Question

Assertion :$f\left(x\right)={lim}_{n\to \infty }{\left(\cos x\right)}^{2n}$, then f is continuous everywhere $\forall xϵ(-\infty ,\infty )$. Reason: $f\left(x\right)=\cos x$ is continuous $\forall xϵR$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

A: At $x\ne n\pi ,nϵI$

$f\left(x\right)={\left(\cos x\right)}^{2n}$

${lim}_{n\to \infty }f\left(x\right)={lim}_{n\to \infty }{\left(\frac{\pm 1}{a}\right)}^{2n}$ (We know that $\left|\cos x\right|<1$

$=\frac{\pm 1}{a^{\infty }}=0$

at $x=n\pi ,nϵI$

$f\left(x\right)={\left(1\right)}^{2n}$

${lim}_{n\to \infty }f\left(x\right)={lim}_{n\to \infty }{1}^{2n}\ne 0$

$\therefore x=n\pi f\left(x\right)$ will be discontinuous

R: for $xϵR$

${lim}_{x\to a}\cos x=f\left(a\right)$$\cos a=\cos a$

$\Rightarrow \cos x$ is continuous on R