KCET 2020 Mathematics Questions with Answers Key Solutions
Solution:
$2^{x}+2^{y}=2^{(x+y)} \ldots . .(1)$
Differentiating both sides w.r.t. $x$, we get
$2^{x} \ln 2+2^{y} y' \ln 2=2^{(x+y)} \ln 2\left(1+y'\right)$
$\Rightarrow 2^{x}+2^{y} y'=2^{(x+y)}\left(1+y'\right)$
$\Rightarrow 2^{x}+2^{y} \cdot y'=2^{(x+y)}+2^{(x+y)} \cdot y'$
$\Rightarrow 2^{x}-2^{(x+y)}=y'\left(2^{(x+y)}-2^{y}\right)$
$\Rightarrow 2^{x}-2^{x}-2^{y}=y'\left(2^{x}+2^{y}-2^{y}\right)[$ From eqn. $(1)]$
$\Rightarrow -2^{y}=y'2^{x}$
$\Rightarrow y'=d y / d x=-2^{y-x}$
Solution:
$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
Put $x=\tan\, \theta$, where $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$f(x)=\sin ^{-1}\left(\frac{2 \tan \,\theta}{1+\tan ^{2} \theta}\right)$
$\Rightarrow f(x)=\sin ^{-1}(\sin 2 \theta)$
$\Rightarrow f(x)=2 \,\theta=2 \tan ^{-1} x \,\,\left(\because \theta=\tan ^{-1} x\right)$
$f'(x)=\frac{2}{1+x^{2}}$
$\therefore f'(\sqrt{3})=\frac{2}{1+3}=\frac{1}{2}$
Solution:
$LHL =\displaystyle\lim _{x \rightarrow 0^{-}}\left(\frac{e^{1 / x}-1}{e^{1 / x}+1}\right)$
$=\frac{e^{-\infty}-1}{e^{-\infty}+1}=\frac{0-1}{0+1}=-1$
Dividing both numerator and denominator by $e^{1 / x}$, we get
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{1-e^{-1 / x}}{1+e^{-1 / x}}\right)$
$RHL =\displaystyle\lim _{x \rightarrow 0^{+}}\left(\frac{1-e^{-1 / x}}{1+e^{-1 / x}}\right)$
$=\frac{1-e^{-\infty}}{1+e^{-\infty}}=\frac{1-0}{1+0}=1$
Solution:
$y=(2 x)^{(n+1)}+(3 x)^{(-n)}$
$\Rightarrow d y / d x=2(n+1) x^{n}-(3 n x)^{(-n-1)}$
$\Rightarrow \left(d^{2} y\right) /\left(d x^{2}\right)=2 n(n+1) x^{(n-1)}+3 n(n+1) x^{(-n-2)}$
$\Rightarrow x^{2}\left(d^{2} y\right) /\left(d x^{2}\right)=n(n+1)\left[(2 x)^{(n+1)}+3 / x^{n}\right]$
$\Rightarrow x^{2}\left(d^{2} y\right) /\left(d x^{2}\right)=n(n+1) y$
Solution:
$2 x=y^{2} \ldots(1)$
$2 x y=K \ldots(2)$
Solving $(1)$ and $(2)$, we get
$( x , y )=\left( K ^{(2 / 3)} / 2, K ^{(1 / 3)}\right)$
Differentiating $(1)$ and $(2)$ w.r.t. $x$
$m _{1}= dy / dx =1 / y \ldots(3)$
$m_{2}=d y / d x=-y / x \ldots(4)$
Both curves intersect each other perpendicularly
$\therefore m _{1} m _{2}=-1$
$\Rightarrow -1 / x=-1$
$\Rightarrow x =1$
$\Rightarrow K ^{(2 / 3)}=2$
$\Rightarrow K ^{2}=8$
Solution:
$(x e)^{y}=e^{x}$
$\Rightarrow y(\log x+1)=x$
$\Rightarrow y=x /(\log x+1)$
$\therefore \left(\frac{dy}{dx} = \frac{log \ x}{(log \ x + 1)^2}\right)$
Solution:
Let one side of the cube be $x$ and surface area be $A$
So, $dx =5 \%=5 x / 100$
Then, $A=(6 x)^{2}$
$\Rightarrow dA / dx =12 x$
$\Rightarrow d A =(12 x ) d x$
$\Rightarrow dA =(12 x )(5 x / 100)$
$\Rightarrow d A =10 A / 100$
$\Rightarrow dA =10 \%$
Solution:
