int\limits_{0}^{1} \frac{x e^{x}}{(2+x)^{3}} d x$ is equal to

A

$\frac{1}{27}, e -\frac{1}{8}$

B

$\frac{1}{9} \cdot e+\frac{1}{4}$

C

$\frac{1}{27} \cdot c+\frac{1}{8}$

D

$\frac{1}{9} \cdot e-\frac{1}{4}$

Solution:

$\int\limits_{0}^{1} e^{x}\left[\frac{1}{(x+2)^{2}}-\frac{2}{(x+2)^{3}}\right] d x$
$=\left[\frac{e^{x}}{(x+2)^{2}}\right]_{0}^{1}=\frac{e}{9}-\frac{1}{4}$

Q34. If $\int \frac{d x}{(x+2)\left(x^{2}+1\right)}=a \log \left|1+x^{2}\right|+b \tan ^{-1} x+\frac{1}{5} \log |x+2|+c$ , then

A

$a =\frac{-1}{10} b =\frac{2}{5}$

B

$a=\frac{-1}{10} b=\frac{-2}{5}$

C

$a=\frac{1}{10} \quad b=\frac{2}{5}$

D

$a=\frac{1}{10} \quad b=\frac{-2}{5}$

Solution:

$\frac{1}{(x+2)\left(x^{2}+1\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+1}$
$A=\frac{1}{5}, B=\frac{-1}{5}, C=\frac{2}{5}$

Q35. Area of the region bounded by the curve $y=\tan x$ , the $x$ -axis and the line $x=\frac{\pi}{3}$ is

A

$\log _{2} \frac{1}{2}$

B

$0$

C

$\log 2$

D

$-\log 2$

Solution:

$A=\int\limits_{0}^{\pi / 3} \tan x d x=\log |\sec x|_{0}^{\pi / 3}=\log 2$

Q36. Evaluate $\int\limits_{2}^{3} x^{2} d x$ as the limit of a sum

A

$\frac{72}{6}$

B

$\frac{25}{7}$

C

$\frac{53}{9}$

D

$\frac{19}{3}$

Solution:

$I=\left[\frac{x^{3}}{3}\right]_{2}^{3}=\frac{1}{3}(27-8)=\frac{19}{3}$

Q37. $\int\limits_{0}^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x$ is equal to

A

$\log 2-1$

B

$-\log 2$

C

$\log 2$

D

$1-\log 2$

Solution:

$\sin x=t$
$\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos x \sin x}{1+\sin x} d x$
$\Rightarrow \int\limits_{0}^{1} \frac{t}{1+t} d t=1-\log 2$

Q38. If $\frac{d y}{d x}+\frac{y}{x}=x^{2}$ , then $2 y(2)-y(1)=$

A

$\frac{11}{4}$

B

$\frac{9}{4}$

C

$\frac{15}{4}$

D

$\frac{13}{4}$

Solution:

$y \cdot x=\frac{x^{4}}{4}+C$
$2 y(2)-y(1)=\frac{15}{4}$

Q39. The solution of the differential equation $\frac{d y}{d x}=(x+y)^{2}$ is

A

$\tan ^{-1}(x+y)=x+c$

B

$\cot ^{-1}(x+y)=c$

C

$\tan ^{-1}(x+y)=0$

D

$\cot ^{-1}(x+y)=x+c$

Solution:

$x+y=z \Rightarrow \frac{d z}{d x}=1+z^{2}$
$\int \frac{1}{1+z^{2}} d z=\int 1 d x$
$\operatorname{Tan}^{-1}(x+y)=x+c$

Q40. If $y(x)$ be the solution of differential equation $x \log x \frac{d y}{d x}+y=2 x \log x, y(e)$ is equal to

A

e

B

2

C

0

D

2 e

Solution:

