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Mathematics · Limits, Continuity and Differentiability

JEE Main 2026 — 22 January, Evening Shift — Question 5

If lim⁡x→0e(a−1)x+2cos⁡bx+(c−2)e−xxcos⁡x−log⁡e(1+x)=2\lim _{x \rightarrow 0} \frac{e^{(a-1) x}+2 \cos b x+(c-2) e^{-x}}{x \cos x-\log _{e}(1+x)}=2, then a2+b2+c2a^{2}+ \mathrm{b}^{2}+\mathrm{c}^{2} is equal to :

  1. Option A:

    55

  2. Option B:

    33

  3. Option C:

    77

    Correct
  4. Option D:

    99

Answer: C

Step-by-step solution

lim⁡x→0(1+(a−1)x+(a−1)2x22!)+2(1−b2x22!)+(c−2)(1−x+x22!)x(1−x22!)−(x−x22…)=2\lim _{x \rightarrow 0} \frac{\left(1+(a-1) x+\frac{(a-1)^{2} x^{2}}{2!}\right)+2\left(1-\frac{b^{2} x^{2}}{2!}\right)+(c-2)\left(1-x+\frac{x^{2}}{2!}\right)}{x\left(1-\frac{x^{2}}{2!}\right)-\left(x-\frac{x^{2}}{2} \ldots\right)}=2

lim⁡x→0(1+2+c−2)+x(a−1−c+2)+x2((a−1)22−b2+(c−22))x22−x32!+…=2\lim _{x \rightarrow 0} \frac{(1+2+c-2)+x(a-1-c+2)+x^{2}\left(\frac{(a-1)^{2}}{2}-b^{2}+\left(\frac{c-2}{2}\right)\right)}{\frac{x^{2}}{2}-\frac{x^{3}}{2!}+\ldots}=2

For which ∵c+1=0⇒c=−1\because c+1=0 \Rightarrow c=-1

∵a−c=−1⇒a=−2\because a-c=-1 \Rightarrow a=-2

∵(a−1)22−b2+(c−22)=1\because \frac{(a-1)^{2}}{2}-b^{2}+\left(\frac{c-2}{2}\right)=1

92−b2−32=1⇒b2=2\frac{9}{2}-b^{2}-\frac{3}{2}=1 \Rightarrow b^{2}=2

a2+b2+c2=4+2+1=7a^{2}+b^{2}+c^{2}=4+2+1=7 .

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Evening Shift
Subject
Mathematics
Chapter
Limits, Continuity and Differentiability
Topic
Evaluation of Limit of Functions

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