Let f- be defined by fx-e3x-3-ln x-tan-1x-1 and g be the inverse function of f- If limx-1x5-5gxg-xsin x-1-pq where p-q- then the least possible value of p-5q is

F(x)=e^3x 3+ln x+tan^ 1(x 1) ⇒ f(1)=1 Since nbsp;g is nbsp;the inverse function of f, ⇒ f(g(x))=x ⋯(1) f (1) = 1⇒ nbsp;g(1) = 1 Now, nbsp;f'(x)=3e^3x ...

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Byju's Answer

Standard XII

Mathematics

Global Minima

Let f:ℝ+→ℝ be...

Question

Let $f : R^{+} \to R$ be defined by $f (x) = e^{3 x - 3} + ln x + {tan}^{- 1} (x - 1)$ and $g$ be the inverse function of $f .$ If $lim x \to 1 \frac{x^{5} - 5 g (x) g^{'} (x)}{sin (x - 1)} = \frac{p}{q}$ where $p, q \in N,$ then the least possible value of $(p - 5 q)$ is

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Solution

$f (x) = e^{3 x - 3} + ln x + {tan}^{- 1} (x - 1)$
$\Rightarrow f (1) = 1$
Since $g$ is the inverse function of $f,$
$\Rightarrow f (g (x)) = x \dots (1)$
$f (1) = 1 \Rightarrow g (1) = 1$
Now, $f^{'} (x) = 3 e^{3 x - 3} + \frac{1}{x} + \frac{1}{1 + (x - 1)^{2}}$
and $f^{''} (x) = 9 e^{3 x - 3} - \frac{1}{x^{2}} - \frac{2 (x - 1)}{{(1 + (x - 1)^{2})}^{2}}$

We have, $g^{'} (1) = \frac{1}{f^{'} (1)}$ [From $(1)$ ]
$\Rightarrow g^{'} (1) = \frac{1}{5}$
and $g^{''} (1) = - \frac{f^{''} (1)}{(f^{'} (1))^{3}} = \frac{- 8}{125}$

$lim x \to 1 \frac{x^{5} - 5 g (x) g^{'} (x)}{sin (x - 1)}$ $(\frac{0}{0} form)$
Using L'Hospitals rule, we get
$lim x \to 1 \frac{5 x^{4} - 5 g (x) g^{''} (x) - 5 (g^{'} (x))^{2}}{cos (x - 1)}$
$= \frac{5 - 5 g (1) g^{''} (1) - 5 (g^{'} (1))^{2}}{cos (0)}$
$= 5 - 5 \times 1 \times (- \frac{8}{125}) - 5 \times \frac{1}{25}$
$= \frac{128}{25} = \frac{p}{q}$
$∴ p - 5 q = 128 - 125 = 3$

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Let f:R+→R be defined by f(x)=e3x−3+lnx+tan−1(x−1) and g be the inverse function of f. If limx→1x5−5g(x)g′(x)sin(x−1)=pq where p,q∈N, then the least possible value of (p−5q) is

Let $f : R^{+} \to R$ be defined by $f (x) = e^{3 x - 3} + ln x + {tan}^{- 1} (x - 1)$ and $g$ be the inverse function of $f .$ If $lim x \to 1 \frac{x^{5} - 5 g (x) g^{'} (x)}{sin (x - 1)} = \frac{p}{q}$ where $p, q \in N,$ then the least possible value of $(p - 5 q)$ is