$I . F=\log x$ ,
$y \log x=2 x(\log x-1)+c$
If $x=e$ then $y=c$ then $y(e)=2 e$

Q41. If $|\vec{a}|=2$ and $|\vec{b}|=3$ and the angle between $\vec{a}$ and $\vec{b}$ is $120^{\circ}$ , then the length of the vector $|\frac{1} {2}^{\vec{a}}-\frac{1}{3}^{\vec{b}}|$ is

A

2

B

$\frac{1}{6}$

C

3

D

1 Q42. If $|\vec{a} \times \vec{b}|+|\vec{a} \cdot \vec{b}|^{2}=36$ and $|\vec{a}|=3$ then $|\vec{b}|$ is equal to

A

9

B

4

C

36

D

2 Q43. If $\vec{\alpha}=\hat{i}-3 \hat{j}, \vec{\beta}=\hat{i}+2 \hat{j}-\hat{k}$ then express $\vec{\beta}$ in the form $\vec{\beta}=\vec{\beta}_{1}+\vec{\beta}_{2}$ where $\vec{\beta}_{1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_{2}$ is perpendicular to $\vec{\alpha}$ then $\vec{\beta}_{1}$ is given by

A

$\frac{5}{8}(\hat{i}-3 \hat{j})$

B

$\hat{i}-3 \hat{j}$

C

$\frac{5}{8}(\hat{i}+3 \hat{j})$

D

$\hat{ i }+3 \hat{ j }$

Solution:

Grace

Q44. The sum of the degree and order of the differential equation $\left(1+y_{1}^{2}\right)^{2 / 3}=y_{2}$ is

A

4

B

5

C

6

D

7

Solution:

$\left(1+ y _{1}^{2}\right)=\left( y _{2}\right)^{3}$
$2+3=5$

Q45. The co-ordinates of foot of the perpendicular drawn from the origin to the plane $2 x-3 y+4 z=29$ are

A

$(2,3,4)$

B

$(2,-3,4)$

C

$(2,-3,-4)$

D

$(-2,-3,4)$

Solution:

verification $(2,-3,4)$

Q46. The angle between the pair of lines $\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ and $\frac{x+1}{1}=\frac{y-4}{4}=\frac{z-5}{2}$ is

A

$\theta=\cos ^{-1}\left[\frac{27}{5}\right]$

B

$\theta=\cos ^{-1}\left[\frac{19}{21}\right]$

C

$\theta=\cos ^{-1}\left[\frac{8 \sqrt{3}}{15}\right]$

D

$\theta=\cos ^{-1}\left[\frac{5 \sqrt{3}}{16}\right]$

Solution:

$\cos \theta=\frac{3 \times 1+5 \times 4+4 \times 2}{\sqrt{3^{2}+5^{2}+4^{2}} \times \sqrt{1^{2}+4^{2}+2^{2}}}=\frac{31}{5 \sqrt{42}}$
$\theta=\cos ^{-1} \frac{31}{5 \sqrt{42}}$

Q47. The corner points of the feasible region of an LPP are $(0,2),(3,0),(6,0),(6,8)$ and $(0,5)$ , then the minimum value of $z=4 x+6 y$ occurs at

A

finite number of points

B

only one point

C

infinite number of points

D

only two points

Solution:

At $(0,2),(3,0), z=12$
Hence minimum at $2$ points.

Q48. A dietician has to develop a special diet using two foods $X$ and $Y$ . Each packet (containing $30 \,g$ ) of food. $X$ contains $12$ units of calcium, $4$ units of iron, $6$ units of cholesterol and $6$ units of vitamin A. Each packet of the same quantity of food $Y$ contains $3$ units of calcium, $20$ units of iron, $4$ units of cholesterol and $3$ units of vitamin A. The diet requires atleast $240$ units of calcium, atleast $460$ units of iron and atmost $300$ units of cholesterol. The corner points of the feasible region are

A

$(2,72),(40,15),(15,20)$

B

$(0,23),(40,15),(2,72)$

C

$(2,72),(15,20),(0,23)$

D

$(2,72),(40,15),(115,0)$

Q49. The distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$ is

A

$\frac{8}{\sqrt{21}}$

B

$\frac{-8}{\sqrt{21}}$

C

$8 \sqrt{21}$

D

$\frac{-8}{21}$

Solution:

