If the function f(x)=\ array l (2)/(x)\ (k_1+1) x+ (k_2-1) x\ , x 0 4, x=0 (2)/(x) _e((2+k_1 x)/(2+k_2 x)), x 0 array . is continuous at x=0 , then k_1^2+k_2^2 is equal to

Correct answer: D. 10 aligned & _ x arrow 0^ - 2 x \ (k_1+1) x+ (k_2-1) x\ =4 & 2(k_1+1)+2(k_2-1)=4 & k_1+k_2=2 & _ x arrow 0^ + 2 x ( 2+k_1 x 2+k_2 x )=4 & _ x arrow 0^ + 1 x (1+ (k_1-k_2) x 2+k_2 x )=2 & k_1-k_2 2 =2 & k_1-k_2=4 & k_1=3, k_2=-1 & k_1^2+k_2^2=9+1=10 aligned

JEE Main 202523 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Continuity and Differentiability

If the function f(x)=\ array l (2)/(x)\ (k_1+1) x+ (k_2-1) x\ , x 0...

If the function

f(x)=\left\{

\begin{array}{l}\frac{2}{x}\left\{\sin \left(k_1+1\right) x+\sin \left(k_2-1\right) x\right\}, \quad x \lt 0 \\ 4, \quad x=0 \\ \frac{2}{x} \log _e\left(\frac{2+k_1 x}{2+k_2 x}\right), \quad x\gt0\end{array}

\right.

is continuous at

\mathrm{x}=0

, then

\mathrm{k}_1^2+\mathrm{k}_2^2

is equal to

Options

A $20$
B $5$
C $8$
D $10$

Correct answer

D. $10$

Step-by-step solution

\begin{aligned} & \lim _{x \rightarrow 0^{-}} \frac{2}{\mathrm{x}}\left\{\sin \left(\mathrm{k}_1+1\right) \mathrm{x}+\sin \left(\mathrm{k}_2-1\right) \mathrm{x}\right\}=4 \\ & \Rightarrow 2\left(\mathrm{k}_1+1\right)+2\left(\mathrm{k}_2-1\right)=4 \\ & \Rightarrow \mathrm{k}_1+\mathrm{k}_2=2 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{2}{\mathrm{x}} \ln \left(\frac{2+\mathrm{k}_1 \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=4 \\ & \Rightarrow \lim _{\mathrm{x} \rightarrow 0^{+}} \frac{1}{\mathrm{x}} \ln \left(1+\frac{\left(\mathrm{k}_1-\mathrm{k}_2\right) \mathrm{x}}{2+\mathrm{k}_2 \mathrm{x}}\right)=2 \\ & \Rightarrow \frac{\mathrm{k}_1-\mathrm{k}_2}{2}=2 \\ & \Rightarrow \mathrm{k}_1-\mathrm{k}_2=4 \\ & \therefore \mathrm{k}_1=3, \mathrm{k}_2=-1 \\ & \mathrm{k}_1^2+\mathrm{k}_2^2=9+1=10\end{aligned}

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