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Consider we have a function as \[g\left( x \right)=\left\{ \begin{matrix}
   \left\{ \dfrac{\max .\left( f\left( t \right) \right)+\min .\left( f\left( t \right) \right)}{2},0\le t\le x \right\},0\le x\le 4 \\
   \left| x-5 \right|+\left| x-4 \right|,4 < x < 5 \\
   \tan \left( {{\sin }^{-1}}\left( \dfrac{6-x}{\sqrt{{{x}^{2}}-12x+37}} \right) \right),x\ge 5 \\
\end{matrix} \right.\]
Where $f\left( x \right)={{x}^{2}}-4x+3$ .
$\displaystyle \lim_{x \to 4}\dfrac{g\left( x \right)-g\left( 2 \right)}{l\text{n}\left( \cos \left( 4-x \right) \right)}$ is equal to-
(A) 0
(B) 1
(C) 2
(D) Does not exist

Last updated date: 20th Jun 2024
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Hint: First we have to find the value of g(2) and g(4) from the given function. For g(4) we need to be careful about the modulus function. Then substitute the values in $\displaystyle \lim_{x \to 4}\dfrac{g\left( x \right)-g\left( 2 \right)}{l\text{n}\left( \cos \left( 4-x \right) \right)}$. You will get an indeterminate form, so now apply L’ HOSPITAL RULE to get an answer.

Complete step-by-step solution:
We have given that;
\[g\left( x \right)=\left\{ \begin{matrix}
   \left\{ \dfrac{\max .\left( f\left( t \right) \right)+\min .\left( f\left( t \right) \right)}{2},0\le t\le x \right\},0\le x\le 4 \\
   \left| x-5 \right|+\left| x-4 \right|,4 < x < 5 \\
   \tan \left( {{\sin }^{-1}}\left( \dfrac{6-x}{\sqrt{{{x}^{2}}-12x+37}} \right) \right),x\ge 5 \\
\end{matrix} \right.\]
Where $f\left( x \right)={{x}^{2}}-4x+3$ .
We have to calculate $\displaystyle \lim_{x \to 4}\dfrac{g\left( x \right)-g\left( 2 \right)}{l\text{n}\left( \cos \left( 4-x \right) \right)}$which is equal to $\dfrac{g\left( 4 \right)-g\left( 2 \right)}{l\text{n}\left( \cos 0 \right)}$
Hence, we have to calculate g(2) and g(4) to get the limit of the above expression.
For g(2), we need to consider;
\[g\left( x \right)=\left\{ \dfrac{\max .\left( f\left( t \right) \right)+\min .\left( f\left( t \right) \right)}{2},0\le t\le x \right\}0\le x\le 4\]
Where $f\left( x \right)=\left( {{x}^{2}}-4x+3 \right)$ .

