If the function f(x)= \begincases\frac72^x-9^x-8^x+1\sqrt2-\sqrt1+\cos x, & x eq 0 a \log _e 2 \log _e 3 & , x=0\endcases is continuous at x=0 , then the value of a^2 is equal to

Limits Continuity and Differentiability PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q34 with Solution

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Correct answer: C

Condition for continuity at $x=0$

For continuity, we need

\lim_{x\to 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=a(\ln 2)(\ln 3).

So we compute the limit.

Simplify the numerator

Notice that

Hence

So the limit becomes

L=\lim_{x\to 0} \frac{(8^x-1)(9^x-1)}{\sqrt{2}-\sqrt{1+\cos x}}.

Simplify the denominator

Use $1+\cos x=2\cos^2\frac{x}{2}.$ Thus

\sqrt{1+\cos x}=\sqrt{2}\,\cos\frac{x}{2}

for $x$ near $0$ (since $\cos(x/2)>0$ near $0$ ).

Therefore

\sqrt{2}-\sqrt{1+\cos x} =\sqrt{2}\left(1-\cos\frac{x}{2}\right).

L=\lim_{x\to 0} \frac{(8^x-1)(9^x-1)}{\sqrt{2}(1-\cos(x/2))}.

Use standard small-angle / exponential limits

As $x\to 0$ ,

8^x-1 \sim x\ln 8, \qquad 9^x-1 \sim x\ln 9.

Also,

1-\cos\frac{x}{2} \sim \frac{1}{2}\left(\frac{x}{2}\right)^2=\frac{x^2}{8}.

Hence

L=\frac{(\ln 8)(\ln 9)}{\sqrt{2}}\cdot \lim_{x\to 0}\frac{x^2}{x^2/8} =\frac{(\ln 8)(\ln 9)}{\sqrt{2}}\cdot 8.

L=\frac{8(\ln 8)(\ln 9)}{\sqrt{2}}.

Now,

\ln 8=3\ln 2, \qquad \ln 9=2\ln 3.

Therefore

L=\frac{8\cdot 3\ln 2\cdot 2\ln 3}{\sqrt{2}} =\frac{48\ln 2\ln 3}{\sqrt{2}} =24\sqrt{2}\,\ln 2\ln 3.

Match with the given value at $x=0$

Given

f(0)=a\ln 2\ln 3.

For continuity,

a\ln 2\ln 3=24\sqrt{2}\ln 2\ln 3.

Thus

a=24\sqrt{2}.

Hence

a^2=(24\sqrt{2})^2=24^2\cdot 2=576\cdot 2=1152.

Check options

\boxed{a^2=1152}

So the correct option is C.

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: C

Condition for continuity at $x=0$

For continuity, we need

\lim_{x\to 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=a(\ln 2)(\ln 3).

So we compute the limit.

Simplify the numerator

Notice that

Hence

So the limit becomes

L=\lim_{x\to 0} \frac{(8^x-1)(9^x-1)}{\sqrt{2}-\sqrt{1+\cos x}}.

Simplify the denominator

Use $1+\cos x=2\cos^2\frac{x}{2}.$ Thus

\sqrt{1+\cos x}=\sqrt{2}\,\cos\frac{x}{2}

for $x$ near $0$ (since $\cos(x/2)>0$ near $0$ ).

Therefore

\sqrt{2}-\sqrt{1+\cos x} =\sqrt{2}\left(1-\cos\frac{x}{2}\right).

L=\lim_{x\to 0} \frac{(8^x-1)(9^x-1)}{\sqrt{2}(1-\cos(x/2))}.

Use standard small-angle / exponential limits

As $x\to 0$ ,

8^x-1 \sim x\ln 8, \qquad 9^x-1 \sim x\ln 9.

Also,

1-\cos\frac{x}{2} \sim \frac{1}{2}\left(\frac{x}{2}\right)^2=\frac{x^2}{8}.

Hence

L=\frac{(\ln 8)(\ln 9)}{\sqrt{2}}\cdot \lim_{x\to 0}\frac{x^2}{x^2/8} =\frac{(\ln 8)(\ln 9)}{\sqrt{2}}\cdot 8.

L=\frac{8(\ln 8)(\ln 9)}{\sqrt{2}}.

Now,

\ln 8=3\ln 2, \qquad \ln 9=2\ln 3.

Therefore

L=\frac{8\cdot 3\ln 2\cdot 2\ln 3}{\sqrt{2}} =\frac{48\ln 2\ln 3}{\sqrt{2}} =24\sqrt{2}\,\ln 2\ln 3.

Match with the given value at $x=0$

Given

f(0)=a\ln 2\ln 3.

For continuity,

a\ln 2\ln 3=24\sqrt{2}\ln 2\ln 3.

Thus

a=24\sqrt{2}.

Hence

a^2=(24\sqrt{2})^2=24^2\cdot 2=576\cdot 2=1152.

Check options

\boxed{a^2=1152}

So the correct option is C.

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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