Distance $=\frac{|2-2-4-4|}{\sqrt{1+4+16}}=\frac{(8)}{\sqrt{21}}$

Q50. Find the mean number of heads in three tosses of a fair coin :

A

$1.5$

B

$2.5$

C

$4.5$

D

$3.5$

Solution:

X 0 1 2 3

$P(x)$ $\frac{1}{8}$ $\frac{3}{8}$ $\frac{3}{8}$ $\frac{1}{8}$

Mean $=\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=\frac{3}{2}=1.5$

Q51. If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}$ and $P(\frac{A}{B})=\frac{1}{4}$ , then $P\left(A^{\prime} \cap B^{\prime}\right)$ is

A

$\frac{1}{4}$

B

$\frac{1}{12}$

C

$\frac{3}{16}$

D

$\frac{3}{4}$

Solution:

$P ( A )=\frac{1}{2}, P ( B )=\frac{1}{3}, P ( A \cap B )=\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}$
$P ( A \cap B )= P (\overline{ A \cup B })$
$P ( A \cap B )=1- P ( A \cup B )$
$P ( A \cap B )=1-\left[\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\right]$
$P ( A \cap B )=1-\left[\frac{6+4-1}{12}\right]$
$P ( A \cap B )=1-\frac{9}{12}$
$P ( A \cap B )=\frac{1}{4}$

Q52. A pandemic has been spreading all over the world. The probabilities are $0.7$ that there will be a lockdown, $0.8$ that the pandemic is controlled in one month if there is a lockdown and $0.3$ that it is controlled in one month if there is no lockdown. The probability that the pandemic will be controlled in one month is

A

$0.65$

B

$1.46$

C

$1.65$

D

$0.46$

Solution:

$P \left( E _{1}\right)=$ probability of there is lockdown $=0.7$
$P\left(E_{2}\right)=$ probability of there is lockdown $=0.3$
$A$ is the event controlled in one month
$P \left( A / E _{1}\right)=0.8, P \left( A / E _{2}\right)=0.3$
$P(A)=0.7(0.8)+(0.3)(0.3)$
$=0.56+0.09=0.65$

Q53. If $A$ and $B$ are two independent events such that $P(\bar{A})=0.75, P(A \cup B)=0.65$ , and $P ( B )=x$ , then find the value of $x$ :

A

$\frac{5}{14}$

B

$\frac{9}{14}$

C

$\frac{8}{15}$

D

$\frac{7}{15}$

Solution:

$P ( A )=\frac{1}{4}, P ( B )=\frac{3}{4}, P ( A \cup B )=\frac{13}{20}$
$\frac{1}{4}+ x -\frac{1}{4} \cdot x =\frac{13}{20}$
$\frac{3}{4} x-\frac{13}{20}-\frac{5}{20}=\frac{8}{20}$
$x=\frac{8}{20} \times \frac{4}{3}=\frac{8}{15}$

Q54. Suppose that the number of elements in set $A$ is $p$ , the number of elements in set $B$ is $q$ and the number of elements in $A \times B$ is $7$ then $p^{2}+q^{2}=$ _____

A

50

B

42

C

51

D

49

Solution:

$n ( A )= p , n ( B )= q$
$n ( A \times B )=7$
$pq =7$
$p ^{2}+ q ^{2}=7^{2}+1^{2}$ or $1^{2}+7^{2}$
$p ^{2}+ q ^{2}=50$

Q55. The domain of the function $f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$ is

A

$[-2,0) \cap(0,1)$

B

$[-2,0)$

C

$[-2,1)$

D

$[-2,0) \cup(0,1)$

Solution:

