P.S.E.B 12th one marks question solutions

$$\color{Black} 1).\text{If} \;f(x)=\log x\;\text{and}\; g(x)=e^x\; \text{find}\;fog?$$

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- $$\color{Black} \text{For a plane}\; ax+by+cz=d$$ $$\color{Black} \text{Direction ratios of normal are }\;\;a,b,c$$ $$\color{Black} \Rightarrow \color{\black}\text{So the direction ratios of normal is }$$ $$\color{Black} \langle 3,1 -1 \rangle$$

NOTE**-PLEASE ROTATE THE SCREEN IN CASE OF QUESTION IS CUT OFF FROM LEFT SIDE

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