Let fx-sin-1x and gx-x2-x-22x2-x-6- If g2-x- 2lim gx- then the domain of the function fog is -

G(2)=x→ 2lim(x 2)(x+1)(2x+3)(x 2)=37 For domain of f(g(x)) | x^2 x 22x^2 x 6 |≤ 1 (∵ domain of f(x) is [ 1,1]) ⇒ (x+1)^2≤ (2x+3)^2 ⇒ 3x^2+10x+8≥ 0 ⇒ (3x+4)(x+2) ...

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Byju's Answer

Standard XII

Mathematics

Global Minima

Let fx=sin-1x...

Question

Let $f (x) = {sin}^{- 1} x$ and $g (x) = \frac{x^{2} - x - 2}{2 x^{2} - x - 6}$ . If $g (2) = lim x \to 2 g (x)$ , then the domain of the function $f o g$ is :

(- \infty, - 2] \cup [- \frac{4}{3}, \infty)

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(- \infty, - 1] \cup [2, \infty)

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(- \infty, - 2] \cup [- 1, \infty)

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(- \infty, - 2] \cup [- \frac{3}{2}, \infty)

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Solution

The correct option is A $(- \infty, - 2] \cup [- \frac{4}{3}, \infty)$
$g (2) = lim x \to 2 \frac{(x - 2) (x + 1)}{(2 x + 3) (x - 2)} = \frac{3}{7}$
For domain of $f (g (x))$
$∣ ∣ ∣ \frac{x^{2} - x - 2}{2 x^{2} - x - 6} ∣ ∣ ∣ \leq 1$
( $∵$ domain of $f (x)$ is $[- 1, 1]$ )
$\Rightarrow (x + 1)^{2} \leq (2 x + 3)^{2} \Rightarrow 3 x^{2} + 10 x + 8 \geq 0$
$\Rightarrow (3 x + 4) (x + 2) \geq 0$
$x ϵ (- \infty, - 2) \cup [- \frac{4}{3}, \infty]$

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Similar questions

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. Assertion :If

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Let f(x)=sin−1x and g(x)=x2−x−22x2−x−6. If g(2)=limx→2g(x), then the domain of the function fog is :

The correct option is A (−∞,−2]∪[−43,∞)g(2)=limx→2(x−2)(x+1)(2x+3)(x−2)=37 For domain of f(g(x)) ∣∣∣x2−x−22x2−x−6∣∣∣≤1 (∵ domain of f(x) is [−1,1]) ⇒(x+1)2≤(2x+3)2⇒3x2+10x+8≥0 ⇒(3x+4)(x+2)≥0 x ϵ (−∞,−2)∪[−43,∞]

Let $f (x) = {sin}^{- 1} x$ and $g (x) = \frac{x^{2} - x - 2}{2 x^{2} - x - 6}$ . If $g (2) = lim x \to 2 g (x)$ , then the domain of the function $f o g$ is :