JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q65

← JEE Main 2022 · 28 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 65 of 90

The function $f:\mathbf R\to\mathbf R$ defined by $f(x)=\displaystyle\lim_{n\to\infty}\frac{\cos(2\pi x)-x^{2n}\sin(x-1)}{1+x^{2n+1}-x^{2n}}$ is continuous for all $x$ in :

\mathbf R-\{-1\}

\mathbf R-\{-1,1\}

\mathbf R-\{1\}

\mathbf R-\{0\}

Answer and worked solution

Official answer: B

Pieces. For $|x|<1$ , $f=\cos2\pi x$ ; for $|x|>1$ , $f=-\frac{\sin(x-1)}x$ . $f(1)=1$ but the right limit is 0; $f(-1)=-(1+\sin2)$ but the right limit is 1.

Continuous on $\mathbf R-\{-1,1\}$ . (B)

Topics:Continuity Limits

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JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 65