$1 - x > 0 , 1 - x \neq 1$
$x -1<0 \,\,\,\,\,\,x \neq 0 \,\,\,\,\,\,x +2 \geq 0$
$x <1 \,\,\,\,\,\,x \geq-2$
$\therefore x \in[-2,0) \cup(0,1)$

Q56. The trigonometric function $y=\tan x$ in the II quadrant

A

decreases from $0$ to $\infty$

B

increases from $0$ to $\infty$

C

decreases from $-\infty$ to $0$

D

increases from $-\infty$ to $0$

Solution:

By graph

Q57. The degree measure of $\frac{\pi}{32}$ is equal to

A

$5^{\circ} 30^{\prime} 20^{\prime \prime}$

B

$5^{\circ} 37^{\prime} 30^{\prime \prime}$

C

$5^{\circ} 37^{\prime} 20^{\prime \prime}$

D

$4^{\circ} 30^{\prime} 30^{\prime \prime}$

Solution:

$\frac{\pi}{32}=\frac{180^{\circ}}{32}=5^{0} 37^{\prime} 30^{\prime \prime}$

Q58. The value of $\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}$ is

A

0

B

$\frac{1}{2}$

C

1

D

$\frac{1}{4}$

Solution:

$\sin \frac{5 \pi}{12} \cdot \sin \frac{\pi}{12}$
$=\frac{1}{2} \sin \frac{\pi}{6}$
$=\frac{1}{2}+\frac{1}{2}=\frac{1}{4}$

Q59. $\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}=$

A

$\sin 2 \theta$

B

$2 \sin \theta$

C

$2 \cos \theta$

D

$2 \cos \frac{\theta}{2}$

Solution:

$1+\cos \theta=2 \cos ^{2}\left(\frac{\theta}{2}\right)$
$\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 8 \theta}}}=2 \cos \theta$

Q60. If $A=\{1,2,3, \ldots \ldots 10\}$ then number of subsets of $A$ containing only odd numbers is

A

31

B

32

C

27

D

30

Solution:

X	0	1	2	3
$P(x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

KCET 2022 Mathematics Questions with Answers Key Solutions

Q1. If all permutations of the letters of the word MASK are arranged in the order as in dictionary with or without meaning, which one of the following is 19th 19^{\text {th }} word?

KAMS

AKMS

SAKM

AMSK

AKMSA\, K\, M \,S A→3!A \rightarrow 3 ! K→3!K \rightarrow 3 ! M→3!M \rightarrow 3 ! SAKM→119SAKM \rightarrow \frac{1}{19}

Q2. If a1,a2,a3,…,a10a_{1}, a_{2}, a_{3}, \ldots, a_{10} is a geometric progression and a3a1=25\frac{a_{3}}{a_{1}}=25, then a9a5\frac{a_{9}}{a_{5}} equals

3(52)3\left(5^{2}\right)

535^{3}

545^{4}

2(52)2\left(5^{2}\right)

a3a1=25\frac{ a _{3}}{ a _{1}}=25 ar2a=25\frac{ ar ^{2}}{ a }=25 r2=52r ^{2}=5^{2} a4a5=ar8ar4=r4=54\frac{ a _{4}}{ a _{5}}=\frac{ ar ^{8}}{ ar ^{4}}= r ^{4}=5^{4}

Q3. If the straight line 2x−3y+17=02 x-3 y+17=0 is perpendicular to the line passing through the points (7,17)(7,17) and (15,β)(15, \beta), then β\beta equals

−5-5

2929

55

−29-29

Q4. The octant in which the point (2,−4,−7)(2,-4,-7) lies is

Eighth

Fourth

Third

Fift

x^{2}-14 x+49=0x^{2}-14 x+49=0

x^{2}-6 x+9=0x^{2}-6 x+9=0

x^{2}-10 x+21=0x^{2}-10 x+21=0

x^{2}-7 x+8=0x^{2}-7 x+8=0

Q6. If 3 x+ i (4 x-y)=6- i3 x+ i (4 x-y)=6- i where xx and yy are real numbers, then the values of xx and yy are respectively,

