\(\frac { { d }^{ 2 }y }{ { d }x^{ 2 } } =-{ cosec }^{ 2 }t.\frac { 1 }{ a\quad cos\quad t } =-\frac { 1 }{ a } { cosec }^{ 2 }t.sec\ t.\)

\(\frac { { d }^{ 2 }y }{ { d }x^{ 2 } } =\frac { 1 }{ at } { sec }^{ 3 }t\)

\(\Rightarrow \left( { 1+x }^{ 2 } \right) ^{ 2 }\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +2x({ 1+x }^{ 2 })\frac { { d }y }{ d{ x } } =2\)

\(\frac{d y}{d x}=\frac{1}{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \sec ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right) \cdot \frac{1}{2} \)
\(=\frac{1}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}=\frac{1}{\sin \left(\frac{\pi}{2}+x\right)}=\frac{1}{\cos x} \)
\(=\sec x . \left.\frac{d^{2} y}{d x^{2}}=\sec x \tan x, \therefore \frac{d^{2} y}{d x^{2}}\right]_{x=\frac{\pi}{1}}=\sqrt{2}\)

\(\Rightarrow \)x²y'+xy'=-y

\({ y }^{ ' }=Ame^{ mx }+{ Bne }^{ nx }\Rightarrow { y }^{ ' }={ Am }^{ 2 }{ e }^{ mx }+{ Bn }^{ 2 }{ e }^{ nx }\),
substitute for y^'', y^', y in LHS.

\(\Rightarrow { x }^{ 2 }{ y }^{ ' }+xy^{ ' }+y=0\)

\(\frac { dy }{ dx } \)=xx cos
x[cos x - x log x . sin x + logx . cos x ]-\(\frac { 4x }{ \left( { x }^{ 2 }-1 \right) ^{ 2 } } \)

\(y'=\frac { log \ (sin\ y) \ + \ y \ \ tan \ \ x }{ log \ (cos\ x) \ -x \ cot \ \ y } \)

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

\(\Rightarrow \)y^'=\(\frac { 1 }{ 2\sqrt { 1-{ x }^{ 2 } } } \)

= (sin x)^{tan x}{sec²x.log (sin x) +1} + (cos x)^{sec x} {sec x tan x log (cos) - sec x tan x}.

Given function \(f(x)=|x-3|=\left\{\begin{aligned} x-3, & x \geq 3 \\ -x+3, & x<3 \end{aligned}\right.\)
\(\underset{x=3}{\mathrm{LHL}} =\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0}\{-(3-h)+3\} \)
\(=\lim _{h \rightarrow 0} h=0 \)
\(\mathrm{RHL} =\lim _{x=3} f(3+h)=\lim _{h \rightarrow 0}\{(3+h)-3\} \)
\(=\lim _{h \rightarrow 0} h=0 \)
\(f(3)=3-3=0\)
\(\text { As } \underset{x=3}{\mathrm{LHL}}=\underset{x=3}{\mathrm{RHL}}=f(3) \text { , }\)
For sontinuity at x = 3,
Hence, function is continuous at x = 3.
\(\underset{x=3}{\operatorname{LHD}} =\lim _{h \rightarrow 0} \frac{f(3-h)-f(3)}{-h}=\lim _{h \rightarrow 0} \frac{(-3+h+3)-(0)}{-h} \)
\(=\lim _{h \rightarrow 0} \frac{h}{-h}=\lim _{h \rightarrow 0}(-1)=-1 \)
\(\operatorname{RHD}_{x=3} =\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}=\lim _{h-0} \frac{(3+h-3)-(0)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0}(1)=1 \)
For differentiability at x = 3,
As Hence, function is not derivable (differentiable) at x = 3.

If function is continuous at \(x=\frac{\pi}{2}, \text { then }\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\mathrm{RHL}=f\left(\frac{\pi}{2}\right)\)
\(\mathrm{LHL}_{x=\frac{\pi}{2}}=\lim _{h \rightarrow 0} f\left(\frac{\pi}{2}-h\right) \)
\(=\lim _{h \rightarrow 0} \frac{1-\sin ^{3}\left(\frac{\pi}{2}-h\right)}{3 \cos ^{2}\left(\frac{\pi}{2}-h\right)}=\lim _{h-0} \frac{1-\cos ^{3} h}{3 \sin ^{2} h} \)
\(=\lim _{h \rightarrow 0} \frac{(1-\cos h)\left(1+\cos ^{2} h+\cos h\right)}{3(1-\cos h)(1+\cos h)} \)
\(=\lim _{h \rightarrow 0} \frac{1+\cos ^{2} h+\cos h}{3(1+\cos h)}=\frac{1+1+1}{3(1+1)}=\frac{1}{2} \ldots .(i i) \)
\(\operatorname{RHL} =\lim _{h=\frac{\pi}{2}} f\left(\frac{\pi}{2}+h\right)=\lim _{h \rightarrow 0} \frac{b\left\{1-\sin \left(\frac{\pi}{2}+h\right)\right\}}{\left\{\pi-2\left(\frac{\pi}{2}+h\right)\right\}^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{(\pi-\pi-2 h)^{2}}=\lim _{h \rightarrow 0} \frac{b(1-\cos h)}{4 h^{2}} \)
\(=\lim _{h \rightarrow 0} \frac{b \cdot 2 \sin ^{2} \frac{h}{2}}{4 h^{2}}=\lim _{h \rightarrow 0} \frac{b}{8}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)^{2} \)
\(=\frac{b}{8} \times 1=\frac{b}{8} \)
Substituting from (ii) and (iii) in (i), we get
\(\frac{1}{2}=\frac{b}{8}=a \Rightarrow a=\frac{1}{2}, b=4\)
Hence, for \(a=\frac { 1 }{ 2 } ,\quad b=4 \text { function is continuous at} x=\frac { \pi }{ 2 } .\)

For function to be continuous at x = 2, we have
\(\mathrm{LHL}_{x=2}=\mathrm{RHL}_{x=2}=f(2)\)
\(\Rightarrow \lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) \)
\(\Rightarrow \lim _{h \rightarrow 0}\{2(2-h)+1\}=\lim _{h \rightarrow 0}\{3(2+h)-1\}=k \)
\(\Rightarrow 4+1=6-1=k \Rightarrow k=5 . \)
k = 5