(i) f (x) = |x|, continuous in [-1, 1]
(ii) LHD at x = 0 \(\neq \) RHD at x = 0.
\(\therefore \)Not derivable at x = 0. Theorem is not applicable.

\(\Rightarrow (1-{ x }^{ 2 })\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -x\frac { dy }{ dx } -2=0\)

\(\frac { dy }{ dx } =\frac { 1 }{ 4\sqrt { a+x } \sqrt { a+\sqrt { a+x } } } \)

\(\frac { dy }{ dx } ={ 6e }^{ 2x }+{ { 6e }^{ 3x } },\frac { { d }^{ 2 }y }{ dx^{ 2 } } =12{ e }^{ 2x }+18{ e }^{ 3x }\) 
substitute in L.H.S.

\(\Rightarrow \) (1+x²) y₂ + (2x-1) y₁=0

\(\text { Given }y =\sqrt{\frac{\sec x-1}{\sec x+1}}=\sqrt{\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}} \)
\(=\sqrt{\frac{1-\cos x}{1+\cos x}} \)
\(=\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}=\sqrt{\tan ^{2} \frac{x}{2}}\)
\(\Rightarrow y=\tan \frac{x}{2} \Rightarrow \frac{d y}{d x}=\frac{1}{2} \sec ^{2} \frac{x}{2}\)
On simplifying we get y= tan \(\frac { x }{ 2 } \)\(\Rightarrow \)y'=\(\frac { 1 }{ 2 } \)sec²\(\frac { x }{ 2 } \)

Consider \(\lim _{x \rightarrow 0} f(x) =\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{x^{2}} \)
\(=\lim _{x \rightarrow 0} \frac{e^{2 x}+1-2 e^{x}}{e^{x} \cdot x^{2}} \)
\(=\lim _{x \rightarrow 0} \frac{1}{e^{x}}\left(\frac{e^{x}-1}{x}\right)^{2} \)
\(=\frac{1}{1} \times(1)^{2}=1 \)
\(f(0) =4 k\)
If continuous then 1 = 4 \(k\Rightarrow k=\frac { 1 }{ 4 } \)

Here, \(f(x)=\left\{\begin{array}{cl} \lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0 \end{array}\right.\) 
At \(x=0, \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right)\) 
\(\therefore \mathrm{LHL}=\lim _{h \rightarrow 0} \lambda\left[(0-h)^{2}-2(0-h)\right]=\lim _{h \rightarrow 0}\left[\lambda\left(h^{2}+2 h\right)\right]=0\) 
\(\mathrm{RHL}=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}(4 x+1)\) 
\(\therefore \mathrm{RHL}=\lim _{h \rightarrow 0}[4(0+h)+1]=\lim _{h \rightarrow 0}[4 h+1]=0+1=1\)
\(\text { [put } x=0+h \text { ; when } x \rightarrow 0^{+} \text {, then } \left.h \rightarrow 0\right] \) 
\(\therefore \mathrm{LHL} \neq \mathrm{RHL}\) 
Thus, f(x) is not continuous at x = 0 for any value of λ.
At x = 1,
 \( \mathrm{LHL} =\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(4 x+1) \)
\(\therefore \mathrm{LHL} =\lim _{h \rightarrow 0}[4(1-h)+1]=\lim _{h \rightarrow 0}[5-4 h]=5-0=5 \)
[put x=1−h; when x→1−,then h→0]   
\( \mathrm{RHL}=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(4 x+1) \)
\(\therefore \lim _{h \rightarrow 0}[4(1+h)+1]=\lim _{h \rightarrow 0}(5+4 h)=5+0=5 \)
[put x=1+h; when x→1, then h→0]   
Also, f(1)=4×1+1=5
\([\because f(x)=4 x+1]\)
Thus f(x) is continuous at x=1 for all values of λ.

For continuity at x - 1
\(\lim _{1} f(x)=\lim _{1} f(x)=f(1)\)
\(\Rightarrow \lim _{x \rightarrow 1}(4)=\lim _{x \rightarrow 1}\left(k x^{2}\right)=k \)
\(\Rightarrow 4=k=k \Rightarrow k=4 \)
k = 4