If gx-1-x and hx - xx-1- then ghxhgx is

G(x) = 1 x, h(x)=xx 1 g(h(x))h(g(x)) = g ( xx 1 )h(1 x) = 1 xx 11 x1 x 1 =x 1 xx 11 x x = 1x 11 x x =x(x 1)(1 x) = x(1 x)^2 h(x)g(x) = xx 11 x =x(1 x)(x 1) = ...

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Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

If gx=1-x and...

Question

If $g (x) = 1 - x$ and $h (x) = \frac{x}{x - 1}$ , then $\frac{g (h (x))}{h (g (x))}$ is

\frac{h (x)}{g (x)}

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\frac{- 1}{x}

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\frac{g (x)}{h (x)}

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\frac{x}{(1 - x)^{2}}

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Solution

The correct option is A $\frac{h (x)}{g (x)}$
$g (x) = 1 - x, h (x) = \frac{x}{x - 1}$
$\frac{g (h (x))}{h (g (x))} = \frac{g (\frac{x}{x - 1})}{h (1 - x)} = \frac{1 - \frac{x}{x - 1}}{\frac{1 - x}{1 - x - 1}}$
$= \frac{\frac{x - 1 - x}{x - 1}}{\frac{1 - x}{- x}} = \frac{\frac{- 1}{x - 1}}{\frac{1 - x}{- x}}$
$= \frac{x}{(x - 1) (1 - x)} = \frac{- x}{(1 - x)^{2}}$
$\frac{h (x)}{g (x)} = \frac{\frac{x}{x - 1}}{1 - x}$
$= \frac{x}{(1 - x) (x - 1)} = \frac{- x}{(1 - x)^{2}}$
$∴ \frac{g (h (x))}{h (g (x))} = \frac{h (x)}{g (x)}$

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If g(x)=1−x and h(x)=xx−1, then g(h(x))h(g(x)) is

The correct option is A h(x)g(x)g(x)=1−x, h(x)=xx−1 g(h(x))h(g(x))=g(xx−1)h(1−x)=1−xx−11−x1−x−1 =x−1−xx−11−x−x=−1x−11−x−x =x(x−1)(1−x)=−x(1−x)2 h(x)g(x)=xx−11−x =x(1−x)(x−1)=−x(1−x)2 ∴g(h(x))h(g(x))=h(x)g(x)

If $g (x) = 1 - x$ and $h (x) = \frac{x}{x - 1}$ , then $\frac{g (h (x))}{h (g (x))}$ is