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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?

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

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

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

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

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,

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

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

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

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

Solution:

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

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

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

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

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

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 $2$ )

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

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

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

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

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

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

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

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

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

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 $\left|A_{1}\right|+\left|A_{2}\right|+\ldots+\left|A_{2021}\right|=$

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

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

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

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

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

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

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

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

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

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

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

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

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 $2 y(2)-y(1)=$

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

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

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

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

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

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

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

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

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

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

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

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 :

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

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

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 $x$ :

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 $p^{2}+q^{2}=$_____

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

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

Solution:

By graph

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

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

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

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

Solution:

Odd number $=\{1,3,5,7,9\}$
No. of sub sets $=2^{5}=32$