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]

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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.$Statement-1 : F(x) is continuous on IR.Statement-2 : f1(x) and f2(x) are continuous on IR.(1) Statemen-1 is false, statement-2 is true.(2) Statemen-1 is true,statement-2 is true;Statement-2 is correct explanation forstatement1.(3) Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation for statement1(4) Statement-1 is true, statement-2 is false [AIEEE 2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4) Q. If$\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)$is equal to:(1) 9/2 (2) 0 (3) 2/9 (4) 8/9 [JEE Mains Offline-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)$\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}$(3) 2(4) 0 [JEE Mains Offline-2014] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)  Q. If the function$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)$(2) (3, 2)(3) (2, 1)(4) (3, 1) [JEE Mains-2018] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (4) Q. Let$\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]

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Sol. (1)

• March 15, 2021 at 8:32 pm

The level of questions tends to easy
Tq

• October 12, 2020 at 12:52 pm

More questions 🙏🏼

• October 12, 2020 at 12:44 pm

Pls more questions 🙏🏼

• September 1, 2020 at 7:38 am

Thanks

• December 2, 2020 at 9:23 pm

thankyou very much

• August 7, 2020 at 7:14 pm

Some solutions are not there

• August 1, 2020 at 9:56 am

Some questions are not explained with perfect solutions…inspite of it its superb…

• June 25, 2020 at 9:32 pm

Superb illustration

• May 20, 2020 at 10:14 pm

Chill bro