Continuity and Differentiability

\(
f(x)= \begin{cases}-x ; & x \in(-2,-1) \\ 1-x ; & x \in[-1,0) \\ 2-x ; & x \in[0,1) \\ 3-x ; & x \in[1,2) \\ x ; & x \in[2,3)\end{cases}
\)

As we can see that function is discontinuous at \(-1,0,1\) and \(2\) . Therefore, function is non-differentiable at \(-1,0,1\) and 2.
Therefore, \(\mathrm{p}=4\) and \(\mathrm{q}=4\).
Then, \(\mathrm{p}+\mathrm{q}=8\).

$L=\underset{x\to 0}{\lim }\frac{{\log }_e\left(\sec \left(ex\right)\cdot \sec \left(e^{2}x\right)\cdot \dots \cdot \sec \left(e^{10}x\right)\right)}{e^2-e^{2\cos x}}$

$=\underset{x\to 0}{\lim }\frac{\ln \left(\sec \left(ex\right)\right)+\ln \left(\sec \left(e^{2}x\right)\right)+\dots \mathrm{..}+\ln \left(\sec \left(e^{10}x\right)\right)}{e^{2\cos x}\left(\frac{e^{2-2\cos x}-1}{2-2\cos x}\right)\times \frac{2-2\cos x}{x^2}\times x^2}$

$=\underset{x\to 0}{\lim }\frac{\ln \left(\sec \left(ex\right)\right)+\ln \left(\sec \left(e^{2}x\right)\right)+\dots \dots +\ln \left(\sec \left(e^{10}x\right)\right)}{e^2{x}^2}$

Using L'H rule

$=\underset{x\to 0}{\lim }\frac{e\tan \left(ex\right)+e^2\tan \left(e^{2}x\right)+\dots \mathrm{..}+e^{10}\tan \left(e^{10}x\right)}{2e^{2}x}$

$=\frac{1}{2e^2}\left[e^2+e^4+e^6+\dots .+e^{20}\right]$

$=\frac{1}{2}\frac{e^2\left({\left(e^2\right)}^{10}-1\right)}{e^2\left(e^2-1\right)}$

$=\frac{1}{2}\frac{\left(e^{20}-1\right)}{\left(e^2-1\right)}$

Using \(\cos \theta=1-\frac{\theta^2}{2}+\ldots\)

\(\cos (\alpha x) \cos ((\alpha+1) x) \cos ((\alpha+2) x) =1-\frac{\alpha^2 x^2+(\alpha+1)^2 x^2+(\alpha+2)^2 x^2}{2}+\ldots\)

So, \(1-\cos (\alpha x) \cos ((\alpha+1) x) \cos ((\alpha+2) x) =\frac{\alpha^2+(\alpha+1)^2+(\alpha+2)^2}{2} x^2\)

Also, \(\sin ^2((\alpha+1) x)=(\alpha+1)^2 x^2\)

\( \therefore \frac{1}{2(\alpha+1)^2}\left[\alpha^2+(\alpha+1)^2+(\alpha+2)^2\right]=2\)
\(\alpha^2+(\alpha+1)^2+(\alpha+2)^2=4(\alpha+1)^2 \)
\( \alpha^2+(\alpha+2)^2=3(\alpha+1)^2 \)
\( \Rightarrow \alpha^2+2 \alpha-1=0\)
Product \(=-1\)

\(\lim_{t\to x}\frac{10t^9f(x)-f'(t)x^{10}}{9t^8}=1\)

\(\Rightarrow 10x^9f(x)-f'(x)x^{10}=9x^8\)

\(\Rightarrow f'(x)-\frac{10}{x}f(x)=-\frac{9}{x^2}\)

Let \(y=f(x)\).

\(IF=e^{-\int \frac{10}{x}\,dx}\)

\(IF=\frac{1}{x^{10}}\)

\(\frac{y}{x^{10}}=\int -\frac{9}{x^2}\cdot \frac{1}{x^{10}}\,dx\)

\(\frac{y}{x^{10}}=-9\int x^{-12}\,dx\)

\(\frac{y}{x^{10}}=\frac{9}{11}x^{-11}+C\)

\(\because y(1)=2\Rightarrow C=\frac{13}{11}\)

\(\Rightarrow y=\frac{9}{11x}+\frac{13}{11}x^{10}\)

Use the Taylor expansions around $x=0$:

$$\cos(2x) = 1 - 2x^2 + \frac{2}{3}x^4 + O(x^6)$$

$$\cos(4x) = 1 - 8x^2 + \frac{32}{3}x^4 + O(x^6)$$

$$\cos(2x) + a\cos(4x) - b = (1+a-b) + (-2 - 8a)x^2 + \left(\frac{2}{3} + \frac{32a}{3}\right)x^4 + O(x^6)$$

For

$$\lim_{x\to 0} \frac{\cos(2x) + a\cos(4x) - b}{x^4}$$

to be finite, the numerator must be $O(x^4)$. Thus the constant and $x^2$ terms must vanish:

$$1 + a - b = 0 \quad\text{(i)}$$

$$-2 - 8a = 0 \quad\text{(ii)}$$

From (ii): $a = -\frac14$.

Substitute into (i):

$$1 - \frac14 - b = 0 \Rightarrow b = \frac34$$

So

$$a+b=-\frac{1}{4}+\frac{3}{4}=\frac{1}{2}\mathrm{.}$$

\(
f(x)= \begin{cases}-x ; & x \in(-2,-1) \\ 1-x ; & x \in[-1,0) \\ 2-x ; & x \in[0,1) \\ 3-x ; & x \in[1,2) \\ x ; & x \in[2,3)\end{cases}
\)

As we can see that function is discontinuous at \(-1,0,1\) and \(2\) . Therefore, function is non-differentiable at \(-1,0,1\) and 2.
Therefore, \(\mathrm{p}=4\) and \(\mathrm{q}=4\).
Then, \(\mathrm{p}+\mathrm{q}=8\).

