f(x) = x + x²
f(-x) = (-x) + (-x)² 
f(-x) ≠ f(x) and f(-x) ≠ -f(x)
\(\therefore\) f is neither even nor odd function.

Given e^y(x+1)=1    ....(1)
Differentiating with respect to 'x' we get,
\({ e }^{ y }(1)+(x+1){ e }^{ y }\frac { dy }{ dx } =0\)   [product rule]
\(\Rightarrow { e }^{ y }+(1)\frac { dy }{ dx } =0\quad [using\quad (1)]\)
\(\Rightarrow \frac { dy }{ dx } =-{ e }^{ y }\)...(2)
Differentiating again with respect to 'x' we get,
\(\frac { d }{ dx } \left( \frac { dy }{ dx } \right) =\frac { d }{ dx } \left( -{ e }^{ y } \right) \)
\(\Rightarrow \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-{ e }^{ y }.\frac { dy }{ dx } \)
\(=\left( \frac { dy }{ dx } \right) \left( \frac { dy }{ dx } \right) \) [using (2)]
\(={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }\)
Hence proved.

Given x = 15(t - sin t)
Differentiating with respect to 't' we get,
\(\frac{dy}{dx}\) = 15(1 - cos t) Also y = 18(1 - cos t)
\(\frac{dy}{dx}\) = 18(sin t)
Now  \(\frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\frac { 18\sin { t } }{ 15\left( 1-\cos { t } \right) } =\frac { 6\sin { t } }{ 5\left( 1-\cos { t } \right) } \)
\(=\frac { 6\times 2\sin { \frac { t }{ 2 } } \cos { \frac { t }{ 2 } } }{ 5\times 2\sin ^{ 2 }{ \frac { t }{ 2 } } } \)      \(\left[ \because sin2A=2sinAcosA\ \ and\ 1-cosA=2{ sin }^{ 2 }\frac { A }{ 2 } \right] \quad \quad \)
\(=\frac { 6 }{ 5 } \cot { \left( \frac { t }{ 2 } \right) } \)

x = a cos \(\theta\), y = a sin \(\theta\)
\(\frac {dx }{ d\theta } =-a\sin { \theta } \) ; \(\frac {dy }{ d\theta } =a\cos { \theta } \)    
\(= { \frac { dy }{ dx} } =\frac { a\cos { \theta } }{ -a\sin { \theta } } =-\cot { \theta } \)
\(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(-\cot \theta)\)
\(=\frac{d}{d \theta}(-\cot \theta) \frac{d \theta}{d x}\)
\(=\operatorname{cosec}^2 \theta\left(\frac{-1}{a \sin \theta}\right)=\frac{-1}{a} \operatorname{cosec}^3 \theta\)

\(\lim _{ x\rightarrow \infty }{ \frac { 2x+5 }{ { x }^{ 2 }+3x+9 } } \)
= \(\lim _{x \rightarrow \infty}{ \frac { x\left( 2+\frac { 5 }{ x } \right) }{ { x }^{ 2 }\left( 1+\frac { 3 }{ x } +\frac { 9 }{ { x }^{ 2 } } \right) } } \)
= \(\lim _{x \rightarrow \infty}{ \frac { 2+\frac { 5 }{ x } }{ { x }^{ 2 }\left( 1+\frac { 3 }{ x } +\frac { 9 }{ { x }^{ 2 } } \right) } } \) \(\left(\because x \rightarrow \infty \frac{1}{x} \rightarrow 0\right)\)

If x = 0, y = 1. The curve cuts the y axis at (0,1)
The curve will not meet the x axis for all real values of x
 

y = e^2x
If x = 0, y = 1. The curve cuts the y axis at (0, 1)
For no real value of x, f(x) equals 0.
Thus it does not meet the x axis for all real values of x.
 

u = sin² x and v = x²
\(\frac { du }{ dx } =2 \sin x \cos x;\quad \frac { dv }{ dx } =2x\)
= 2 sinx 
\(\therefore \frac { du }{ dv } =\frac { \frac { du }{ dx } }{ \frac { dv }{ dx } } =\frac { sin2x }{ 2x } \)

\(f(x)=\frac{x-1}{3 x+1} \)
\(f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{3}{x}+1}=\frac{1-x}{3+x} \)
\(\frac{1}{f(x)}=\frac{3 x+1}{x-1} \)
 

Given f(x) = x³ - k x² + 2x
\(\therefore\) f is an odd function,
f(-x) = -f(x)
= (-x)³ - k(-x)² + 2(-x) = -x³ + kx² - 2x
-x³ - kx² - 2x = -[x³ - kx² + 2x]
\(-2 k x^2=0 \Rightarrow k=0\)