$ \int \frac{1+x^{4}}{1+x^{6}} d x =\int \frac{1+x^{4}-x^{2}+x^{2}}{\left(1+x^{2}\right)\left(1-x^{2}+x^{4}\right)} d x$
$=\int\left(\frac{1-x^{2}+x^{4}}{\left(1+x^{2}\right)\left(1-x^{2}+x^{4}\right)}+\frac{x^{2}}{\left(1+x^{2}\right)\left(1-x^{2}+x^{4}\right)}\right) d x $
$=\int\left(\frac{1}{1+x^{2}}+\frac{1}{3} \cdot \frac{3 x^{2}}{1+\left(x^{3}\right)^{2}}\right) d x $
$=\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^{3}+C $
Solution:
$(y = \frac{log_e \ x}{x})$
$⇒ \left(\frac{dy}{dx}=\frac{1 - log_e x}{x^2}\right)$
For maxima, $dy/dx = 0$
$⇒ 1 - log_e x = 0$
$⇒ x = e$
$dy/dx$ changes sign from positive to negative at $x = e$
$∴ y_{\text{max}} = 1/e$
Solution:
$\left(I = \int e^{sin\,x} \ sin \,2x \ dx = 2 \int e^{sin\,x} \ sin \,x \ cos \,x \ dx\right)$
Let $t=\sin\, x$
$\Rightarrow d t=\cos \,x d x$
Therefore, $I=2 \int {te}^{t} d t$
$=2\left(t e^{t}-e^{t}\right)+C$
$=2(\sin \,x-1) e^{\sin\, x}+C$
Solution:
$\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \cos ^{-1} x d x$
$=\left|x \cos ^{-1} x\right|_{-\frac{1}{2}}^{\frac{1}{2}}+\int\limits_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x}{\sqrt{1-x^{2}}} d x \quad$ [Using integration by parts]
$=\frac{1}{2} \cos ^{-1}\left(\frac{1}{2}\right)+\frac{1}{2} \cos ^{-1}\left(-\frac{1}{2}\right)-\left[\left(\frac{1}{2}\right) \cdot 2 \sqrt{1-x^{2}}\right]_{-\frac{1}{2}}^{\frac{1}{2}}$
$=\frac{1}{2} \cdot \frac{\pi}{3}+\frac{1}{2} \cdot \frac{2 \pi}{3}-0$
$=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$
Solution:
Let $\frac{3 x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\dots$(i)
$\Rightarrow \int \frac{3 x+1}{(x-1)(x-2)(x-3)} d x$
$=A \log |x-1|+B \log |x-2|+C \log |x-3|+C$
Now, $3 x+1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)$ [From eqn. (1)]
Putting $x=1, x=2, x=3$ in the above equation one at a time, we get
$A=2, B=-7, C=5$
Solution:
$I=\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$
Let $x=\tan \theta$
$\Rightarrow d x=\sec ^{2} \theta d \theta$
$I=\int_{0}^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta$
$=\int_{0}^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta$
$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan \theta}\right) d \theta$
$=\int_{0}^{\frac{\pi}{4}} \log 2 d \theta-\int_{0}^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$\Rightarrow 2 I=\frac{\pi}{4} \log 2$
$\Rightarrow I=\frac{\pi}{8} \log 2$
Solution:
Solving $y^2= 8x$ and $y = 2x$, we get
$(x, y) = (0, 0), (0, 4)$
So, area bounded by the curve is
$\int\limits_{0}^{2}\left(2 \sqrt{2} x^{1 / 2}-2 x\right) d x$
$=\left(\frac{4 \sqrt{2}}{3} x^{3 / 2}-x^{2}\right)_{0}^{2}=\frac{4}{3}$ sq. units
Solution:
$I=\int\limits_{-\frac{\pi}{2}^{\frac{\pi}{2}} \frac{\cos x}{1+e^{x}}} d x \dots$(1)
