If fx-x3-1- x-0 x2-1- x- 0- - gx-x-11-3- x-1 x-11-2- x- 1- then gof x is equal to

Let 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)(x)=g(f(x))=g(x^2+1)=((x^2+1) 1) ...

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Byju's Answer

Standard XI

Mathematics

Composite Function

If fx=x3+1, ...

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. Let f(x) =

{\begin{matrix} 1 + x^{3}, x < 0 x^{2} - 1, x \geq 0 \end{matrix}

; g(x) =

{\begin{matrix} (x - 1)^{1 / 3}, x < 0 (x + 1)^{1 / 2}, x \geq 0 \end{matrix}

. Discuss the continuity of g(f(x))

Q. Let two functions are defined as

g (x) = {\begin{matrix} x^{2}, & - 1 \leq x \leq 2 x + 2, & 2 \leq x \leq 3 \end{matrix},

and

f (x) = {\begin{matrix} x + 1 & x \leq 1 2 x + 1 & 1 < x \leq 2, \end{matrix}

then find

g o f

(1)

Q. Find fog and gof if
(i)

f (x) = e^{x}, g (x) = \log_{e} x

(ii)

f (x) = x^{2}, g (x) = \cos x

(iii)

f (x) = | x |, g (x) = \sin x

(iv)

f (x) = x + 1, g (x) = e^{x}

(v)

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

(vi)

f (x) = x + 1, g (x) = \sin x

(vii)

f (x) = x + 1, g (x) = 2 x + 3

(viii)

f (x) = c, c \in R, g (x) = \sin x^{2}

(ix)

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

Q. If

f (x) = {\begin{matrix} \begin{matrix} x - 3, & x < 0 \end{matrix} \begin{matrix} x^{2} - 3 x + 2, & x \geq 0 \end{matrix} \end{matrix}

and

g (x) = f (| x |) + | f (x) |

, then

g (x)

Q. Let

g (x) = 1 + x - [x]

and

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1 & x < 0 0, & x = 0 1, & x > 0 \end{matrix}

then for all

x

f [g (x)]

is equal to:

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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