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{\left(2-\frac{3}{x}+\frac{5}{x^2}\right){\left(1-\frac{1}{3x}\right)}^{x/2}}{\left(3+\frac{5}{x}+\frac{4}{x^2}\right){\left(1+\frac{2}{3x}\right)}^{x/2}} \\ \underset{x\to \infty }{\lim }\frac{\left(2-\frac{3}{x}+\frac{5}{x^2}\right)}{\left(3+\frac{5}{x}+\frac{4}{x^2}\right)}\underset{x\to \infty }{\lim }\frac{{\left(1-\frac{1}{3x}\right)}^{x/2}}{{\left(1+\frac{2}{3x}\right)}^{x/2}} \\ =\frac{2}{3}\underset{x\to \infty }{\lim }{\left(\frac{\left(1-\frac{1}{3x}\right)}{\left(1+\frac{2}{3x}\right)}\right)}^{x/2} \\ =\frac{2}{3}·\frac{e^{\frac{x}{2}\left(1-\frac{1}{3x}-1\right)}}{e^{\frac{x}{2}\left(1+\frac{2}{3x}-1\right)}} \\ =\frac{2}{3}·\frac{{\mathrm{e}}^{-\frac{1}{6}}}{{\mathrm{e}}^{1/3}}=\frac{2}{3}{\mathrm{e}}^{-\frac{1}{2}} \end{gathered}$$

\(\lim _{x \rightarrow \infty} \frac{\left(2 x^2-3 x+5\right)(3 x-1)^{\frac{x}{2}}}{\left(3 x^2+5 x+4\right) \sqrt{(3 x+2)^x}}\)

$=\underset{x\to \infty }{\lim }\frac{\left(2-\frac{3}{x}+\frac{5}{x^2}\right){\left(1-\frac{1}{3x}\right)}^{x/2}}{\left(3+\frac{5}{x}+\frac{4}{x^2}\right){\left(1+\frac{2}{3x}\right)}^{x/2}}$

$=\underset{x\to \infty }{\lim }\frac{\left(2-\frac{3}{x}+\frac{5}{x^2}\right)}{\left(3+\frac{5}{x}+\frac{4}{x^2}\right)}\underset{x\to \infty }{\lim }\frac{{\left(1-\frac{1}{3x}\right)}^{x/2}}{{\left(1+\frac{2}{3x}\right)}^{x/2}}$

$=\frac{2}{3}\underset{x\to \infty }{\lim }{\left(\frac{\left(1-\frac{1}{3x}\right)}{\left(1+\frac{2}{3x}\right)}\right)}^{x/2}$

$=\frac{2}{3}·\frac{e^{\frac{x}{2}\left(1-\frac{1}{3x}-1\right)}}{e^{\frac{x}{2}\left(1+\frac{2}{3x}-1\right)}}$

$=\frac{2}{3}·\frac{{\mathrm{e}}^{-\frac{1}{6}}}{{\mathrm{e}}^{1/3}}=\frac{2}{3}{\mathrm{e}}^{-\frac{1}{2}}$

I. Wrong by definition.

II. It's not necessary.

For example:

$f\left(x\right)=x$ is continuous at $x=0$

But $\frac{1}{f\left(x\right)}=\frac{1}{x}$ is not continuous at $x=0$.

Doubtful points: \(-1,0,1,\sqrt{2},\sqrt{3},2\)

At \(x=\sqrt{2},\sqrt{3}\),

\(f(x)=(2x^2+x-[x])+[x^2]\)

Here \(2x^2+x-[x]\) is continuous, but \([x^2]\) is discontinuous.

So, \(f(x)\) is discontinuous at \(x=\sqrt{2},\sqrt{3}\).

At \(x=-1\):

R.H.L. \(\Rightarrow f(-1^+)=2-1-(-1)+0=2\)

\(f(-1)=2-1-(-1)+1=3\)

So, \(f(x)\) is discontinuous at \(x=-1\).

At \(x=2\):

L.H.L. \(\Rightarrow f(2^-)=8+2-1+3=12\)

\(f(2)=8+2-2+4=12\)

R.H.L. \(\Rightarrow f(2^+)=8+2-2+4=12\)

So, \(f(x)\) is continuous at \(x=2\).

At \(x=0\):

L.H.L. \(\Rightarrow f(0^-)=0+0-(-1)+0=1\)

\(f(0)=0\)

So, \(f(x)\) is discontinuous at \(x=0\).

At \(x=1\):

L.H.L. \(\Rightarrow f(1^-)=2+1-0+0=3\)

\(f(1)=2+1-1+1=3\)

R.H.L. \(\Rightarrow f(1^+)=2+1-1+1=3\)

So, \(f(x)\) is continuous at \(x=1\).

Given \(f'(x)=3x^2\sin\left(\dfrac{1}{x}\right)-x\cos\left(\dfrac{1}{x}\right)\).

Differentiating again,

\(f''(x)=6x\sin\left(\dfrac{1}{x}\right)-3\cos\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)-\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}\)

\(\Rightarrow f''(x)=6x\sin\left(\dfrac{1}{x}\right)-4\cos\left(\dfrac{1}{x}\right)-\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}\)

At \(x=\dfrac{2}{\pi}\),

\(f''\left(\dfrac{2}{\pi}\right)=6\cdot\dfrac{2}{\pi}\sin\dfrac{\pi}{2}-4\cos\dfrac{\pi}{2}-\dfrac{\sin\dfrac{\pi}{2}}{\dfrac{2}{\pi}}\)

\(\Rightarrow f''\left(\dfrac{2}{\pi}\right)=\dfrac{12}{\pi}-0-\dfrac{\pi}{2}\)

\(\Rightarrow f''\left(\dfrac{2}{\pi}\right)=\dfrac{24-\pi^2}{2\pi}\)

Given function \(f(x) = ax^3 + bx^2 + cx + 41\)
$f^{\text{'}}\left(x\right)=3a{x}^2+2bx+c$
$\Rightarrow f^{\text{'}}\left(1\right)=3a+2b+c=2\dots \left(i\right)$
$f^{"}\left(x\right)=6ax+2b$
\(\Rightarrow f^{\prime \prime}(1)=6 a+2 b=4\)

\(3 a+b=2 \ldots(i i)\)

\((i)-(i i) \Rightarrow b+c=0 \ldots(i i i)\)

\(f(1)=40 \Rightarrow a+b+c+41=40\)
use (iii) use (iii) \(\Rightarrow a+41=40 \Rightarrow a=-1\)
by (ii) \(\Rightarrow-3+b=2 \Rightarrow b=5\) and \( c=-5\)
\(a^2 + b^2 + c^2 = 1 + 25 + 25 = 51\)

Given \(f(x)\) is continuous at \(x=0\)

$f\left(0\right)=a$

$$\begin{gathered} \mathrm{RHL}=\underset{x\to 0^{+}}{\lim }\frac{\sin x(1-\cos x)}{x^3} \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{x^2}\right) \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{LHL}=\underset{x\to 0^{-}}{\lim }\left(b^2\sin \left(\frac{\pi }{2}\left[\frac{\pi }{2}(\sin x+\cos x)\cos x\right]\right)\right) \\ =b^2 \end{gathered}$$

$\therefore a=\frac{1}{2}$

$b^2=\frac{1}{2}$

So, $\left(a^2+b^2\right)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$

(B)

Let \(g(x)=ax+b\).

