If the function f ( x )= \ begin array 35 l frac 72 x - 9 x - 8 x +1 √(2)-sqrt 1+ cos x & ,xne 0 \\ lo g e 2 lo g e 3 & ,x=0 \\end array . is continuous at x=0 , then the value of a 2 is equal to :

Mathematics · Limits, Continuity and Differentiability

JEE Main 2024 — 4 April, Shift 2 — Question 1

If the function $f\left( x \right)=\left\{ \begin{array}{*{35}{l}}\frac{{{72}^{x}}-{{9}^{x}}-{{8}^{x}}+1}{\sqrt{2}-\sqrt{1+\text{cos}x}} & ,x\ne 0 \\\text{lo}{{\text{g}}_{e}}2\text{lo}{{\text{g}}_{e}}3 & ,x=0 \\\end{array} \right.$ is continuous at $x=0$ , then the value of $a^{2}$ is equal to :

Option A:
968
Option B:
1152
Correct
Option C:
746
Option D:
1250

Answer: B

Step-by-step solution

$\lim _{x \rightarrow 0} f(x)=a \ln 2 \ln 3$

$\lim _{n \rightarrow 0} \frac{72^{x}-9^{x}-8^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}}=\lim _{x \rightarrow 0} \frac{\left(8^{x}-1\right)\left(9^{x}-1\right)}{\sqrt{2}-\sqrt{1+\cos x}}$

$\lim _{n \rightarrow 0}\left(\frac{8^{x}-1}{x}\right)\left(\frac{9^{x}-1}{x}\right)\left(\frac{x^{2}}{1-\cos x}\right)(\sqrt{2}+\sqrt{1+\cos x})$

$\therefore \ln 8 \times \ln 9 \times 2 \times 2 \sqrt{2}=24 \sqrt{2} \ln 2 \ln 3$

$\therefore \mathrm{a}=24 \sqrt{2}, \mathrm{a}^{2}=576 \times 2=1152$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Continuity