Continuity and Differentiability

$LHS=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to \infty }{lim}\frac{-1}{h}ln\left(\frac{1-\frac{h}{a}}{1+\frac{h}{b}}\right)=\underset{h\to 0}{lim}\frac{ln\left(1-\frac{h}{a}\right)}{a\left(\frac{-h}{a}\right)}+\underset{h\to 0}{lim}\frac{ln\left(1+\frac{h}{b}\right)}{b\left(\frac{h}{b}\right)}=\frac{1}{a}+\frac{1}{b}.............\left(i\right)$

$RHS=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{cos2x-1}{x^2+1-1}\left(\sqrt[]{x^2+1}+1\right)=\underset{x\to 0}{lim}\frac{-2si{n}^{x}x}{x^2}*2=-4...............\left(ii\right)$

$and\underset{x\to 0}{lim}f\left(x\right)=k.............\left(iii\right)$

$f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$

$\Rightarrow \frac{1}{a}+\frac{1}{b}=-4=k$

From (i), (ii) and (iii) we get $\therefore \frac{1}{a}+\frac{1}{b}+\frac{4}{k}=k+\frac{4}{k}=-4-1=-5$

$f\left(x\right)=\left|x^2-2x-3\right|e^{\left|9x^2-12x+4\right|}$

$\Rightarrow f\left(x\right)=\{\begin{array}{l}\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x<-1\\ -\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},-1\le x<3\\ \left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x\ge 3\end{array}$

$f\text{'}\left(-1^{-}\right)\ne f\left(-1^{+}\right),$

$f\text{'}\left(3^{-}\right)\ne f\left(3^{+}\right)$

Number of non differential points is 2 at x = -1, 3.

Let f (x) = 3x⁴ + 4x³ – 12x² + 4

So, f' (x) = 12x³ + 12x² – 24x = 12x (x² + x – 2)

=12x (x + 2) (x – 1)

So, number of distinct real roots = 4

$f\left(0^{-}\right)=f\left(0\right)=-a$

$f\left(0^{+}\right)=\underset{x\to 0^{+}}{lim}\frac{sin2x\left(\frac{1}{cos2x}-1\right)}{b{x}^3}$ $=\frac{4}{b}$

ab = 4 10 – ab = 14

$f\left(x\right)=\left\{x\right\},g\left(x\right)=1?\left\{x\right\}$

Given,

$f\left(x\right)=\underset{f_1\left(x\right)}{\underset{?}{\left(x^2-2x+7\right)}}\underset{f_2\left(x\right)}{\underset{?}{\left(e^{\left(4x^3-12x^2-180x+31\right)}\right)}}$        

f₁ (x) = x² – 2x + 7

So f (x) is decreasing in [-3, 0]

and positive also

$\therefore$ absolute maximum value of f (x) occurs at x = -3

$\therefore \alpha =-3$      

  $f\left(x\right)\{\begin{array}{l}{\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{x}{\int }}\left(8-t\right)dt;x>4}}\\ x^2+bx;x\le 4\end{array}$

f(x) is continuous at x = 4

$16+4b={\displaystyle \underset{0}{\overset{3}{\int }}\left(2-t\right)dt+{\displaystyle \underset{3}{\overset{4}{\int }}\left(8-t\right)dt}}$           

16 + 4b = 15

$\Rightarrow b=\frac{-1}{4}$               

$forx\le 4,$ $\begin{array}{l}f\left(x\right)=x^2-\frac{x}{4}\\ f\text{'}\left(x\right)=2x-\frac{1}{4}\end{array}$                 

f(x) is increasing in  $\left(\frac{1}{8},\infty \right)$  

 rate change at $x=\frac{1}{8},$  from -ve to +ve. So minima occurs.

Find RHL, x = -1 + h

$\underset{h\to 0}{lim}a.sin\left(\frac{\pi \left[-1+h\right]}{2}\right)+\left[2+1-h\right]=-a+2$               

 Find LHL, x = -1 – h

$\underset{h\to 0}{lim}asin\frac{\left(\pi \left[-1-h\right]\right)}{2}=\left[2+1+h\right]=3$               

${\displaystyle \underset{0}{\overset{4}{\int }}f\left(x\right)dx={\displaystyle \underset{0}{\overset{1}{\int }}1dx+{\displaystyle \underset{1}{\overset{2}{\int }}\left(-1\right)dx+{\displaystyle \underset{2}{\overset{3}{\int }}\left(-1\right)dx+{\displaystyle \underset{3}{\overset{4}{\int }}\left(-1\right)dx=-2}}}}}$               

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

$\therefore \frac{f\left(x\right)+f\left(1-x\right)}{2}=1$

i.e. f (x) + f (1 – x) = 2

$\therefore f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+....+f\left(\frac{99}{100}\right)$

${\displaystyle \sum _{x=1}^{49}f\left(\frac{x}{100}\right)+f\left(1-\frac{x}{100}\right)+f\left(\frac{1}{2}\right)}$

= 49 * 2 + 1 = 99