$\Rightarrow I=\int\limits_{-\frac{\pi}{2}^{\frac{\pi}{2}} \frac{\cos \left(\frac{\pi}{2}-\frac{\pi}{2}-x\right)}{1+e^{\frac{\pi}{2}-\frac{\pi}{2}-x}} d x}$
$\Rightarrow I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos\, x}{1+e^{-x}} d x$
$=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{x} \cos x}{1+e^{x}} d x \cdots$(2)
From (1) and (2), we have
$2 I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(e^{x}+1\right) \cos\, x}{e^{x}+1} d x$
$=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos \,x\, d x$
$\Rightarrow 2 I=\left.\sin\, x\right|_{-\frac{\pi}{2}} ^{\frac{\pi}{2}} $
$\Rightarrow I=1$
Solution:
$\left(y = [(\frac{C_2 + C_3}{C_1})e^{C_4}]e^x = A e^x,\right)$
where $A = \left((\frac{C_2 + C_3}{C_1})e^{C_4}\right)$
Order = Number of independent arbitrary constants $= 1$
Solution:
$x^{2} d y-2 x y d x=x^{4} \cos\, x d x$
$\left(\frac{dy}{dx} = \frac{x^4 \ cos\, x+2xy}{x^2}\right)$
$\Rightarrow d y / d x-2 y / x=x^{2} \cos \,x$
I.F. $= e ^{\int-2 / x d x}=e^{-2 \log x}=1 / x^{2}$
Therefore, the general solution is
$\left(y(\frac{1}{x^2}) = \int \frac{1}{x^2}(x^2 \ cos\, x)dx = sin\,x + c\right)$
$\therefore y = x ^{2}(\sin\, x + c )$
$=x^{2} \sin\, x+c x^{2}$
Solution:
Area bounded by $y = 2x+1$ with $x$- axis
$= (1/2) (1/2)(1) +(1/2) (3/2)(3) $
$= 5/2$ sq. units.
Solution:
$\overrightarrow{A B}=(1,1,1), \overrightarrow{A C}=(1,3,5)$
$\Rightarrow \overrightarrow{B C}=\overrightarrow{A C}-\overrightarrow{A B}=(0,2,4)$
Since $D$ is the mid-point of $\overrightarrow{B C}$,
$ \therefore \overrightarrow{B D}=(0,1,2)$
$\overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B D}$
$=(1,2,3)$
$\therefore |\overrightarrow{A D}| =\sqrt{1+4+9} $
$=\sqrt{14}$
Solution:
$|\vec{a}-\vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}||\vec{b}| \cos \theta$
$=2(1-\cos \theta) \,\,\,(\because|\vec{a}|=|\vec{b}|=1)$
$\Rightarrow |\vec{a}-\vec{b}|^{2}=2\left(2 \sin ^{2} \frac{\theta}{2}\right)$
$\therefore \sin \frac{\theta}{2}=\frac{|\vec{a}-\vec{b}|}{2}$
Solution:
Given, $\left(\frac{dy}{dx}= \frac{2x}{y}\right)$
$\Rightarrow y d y=2 x d x$
$\Rightarrow \int y d y=\int 2 x d x$
$\Rightarrow y^{2} / 2=x^{2}+A$, where $A$ is a constant.
The above equation represents a hyperbola.
Solution:
$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144$
$\Rightarrow |\vec{a}|^{2}|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2}|\vec{b}|^{2} \cos ^{2} \theta=144$
$\Rightarrow |\vec{a}|^{2}|\vec{b}|^{2}=144$
$\Rightarrow |\vec{b}|^{2}=\frac{144}{36}=4 \,\,\,(\because|\vec{a}|=6)$
$\Rightarrow |\vec{b}|=2$
Solution:
Signs of $x$-coordinate, $y$-coordinate and $z$-coordinate are $+,-,+ $ respectively.
$∴(1,-3,4)$ lies in the fourth octant.
Solution:
Given vectors are coplanar.