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

f(5) = 5² - 4(5) = 5

\(x=a\left( t+{{1}\over{}t} \right)\)
Differentiating with respect to 't' we get,
\({{dx}\over{dt}}=a\left( 1-{{1}\over{t^2}} \right)=a\left( {{t^2-1}\over{t^2}}\right)\)
Also  \(y=a\left( t={{1}\over{t}}\right)\) ....(1)
Differentiating with respect to 't' we get,
\({{dy}\over{dt}}=a\left( 1+{{1}\over{t^2}}\right)=a\left( {{t^2+1}\over{t^2}} \right)\)  ...(2)
Now, \({{{{dy}\over{dx}}={{dy}\over{dt}}}\over{{{dy}\over{dt}}}}=\frac { a({ t }^{ 2 }+1) }{ \frac { { t }^{ 2 } }{ \frac { a({ t }^{ 2 }-1) }{ { t }^{ 2 } } } } ={{a(t^2+1)}\over{t^2}}\times{{t^2}\over{a(t^2-1)}}={{t^2+1}\over{t^2-1}}\) ......(3)
Also, \({{x}\over{y}}={{a\left( t+{{1}\over{t}} \right)}\over{a\left( t-{{1}\over{t}} \right)}}={{{{r^2+1}\over{t}}}\over{{{r^2-1}\over{t}}}}={{t^2+1}\over{}t}\times{{t}\over{t^2-1}}\)
\({{x}\over{y}}={{t^2+1}\over{t^2-1}}\)    ....(4)
From (3) and (4), \({{dy}\over{dx}}={{x}\over{y}}\)

Let y = (sec x -1) (sec x +1)
y = sec² x -1    \(\left[ \therefore (a+b)(a-b)={ a }^{ 2 }-{ b }^{ 2 } \right] \)
\(\therefore\)y =tan²x
Differentiating with respect to 'x' we get 
\(\frac { dy }{ dx } =2tanx\frac { d }{ dx } (tan x)\) \([\therefore 1+tan^{ 2 }x=sec^{ 2 }x]\)
=2tan x.sec² x

\(\underset { h\rightarrow 0 }{ lim } \frac { \sqrt { x+h } -\sqrt { x } }{ h } =\underset { h\rightarrow 0 }{ lim } \frac { (x+h)^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 1 }{ 2 } } }{ (x+h)-x } \)  [Adding and subtracting x in the denominator]
= \(\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } -1 }\left[ \therefore \underset { x-\rightarrow a }{ lim } \frac { { x }^{ n }-{ a }^{ n } }{ z-a } =n.a^{ n-1 } \right] \)
= \(\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } }=\frac { 1 }{ 2\sqrt { x } } \)

\(\begin{matrix} \underset { x\rightarrow 1 }{ lim } & \frac { { x }^{ 4 }-1 }{ x-1 } \end{matrix}=\begin{matrix} \underset { x\rightarrow k }{ lim } & \frac { { x }^{ 3 }-{ k }^{ 3 } }{ { x }^{ 2 }-{ k }^{ 2 } } \end{matrix}\)
\(=\begin{matrix} \underset { x\rightarrow 1 }{ lim } & \frac { { x }^{ 4 }-1 }{ x-1 } \end{matrix}\)
\(=\begin{matrix} \underset { x\rightarrow k }{ lim } & \frac { { x }^{ 3 }-{ k }^{ 3 } }{ { x }-{ k } } \end{matrix}\times \frac { x-k }{ { x }^{ 2 }-{ k }^{ 2 } } \)       [multiplying the numerator and denominator ]
\(\Rightarrow { 4.1 }^{ 4-1 }=\begin{matrix} \underset { x\rightarrow k }{ lim } & \frac { { x }^{ 3 }-{ k }^{ 3 } }{ { x }-{ k } } \end{matrix}\times \frac { 1 }{ \begin{matrix} \underset { x\rightarrow k }{ lim } & \frac { { x }^{ 2 }-{ k }^{ 2 } }{ { x }-{ k } } \end{matrix} } \Rightarrow 4=\frac { 3.{ k }^{ 2 } }{ 2.k } \)   
\(=\left[ \because \underset{ x\rightarrow a } {lim}\frac { { x }^{ n }-{ a }^{ n } }{ x-a } =n.{ a }^{ n-1 } \right] \)
\(\Rightarrow 4=\frac { 3k }{ 2 } \Rightarrow \frac { 4\times 2 }{ 3 } =k\Rightarrow \frac { 8 }{ 3 } \)