$f(x)=\frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}, \quad \mathrm{x} \neq 2$
$\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2$
where $P(x)$ is a polynomial such that $P^{\prime \prime}(x)$ is always a constant and $P(3)=9$. If $f(x)$ is continuous at $x=2$, then $P(5)$ is equal to $.....$
$\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2$
$\mathrm{P}^{\prime \prime}(\mathrm{x})=$ const. $\Rightarrow \mathrm{P}(\mathrm{x})$ is a 2 degree polynomial
$f(x)$ is cont. at $x=2$
$f\left(2^{+}\right)=f\left(2^{-}\right)$
$\lim _{x \rightarrow 2^{+}} \frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}=7$
$\lim _{x \rightarrow 2^{+}} \frac{(x-2)(a x+b)}{\sin (x-2)}=7 \Rightarrow 2 a+b=7$
$P(x)=(x-2)(a x+b)$
$P(3)=(3-2)(3 a+b)=9 \Rightarrow 3 a+b=9$
$a=2, b=3$
$P(5)=(5-2)(2.5+3)=3.13=39$
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${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then
$\pi$
$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$