Question

If the function $f\left(x\right)=\lambda \left|\sin x\right|+{\lambda }^2\left|\cos x\right|+g\left(\lambda \right),\lambda \in R$ is periodic with fundamental period $\frac{\pi }{2},$ then

• A. $\lambda =0,1$
• B. $\lambda =1$
• C. $\lambda =0$
• D. $\lambda =-1$

Answer 1

The correct option is B $\lambda =1$

As, period of $f\left(x\right)$ is $\frac{\pi }{2}$
$\Rightarrow f(\frac{\pi }{2}+x)=f\left(x\right)\forall x\in R$
$\Rightarrow \lambda \left|\cos x\right|+{\lambda }^2\left|\sin x\right|+g\left(\lambda \right)=\lambda \left|\sin x\right|+{\lambda }^2\left|\cos x\right|+g\left(\lambda \right)$
$\Rightarrow (\lambda -{\lambda }^2)\left|\cos x\right|+({\lambda }^2-\lambda )\left|\sin x\right|=0\forall x\in R$
$\Rightarrow \lambda -{\lambda }^2=0\text{or}|\sin x|=|\cos x|$
$\Rightarrow \lambda =0,1\text{or}x=\frac{(2n+1)\pi }{4},n\in Z$
But at $\lambda =0,f\left(x\right)$ becomes a constant function, so $\lambda \ne 0$
$\Rightarrow \lambda =1\text{or}x=\frac{(2n+1)\pi }{4},n\in Z$