Now function \(f(x)\) is continuous at \(x=0\).

\(\therefore \lim_{x\to 0^+}f(x)=f(0)\)

\(\Rightarrow \lim_{x\to 0^+}\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}=b\)

\(\Rightarrow b=0\)

\(\therefore g(x)=ax\)

Now, for \(x>0\),

\(f(x)=\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}\)

Taking log on both sides,

\(\ln f(x)=\frac{1}{x}\ln\left(\frac{1+x}{2+x}\right)\)

Differentiating both sides,

\(\frac{f'(x)}{f(x)}=\frac{1}{x}\cdot\frac{1}{(1+x)(2+x)}-\frac{1}{x^2}\ln\left(\frac{1+x}{2+x}\right)\)

\(\Rightarrow f'(x)=\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}\left[\frac{1}{x(1+x)(2+x)}-\frac{1}{x^2}\ln\left(\frac{1+x}{2+x}\right)\right]\)

\(\therefore f'(1)=\frac{2}{3}\left[\frac{1}{6}-\ln\left(\frac{2}{3}\right)\right]\)

\(\Rightarrow f'(1)=\frac{1}{9}-\frac{2}{3}\ln\left(\frac{2}{3}\right)\)

And \(f(-1)=g(-1)=-a\).

\(\therefore -a=\frac{1}{9}-\frac{2}{3}\ln\left(\frac{2}{3}\right)\)

\(\Rightarrow a=\frac{2}{3}\ln\left(\frac{2}{3}\right)-\frac{1}{9}\)

\(\therefore g(3)=3a\)

\(=2\ln\left(\frac{2}{3}\right)-\frac{1}{3}\)

\(=\ln\left(\frac{4}{9}\right)-\ln e^{\frac{1}{3}}\)

\(=\ln\left(\frac{4}{9e^{\frac{1}{3}}}\right)\)

Given: $y\left(\theta \right)=\frac{2\cos \theta +\cos 2\theta }{\cos 3\theta +4\cos 2\theta +5\cos \theta +2}$

We know that, $\cos 2\theta =2{\cos }^2\theta -1$

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$y\left(\theta \right)=\frac{2\cos \theta +2{\cos }^2\theta -1}{4{\cos }^3\theta -3\cos \theta +8{\cos }^2\theta -4+5\cos \theta +2}$

$y\left(\theta \right)=\frac{2{\cos }^2\theta +2\cos \theta -1}{4{\cos }^3\theta +8{\cos }^2\theta +2\cos \theta -2}$

$y\left(\theta \right)=\frac{2{\cos }^2\theta +2\cos \theta -1}{(2\cos \theta +2)(2{\cos }^2\theta +2\cos \theta -1)}$

$y\left(\theta \right)=\frac{1}{2(\cos \theta +1)}$

$y\left(\theta \right)=\frac{1}{2·2{\cos }^2\frac{\theta }{2}}$

$y\left(\theta \right)=\frac{1}{4}{\sec }^2\frac{\theta }{2}$

$y\text{'}\left(\theta \right)=\frac{1}{4}·2\sec \frac{\theta }{2}\left(\sec \frac{\theta }{2}\tan \frac{\theta }{2}\right)\frac{1}{2}$

$y\text{'}\left(\theta \right)=\frac{1}{4}{\sec }^2\frac{\theta }{2}\tan \frac{\theta }{2}$

$y\text{'}\text{'}\left(\theta \right)=\frac{1}{4}\left({\sec }^4\frac{\theta }{2}\times \frac{1}{2}+{\sec }^2\frac{\theta }{2}{\tan }^2\frac{\theta }{2}\right)$

$y\text{'}\text{'}\left(\theta \right)=\frac{1}{8}{\sec }^4\frac{\theta }{2}+\frac{1}{4}{\sec }^2\frac{\theta }{2}{\tan }^2\frac{\theta }{2}$

$$\begin{gathered} y\text{'}\text{'}\left(\theta \right)+y\text{'}\left(\theta \right)+y\left(\theta \right)=\frac{1}{8}{\sec }^4\frac{\theta }{2}+\frac{1}{4}{\sec }^2 \\ \frac{\theta }{2}{\tan }^2\frac{\theta }{2}+\frac{1}{4}{\sec }^2\frac{\theta }{2}\tan \frac{\theta }{2}+\frac{1}{4}{\sec }^2\frac{\theta }{2} \end{gathered}$$

$\text{at }\theta =\frac{\pi }{2},$

$y\text{'}\text{'}\left(\frac{\pi }{2}\right)+y\text{'}\left(\frac{\pi }{2}\right)+y\text{'}\left(\frac{\pi }{2}\right)=\frac{1}{8}\times 4+\frac{1}{4}\times 2\times 1+\frac{1}{4}\times 2\times 1+\frac{1}{4}\times 2$

$=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2$

\(\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0^{+}} f(x)=\left|\begin{array}{ccc}
a+1 & 1 & b \\
a & 1+1 & b \\
a & 1 & b+1
\end{array}\right|=\mu+\alpha a+\beta b \\
& R_1 \rightarrow R_1-R_2 \\
& R_2 \rightarrow R_2-R_3 \\
& \left|\begin{array}{ccc}
1 & -1 & 0 \\
0 & 1 & -1 \\
a & 1 & b+1
\end{array}\right|=\mu+\alpha a+\beta b \\
& C_2 \rightarrow C_2+C_1 \\
& \left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & -1 \\
a & a+1 & b+1
\end{array}\right|=\mu+\alpha a+\beta b \\
& a+b+2=\mu+\alpha a+\beta b \\
& \quad \text { So } \alpha=1, \mu=2, \beta=1
\end{aligned}\\
&\text { Now, } \quad(\mu+\alpha+\beta)^2=16
\end{aligned}\)

$\underset{x\to 0}{\lim }cosecx\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)$

$=\underset{x\to 0}{\lim }\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sin x\left(\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x+4}\right)}$

$=\underset{x\to 0}{\lim }\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x(\sqrt{5}+\sqrt{5})}$

on applying, L'Hospitals rule,

$=\underset{x\to 0}{\lim }\frac{-2\cos x\sin x-3\sin x-\cos x}{\cos x·2\sqrt{5}}=-\frac{1}{2\sqrt{5}}$

(A) RHD at \(x=0: \lim _{h \rightarrow 0} \frac{\left(2-2 h^2-h^2 \sin \frac{1}{h}\right)-2}{h}=0\)

Similarly LHD at \(\mathrm{x}=0\) is also equal to 0 .

