Question

Consider the curves $C_1$ and $C_2$ given by $C_1:y=1+\cos x$ and $C_2:y=1+\cos (x–\alpha )$ for $\alpha \in (0,\frac{\pi }{2}),x\in [0,\pi ]$
The value of ${}^{\prime }{\alpha }^{\prime }$ for which the area of the figure bounded by the curves $C_1,C_2$ and $x=0$ is the same as that of the figure bounded by $c_2,y=1$ and $x=\pi$ is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{5\pi }{12}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $\frac{\pi }{3}$


As $A_1=A_4$
${\int }_0^{\beta }(\cos x-\cos (x-\alpha \left)\right)dx={\int }_{\frac{\pi }{2}+\alpha }^{\pi }1-(1+\cos (x-\alpha \left)\right)dx$
as $1+\cos \beta =1+\cos (\beta -\alpha )\Rightarrow \beta =\frac{\alpha }{2}$
$\Rightarrow \sin \frac{\alpha }{2}=\frac{l}{2}\Rightarrow \alpha =\frac{\pi }{3}$