Question

The function $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\lambda }\left|\sin \left(x\right)\right|+{\mathrm{\lambda }}^2\left|\cos \left(x\right)\right|+\mathrm{g}\left(\mathrm{\lambda }\right)$ has period equal to $\frac{\mathrm{\pi }}{2}$ if $\mathrm{\lambda }$ is

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$1$

Explanation for correct option

Compute the value of $\mathrm{\lambda }$.

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{\lambda }\left|\sin \left(x\right)\right|+{\mathrm{\lambda }}^2\left|\cos \left(x\right)\right|+\mathrm{g}\left(\mathrm{\lambda }\right)$ which has period equal to $\frac{\mathrm{\pi }}{2}$.

A function $\mathrm{f}$ is said to be periodic if it repeats its nature after specific interval and that interval is called as the period of the function.

We will use the condition if $\mathrm{f}\left(\mathrm{x}\right)$ is periodic function with period is equal to $\mathrm{a}$ that is, $\mathrm{f}\left(\mathrm{x}+\mathrm{a}\right)=\mathrm{f}\left(\mathrm{x}\right)$.

Here, $\mathrm{a}=\frac{\mathrm{\pi }}{2}$.

So,

$\Rightarrow$ $\mathrm{f}\left(\mathrm{x}+\frac{\mathrm{\pi }}{2}\right)=\mathrm{f}\left(\mathrm{x}\right)$

$\Rightarrow$ $\mathrm{\lambda }\left|\left(\sin \left(\mathrm{x}+\frac{\mathrm{\pi }}{2}\right)\right)\right|+{\mathrm{\lambda }}^2\left|\cos \left(\mathrm{x}+\frac{\mathrm{\pi }}{2}\right)\right|+\mathrm{g}\left(\mathrm{\lambda }\right)=\mathrm{\lambda }\left|\sin \mathrm{x}\right|+{\mathrm{\lambda }}^2\left|\cos \mathrm{x}\right|+\mathrm{g}\left(\mathrm{\lambda }\right)$

$\Rightarrow$ $\mathrm{\lambda }\left|\cos \mathrm{x}\right|+{\mathrm{\lambda }}^2\left(-\sin \mathrm{x}\right)=\mathrm{\lambda }\left|\sin \mathrm{x}\right|+{\mathrm{\lambda }}^2\left|\cos \mathrm{x}\right|$

$\Rightarrow$ $\mathrm{\lambda }\left|\cos \mathrm{x}\right|-\left|\sin \mathrm{x}\right|{\mathrm{\lambda }}^2\left(-\sin \mathrm{x}-\left|\cos \mathrm{x}\right|\right)=0$

$\Rightarrow$ $\left({\mathrm{\lambda }}^2-\mathrm{\lambda }\right)\left(\left|\cos x\right|-\left|\sin x\right|\right)=0$

$\Rightarrow$ ${\mathrm{\lambda }}^2-\mathrm{\lambda }=0$

$\Rightarrow$ $\mathrm{\lambda }\left(\mathrm{\lambda }-1\right)=0$

$\Rightarrow$ $\mathrm{\lambda }=0,1$

The value of $\mathrm{\lambda }$ is $0$ or $1$.

If we take $\mathrm{\lambda }=0$, then we get,

$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{\lambda }\right)$

Here, $\mathrm{f}\left(\mathrm{x}\right)$ is a constant function, and constant functions can not be periodic. So, $\mathrm{\lambda }$ can never be $0$.

Therefore, the value of $\mathrm{\lambda }$ is $1$.

Hence, the Option $\left(\mathrm{B}\right)$ is correct.