Limits, Continuity & Differentiability · JEE Main previous year questions

2022 28 questions

1. Q222022 · 25 Jun 2022 · Shift 2· Numerical answer

   Let $f(x)=[2x^2+1]$ and $g(x)=\{\begin{array}{ll}2x-3,& x<0\\ 2x+3,& x\ge 0\end{array}$, where $[t]$ is the greatest integer $\le t$. Then, in the open interval $(-1,1)$, the number of points where $f\circ g$ is discontinuous is equal to $\underline{}$.

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   Answer: 62

2. Q52022 · 24 Jun 2022 · Shift 2

   Let $$f(x)=\{\begin{array}{ll}\frac{\sin (x-[x])}{x-[x]}& ,x\in (-2,-1)\\ \max \{2x,3[|x|]\}& ,|x|<1\\ 1& ,otherwise\end{array}$$ where [t] denotes greatest integer $\le t$. If m is the number of points where $f$ is not continuous and n is the number of points where $f$ is not differentiable, then the ordered pair (m, n) is :

   1. A.(3, 3)
   2. B.(2, 4)
   3. C.(2, 3)
   4. D.(3, 4)
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   Answer: C

3. Q162022 · 24 Jun 2022 · Shift 2

   ${\lim }_{n\to \infty }\left(\frac{n^2}{(n^2+1)(n+1)}+\frac{n^2}{(n^2+4)(n+2)}+\frac{n^2}{(n^2+9)(n+3)}+\cdots +\frac{n^2}{(n^2+n^2)(n+n)}\right)$ is equal to :

   1. A.$\frac{\pi }{8}+\frac{1}{4}{\log }_{e}2$
   2. B.$\frac{\pi }{4}+\frac{1}{8}{\log }_{e}2$
   3. C.$\frac{\pi }{4}-\frac{1}{8}{\log }_{e}2$
   4. D.$\frac{\pi }{8}+{\log }_e\sqrt{2}$
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   Answer: A