∴ Differentiable at \(\mathrm{x}=0\)

(D)
$$\begin{gathered} \mathrm{f}\left(0\right)=2, \\ \mathrm{f}(0+\mathrm{h})<2 \\ \mathrm{f}(0-\mathrm{h})<2 \end{gathered}$$
therefore \(x=0\) is a point of local maxima

(B) Now,

$f^{\text{'}}\left(x\right)=-\left(4x+2x\sin \frac{1}{x}\right)+\cos \frac{1}{x}$

for any \(\delta>0\), choose \(x=\frac{1}{2 \pi n} \in(0, \delta)\) with \(n\) large.

Then \(\sin \frac{1}{x}=0, \cos \frac{1}{x}=1\), so

\(f^{\prime}(x)=1-4 x>0\)

A decreasing function cannot have \(f^{\prime}(x)>0\) at a point where derivative exists.

(C) for any \(\delta>0\), choose \(x=-\frac{1}{(2 n+1) \pi} \in(-\delta, 0)\)

with \(n\) large. Then \(\cos \frac{1}{x}=-1\) and

\(f^{\prime}(x)=-4 x-2 x \sin \frac{1}{x}-1 \leq 6|x|-1<0\)

for sufficiently small \(|x|\)

Hence \(f\) cannot be increasing on ( \(-\delta, 0\) )

Let degree of \(f(x)\) be \(n\)

Since \(f(x)=f^{\prime}(x) f^{\prime \prime}(x)\),

comparing degrees gives \(n=(n-1)+(n-2)\), so \(n=3\)

Let \(f(x)=a x^3+b x^2+c x+d\)
Since \(f(0)=0, d=0\), so \(f(x)=a x^3+b x^2+c x\)
Now \(f^{\prime}(x)=3 a x^2+2 b x+c\) and \(f^{\prime \prime}(x)=6 a x+2 b\)
Using \(f(x)=f^{\prime}(x) f^{\prime \prime}(x)\):

\(a x^3+b x^2+c x=\left(3 a x^2+2 b x+c\right)(6 a x+2 b)\)
So \(a x^3+b x^2+c x=18 a^2 x^3+18 a b x^2+\left(6 a c+4 b^2\right) x+2 b c\)
Comparing coefficients:

\(a=18 a^2\)

Hence \(a=\frac{1}{18}\)

Also \(b c=0\) and \(c=6 a c+4 b^2\)

Using \(a=\frac{1}{18}, c=\frac{c}{3}+4 b^2\), so \(c=6 b^2\)

Since \(b c=0\), we get \(b=0\), hence \(c=0\)

Therefore \(f(x)=\frac{x^3}{18}\)

So \(f^{\prime}(x)=\frac{x^2}{6}\) and \(f^{\prime \prime}(x)=\frac{x}{3}\)

Hence \(f^{\prime}(2)=\frac{2}{3}, f^{\prime \prime}(2)=\frac{2}{3}\),

and \(\int_0^2 f(x) d x=\int_0^2 \frac{x^3}{18} d x=\left.\frac{x^4}{72}\right|_0 ^2=\frac{2}{9}\)

Therefore \(36\left(f^{\prime}(2)+f^{\prime \prime}(2)+\int_0^2 f(x) d x\right)=36\left(\frac{2}{3}+\frac{2}{3}+\frac{2}{9}\right)=56\)

$y\left(x\right)=\left|\begin{array}{ccc}\sin x& \cos x& \sin x+\cos x+1\\ 27& 28& 27\\ 1& 1& 1\end{array}\right|$

$C_3\to C_3-C_1$

$y\left(x\right)=\left|\begin{array}{ccc}\sin x& \cos x& \cos x+1\\ 27& 28& 0\\ 1& 1& 0\end{array}\right|$

$$\begin{gathered} y=-\cos x-1 \\ \Rightarrow \frac{dy}{dx}=\sin x \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\cos x \end{gathered}$$

Now, $\frac{d^{2}y}{d{x}^2}+y=\cos x+\left(-\cos x-1\right)=-1$

\(f(x)=x^5+2x^3+3x+1\)

\(\Rightarrow f'(x)=5x^4+6x^2+3\)

\(\Rightarrow f'(1)=5+6+3=14\)

Given \(g(f(x))=x\).

Differentiating with respect to \(x\),

\(g'(f(x))\cdot f'(x)=1\)

For \(f(x)=7\),

\(x^5+2x^3+3x+1=7\)

\(\Rightarrow x=1\)

So, \(g(7)=1\).

Now,

\(g'(f(1))\cdot f'(1)=1\)

\(\Rightarrow g'(7)\cdot14=1\)

\(\Rightarrow g'(7)=\dfrac{1}{14}\)

Therefore,

\(\dfrac{g(7)}{g'(7)}=\dfrac{1}{1/14}\)

\(\Rightarrow \dfrac{g(7)}{g'(7)}=14\)

$$\begin{gathered} \frac{d}{dx}\left({\tan }^{-1}\left(\cos \sqrt{x}\right)+{\sec }^{-1}\left(e^x\right)\right)= \\ \frac{d}{dx}\left({\tan }^{-1}\left(\cos \sqrt{x}\right)+{\cos }^{-1}\left(\frac{1}{e^x}\right)\right)= \\ \{ hint \frac{d}{dx}\left({\tan }^{-1}\left(x\right)\right)=\frac{1}{x^2+1} , \frac{d}{dx}\left({\cos }^{-1}\left(x\right)\right)=\frac{-1 }{\sqrt{1-x^2}} \} \\ =\frac{1}{{\cos }^2\left(\sqrt{x}\right)+1}·\frac{ \mathrm{d}}{ \mathrm{d}x}\left[\cos \right(\sqrt{x}\left)\right]-\frac{1 }{\sqrt{1-\frac{1}{{\left({\mathrm{e}}^x\right)}^2}}{\left({\mathrm{e}}^x\right)}^2}\frac{-1}{{\left({\mathrm{e}}^x\right)}^2}·\frac{ \mathrm{d}}{ \mathrm{d}x}\left[{\mathrm{e}}^x\right] \{apply chain rule \} \\ =\frac{(-\sin (\sqrt{x}\left)\right)·\frac{\mathrm{d}}{\mathrm{d}x}\left[\sqrt{x}\right]}{{\cos }^2\left(\sqrt{x}\right)+1}+\frac{{\mathrm{e}}^x{\mathrm{e}}^{-2x}}{\sqrt{1-{\mathrm{e}}^{-2x}}} \\ =-\frac{\frac{1}{2}{x}^{\frac{1}{2}-1}\sin \left(\sqrt{x}\right)}{{\cos }^2\left(\sqrt{x}\right)+1}+\frac{{\mathrm{e}}^{-x}}{\sqrt{1-{\mathrm{e}}^{-2x}}} \\ -\frac{\sin \left(\sqrt{x}\right)}{2\sqrt{x}\left({\cos }^2\left(\sqrt{x}\right)+1\right)}+\frac{1}{\sqrt{e^{2x}-1}} \\ \mathrm{now} \mathrm{find} \mathrm{value} \mathrm{of} -\frac{\sin \left(\sqrt{\mathrm{x}}\right)}{2\sqrt{\mathrm{x}}\left({\cos }^2\left(\sqrt{\mathrm{x}}\right)+1\right)}+\frac{1}{\sqrt{{\mathrm{e}}^{2\mathrm{x}}-1}} \mathrm{at} \mathrm{x}=\frac{{\mathrm{\pi }}^2}{4} \\ \mathrm{we} \mathrm{get} \\ -\frac{1}{\pi }+\frac{1}{\sqrt{e^{\frac{{\pi }^2}{2}}-1}} \end{gathered}$$