$⇒ \left(\begin{vmatrix} 2 &-3 &4 \\ 2& 1 &-1 \\ \lambda & -1 &2 \end{vmatrix} = 0\right)$
$\Rightarrow 2(1)+3(4+\lambda)+4(-2-\lambda)=0$
$\Rightarrow 2+12+3 \lambda-8-4 \lambda=0$
$\Rightarrow \lambda=6$
Solution:
Let $\left(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}=t\right)$
$\Rightarrow (x, y, z)=(2 t+3,3 t+3,6 t-5)$
$\therefore $ d.r.’s of the line perpendicular to
$\left(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}\right)$ and
joining $(2 t+3,3 t+3,6 t-5)$
and $(1,2,-4)$ is $(2 t+2,3 t+1,6 t-1)$
$\therefore 2(2 t+2)+3(3 t+1)+6(6 t-1)=0$
$\Rightarrow t =-1 / 49$
$\therefore $ Distance $=\sqrt{(2t+2)^2+(3t+1)^2+(6t-1)^2}$
$ = \sqrt{49t^2+2t+6}$
$= \sqrt{\frac{1}{49}-\frac{2}{49}+6}=\frac{\sqrt{293}}{7}$
Solution:
Given line is $ \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
and plane is $2x-2y+z = 5$
So, $\sin\, \theta=\left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right|$
$\Rightarrow \sin\, \theta=\left|\frac{(3 \hat{\imath}+4 \hat{j}+5 \hat{k}) \cdot(2 \hat{\imath}-2 \hat{\jmath}+\hat{k})}{\sqrt{9+16+25} \sqrt{4+4+1}}\right|$
$=\frac{6-8+5}{\sqrt{50} \sqrt{9}}=\frac{1}{5 \sqrt{2}}$
$=\frac{1}{5 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{10}$
Solution:
Given $\alpha = \beta = \pi/3$
Let acute angle made by $z$ -axis be $y$
Then, $cos^2 \alpha + cos^2 \beta + cos^2 \gamma$
$\Rightarrow (\frac{1}{2})^2+(\frac{1}{2})^2 + cos^2 \gamma = 1$
$\Rightarrow \frac{1}{4}+\frac{1}{4} + cos^2 \gamma = 1 $
$\Rightarrow cos^2 \gamma = \frac{1}{2}$
$\Rightarrow cos \gamma = \pm \frac{1}{\sqrt 2}$
$\Rightarrow \gamma = \frac{\pi}{4}$
$[\therefore y $ is acute]
Solution:
Given corner points are $(0,3),(1,1),(3,0)$
$z=p x+q y$
At $(3,0), z=3 p$
At $(1,1), z=p+q$
It is given that the minimum of $z$ occurs at $(3,0)$ and $(1,1)$
$\Rightarrow 3 p=p+q$
$\Rightarrow 2 p = q$
$\Rightarrow p = q / 2$
Solution:
$y$ Intercept of $x+y=5$ is $(0,5)$
$y$ -intercept of $x+3 y=9$ is $(0,3)$
The intersection point of $x+y=5$ and $x+3 y=9$ 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, $Z_{\max }=47$ at $(3,2)$.
Q30. A die is thrown $10$ times, the probability that an odd number will come up at least one time is
Solution:
Given $n =10$
Probability of odd number, $p =1 / 2$
$\therefore q =1 / 2$
Required probability $= P ( X \geq 1)$
$=1- P ( X =0)$
$= 1- \,{}^{10}C_0 (1/2)\,{}^{10-0}(1/2)^0$
$=1-1 / 2^{10}$
$=1-1 / 1024$
$=1023 / 1024$
Solution:
Given $P(A)=1 / 3, P(B)=1 / 2, P(A \cap B)=1 / 6$
So, $P\left(A' | B\right)=1-P(A | B)$
$= 1 - \frac{P(A \cap B)}{P(B)}$
$=1-1 / 3$
$=2 / 3$
Solution:
Let $P\left(A \mid E_{1}\right)=x$
By Bayes' theorem,
$P\left(E_{2} |A\right)=\frac{P\left(E_{2}\right) P\left(A |E_{2}\right)}{P\left(E_{1}\right) P\left(A |E_{1}\right)+P\left(E_{2}\right) P\left(A |E_{2}\right)} $
