Question

Assertion :Statement -1: If $f\left(x\right)=\{\begin{array}{cc}x\cos x.\sin \left(\frac{1}{x\cos x}\right),& \text{whenever defined}\\ 0& \text{otherwise}\end{array}$, then $f\left(x\right)$ is continuous Reason: Statement -2 : $\underset{x\to \infty }{lim}\frac{\sin x}{x}=0$

• 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 A Statement -1 is True, Statement -2 is True ; Statement -2 is a correct explanation for Statement -1

$xcosxsin\left(\frac{1}{xcosx}\right)$ is undefined for $x=0andx=kwhere$
$k=(2n+1)\frac{\pi }{2},n\in Z$
$\therefore f\left(0\right)=0,f\left(k\right)=0$
Now ${lim}_{x\to 0}f\left(x\right)=\left(0\right)\left(1\right)(finitevaluebetween-1and1)=0=f\left(0\right)$
$\therefore f\left(x\right)$ is continuous at $x=0$
Also $li{m}_{x\to k}f\left(x\right)=\left(k\right)\left(0\right)(finitevaluebetween-1and1)=0=f\left(k\right)$
$\therefore f\left(x\right)$ is continuous at $x=k$
Hence, the assertion is true. The reason is also true and it explains the Assertion.