\(
\begin{aligned}
&\lim_{x\to 0}
\frac{\cos(2x)+a\cos(4x)-b}{x^4}
\end{aligned}
\)

\(
\begin{aligned}
&=
\lim_{x\to 0}
\frac{
\left(
1-\frac{(2x)^2}{2!}
+\frac{(2x)^4}{4!}
-\cdots
\right)
+
a\left(
1-\frac{(4x)^2}{2!}
+\frac{(4x)^4}{4!}
-\cdots
\right)
-b
}
{x^4}
\end{aligned}
\)

For the limit to be finite,

\(
\begin{aligned}
&1+a-b=0
\end{aligned}
\)

and

\(
\begin{aligned}
&-2-8a=0\\
&\Rightarrow 8a=-2\\
&\Rightarrow a=-\frac{1}{4}
\end{aligned}
\)

\(
\begin{aligned}
&\Rightarrow b=1+a\\
&=1-\frac{1}{4}\\
&=\frac{3}{4}
\end{aligned}
\)

\(
\begin{aligned}
&a+b\\
&=-\frac{1}{4}+\frac{3}{4}\\
&=\frac{1}{2}
\end{aligned}
\)

$$\begin{gathered} y={\tan }^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3/2}}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{x}-x}{1+\sqrt{x}·x}\right) \\ ={\tan }^{-1}\left(\sqrt{x}\right)-{\tan }^{-1}\left(x\right) \end{gathered}$$

Differentiating w.r.t. x, we get

$$\begin{gathered} y^{\text{'}}\left(x\right)=\frac{1}{1+x}·\frac{1}{2\sqrt{x}}-\frac{1}{1+x^2} \\ \Rightarrow y^{\text{'}}\left(1\right)=\frac{1}{2}·\frac{1}{2}-\frac{1}{2} \\ \Rightarrow y^{\text{'}}\left(1\right)=-\frac{1}{4} \end{gathered}$$

$$\begin{gathered} \alpha =\underset{x\to \infty }{\lim }{\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)}^x \left(1^{\infty }\text{ form }\right) \\ \mathrm{\alpha }=e^{\underset{\mathrm{x}\to \infty }{\lim }\mathrm{x}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}\right)-1\right)} \\ \Rightarrow \mathrm{\alpha }=e^{\underset{\mathrm{x}\to \infty }{\lim }\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\mathrm{x}\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}-\left(\frac{1-\mathrm{e}}{\mathrm{e}}\right) \right)} \\ \Rightarrow \mathrm{\alpha }=e^{\frac{\mathrm{e}}{1-\mathrm{e}}\underset{\mathrm{x}\to \infty }{\lim }\mathrm{x}\left(1-\frac{\mathrm{x}}{1+\mathrm{x}}\right)} \\ \Rightarrow \mathrm{\alpha }=e^{\frac{\mathrm{e}}{1-\mathrm{e}}.1} \\ \therefore \mathrm{\alpha }={\mathrm{e}}^{\frac{\mathrm{e}}{1-\mathrm{e}}}\Rightarrow \mathrm{log\alpha }=\frac{\mathrm{e}}{1-\mathrm{e}} \\ \therefore \text{ Required value }=\frac{\frac{\mathrm{e}}{1-\mathrm{e}}}{1+\frac{\mathrm{e}}{1-\mathrm{e}}}=\mathrm{e} \end{gathered}$$

Given \(f(x)\) is continuous at \(x=0\)

$f\left(0\right)=a$

$$\begin{gathered} \mathrm{RHL}=\underset{x\to 0^{+}}{\lim }\frac{\sin x(1-\cos x)}{x^3} \\ =\underset{x\to 0^{+}}{\lim }\left(\frac{\sin x}{x}\right)\left(\frac{1-\cos x}{x^2}\right) \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{LHL}=\underset{x\to 0^{-}}{\lim }\left(b^2\sin \left(\frac{\pi }{2}\left[\frac{\pi }{2}(\sin x+\cos x)\cos x\right]\right)\right) \\ =b^2 \end{gathered}$$

$\therefore a=\frac{1}{2}$

$b^2=\frac{1}{2}$

So, $\left(a^2+b^2\right)=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$

Given \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right)\)

\(y=\ln \left(1-x^2\right)-\ln \left(1+x^2\right)\)

Differentiating both sides with respect to \(x\) :

\(y^{\prime}=\frac{-2 x}{1-x^2}-\frac{2 x}{1+x^2}\)

\(=\frac{-2 x-2 x^3-2 x+2 x^3}{1-x^4}=\frac{-4 x}{1-x^4}\)

Now, \(y^{\prime \prime}=\frac{-4\left(1-x^4\right)-(-4 x)\left(-4 x^3\right)}{\left(1-x^4\right)^2}\)

\(y^{\prime \prime}=\frac{-4+4 x^4-16 x^4}{\left(1-x^4\right)^2}=\frac{-4-12 x^4}{\left(1-x^4\right)^2}\)

Now, \(y^{\prime}\left(\frac{1}{2}\right)=\frac{-4(1 / 2)}{1-1 / 16}=\frac{-2}{15 / 16}=-\frac{32}{15}\)

