If fx-1x- gx-1x2 and hx-x2- then which of the following is always true-

The correct option is B fog(x)=x^2 ∀ x ≠ 0, h(g(x))=(g(x))^2 ∀ x ≠ 0f(x)=1x, g(x)=1x^2 and h(x)=x^2 f(g(x))=x^2, ∀ x ≠ 0 h(g(x))=1x^4=g((x))^2, ∀ x ≠ 0 ...

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Byju's Answer

Standard XI

Mathematics

AM,GM,HM Inequality

If fx=1x, gx=...

Question

If $f (x) = \frac{1}{x}, g (x) = \frac{1}{x^{2}}$ and $h (x) = x^{2}$ , then which of the following is always true?

f o g (x) = x^{2} \forall x \neq 0, h (g (x)) = \frac{1}{x^{2}}

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h (g (x)) = \frac{1}{x^{2}} \forall x \neq 0, f o g (x) = x^{2}

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f o g (x) = x^{2} \forall x \neq 0, h (g (x)) = (g (x))^{2} \forall x \neq 0

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None of these

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Solution

The correct option is C $f o g (x) = x^{2} \forall x \neq 0, h (g (x)) = (g (x))^{2} \forall x \neq 0$
$f (x) = \frac{1}{x}, g (x) = \frac{1}{x^{2}}$ and $h (x) = x^{2}$
$f (g (x)) = x^{2}, \forall x \neq 0$
$h (g (x)) = \frac{1}{x^{4}} = g ((x))^{2}, \forall x \neq 0$

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Similar questions

Q. If

f (x) = \frac{x}{\sqrt{1 - x^{2}}}, g (x) = \frac{x}{\sqrt{1 + x^{2}}}

, then

(f o g) (x) =

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f (x) = \frac{x}{\sqrt{1 - x^{2}}}

and g(x) =

f (x) = \frac{x}{\sqrt{1 + x^{2}}}

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g (x) = x (1 - x^{2}),

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f o g (x)

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f o g (x) = ⎧ ⎨ ⎩ \begin{matrix} \begin{matrix} - 1, & x \in (- 1, 0) \cup (1, \infty) \end{matrix} \begin{matrix} 0, \end{matrix} x \in {- 1, 0, 1} \begin{matrix} 1, & x \in (- \infty, - 1) \cup (0, 1) \end{matrix} \end{matrix}

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If $f (x) = \sqrt{x + 3}$ and $g (x) = 1 + x^{2},$ then $f o g (x) =$ ____.

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If f(x)=1x, g(x)=1x2 and h(x)=x2, then which of the following is always true?

The correct option is C fog(x)=x2 ∀ x≠0, h(g(x))=(g(x))2 ∀ x≠0f(x)=1x, g(x)=1x2 and h(x)=x2 f(g(x))=x2, ∀ x≠0 h(g(x))=1x4=g((x))2, ∀ x≠0

If $f (x) = \frac{1}{x}, g (x) = \frac{1}{x^{2}}$ and $h (x) = x^{2}$ , then which of the following is always true?

The correct option is C $f o g (x) = x^{2} \forall x \neq 0, h (g (x)) = (g (x))^{2} \forall x \neq 0$
$f (x) = \frac{1}{x}, g (x) = \frac{1}{x^{2}}$ and $h (x) = x^{2}$
$f (g (x)) = x^{2}, \forall x \neq 0$
$h (g (x)) = \frac{1}{x^{4}} = g ((x))^{2}, \forall x \neq 0$