CUET Preparation Today
The value of $\frac{\left\{\int\limits_{-1 / 2}^{1 / 2} \cos 2 x . \log \left(\frac{1+x}{1-x}\right) d x\right\}}{\left\{\int\limits_0^{1 / 2} \cos 2 x . \log \left(\frac{1+x}{1-x}\right) d x\right\}}$ is |
0 4 2 8 |
0 |
Since $f(x)=\cos 2 x . \log \left(\frac{1+x}{1-x}\right)$ is an odd function. ∴ $\int\limits_{-1 / 2}^{1 / 2} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x=0$ Hence, $\frac{\int\limits_{-1 / 2}^{1 / 2} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x}{\int\limits_0^{1 / 2} \cos 2 x \log \left(\frac{1+x}{1-x}\right) d x}=0$ |