$\Rightarrow \frac{1}{2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{2}\right) x+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}$
$\Rightarrow x=\frac{2}{3} $
$\therefore P\left(E_{1} |A\right)=\frac{P\left(E_{1}\right) P\left(A |E_{1}\right)}{P\left(E_{1}\right) P\left(A | E_{1}\right)+P\left(E_{2}\right) P\left(A | E_{2}\right)} $
$=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}=\frac{1}{2}$
Solution:
Required probability $= P \left( A ' BC \right)+ P \left( AB ' C \right)+ P \left( ABC '\right)$
$=1 / 2 \times 1 / 4 \times 1 / 3+1 / 2 \times 3 / 4 \times 1 / 3+1 / 2 \times 1 / 4 \times 2 / 3$
$=1 / 24+1 / 8+1 / 12$
$=(1+3+2) / 24$
$=1 / 4$
Solution:
$n(A)=2$
Given, $2^{(n(A) \cdot n(B))}=1024$
$\Rightarrow (2)^{(2 \cdot n(B))}=(2)^{10}$
$\Rightarrow 2 \cdot n(B)=10$
$\Rightarrow n(B)=5$
Solution:
$\sin ^{2} 51^{\circ}+\sin ^{2} 39^{\circ}$
$=\cos ^{2} 39^{\circ}+\sin ^{2} 39^{\circ}$
$=1$
Solution:
$\tan \,A+\cot \,A=2$
$\Rightarrow (\tan \,A+\cot \,A)^{2}=4$
$\Rightarrow \tan ^{2} A+\cot ^{2} A+2 \tan \,A \cot\, A=4$
$\Rightarrow \tan ^{2} A+\cot ^{2} A=2$
$\Rightarrow \left(\tan ^{2} A+\cot ^{2} A\right)^{2}=4$
$\Rightarrow \tan ^{4} A+\cot ^{4} A+2 \tan ^{2} A \cot ^{2} A=4$
$\Rightarrow \tan ^{4} A+\cot ^{4} A=2$
Solution:
Total number of subsets of $A$ is $2^{ n (A)}=2^{6}=64$
Number of subsets of A which contain at least two elements is
$64- \left({ }^{6} C _{0}+{ }^{6} C _{1}\right)$
$=64-(1+6)$
$=57$
Solution:
$|z+1|=|z-1|$
Let $z=x+i y$
Then, $|x+i y+1|=|x+i y-1|$
$\Rightarrow \sqrt{(x+1)^{2}+y^{2}}=\sqrt{(x-1)^{2}+y^{2}}$
$\Rightarrow (x+1)^{2}+y^{2}=(x-1)^{2}+y^{2}$
$\Rightarrow x^{2}+2 x+1+y^{2}=x^{2}+1-2 x+y^{2}$
$\Rightarrow 4 x=0$
$\Rightarrow x=0$ which represents $y-$ axis.
Solution:
${ }^{16} C _{9}+{ }^{16} C _{10^{-}}{ }^{16} C _{6^{-}}{ }^{16} C _{7}$
$={ }^{16} C _{9}+{ }^{16} C _{10^{-}}{ }^{16} C _{10^{-}}{ }^{16} C _{9}=0$
$\left(\because{ }^{ n } C _{ r }={ }^{ n } C _{( n - r )}\right)$
Solution:
Number of terms in the expansion of
$( x + y + z )^{10}$ is ${ }^{(10+3-1)} C _{10}$
$={ }^{12} C _{10}=12 ! /(2 ! 10 !)=66$
Solution:
For $n =1,2,3,2^{ n }> n !$
$P(4): 2^{4}<4 !$
So, smallest positive integer, $n =4$.
Solution:
Product of slopes $= -1$
$⇒ ll'' + mm' = 0$
Q43. If the parabola $x^2=4ay$ passes through the point $(2, 1)$, then the length of the latus rectum is
Solution:
$x^{2}=4 a y$
Given parabola passes through $(2,1)$.
$\Rightarrow 2^{2}=4 a$
$\Rightarrow a=1$
Length of latus rectum= $4 a=4$.
Solution:
$S _{ n }= n ^{2}+ n$
$S _{1}=1+1=2= T _{1}$
$S _{2}=2^{2}+2=6= T _{1}+ T _{2}$
$\therefore T _{2}= S _{2}- S _{1}=4$
Common difference, $d = T _{2}- T _{1}$
$=4-2$
$=2$
Solution:
Negation: For some real numbers $x$ and $y, x + y ≠ y + x$.