\(y^{\prime \prime}\left(\frac{1}{2}\right)=\frac{-19 / 4}{225 / 256}=-\frac{19}{4} \cdot \frac{256}{225}=-\frac{1216}{225}\)

\(225\left(y^{\prime}-y^{\prime \prime}\right)=225\left(-\frac{32 \cdot 15}{225}+\frac{1216}{225}\right)\)

\(225\left(y^{\prime}-y^{\prime \prime}\right)=-480+1216=736\)

Given \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right)\)

\(y=\ln \left(1-x^2\right)-\ln \left(1+x^2\right)\)

Differentiating both sides with respect to \(x\) :

\(y^{\prime}=\frac{-2 x}{1-x^2}-\frac{2 x}{1+x^2}\)

\(=\frac{-2 x-2 x^3-2 x+2 x^3}{1-x^4}=\frac{-4 x}{1-x^4}\)

Now, \(y^{\prime \prime}=\frac{-4\left(1-x^4\right)-(-4 x)\left(-4 x^3\right)}{\left(1-x^4\right)^2}\)

\(y^{\prime \prime}=\frac{-4+4 x^4-16 x^4}{\left(1-x^4\right)^2}=\frac{-4-12 x^4}{\left(1-x^4\right)^2}\)

Now, \(y^{\prime}\left(\frac{1}{2}\right)=\frac{-4(1 / 2)}{1-1 / 16}=\frac{-2}{15 / 16}=-\frac{32}{15}\)

\(y^{\prime \prime}\left(\frac{1}{2}\right)=\frac{-19 / 4}{225 / 256}=-\frac{19}{4} \cdot \frac{256}{225}=-\frac{1216}{225}\)

\(225\left(y^{\prime}-y^{\prime \prime}\right)=225\left(-\frac{32 \cdot 15}{225}+\frac{1216}{225}\right)\)

\(225\left(y^{\prime}-y^{\prime \prime}\right)=-480+1216=736\)

We have, \(f(x)\) is continuous at \(x=0\).

\(\therefore f(0^-)=f(0)=f(0^+)\)

\(\therefore \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}\frac{\tan((a+1)x)+b\tan x}{x}\)

\(=\lim_{x\to 0^-}\frac{\tan((a+1)x)}{(a+1)x}\cdot(a+1)+b\lim_{x\to 0^-}\frac{\tan x}{x}\)

\(=a+1+b=3\), because \(f(0)=3\).

And,

\(\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{\sqrt{ax+b^2x^2}-\sqrt{ax}}{b\sqrt{a}\,x\sqrt{x}}\)

\(=\lim_{x\to 0^+}\frac{\sqrt{x}\left(\sqrt{a+b^2x}-\sqrt{a}\right)}{b\sqrt{a}\,x\sqrt{x}}\)

\(=\lim_{x\to 0^+}\frac{\sqrt{a+b^2x}-\sqrt{a}}{b\sqrt{a}\,x}\)

\(=\lim_{x\to 0^+}\frac{\sqrt{a+b^2x}-\sqrt{a}}{b\sqrt{a}\,x}\cdot\frac{\sqrt{a+b^2x}+\sqrt{a}}{\sqrt{a+b^2x}+\sqrt{a}}\)

\(=\lim_{x\to 0^+}\frac{b^2x}{b\sqrt{a}\,x\left(\sqrt{a+b^2x}+\sqrt{a}\right)}\)

\(=\lim_{x\to 0^+}\frac{b^2}{b\sqrt{a}\left(\sqrt{a+b^2x}+\sqrt{a}\right)}\)

\(=\frac{b^2}{b\sqrt{a}\left(\sqrt{a}+\sqrt{a}\right)}\)

\(=\frac{b}{2a}=3\)

\(\therefore a+1+b=3\) and \(\frac{b}{2a}=3\)

\(\Rightarrow \frac{b}{a}=6\)

We have, \(f(x)\) is continuous at \(x=0\).

\(\therefore f(0^-)=f(0)=f(0^+)\)

\(\therefore \lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}\frac{\tan((a+1)x)+b\tan x}{x}\)

\(=\lim_{x\to 0^-}\frac{\tan((a+1)x)}{(a+1)x}\cdot(a+1)+b\lim_{x\to 0^-}\frac{\tan x}{x}\)

\(=a+1+b=3\), because \(f(0)=3\).

And,

\(\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{\sqrt{ax+b^2x^2}-\sqrt{ax}}{b\sqrt{a}\,x\sqrt{x}}\)

\(=\lim_{x\to 0^+}\frac{\sqrt{x}\left(\sqrt{a+b^2x}-\sqrt{a}\right)}{b\sqrt{a}\,x\sqrt{x}}\)

\(=\lim_{x\to 0^+}\frac{\sqrt{a+b^2x}-\sqrt{a}}{b\sqrt{a}\,x}\)

\(=\lim_{x\to 0^+}\frac{\sqrt{a+b^2x}-\sqrt{a}}{b\sqrt{a}\,x}\cdot\frac{\sqrt{a+b^2x}+\sqrt{a}}{\sqrt{a+b^2x}+\sqrt{a}}\)

\(=\lim_{x\to 0^+}\frac{b^2x}{b\sqrt{a}\,x\left(\sqrt{a+b^2x}+\sqrt{a}\right)}\)

\(=\lim_{x\to 0^+}\frac{b^2}{b\sqrt{a}\left(\sqrt{a+b^2x}+\sqrt{a}\right)}\)

\(=\frac{b^2}{b\sqrt{a}\left(\sqrt{a}+\sqrt{a}\right)}\)

\(=\frac{b}{2a}=3\)

\(\therefore a+1+b=3\) and \(\frac{b}{2a}=3\)

\(\Rightarrow \frac{b}{a}=6\)

Let$t=x-1$. As $x\to 1^{+},t\to 0^{+}$.
Limit is $\underset{t\to 0}{\lim }\frac{t(6+\lambda \cos t)-\mu \sin t}{t^3}=-1$.
Using series expansions:
Numerator $=t\left(6+\lambda \left(1-\frac{t^2}{2}\right)\right)-\mu \left(t-\frac{t^3}{6}\right)=t(6+\lambda -\mu )+t^3\left(-\frac{\lambda }{2}+\frac{\mu }{6}\right)+\dots$
For limit to be finite, coeff. of $t=0⟹6+\lambda -\mu =0⟹\mu -\lambda =6$.
Limit value is coeff. of $t^3:-\frac{\lambda }{2}+\frac{\mu }{6}=-1⟹-3\lambda +\mu =-6$.
Solving the system: $(\lambda +6)-3\lambda =-6⟹-2\lambda =-12⟹\lambda =6,\mu =12$.
Value of $\lambda +\mu =6+12=18$.