3,9

2,9

2,4

3,4

3 x=63 x=6 x=2 x=2 4 x-y=-14 x-y=-1 8-y=-18-y=-1 9= y9= y

Q7. If the standard deviation of the numbers -1,0,1, k-1,0,1, k is \sqrt{5}\sqrt{5} where k > 0k > 0, then kk is equal to

4 \sqrt{\frac{5}{3}}4 \sqrt{\frac{5}{3}}

2 \sqrt{\frac{10}{3}}2 \sqrt{\frac{10}{3}}

\sqrt{6}\sqrt{6}

2 \sqrt{6}2 \sqrt{6}

Q8. If the set xx contains 77 elements and set yy contains 8 elements, then the number of bijections from xx to yy is

00

7!7!

8 P_{7}8 P_{7}

8 !8 !

n ( A ) \neq n ( B )n ( A ) \neq n ( B ) Number of bijections is zero

Q9. If f: R \rightarrow Rf: R \rightarrow R be defined by f(x)=\begin{cases}2 x & : & x>3 \\x^{2} & : & 1 < x \leq 3 \\3 x & : & x \leq 1\end{cases}f(x)=\begin{cases}2 x & : & x>3 \\x^{2} & : & 1 < x \leq 3 \\3 x & : & x \leq 1\end{cases} then f(-1)+f(2)+f(4)f(-1)+f(2)+f(4) is

5

9

10

14

f(-1)=3(-1)=-3f(-1)=3(-1)=-3 f(2)=2^{2}=4f(2)=2^{2}=4 f(4)=2(4)=8f(4)=2(4)=8 f(-1)+f(2)+f(4)f(-1)+f(2)+f(4) =-3+4+8=9=-3+4+8=9

Q10. Let the relation RR is defined in NN by a R ba R b, if 3 a+2 b=273 a+2 b=27 then RR is

\{(1,12)(3,9)(5,6)(7,3)\}\{(1,12)(3,9)(5,6)(7,3)\}

\{(1,12)(3,9)(5,6)(7,3)(9,0)\}\{(1,12)(3,9)(5,6)(7,3)(9,0)\}

\left\{\left(0, \frac{27}{2}\right)(1,12)(3,9)(5,6)(7,3)\right\}\left\{\left(0, \frac{27}{2}\right)(1,12)(3,9)(5,6)(7,3)\right\}

\{(2,1)(9,3)(6,5)(3,7)\}\{(2,1)(9,3)(6,5)(3,7)\}

2 b =27-3 a2 b =27-3 a b =\frac{27-3 a }{2}b =\frac{27-3 a }{2} R =\{(1,2),(3,9),(5,6),(7,3)\}R =\{(1,2),(3,9),(5,6),(7,3)\}

Q11. \displaystyle\lim _{y \rightarrow 0} \frac{\sqrt{3+y^{3}}-\sqrt{3}}{y^{3}}=\displaystyle\lim _{y \rightarrow 0} \frac{\sqrt{3+y^{3}}-\sqrt{3}}{y^{3}}=

\frac{1}{2 \sqrt{3}}\frac{1}{2 \sqrt{3}}

2 \sqrt{3}2 \sqrt{3}

\frac{1}{3 \sqrt{2}}\frac{1}{3 \sqrt{2}}

3 \sqrt{2}3 \sqrt{2}

\displaystyle\lim _{y \rightarrow 0} \frac{3+y^{3}-3}{y\left(\sqrt{3+y^{3}}+\sqrt{3}\right)}=\frac{1}{2 \sqrt{3}}\displaystyle\lim _{y \rightarrow 0} \frac{3+y^{3}-3}{y\left(\sqrt{3+y^{3}}+\sqrt{3}\right)}=\frac{1}{2 \sqrt{3}}

Q12. If AA is a matrix of order 3 \times 33 \times 3, then \left(A^{2}\right)^{-1}\left(A^{2}\right)^{-1} is equal to