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, $\sigma = \sqrt{\frac{1}{n} \sum(x_i - \bar{x})^2}$
$\Rightarrow \sigma = \sqrt{\frac{10}{5}} = \sqrt{2}$
Solution:
$\displaystyle\lim_{x \rightarrow 0}\left(\frac{\tan \,x}{\sqrt{2 x+4}-2}\right) \,\,\,\left[\frac{0}{0}\right]$ form
Applying L' Hopitals' rule, we get
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{\sec ^{2} x}{\frac{2}{2x\sqrt{2x +4}}-0}\right)$
$=\displaystyle\lim _{x \rightarrow 0}\left((\sqrt{2 x+4}) \sec ^{2} x\right)$
$=\sqrt{2 \times 0+4} \times 1=2$
Solution:
$R= \{(1,1)\}$ on set $\{1,2,3\}$
Clearly, $R$ is symmetric and transitive.
Q49. Let $f:[2,\infty) \to R$ be the function defined $f(x) = x^2 - 4x + 5$ , then the range of f is
Solution:
$f(x)=(x-2)^{2}+1 \geq 1, \forall x \in[2, \infty)$
$f_{\min }=1$ at $x=2$
$\therefore $ Range of $f$ is $[1, \infty)$
Solution:
Given, $P(A)=2 P(B)=3 P(C)$
$\Rightarrow P ( C )=2 / 3 P ( B )$
Since $A , B , C$ are three mutually exclusive and exhaustive events
$\therefore P(A)+P(B)+P(C)=1$
$\Rightarrow P ( B )=3 / 11$
Solution:
For $f$ to be defined,
$x-1 \geq 0$ and $-1 \leq \sqrt{(}(x-1) \leq 1$
$\Rightarrow x \geq 1$ and $0 \leq x -1 \leq 1$
$\Rightarrow x \geq 1$ and $1 \leq x \leq 2$
$\Rightarrow x \in[1,2]$
Hence, domain of $f$ is $[1,2]$.
Solution:
$\pi / 3 \notin[-1,1]$ which is the domain of $\sin ^{-1} x, \cos ^{-1} x$
So, $\cos \left(\sin ^{-1} \pi / 3+\cos ^{-1} \pi / 3\right)$ does not exist.
Solution:
$A^{2}=\begin{pmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{pmatrix}\begin{pmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{pmatrix}=I$
$\therefore A^{4}=\left(A^{2}\right)^{2}=I^{2}=I$
Solution:
$A=\{a, b, c\}$
$n(A)=3$
Number of binary operations is
$n ( A )^{( n ( A ))^ 2}=3^{\left(3^ 2\right)}=3^{9}$
Solution:
We know that if $B A=A B=I$, then $A$ and $B$ are inverse of each other. Let $B=\begin{pmatrix}2 & 1 \\ 3 & 2\end{pmatrix}$
$ \Rightarrow |B|=1$
$adj(B)=\begin{pmatrix}2 & -1 \\ -3 & 2\end{pmatrix}$
$B^{-1}=\begin{pmatrix}2 & -1 \\ -3 & 2\end{pmatrix}$
$\therefore A=B^{-1}=\begin{pmatrix}2 & -1 \\ -3 & 2\end{pmatrix}$
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).
$∴f(0) = 0.$
Q57. If $A$ and $B$ are square matrices of same order and $B$ is a skew symmetric matrix, then $A'BA$ is
Solution:
$B$ is a skew symmetric matrix.
$\Rightarrow B'=- B$
Now, $\left(A' B A\right)'=A' B'\left(A'\right)'$
$=A'(-B) A$
$=-\left(A'B A\right)$
Hence, $A'BA$ is a skew symmetric matrix.
Solution:
$|A$ adj $A|=| A|| $adj $A |$
$=|A||A|^{3-1}$
$=5\cdot 5^{2}$
$=125$
Solution:
Given, $f$ is continuous at $x=0$.
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} f(x)=f(0) $
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{1-\cos\, K x}{x\, \sin\, x}=\frac{1}{2}$
Applying L' Hopitals' rule, we get
$\displaystyle\lim _{x \rightarrow 0} \frac{K\, \sin\, K x}{\sin\, x+x\, \cos\, x}=\frac{1}{2}$
Again, by L' Hopitals' rule,
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{K^{2} \cos \,K x}{\cos \,x-x\, \sin\, x+\cos \,x}\right)=\frac{1}{2} $
$\Rightarrow \frac{K^{2}}{1-0+1}=\frac{1}{2}$
$\Rightarrow K=\pm 1$
Solution: