$$\color{Black} 1).\text{If} \;f(x)=\log x\;\text{and}\; g(x)=e^x\;
\text{find}\;fog?$$
SOLUTION-$$\color{Black} fog= f(g(x))= f(e^x)\;\;\;\;\;\left ( \because
g(x)=e^x \right )$$$$\color{Black} \Rightarrow fog=\log(e^x)= x \;\;\; \;\;\;\;\;\left ( \because \ln(e^x)=x \right )$$
$$\color{Black} \Rightarrow \text{x is right answer}$$
$$\color{Black}2)\;\color{Black} .A\;\text{is a square matrix of order
}3\times 3\;\;\;$$ $$\color{Black} \text{and}\;|A|=5\;\;\;\; \text{then find}\;\;
|adj(A)|$$
SOLUTION- $$\color{Black} |adjA|=|A|^{n-1} \text{Where n is order of matrix} $$
$$\color{Black} \Rightarrow |adj A|=(5)^2=25$$
$$\color{Black}\Rightarrow \text{so 25 is answer} \;\;\;\;\;\;\;\;\;\;\;$$ $$\color{Black}3)\; \text{Principle value of} \sin^{-1}(\frac{1}{2}) $$
SOLUTION-$$\color{Black} \text{we know that}
\sin(\frac{\pi}{6})=\frac{1}{2}$$ $$\color{Black} \Rightarrow
\sin^{-1}(\frac{1}{2})=\sin^{-1}(\sin(\frac{\pi}{6})$$ $$\color{Black}
\Rightarrow \sin^{-1}(\sin(\frac{\pi}{6})=\frac{\pi}{6}$$ $$\color{Black}
\Rightarrow \text{So the answer is }\;\; \frac{\pi}{6}$$$$\color{Black} 4)\;\; \text{if}\;f(x)=\begin{cases} \displaystyle
\frac{x^2-9}{x-3} & \text{ if } x\neq 3 \\ m & \text{ if } x= 3
\end{cases}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$ $$\color{Black} \text{is continuous at 3 then find m}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$
SOLUTION-
$$ \color{Black} \Rightarrow \;\;\lim_{x \to 3^{-}}\frac{x^2-9}{(x-3)}=\lim_{n \to 3^{-}}\frac{\cancel{(x-3)}(x+3)}{\cancel{(x-3)}}$$
$$ \color{Black} \Rightarrow \lim_{x \to 3^{-}} (x+3)=3+3=6\;\;\;\;...(1)$$
$$\color{Black} \text{ If function is continous at 'a' then }$$ $$\color{Black} \;\lim_{x\to {a^{-}}} f(x)=\lim_{x\to {a^{+}}} f(x)=f(a)$$
$$ \color{Black} \Rightarrow f(3)=\lim_{x \to 3^{-}} (x+3) $$
$$\color{Black}\text{ we have f(3)=m given in question}$$
$$\color{Black} \Rightarrow m=3+3=6 $$
$$\color{Black} \text{so} \;\;6=m$$$$\color{Black}5).\text{if} \;\; y=\log(\tan x) \; \;\text{then} \; \frac{dy}{dx}\; \text{is} $$
SOLUTION$$\color{Black} \text{we know that} \;\;\frac{d}{dx}\log x=\frac{1}{x}\times \frac{d}{dx}.x$$
$$\color{Black} \Rightarrow \frac{d}{dx}\log(\tan x)=\frac{1}{\tan x}\times \frac{d}{dx}\tan x$$
$$ \color{Black} \Rightarrow \frac{1}{\tan x}\times \frac{d}{dx}\tan x$$
$$\color{Black} \Rightarrow \frac{1}{\tan x}\times \sec^2 x= \frac{\sec^2 x}{\tan x}$$
$$\color{Black} \text{Answer is }\;\;\frac{\sec^2 x}{\tan x} $$$$\color{Black} 6) .\text{Find}\; \int_{0}^{\pi/2} \frac{\sin^{3/2}x}{\sin^{3/2}x +\cos^{3/2}x}$$
SOLUTION
$$\color{Black} \text{We have a formula}$$
$$\color{Black} \int_{0}^{a}f(x)\;dx=\int_{0}^{a}f(a-x)\;dx$$
$$\color{Black} \text{we will use this formula in question}$$
$$\color{Black} \text{Let}\;\; I=\int_{0}^{\pi/2}\frac{\sin^{3/2} x}{\sin^{3/2} x+\cos^{3/2} x} \;$$
$$\color{Black} \text{(Now according to formula)} $$
$$\color{Black} =\int_{0}^{\pi/2} \frac{\sin^{3/2} (\frac{\pi}{2}-x)}{\sin^{3/2} (\frac{\pi}{2}-x))+\cos^{3/2} (\frac{\pi}{2}-x)} \;\;\;\;..1)$$
$$\color{Black} \Rightarrow I=\displaystyle \int_{0}^{\pi/2} \frac{\cos^{3/2} (x)}{\cos^{3/2} x+\sin^{3/2} x} \;\;\;\;......2) $$
$$\textbf{Add (1) and (2) we get}$$
$$\color{Black} 2I=\displaystyle \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2}x+\cos^{3/2} x}+\displaystyle \int_{0}^{\pi/2} \frac{\cos^{3/2} (x)}{\cos^{3/2} x+\sin^{3/2} x}$$
$$\color{Black} 2I=\displaystyle \int_{0}^{\pi/2} \frac{\sin^{3/2} x+\cos^{3/2} (x)}{\sin^{3/2}x+\cos^{3/2} x}$$
$$\color{Black} 2I=\displaystyle \int_{0}^{\pi/2} \frac{\bcancel{\sin^{3/2} x+\cos^{3/2} x}}{\bcancel{\sin^{3/2}x+\cos^{3/2} x}}=\int_{0}^{\pi/2}1\;dx$$
$$\color{Black} \Rightarrow 2I=|x|_0^{\pi/2}=\frac{\pi}{2}$$
$$\color{Black} \Rightarrow 2I=\frac{\pi}{2}$$
$$\color{Black} \text{Answer is }\;\; \frac{\pi}{4}$$
$$\color{Black} 7).\text{The degree of differential equation}$$
$$\color{black}\displaystyle \frac{d^2y}{dx^2}+\left( \frac{dy}{dx}\right)^3+y=0$$
SOLUTION
$$\color{Black}\text{(we know that degree of a given differential}$$ $$\color{black}\text{equation is power of highest order derivative)}$$
$$\color{Black}\text{Here maximum derivative is }\;\;\frac{d^2y}{dx^2} \;\;\;\text{and it}$$ $$\color{black}\text{has power 1}$$
$$\color{Black}\text{Answer is 1}$$$$\color{Black}8).\text{If}\; \vec{a}=2\hat{i}+3\hat{j}-\hat{k}\;\;\text{then find} \displaystyle \;\;|\vec{a}|$$
SOLUTION
$$\color{Black}|\vec{a}|=\sqrt{2^2+3^2+(-1)^2}=\sqrt{4+9+1}=\sqrt{14}$$
$$\color{Black}\text{Answer is}\;\sqrt{14}$$$$\color{Black} 9).\text{Find the direction ratios of normal}$$ $$\color{Black} \text{to the plane which is parallel to}$$
$$\color{Black} \;3x+y-z=11$$
SOLUTION-
$$\color{Black} \text{For a plane}\; ax+by+cz=d$$
$$\color{Black} \text{Direction ratios of normal are }\;\;a,b,c$$
$$\color{Black} \Rightarrow \color{\black}\text{So the direction ratios of normal is }$$
$$\color{Black} \langle 3,1 -1 \rangle $$
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