If $f: R \rightarrow R, g: R \rightarrow R$ are defined by
$f(x)=5\, x-3, g(x)=x^{2}+3$, then $g o f^{-1}(3)$ is equal to
- $\frac{25}{3}$
- $\frac{111}{25}$
- $\frac{9}{25}$
- $\frac{25}{111}$
The Correct Option is B
Solution and Explanation
Let $ y=5 x-3 $
$ \Rightarrow y+3=5 x $
$ \Rightarrow x=\frac{y+3}{5} $
$ \therefore f^{-1}(y)=\frac{y+3}{5} $
$\Rightarrow f^{-1}(x)=\frac{x+3}{5}$
and $g(x)=x^{2}+3$
Now, $g o f^{-1}(3)=g\left[f^{-1}(3)\right]$
$=g\left(\frac{3+3}{5}\right)=g\left(\frac{6}{5}\right) $
$ \Rightarrow g\left(\frac{6}{5}\right)=\frac{(6)^{2}}{(5)^{2}}+3 =\frac{36}{25}+3=\frac{111}{25} $
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Concepts Used:
Differentiability
Differentiability of a function A function f(x) is said to be differentiable at a point of its domain if it has a finite derivative at that point. Thus f(x) is differentiable at x = a
\(\frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
⇒ f'(a – 0) = f'(a + 0)
⇒ left-hand derivative = right-hand derivative.
Thus function f is said to be differentiable if left hand derivative & right hand derivative both exist finitely and are equal.
If f(x) is differentiable then its graph must be smooth i.e. there should be no break or corner.
Note:
(i) Every differentiable function is necessarily continuous but every continuous function is not necessarily differentiable i.e. Differentiability ⇒ continuity but continuity ⇏ differentiability
(ii) For any curve y = f(x), if at any point \(\frac{d y}{d x}\) = 0 or does not exist then, the point is called “critical point”.
3. Differentiability in an interval
(a) A function fx) is said to be differentiable in an open interval (a, b), if it is differentiable at every point of the interval.
(b) A function f(x) is differentiable in a closed interval [a, b] if it is
- Differentiable at every point of interval (a, b)
- Right derivative exists at x = a
- Left derivative exists at x = b.
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