f(x) = 2^x (Given)
\(\therefore \) f(x + y) = 2^x+y
= 2^x.2^y
= f(x).f(y)

y = 9 - x² 

\(f(x)=x^2-|x|\)
\(f(-x)=(-x)^2-|-x|=x^2-|x|\)
\(=f(x)\)
\(\therefore\) f is an even function.

\(\underset { x\rightarrow a }{ lim } \cfrac { { x }^{ \frac { 3 }{ 5 } }-{ a }^{ \frac { 3 }{ 5 } } }{ { x }^{ \frac { 1 }{ 5 } }-{ a }^{ \frac { 1 }{ 5 } } } =\underset { x\rightarrow a }{ lim } \cfrac { { x }^{ \frac { 3 }{ 5 } }-{ a }^{ \frac { 3 }{ 5 } } }{ \frac { { x }^{ \frac { 4 }{ 5 } }-{ a }^{ \frac { 1 }{ 5 } } }{ x-a } } \) 
= \(\cfrac { \frac { 3 }{ 5 } \left( a \right) ^{ -\frac { 2 }{ 5 } } }{ \frac { 1 }{ 5 } \left( a \right) ^{ -\frac { 4 }{ 5 } } } =3{ a }^{ \frac { -2 }{ 5 } +\frac { 4 }{ 5 } }=3a^{ \frac { 2 }{ 5 } }\)

We know that, domain set is (0, \(\infty\)), range set is R and the curve passing the point (1, 0)
Case (i) when a > 1 and x > 0
\((f(x)=\log _a x= \begin{cases} < 0 & \text { if } 0 < x <1 \\ =0 & \text { if } x=1 \\ > 0 & \text { if } x > 1\end{cases} \)
We noticed that as x increases, f(x) is also increases.
\(\therefore \) The graph is as shown in the figure.

Case (ii) when 0 < a < 1 and x > 0
\((f(x)=\log _a x= \begin{cases} > 0 & \text { if } 0 < x < 1 \\ =0 & \text { if } x=1 \\ < 0 & \text { if } x > 1\end{cases} \)
We noticed that x increases, the value of f(x) decreases.
\(\therefore \) The graph is as shown in the figure

\(\lim _{ x\leftarrow 2 }{ \frac { { x }^{ n }-2^{ n } }{ x-2 } =448 } \) 
n2^n-1 = \(448=7 \times 2^6\)
n2^n-1 \(=7 \times 2^{7-1}\)
n = 7

\(f(x)=\frac { |x| }{ x } \)
\(y=\frac{x}{x}=1 \text { if } x>0\)
\(y=\frac{-x}{x}=-1 \text { if } x<0\)

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

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

\(f(x+1) =(x+1)^2-(x+1)+1 \)
\(=x^2+2 x+1-x-1+1 \)
\(=x^2+x+1 \)