\[g\left( 2 \right)=\left\{ \dfrac{\max .f\left( t \right)+\min .f\left( t \right)}{2},0\le t\le 2 \right\}\]
$f\left( x \right)={{x}^{2}}-4x+3$
$f\left( x \right)={{x}^{2}}-4x+3$ can also be written as;
  & {{\left( x-2 \right)}^{2}}+3-4={{\left( x-2 \right)}^{2}}-1 \\
 & OR \\
 & f\left( t \right)={{\left( t-2 \right)}^{2}}-1 \\
Let us calculate max (f(t)) and min (f(t)) for \[0\le t\le 2\],
min f(t)= -1 (at t=2)
max f(t)= 3 (at t=0)
$g\left( 2 \right)=\dfrac{\min f\left( t \right)+\max f\left( t \right)}{2}$can be given by;
$g\left( 2 \right)=\dfrac{3-1}{2}=1$
Now, let us calculate g (4) from the given function.
For g (4) we have to consider two functions of g (x). As for ${{4}^{+}}$, we have to consider $\left| x-5 \right|+\left| x-4 \right|$ and for ${{4}^{-}},\dfrac{\max f\left( t \right)+\min f\left( t \right)}{2},0 < t\le x$ for 4 < x < 5 and $0\le x\le 4$ respectively.
For left-hand limit, we have function;
$g\left( {{4}^{-}} \right)=\left\{ \dfrac{\max f\left( t \right)+\min f\left( t \right)}{2} \right\},0\le t < 4$
Where $f\left( x \right)={{x}^{2}}-4x+3$ or
$f\left( x \right)={{\left( x-2 \right)}^{2}}-1$
The minimum value of f (t) will be at t=2 and the maximum may occur at endpoints (0 or 4) from the graph.
  & f\left( 0 \right)=4-1=3 \\
 & f\left( 4 \right)={{2}^{2}}-1=3 \\
Hence, f (0) = f(4) = maximum value of f(t) = 3 for $0\le t\le 4$
Hence, we can get
$g\left( {{4}^{-}} \right)=\dfrac{3-1}{2}=1$
Now, let us calculate RHL, for which we have to calculate
$g\left( x \right)=g\left( {{4}^{+}} \right)=\left| x-5 \right|+\left| x-4 \right|\text{ 4x5}$
Here, we need to define modules function as expressed below;
\[\left| x \right|=\left\{ \begin{matrix}
   x \\
   -x \\
   0 \\
\end{matrix}\text{ }\begin{matrix}
   x > 0 \\
   n < 0 \\
   x=0 \\
\end{matrix} \right.\]
  & \left| x-5 \right|+\left| x-4 \right|=\left\{ \begin{matrix}
   -\left( x-5 \right)-\left( x-4 \right), \\
   -\left( x-5 \right)+\left( x-4 \right), \\
   x-5+x-4, \\
\end{matrix}\text{ }\begin{matrix}
   x < 5 \\
   4 < x < 5 \\
   x > 5 \\
\end{matrix} \right. \\
 & Or \\
 & \left| x-5 \right|+\left| x-4 \right|=\left\{ \begin{matrix}
   -2x+9, \\
   1, \\
   2x-9, \\
\end{matrix}\text{ }\begin{matrix}
   x < 5 \\
   4 < x < 5 \\
   x > 5 \\
\end{matrix} \right. \\
Hence, for, \[4 < x < 5,\left| x-5 \right|+\left| x+4 \right|\] can be replaced by 1.
Hence, $g\left( {{4}^{+}} \right)=1$ (constant term)
As $g\left( {{4}^{-}} \right)=g\left( {{4}^{+}} \right)$, therefore the limit will exist and equals 1.
Now, coming to the question part;
$\dfrac{g\left( 4 \right)-g\left( 2 \right)}{\ln \left( \cos 0{}^\circ \right)}=\dfrac{1-1}{0}=\dfrac{0}{0}$
which is an indeterminate form of limit.
As we know, we need to apply L’ HOSPITAL Theorem to solve the limit which is defined as;
If any limit is of the form $\displaystyle \lim_{x \to c}\dfrac{f\left( x \right)}{g\left( x \right)}$ and giving an indeterminate form of $\dfrac{0}{0}\text{ or }\dfrac{\infty }{\infty }$ then we have to differentiate numerator and denominator individually i.e.
$\displaystyle \lim_{x \to c}\dfrac{f'\left( x \right)}{g'\left( x \right)}$and then solve it.
Now, differentiating the given function in question;
$\displaystyle \lim_{x \to 4}\dfrac{g\left( x \right)-g\left( 2 \right)}{\ln \left( \cos \left( 4-x \right) \right)}=\displaystyle \lim_{x \to 4}\dfrac{\left( g'\left( x \right)-0 \right)}{\dfrac{1}{\cos \left( 4-x \right)}\sin \left( 4-x \right)}$
  & \dfrac{d}{dx}\ln x=\dfrac{1}{x} \\
 & \dfrac{d}{dx}\cos x=-\sin x \\
 & As:\dfrac{\sin x}{\cos x}=\tan x \\
Hence, we have to calculate
$\displaystyle \lim_{x \to 4}\dfrac{g'\left( x \right)}{\tan \left( 4-x \right)}$………………. (1)
Here, we have to get g’(4) i.e. $g'\left( {{4}^{+}} \right)\And g'\left( {{4}^{-}} \right)$ both. As, we have already defined g(x) for ${{4}^{+}}\And {{4}^{-}}$ as below;
For $g\left( {{4}^{-}} \right)$ we have function;
  & g\left( x \right)=\dfrac{\max .f\left( t \right)+\min .f\left( t \right)}{2},0\le t\le 4 \\
 & Where\text{ }f\left( t \right)={{t}^{2}}-4t+3 \\
We have calculated the value of g(4) already above that is 1.
Hence, $g\left( {{4}^{-}} \right)=1$
And for ${{4}^{+}}$, we have
$g\left( x \right)=1\text{ }4 < x < 5$
Now $g\left( x \right)$ is a constant function, so
$g'\left( x \right)=0$ for $x=4\left( {{4}^{-}}\And {{4}^{+}}both \right)$;
Hence, from equation (1), we have;
$\displaystyle \lim_{x \to 4}\dfrac{g'\left( x \right)}{\tan \left( 4-x \right)}=\dfrac{0}{0}$
As, we know $\displaystyle \lim_{h\to 0}\left( \dfrac{\tanh }{h} \right)=1$, where we find $\left( \dfrac{0}{0} \right)$ a form of limit.
So by using the above result we can write the above limit as;
$\displaystyle \lim_{x \to 4}\dfrac{g'\left( x \right)}{\tan \left( 4-x \right)}=1$ where $g'\left( 4 \right)\to 0\text{ and }\tan \left( 4-x \right)$ is tending to zero as well.
Hence, ‘B’ is the correct answer.

Note: We don’t need to calculate any limit for
$\tan \left( {{\sin }^{-1}}\dfrac{6-x}{\sqrt{{{x}^{2}}-12x+37}} \right)$
One can apply LHD(Left-hand derivative) and RHD(Right-Hand derivative) by formula
  & \displaystyle \lim_{x \to C}\dfrac{f\left( x \right)-f\left( c \right)}{x-c} \\
 & LHD=\displaystyle \lim_{h\to 0}\dfrac{f\left( c-h \right)-f\left( c \right)}{-h} \\
 & RHD=\displaystyle \lim_{h\to 0}\dfrac{f\left( c+h \right)-f\left( c \right)}{h} \\
In solution, we have calculated g(x) then directly differentiated it to get g’(x) to put direct limits too $\dfrac{g'\left( x \right)}{\tan \left( 4-x \right)}$. As the above process is longer than given in solution.
At last, we need to compare $\displaystyle \lim_{x \to 4}\dfrac{g'\left( x \right)}{\tan \left( 4-x \right)}$ to $\displaystyle \lim_{x \to 0}\dfrac{x}{\tan x}=1$ to get an answer.