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Functions question

2020 · 8 Jan · Shift 1 · Q26
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Functions question

2020 · 8 Jan · Shift 1 · Q26

JEE MainMathematicsFunctionsMCQ+4 / −1
The inverse function of f(x) = 82x−8−2x82x+8−2x{{{8^{2x}} - {8^{ - 2x}}} \over {{8^{2x}} + {8^{ - 2x}}}}82x+8−2x82x−8−2x​, x ∈\in∈ (-1, 1), is :
  1. A
    14log⁡e(1−x1+x){1 \over 4}{\log _e}\left( {{{1 - x} \over {1 + x}}} \right)41​loge​(1+x1−x​)
  2. B
    14(log⁡8e)log⁡e(1−x1+x){1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 - x} \over {1 + x}}} \right)41​(log8​e)loge​(1+x1−x​)
  3. C
    14(log⁡8e)log⁡e(1+x1−x){1 \over 4}\left( {{{\log }_8}e} \right){\log _e}\left( {{{1 + x} \over {1 - x}}} \right)41​(log8​e)loge​(1−x1+x​)
  4. D
    14log⁡e(1+x1−x){1 \over 4}{\log _e}\left( {{{1 + x} \over {1 - x}}} \right)41​loge​(1−x1+x​)
View written solutionFree

Correct answer: C

  1. Given function

We need the inverse of

f(x)=82x−8−2x82x+8−2x,x∈(−1,1).f(x)=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}}, \quad x\in(-1,1).f(x)=82x+8−2x82x−8−2x​,x∈(−1,1).

Let

y=82x−8−2x82x+8−2x.y=\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}}.y=82x+8−2x82x−8−2x​.

We now solve for xxx in terms of yyy.

  1. Simplify the expression

Multiply numerator and denominator by 82x8^{2x}82x:

y=84x−184x+1.y=\frac{8^{4x}-1}{8^{4x}+1}.y=84x+184x−1​.

Let

t=84x.t=8^{4x}.t=84x.

Then

y=t−1t+1.y=\frac{t-1}{t+1}.y=t+1t−1​.
  1. Solve for ttt

From

y=t−1t+1,y=\frac{t-1}{t+1},y=t+1t−1​,

we get

y(t+1)=t−1.y(t+1)=t-1.y(t+1)=t−1.

So,

yt+y=t−1.yt+y=t-1.yt+y=t−1.

Rearrange:

t−yt=y+1,t-yt=y+1,t−yt=y+1, t(1−y)=1+y.t(1-y)=1+y.t(1−y)=1+y.

Hence,

t=1+y1−y.t=\frac{1+y}{1-y}.t=1−y1+y​.

Since t=84xt=8^{4x}t=84x,

84x=1+y1−y.8^{4x}=\frac{1+y}{1-y}.84x=1−y1+y​.
  1. Take logarithm

Taking natural log on both sides,

4xln⁡8=ln⁡(1+y1−y).4x\ln 8=\ln\left(\frac{1+y}{1-y}\right).4xln8=ln(1−y1+y​).

Thus,

x=14ln⁡8ln⁡(1+y1−y).x=\frac{1}{4\ln 8}\ln\left(\frac{1+y}{1-y}\right).x=4ln81​ln(1−y1+y​).

Now,

1ln⁡8=log⁡8e.\frac{1}{\ln 8}=\log_8 e.ln81​=log8​e.

Therefore,

x=14(log⁡8e)ln⁡(1+y1−y).x=\frac{1}{4}(\log_8 e)\ln\left(\frac{1+y}{1-y}\right).x=41​(log8​e)ln(1−y1+y​).

Replacing yyy by xxx for the inverse function variable,

f−1(x)=14(log⁡8e)ln⁡(1+x1−x).f^{-1}(x)=\frac{1}{4}(\log_8 e)\ln\left(\frac{1+x}{1-x}\right).f−1(x)=41​(log8​e)ln(1−x1+x​).
  1. Match with the options

This is exactly Option C:

14((log⁡8e))log⁡e(1+x1−x).\frac{1}{4}\left((\log_8 e)\right)\log_e\left(\frac{1+x}{1-x}\right).41​((log8​e))loge​(1−x1+x​).
  1. Check monotonicity / validity

The function is of the form

a−a−1a+a−1\frac{a-a^{-1}}{a+a^{-1}}a+a−1a−a−1​

with a=82x>0a=8^{2x}>0a=82x>0, so it is essentially a hyperbolic tangent type expression and is strictly increasing. Hence inverse exists on the given interval.

Therefore the inverse found is valid.

Final Answer: Option C

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