\(f(x)=\left\{\begin{array}{cc}\frac{a|x|+x^2-2(\sin |x|)(\cos |x|)}{x} & , x \neq 0 \\ b & , x=0\end{array}\right.\)

for continuity

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)=f\left(0\right)$

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\frac{-ax+x^2+2\left(\sin x\right)\cos x}{x} \\ =\underset{x\to 0^{-}}{\lim }\frac{-ax}{x}+\frac{x^2}{x}+\frac{2\sin \left(2x\right)}{2x}=-a+2 \end{gathered}$$

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{ax+x^2-2\left(\sin x\right)\cos x}{x} \\ =\underset{x\to 0^{+}}{\lim }\frac{ax}{x}+\frac{x^2}{x}-\frac{2\sin \left(2x\right)}{2x}=a-2 \end{gathered}$$

So we get

\( -a+2=a-2=b \)

\(\Rightarrow 2 a=4 \)

\(\Rightarrow a=2, b=0 \)

\( \therefore a+b=2\)

\(\begin{aligned}
& e^{x_0}+x_0=0 \\
& g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)} \\
& g(x)=x-\frac{\alpha\left(e^x+x\right)}{3\left(e^x+1\right)}, g(x)+e^{x_0}=x+e^{x_0}-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \\
& g(x)+e^{x_0}=\left(x-x_0\right)-\frac{\alpha}{3}\left(\frac{e^x+x}{e^x+1}\right) \\
& \Rightarrow \text { As very clear } \mathrm{g}\left(\mathrm{x}_0\right)+\mathrm{x}_0 \Rightarrow 0 \\
& \lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| \frac{0 \uparrow}{0 \uparrow}=\lim _{x \rightarrow x_0}\left|\frac{g^{\prime}(x)}{1}\right|=\left|g^{\prime}\left(x_0\right)\right|
\end{aligned}\)

\(\begin{aligned}
& g^{\prime}(x)=1-\frac{\alpha}{3}\left(\frac{\left(e^x+1\right)\left(e^x+1\right)-\left(e^x+x\right) e^x}{\left(e^x+1\right)^2}\right) \\
& \mathrm{g}^{\prime}\left(\mathrm{x}_0\right)=1-\frac{\alpha}{3}\left(\frac{\left(\mathrm{e}^{x_0}+1\right)^2}{\left(\mathrm{e}^{x_0}+1\right)^2}\right)=1-\frac{\alpha}{3} \\
& \Rightarrow \lim _{x \rightarrow x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|g^{\prime}\left(x_0\right)\right|=\left|1-\frac{\alpha}{3}\right|
\end{aligned}\)

\((\sin x\cos y)\left(f(2x+2y)-f(2x-2y)\right)=(\cos x\sin y)\left(f(2x+2y)+f(2x-2y)\right)\)

\(\Rightarrow f(2x+2y)(\sin x\cos y-\cos x\sin y)=f(2x-2y)(\sin x\cos y+\cos x\sin y)\)

\(\Rightarrow f(2x+2y)\sin(x-y)=f(2x-2y)\sin(x+y)\)

\(\Rightarrow \dfrac{f(2x+2y)}{\sin(x+y)}=\dfrac{f(2x-2y)}{\sin(x-y)}\)

Put \(2x+2y=m\) and \(2x-2y=n\).

\(\Rightarrow \dfrac{f(m)}{\sin\left(\dfrac{m}{2}\right)}=\dfrac{f(n)}{\sin\left(\dfrac{n}{2}\right)}=K\)

\(\Rightarrow f(m)=K\sin\left(\dfrac{m}{2}\right)\)

Therefore, \(f(x)=K\sin\left(\dfrac{x}{2}\right)\)

\(\Rightarrow f'(x)=\dfrac{K}{2}\cos\left(\dfrac{x}{2}\right)\)

Put \(x=0\).

\(\dfrac{1}{2}=\dfrac{K}{2}\)

\(\Rightarrow K=1\)

So,

\(f'(x)=\dfrac{1}{2}\cos\left(\dfrac{x}{2}\right)\)

\(\Rightarrow f''(x)=-\dfrac{1}{4}\sin\left(\dfrac{x}{2}\right)\)

Now,

\(f''\left(\dfrac{5\pi}{3}\right)=-\dfrac{1}{4}\sin\left(\dfrac{5\pi}{6}\right)\)

\(\Rightarrow f''\left(\dfrac{5\pi}{3}\right)=-\dfrac{1}{8}\)

If the required value is \(24f''\left(\dfrac{5\pi}{3}\right)\), then

\(24f''\left(\dfrac{5\pi}{3}\right)=24\left(-\dfrac{1}{8}\right)\)

\(\Rightarrow 24f''\left(\dfrac{5\pi}{3}\right)=-3\)

If the required value is \(4f''\left(\dfrac{5\pi}{3}\right)\), then

\(4f''\left(\dfrac{5\pi}{3}\right)=4\left(-\dfrac{1}{8}\right)\)

\(\Rightarrow 4f''\left(\dfrac{5\pi}{3}\right)=-\dfrac{1}{2}\)

\(\alpha=\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x\left(1^{\infty}\right.\) form \()\)

$\mathrm{\alpha }=e^{\underset{\mathrm{x}\to \infty }{\lim }\mathrm{x}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}\right)-1\right)}$

$\Rightarrow \mathrm{\alpha }=e^{\underset{\mathrm{x}\to \infty }{\lim }\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\mathrm{x}\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}-\left(\frac{1-\mathrm{e}}{\mathrm{e}}\right) \right)}$

$\Rightarrow \mathrm{\alpha }=e^{\frac{\mathrm{e}}{1-\mathrm{e}}\underset{\mathrm{x}\to \infty }{\lim }\mathrm{x}\left(1-\frac{\mathrm{x}}{1+\mathrm{x}}\right)}$

$\Rightarrow \mathrm{\alpha }=e^{\frac{\mathrm{e}}{1-\mathrm{e}}.1}$

$\therefore \mathrm{\alpha }={\mathrm{e}}^{\frac{\mathrm{e}}{1-\mathrm{e}}}\Rightarrow \mathrm{log\alpha }=\frac{\mathrm{e}}{1-\mathrm{e}}$

