If fx- left begin matrix x - 3-1-8 x -0- x- 2-1- - x - geq 0 x -1- -frac 13- - x -1 x -1-lfrac12- - x lgeq 1lend matrix right-l then gof x is equal toA- x - - x - RB- x -1- - x - RC- x -1- - x - RD- None of these

The correct option is A x,∀ xϵ RLet x lt; 0. ∴ (gof)(x) = g(f(x)) = g(x^3+1) = [(x^3+1) 1]^1/3 (∵ x lt; 0 ⇒ x^3+1 lt; 1) =(x^3)^1/3=x Let x ≥ 0. ∴ (gof ...

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Byju's Answer

Standard IX

Mathematics

Composite Function

If fx= left b...

Question

If $f (x) = {\begin{matrix} x^{3} + 1, & x < 0 x^{2} + 1, & x \geq 0 \end{matrix}, g (x) = ⎧ ⎨ ⎩ \begin{matrix} (x - 1)^{\frac{1}{3}}, & x < 1 (x - 1)^{\frac{1}{2}}, & x \geq 1 \end{matrix}$ then (gof) (x) is equal to

x, \forall x ϵ R

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x - 1, \forall x ϵ R

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x + 1, \forall x ϵ R

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N o n e o f t h e s e

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Solution

The correct option is A $x, \forall x ϵ R$
Let x < 0.
$∴$ $(g o f) (x) = g (f (x)) = g (x^{3} + 1) = [(x^{3} + 1) - 1]^{\frac{1}{3}}$
$(∵ x < 0 \Rightarrow x^{3} + 1 < 1)$
$= (x^{3})^{\frac{1}{3}} = x$
Let x $\geq$ 0.
$∴ (g o f) (x) = g (f (x)) = g (x^{2} + 1) = ((x^{2} + 1) - 1)^{\frac{1}{2}}$
$(∵ x \geq 0 \Rightarrow x^{2} + 1 \geq 1)$
$= (x^{2})^{\frac{1}{2}} = | x | = x (∵ x \geq 0)$
$∴ (g o f) (x) = x \forall x ϵ R$

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Similar questions

Q. If

f (x) = {\begin{matrix} x^{3} + 1, & x < 0 x^{2} + 1, & x \geq 0 \end{matrix}, g (x) = ⎧ ⎨ ⎩ \begin{matrix} (x - 1)^{\frac{1}{3}}, & x < 1 (x - 1)^{\frac{1}{2}}, & x \geq 1 \end{matrix}

then (gof) (x) is equal to

Find

$lim x \to 1$ f(x) where f(x)= ${\begin{matrix} x^{2} - 1, & x \leq 1 - x^{2} - 1 & x > 1 \end{matrix}$

Q. If

f (x) = {\begin{matrix} x - 1, x \geq 1, 2 x^{2} - 2, x < 1 \end{matrix} g (x) = {\begin{matrix} x + 1, x > 0 - x^{2} + 1, x \leq 0 \end{matrix} and h (x) = | x |, then lim x \to 0 f (g (h (x)))

The function

$f (x) = {\begin{matrix} | x - 3 |, & x \geq 1 \frac{x^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, & x < 1 \end{matrix}$ ,

f (x) = {\begin{matrix} x + 1 & x < 0 x^{2} & x \geq 0 \end{matrix}

and

g (x) = {\begin{matrix} x^{3} & x < 1 2 x - 1 & x \geq 1 \end{matrix}

Then find

f (g (x))

and find its domain and range.

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If f(x)={x3+1,x<0x2+1,x≥0, g(x)=⎧⎨⎩(x−1)13,x<1(x−1)12,x≥1 then (gof) (x) is equal to

The correct option is A x,∀xϵRLet x < 0. ∴ (gof)(x)=g(f(x))=g(x3+1)=[(x3+1)−1]13 (∵x<0⇒x3+1<1) =(x3)13=x Let x ≥ 0. ∴(gof)(x)=g(f(x))=g(x2+1)=((x2+1)−1)12 (∵x≥0⇒x2+1≥1) =(x2)12=|x|=x (∵x≥0) ∴(gof)(x)=x∀xϵR

If $f (x) = {\begin{matrix} x^{3} + 1, & x < 0 x^{2} + 1, & x \geq 0 \end{matrix}, g (x) = ⎧ ⎨ ⎩ \begin{matrix} (x - 1)^{\frac{1}{3}}, & x < 1 (x - 1)^{\frac{1}{2}}, & x \geq 1 \end{matrix}$ then (gof) (x) is equal to