Question

${\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}g\left(\theta \right)d\theta =?$

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is D $-2$

Given $f\left(x\right)+f\left(y\right)=\frac{x+y}{xy}\forall x,yϵR=\left(0\right)$

$\Rightarrow f\left(x\right)=\frac{1}{x}$

Now $g\left(\theta \right)={\int }_{-(\overrightarrow{a}\cdot \overrightarrow{b}{)}^2}^{-\lambda }\frac{1}{x2}dx+{\int }_{\lambda }^{\mid \overrightarrow{a}\times \overrightarrow{b}{\mid }^2}\frac{1}{x^2}dx-\frac{2}{\lambda }$

Here$\left(\right(\overrightarrow{a}\cdot \overrightarrow{b}{)}^2=\mid \overrightarrow{a}{\mid }^2\mid \overrightarrow{b}{\mid }^2{\cos }^2\theta ={\cos }^2\theta ,\mid \overrightarrow{a}\times \overrightarrow{b}{\mid }^2={\sin }^2\theta$

$=[-\frac{1}{x}{]}_{-{\cos }^2\theta }^{-\lambda }+[-\frac{1}{x}{]}_{\lambda }^{{\sin }^2\theta }-\frac{2}{\lambda }=\frac{1}{\lambda }-\frac{1}{{\cos }^2\theta }+\frac{-1}{{\sin }^2\theta }-\left(\frac{-1}{x}\right)-\frac{2}{\lambda }$

$=-\frac{1}{si{n}^2\theta }-\frac{1}{{\cos }^2\theta }=\frac{-1}{{\sin }^2{\cos }^2\theta }=-4{\text{cosec}}^{2}2\theta$

${\int }_{\frac{\pi }{8}}^{\frac{\pi }{4}}-4{\text{cosec}}^{2}2\theta d\theta$

$=-4[\frac{-\cot 2\theta }{2}{]}_{\frac{\pi }{8}}^{\frac{\pi }{4}}=2(0-1)=-2$