Required value \(=\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=e\)

$$\begin{gathered} \cos \left|\mathbf{x}\right|\text{ is always differentiable } \\ \therefore \text{ we have to check only for }\left|{\mathrm{x}}^2-\mathrm{ax}+2\right| \\ \therefore \text{ Not differentiable at } \\ {x}^2-ax+2=0 \\ \text{ One root is given, }\alpha =2 \\ \therefore 4-2a+2=0 \\ a=3 \\ \therefore \text{ other root }\beta =1 \\ \text{ and for }x=1f\left(x\right)\text{ is non differentiable } \\ \text{the distance of the point (2, 1) from} \\ \text{ the line 12 x+5 y+10=0 is} \\ \frac{\left|12\left(2\right)+5\left(1\right)+10\right|}{13}=3 \end{gathered}$$

Here \(f(x)=\mathrm{e}^{\sin |x|}-|x|\)

we need to check differentiability at \(x=0\) only

For \(x>0\) : \(f(x) =e^{\sin x}-x\)

\(f^{\prime}(x)=\cos x\cdot e^{\sin x}-1\)

\(\Rightarrow f^{\prime}\left(0^{+}\right)=e^0 \cos (0)-1=0\)

For \(x<0\) : \(f(x) =e^{-\sin x}+x\)

\(f^{\prime}(x)=-\cos x\cdot e^{-\sin x}+1\)

\(\Rightarrow f^{\prime}\left(0^{-}\right)=-e^0 \cos (0)+1=0\)

Since \(f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right)=0\), the function is differentiable at \(x=0\)

Hence \(f(x)\) is differentiable for all \(x\in R\)

If \(x \in\left(-\pi,-\frac{\pi}{2}\right)\)

Then, \(f(x)=\mathrm{e}^{-\sin x}+x\)

\(f^{\prime}(x)=-e^{-\sin x} \cos x+1\)

for \(x \in\left(-\pi, \frac{-\pi}{2}\right) , \cos x<0\)

\(\therefore f^{\prime}(x)>0\)

\(\therefore f(x)\) is increasing

\( y=\tan ^{-1}\left(\frac{\frac{3}{4}-\tan x}{1+\frac{3}{4} \tan x}\right)+2 \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)\)
For \( 2 \tan ^{-1}\left(\frac{\mathrm{x}}{1+\sqrt{1-\mathrm{x}^2}}\right) \)
put \( \mathrm{x}=\sin \theta \)
\( \therefore 2 \tan ^{-1}\left(\frac{\sin \theta}{1+\cos \theta}\right)=\theta=\sin ^{-1} \mathrm{x}\)
Hence
\( y=\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}(\tan x)+\sin ^{-1} x \)
\( \frac{d y}{d x}=-1+\frac{1}{\sqrt{1-x^2}} \)
\( \mathrm{x}=\frac{\sqrt{3}}{2}\)
\( \frac{\mathrm{dy}}{\mathrm{dx}}=-1+2=1\)

Given \(f(0)=2\).

So, \(g(2)=0\).

Since \(g(x)\) is inverse of \(f(x)\),

\(g(f(x))=x\)

Differentiating with respect to \(x\),

\(g'(f(x))f'(x)=1\)

Putting \(x=0\),

\(g'(f(0))f'(0)=1\)

\(\Rightarrow g'(2)=\dfrac{1}{f'(0)}\)

Now,

\(f'(x)=5x^4+\dfrac{2}{4}e^{x/4}\)

\(\Rightarrow f'(0)=0+\dfrac{1}{2}=\dfrac{1}{2}\)

So,

\(g'(2)=\dfrac{1}{1/2}=2\)

Therefore,

\(8g'(2)=8\cdot2=16\)

Let

$$A=f(2x+2y),\quad B=f(2x-2y).$$

Then

$$\sin x\cos y(A-B)=\cos x\sin y(A+B).$$

Rearrange:

$(\sin x\cos y-\cos x\sin y)A$$=(\sin x\cos y+\cos x\sin y)B\mathrm{.}$

Using $\sin x\cos y-\cos x\sin y=\sin(x-y)$ and
$\sin x\cos y+\cos x\sin y=\sin(x+y)$, we get

$\sin (x-y)f(2x+2y)$$=\sin (x+y)f(2x-2y)\mathrm{.} \left(1\right)$

Now set

$$u=2(x+y),\quad v=2(x-y).$$

Then $x+y=\frac u2,\ x-y=\frac v2$, and (1) becomes

$\begin{array}{c}\sin \frac{v}{2}f\left(u\right)=\sin \frac{u}{2}f\left(v\right) \text{for all }u,v\in \mathrm{ℝ}\mathrm{.} \text{(2)}\end{array}$

From (2), when $\sin\frac u2\neq 0$ and $\sin\frac v2\neq 0$,

$$\frac{f(u)}{\sin(u/2)} = \frac{f(v)}{\sin(v/2)}.$$

Thus there exists a constant $k$ such that

$$f(t)=k\sin\frac t2 \quad\text{for all } t \text{ with }\sin\frac t2\neq 0.$$

If $\sin\frac t2=0$ (i.e. $t=2\pi n$), choose some $v$ with $\sin\frac v2\neq 0$; from (2):

$$\sin\frac v2\, f(t) = \sin\frac t2\, f(v)=0 \Rightarrow f(t)=0,$$

which matches $k\sin \frac t2$.
So for all real $t$,

$$f(t)=k\sin\frac t2.$$

Use $f'(0)=\frac12$:

$f^{\text{'}}\left(t\right)=k\cdot \frac{1}{2}\cos \frac{t}{2}$$\Rightarrow f^{\text{'}}\left(0\right)=k\cdot \frac{1}{2}=\frac{1}{2}\Rightarrow k=1.$

Hence

$$f(t)=\sin\frac t2.$$

$f\left(t\right)=\sin \frac{t}{2},f^{\text{'}}\left(t\right)=\frac{1}{2}\cos \frac{t}{2},$ $f^{\text{'}\text{'}}\left(t\right)=\frac{1}{2}\cdot (-\frac{1}{2}\sin \frac{t}{2})=-\frac{1}{4}\sin \frac{t}{2}\mathrm{.}$

At $t=\frac{5\pi}{3}$:

$f^{\text{'}\text{'}}\left(\frac{5\pi }{3}\right)=-\frac{1}{4}\sin (\frac{5\pi }{3}\cdot \frac{1}{2})$ $=-\frac{1}{4}\sin \left(\frac{5\pi }{6}\right)=-\frac{1}{4}\cdot \frac{1}{2}=-\frac{1}{8}\mathrm{.}$

Therefore,

$$24f^{\mathrm{\prime }\mathrm{\prime }}\left(\frac{5\pi }{3}\right)=24(-\frac{1}{8})=-3.$$