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Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 23 January, Morning Shift — Question 3

If the function

f(x)={2x{sin⁡(k1+1)x+sin⁡(k2−1)x},x<04,x=02xlog⁡e(2+k1x2+k2x),x>0f(x) = \begin{cases} \frac{2}{x} \{\sin(k_1+1)x + \sin(k_2-1)x\}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left(\frac{2+k_1x}{2+k_2x}\right), & x > 0 \end{cases}

is continuous at x=0\mathrm{x}=0, then k12+k22\mathrm{k}_{1}{ }^{2}+\mathrm{k}_{2}{ }^{2} is equal to

  1. Option A:

    8

  2. Option B:

    20

  3. Option C:

    5

  4. Option D:

    10

    Correct

Answer: D

Step-by-step solution

lim⁡x→0−2x{sin⁡(k1+1)x+sin⁡(k2−1)x}=4\lim _{x \rightarrow 0^{-}} \frac{2}{x}\left\{\sin \left(k_{1}+1\right) x+\sin \left(k_{2}-1\right) x\right\}=4

⇒2(k1+1)+2(k2−1)=4\Rightarrow 2\left(\mathrm{k}_{1}+1\right)+2\left(\mathrm{k}_{2}-1\right)=4 ⇒k1+k2=2\Rightarrow \mathrm{k}_{1}+\mathrm{k}_{2}=2

⇒lim⁡x→0+2xln⁡(2+k1x2+k2x)=4\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

⇒lim⁡x→0+1xln⁡(1+(k1−k2)x2+k2x)=2\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

⇒k1−k22=2\Rightarrow \frac{\mathrm{k}_{1}-\mathrm{k}_{2}}{2}=2

⇒k1−k2=4\Rightarrow \mathrm{k}_{1}-\mathrm{k}_{2}=4

∴k1=3,k2=−1\therefore \mathrm{k}_{1}=3, \mathrm{k}_{2}=-1

k12+k22=9+1=10\mathrm{k}_{1}^{2}+\mathrm{k}_{2}^{2}=9+1=10

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Morning Shift
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Continuity at a point & in an interval