Continuity - JEE Main Previous Year Question with Solutions

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Q. The values of p and q for which the function f(x) =$\left\{\begin{aligned} \frac{\sin (\mathrm{p}+1) \mathrm{x}+\sin \mathrm{x}}{\mathrm{x}} &, \quad \mathrm{x}<0 \\ \mathrm{q} &, \quad \mathrm{x}=0 \\ \frac{\sqrt{\mathrm{x}+\mathrm{x}^{2}}-\sqrt{\mathrm{x}}}{\mathrm{x}^{\frac{3}{2}}} &, \quad \mathrm{x}>0 \end{aligned}\right.$ is continuous for all x in R, are :-

(1) $\mathrm{p}=-\frac{3}{2}, \mathrm{q}=\frac{1}{2}$

(2) $\mathrm{p}=\frac{1}{2}, \mathrm{q}=\frac{3}{2}$

(3) $\mathrm{p}=\frac{1}{2}, \mathrm{q}=-\frac{3}{2}$

(4) $\mathrm{p}=\frac{5}{2}, \mathrm{q}=\frac{1}{2}$

[AIEEE 2011]

Sol. (1)

Q. Define $F(x)$ as the product of two real functions $f_{1}(x)=x, x \in \mathbb{R},$ and $f_{2}(x) $=\left\{\begin{array}{ccc}{\sin \frac{1}{x},} & {\text { if }} & {x \neq 0} \\ {0,} & {\text { if }} & {x=0}\end{array}\right.$ as follows : $\mathrm{F}(\mathrm{x})=\left\{\begin{array}{cc}{\mathrm{f}_{1}(\mathrm{x}) \cdot \mathrm{f}_{2}(\mathrm{x})} & {\text { if } \quad \mathrm{x} \neq 0} \\ {0,} & {\text { if } \quad \mathrm{x}=0}\end{array}\right.$

Statement-1 : F(x) is continuous on IR.

Statement-2 : f1(x) and f2(x) are continuous on IR.

(1) Statemen-1 is false, statement-2 is true.

(2) Statemen-1 is true,statement-2 is true;Statement-2 is correct explanation for

statement1.

(3) Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement1

(4) Statement-1 is true, statement-2 is false

[AIEEE 2011]

Sol. (4)

Q. If $\mathrm{f}(\mathrm{x})$ is continuous and $\mathrm{f}(9 / 2)=2 / 9,$ then $\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}\left(\frac{1-\cos 3 \mathrm{x}}{\mathrm{x}^{2}}\right)$ is equal to:

(1) 9/2 (2) 0 (3) 2/9 (4) 8/9

[JEE Mains Offline-2014]

Sol. (3)

$\mathrm{F}\left(\frac{9}{2}\right)=\frac{2}{9} \quad \lim _{x \rightarrow 0} \mathrm{F}\left[\frac{1-\cos 3 \mathrm{x}}{\mathrm{x}^{2}}\right]$

F(x) is continues and well defined than we can take limit inside.

$\Rightarrow \mathrm{F}\left[\lim _{x \rightarrow 0} \frac{1-\cos 3 x}{x^{2}}\right]$ use lopital

$\Rightarrow \mathrm{F}\left[\lim _{x \rightarrow 0} \frac{+3 \sin 3 \mathrm{x}}{2 \mathrm{x}}\right]=\operatorname{again}$

$\Rightarrow \mathrm{F}\left[\lim _{x \rightarrow 0} \frac{+9 \cos 3 x}{2}\right]=\mathrm{F}\left[\frac{9}{2}\right]=\frac{2}{9}$

Q. If the function $f(x)=\left\{\begin{array}{ll}{\frac{\sqrt{2+\cos x}-1}{(\pi-x)^{2}},} & {x \neq \pi} \\ {k} & {, x=\pi}\end{array}\right.$ is continuous at $x=\pi,$ then $k$ equals:-

(1) $\frac{1}{4}$

( 2)$\frac{1}{2}$

(3) 2

(4) 0

[JEE Mains Offline-2014]

Sol. (1)

Q. If the function $f$ defined as $f(x)=\frac{1}{x}-\frac{k-1}{e^{2 x}-1}, x \neq 0,$ is continuous at $x=0,$ then the ordered pair $(k, f(0))$ is equal to :

( 1)$\left(\frac{1}{3}, 2\right)$

(2) (3, 2)

(3) (2, 1)

(4) (3, 1)

[JEE Mains-2018]

Sol. (4)

Q. Let $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}{(\mathrm{x}-1)^{\frac{1}{2-\mathrm{x}}},} & {\mathrm{x}>1, \mathrm{x} \neq 2} \\ {\mathrm{k}} & {, \mathrm{x}=2}\end{array}\right.$

The value of k for which f is continuous at x = 2 is :

(1) $\mathrm{e}^{-1}$

(2) e

(3) $\mathrm{e}^{-2}$

(4) 1

[JEE Mains-2018]

Sol. (1)

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