\left(-A^{2}\right)^{2}\left(-A^{2}\right)^{2}

A ^{2}A ^{2}

\left(A^{-1}\right)^{2}\left(A^{-1}\right)^{2}

(-A)^{-2}(-A)^{-2}

\left( A ^{2}\right)^{-1}=\left( A ^{-1}\right)^{2}\left( A ^{2}\right)^{-1}=\left( A ^{-1}\right)^{2}

Q13. If A=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}A=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}, then the inverse of the matrix A^{3}A^{3} is

AA

11

-1-1

- A- A

Q14. If AA is a skew symmetric matrix, then A^{2021}A^{2021} is

Row matrix

Symmetric matrix

Column matrix

Skew symmetric matrix

A ^{ T }=- AA ^{ T }=- A or A ^{ n }A ^{ n } is slow symmetric if nn is odd P = A ^{2021}P = A ^{2021} P ^{ T }=\left[ A ^{2021}\right]^{ T }=\left[ A ^{ T }\right]^{2021}=(- A )^{2021}=- PP ^{ T }=\left[ A ^{2021}\right]^{ T }=\left[ A ^{ T }\right]^{2021}=(- A )^{2021}=- P

Q15. If A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} then (a I+b A)^{n}(a I+b A)^{n} is (where II is the identity matrix of order 22 )

Q1. If all permutations of the letters of the word MASK are arranged in the order as in dictionary with or without meaning, which one of the following is $19^{\text {th }}$ word?

$A\, K\, M \,S$
$A \rightarrow 3 !$
$K \rightarrow 3 !$
$M \rightarrow 3 !$
$SAKM \rightarrow \frac{1}{19}$

Q2. If $a_{1}, a_{2}, a_{3}, \ldots, a_{10}$ is a geometric progression and $\frac{a_{3}}{a_{1}}=25$ , then $\frac{a_{9}}{a_{5}}$ equals

$3\left(5^{2}\right)$

$5^{3}$

$5^{4}$

$2\left(5^{2}\right)$

$\frac{ a _{3}}{ a _{1}}=25$
$\frac{ ar ^{2}}{ a }=25$
$r ^{2}=5^{2}$
$\frac{ a _{4}}{ a _{5}}=\frac{ ar ^{8}}{ ar ^{4}}= r ^{4}=5^{4}$

Q3. If the straight line $2 x-3 y+17=0$ is perpendicular to the line passing through the points $(7,17)$ and $(15, \beta)$ , then $\beta$ equals

$-5$

$29$

$5$

$-29$

Q4. The octant in which the point $(2,-4,-7)$ lies is

$x^{2}-14 x+49=0$

$x^{2}-6 x+9=0$

$x^{2}-10 x+21=0$

$x^{2}-7 x+8=0$

Q6. If $3 x+ i (4 x-y)=6- i$ where $x$ and $y$ are real numbers, then the values of $x$ and $y$ are respectively,

$3 x=6$
$x=2$
$4 x-y=-1$
$8-y=-1$
$9= y$

Q7. If the standard deviation of the numbers $-1,0,1, k$ is $\sqrt{5}$ where $k > 0$ , then $k$ is equal to

$4 \sqrt{\frac{5}{3}}$

$2 \sqrt{\frac{10}{3}}$

$\sqrt{6}$

$2 \sqrt{6}$

Q8. If the set $x$ contains $7$ elements and set $y$ contains 8 elements, then the number of bijections from $x$ to $y$ is

$0$

$7!$

$8 P_{7}$

$8 !$

$n ( A ) \neq n ( B )$
Number of bijections is zero

Q9. If $f: R \rightarrow R$ be defined by
$f(x)=\begin{cases}2 x & : & x>3 \\x^{2} & : & 1 < x \leq 3 \\3 x & : & x \leq 1\end{cases}$
then $f(-1)+f(2)+f(4)$ is

