Question

Assertion :$f\left(x\right)=\frac{2}{\pi }x\sin x+x^3$, where $x\in \left[\begin{array}{c}0,\frac{\pi }{2}\end{array}\right]$.

Statement-1 : $f\left(x\right)=\frac{\pi }{2}$ has exactly one solution in $x\in \left[\begin{array}{c}0,\frac{\pi }{2}\end{array}\right]$
and Reason: Statement-2 : $f\left(x\right)\ge 0$ for all $x$ in $\left[\begin{array}{c}0,\frac{\pi }{2}\end{array}\right]$

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
• C. Statement-1 is True, Statement-2 is False.
• D. Statement-1 is False, Statement-2 is True.

Answer 1

The correct option is B Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

Given, $f\left(x\right)=\frac{2}{\pi }x\sin x+x^3$

For $x>0;f\left(x\right)>0$ for $x\in [0,\frac{\pi }{2}]$

$($ as $\sin x$ is increasing in $x\in [0,\frac{\pi }{2}])$

For $x<0;f\left(x\right)<0$

$($ as $\sin x<0$ for $x\in [0,\frac{\pi }{2}])$

And for $x=0;f\left(x\right)=0$

$\Rightarrow f\left(x\right)=0$ has at least one root in $x\in [0,\frac{\pi }{2}]$

Also $f^{\prime }\left(x\right)=\frac{2}{\pi }\sin x+\frac{2}{\pi }x\cos x+3x^2$

And $\sin x\ge 0$ for $x\in [0,\frac{\pi }{2}]$

$x\cos x\ge 0$ for $x\in [0,\frac{\pi }{2}]$

$x^2\ge 0$ for $x\in R$

$\therefore f^{\prime }\left(x\right)\ge 0$ for $x\in [0,\frac{\pi }{2}]$

$\therefore f\left(x\right)\ge 0$ for $x\in [0,\frac{\pi }{2}]$