Question

JEE(mains) 2020 (Phase 1) maths function

Answer 1

EXPLANATION.

$\sf (1) = f(x) = \dfrac{8^{2x} - 8^{-2x} }{8^{2x} + 8^{-2x} } \in (-1,1)$

As we know that,

We can write whole function = y.

Using componendo and dividendo, we get.

$\sf \implies \dfrac{y + 1}{y - 1} = \dfrac{\bigg(8^{2x} - 8^{-2x} \bigg) + \bigg(8^{2x} + 8^{-2x} \bigg)}{\bigg(8^{2x}- 8^{-2x} \bigg) - \bigg(8^{2x} + 8^{-2x}\bigg)_ }$

$\sf \implies \dfrac{y + 1}{y - 1} = \dfrac{2 \times 8^{2x} }{- 2 \times 8^{-2x} }$

$\sf \implies \dfrac{1 + y}{1 - y} = 8^{4x}$

$\sf \implies 4x = log_{8} \bigg( \dfrac{1 + y}{1 - y} \bigg)$

$\sf \implies x = \dfrac{1}{4} log_{8} \bigg(\dfrac{1 + y}{1 - y} \bigg)$

$\sf \implies f^{-1} (x) = \dfrac{1}{4} log_{8} \bigg( \dfrac{1 + x}{1 - x} \bigg)$

Option [A] is correct answer.