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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Continuity and Differentiability

If the function f(x)= cases 72^x-9^x-8^x+1 2 - 1+ x , & x ≠ 0 a _e...

If the function f(x)=f(x)=\begin{cases}\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x \neq 0 \\ a \log _e 2 \log _e 3 & , x=0\end{cases}$iscontinuousatis continuous atx=0,thenthevalueof, then the value ofa^2$ is equal to

Options

  1. A968
  2. B1152
  3. C746
  4. D1250

Correct answer

B. 1152

Step-by-step solution

lim⁡x→0f(x)=aln⁡2ln⁡3lim⁡n→072x−9x−8x+12−1+cos⁡x=lim⁡x→0(8x−1)(9x−1)2−1+cos⁡xlim⁡n→0(8x−1x)(9x−1x)(x21−cos⁡x)(2+1+cos⁡x)∴ln⁡8×ln⁡9×2×22=242ln⁡2ln⁡3∴a=242,a2=576×2=1152\begin{aligned} & \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 a=24 \sqrt{2}, a^2=576 \times 2=1152\end{aligned}
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