$f(-1)=3(-1)=-3$
$f(2)=2^{2}=4$
$f(4)=2(4)=8$
$f(-1)+f(2)+f(4)$
$=-3+4+8=9$

Q10. Let the relation $R$ is defined in $N$ by $a R b$ , if $3 a+2 b=27$ then $R$ is

$\{(1,12)(3,9)(5,6)(7,3)\}$

$\{(1,12)(3,9)(5,6)(7,3)(9,0)\}$

$\left\{\left(0, \frac{27}{2}\right)(1,12)(3,9)(5,6)(7,3)\right\}$

$\{(2,1)(9,3)(6,5)(3,7)\}$

$2 b =27-3 a$
$b =\frac{27-3 a }{2}$
$R =\{(1,2),(3,9),(5,6),(7,3)\}$

Q11. $\displaystyle\lim _{y \rightarrow 0} \frac{\sqrt{3+y^{3}}-\sqrt{3}}{y^{3}}=$

$\frac{1}{2 \sqrt{3}}$

$2 \sqrt{3}$

$\frac{1}{3 \sqrt{2}}$

$3 \sqrt{2}$

$\displaystyle\lim _{y \rightarrow 0} \frac{3+y^{3}-3}{y\left(\sqrt{3+y^{3}}+\sqrt{3}\right)}=\frac{1}{2 \sqrt{3}}$

Q12. If $A$ is a matrix of order $3 \times 3$ , then $\left(A^{2}\right)^{-1}$ is equal to

$\left(-A^{2}\right)^{2}$

$A ^{2}$

$\left(A^{-1}\right)^{2}$

$(-A)^{-2}$

$\left( A ^{2}\right)^{-1}=\left( A ^{-1}\right)^{2}$

Q13. If $A=\begin{bmatrix}2 & -1 \\ 3 & -2\end{bmatrix}$ , then the inverse of the matrix $A^{3}$ is

$A$

$1$

$-1$

$- A$

Q14. If $A$ is a skew symmetric matrix, then $A^{2021}$ is

$A ^{ T }=- A$ or $A ^{ n }$ is slow symmetric if $n$ is odd
$P = A ^{2021}$
$P ^{ T }=\left[ A ^{2021}\right]^{ T }=\left[ A ^{ T }\right]^{2021}=(- A )^{2021}=- P$

Q15. If $A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ then $(a I+b A)^{n}$ is (where $I$ is the identity matrix of order $2$ )

$a ^{2} I+a^{n-1} b \cdot A$

$a^{n} I+n a^{n} b A$

$a^{n} I+n \cdot a^{n-1} b \cdot A$

$a^{n} I+b^{n} A$

Q16. If $A$ is a $3 \times 3$ matrix such that $\mid 5 \cdot$ adj $A \mid=5$ then $|A|$ is equal to

$\pm 1$

$\pm 1 / 5$

$\pm 1 / 25$

$\pm 5$

$A_{3 \times 3}$ matrix $|5 . \operatorname{AdjA}|=5$
$\Rightarrow 5^{3}| A |^{2}=5$
$\Rightarrow\left| A ^{2}\right|=\frac{1}{5^{2}}$
$| A |=\pm \frac{1}{5}$

Q17. If there are two values of ' $a$ ' which makes determinant
$\Delta=\begin{vmatrix}1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a\end{vmatrix}=86$
Then the sum of these numbers is

$-4$

Q18. If the vertices of a triangle are $(-2,6)(3,-6)$ and $(1,5)$ , then the area of the triangle is

Q19. Domain of $\cos ^{-1}[x]$ is, where $[ . ]$ denotes a greatest integer function

$(-1,2]$

$[-1,2]$

$(-1,2)$

$[-1,2)$

$\cos ^{-1}[x]$
$-1 \leq[x] \leq 1$
$\Rightarrow[x]=\{-1,0,1\}$
$x \in[-1,2)$

Q20. If $y=\left(1+x^{2}\right) \tan ^{-1} x-x$ then $\frac{d y}{d x}$ is

$2 x \tan ^{-1} x$