Continuity and Differentiability

$f\left(x\right)=4\left|2x+3\right|+9\left[x+\frac{1}{2}\right]-12\left[x+20\right],-20<x<20$ doubtful points for differentiability : $x=\frac{-3}{2},$

$f\left(x\right)=4\left(2x+3\right)+9\left(-1\right)-12\left(-2\right)-240=8x-213forx=\frac{-3}{2}+h$

= $-8x-230$ for x = $\frac{-3}{2}-h$

Not diff. at x = $\frac{-3}{2}$

other doubtful points : $x+\frac{1}{2}=integer$

$-20+\frac{1}{2}<x+\frac{1}{2}<20+\frac{1}{2}$

$x+\frac{1}{2}=-19,-18,....,19,20$

$x=-19.5,-18.5,-17.5,.......,18.5,19.5\to$ total 40 numbers.

No. of number = 19.5 – $\left(-19.5\right)+1=40\left(-1.5\right)included$

$-20<x<90\Rightarrow x=-19,-18,.....,18,15\to 39points$

No. of number = 19 (19) + 1 = 39

Total : 40 + 39 = 79

g(1) = ?

    $=f\left(f\left(f\left(1\right)\right)\right)+f\left(f\left(1\right)\right)$            

$f\left(1\right)={\left(2\left(1-\frac{1}{2}\right)\left(2+1\right)\right)}^{\frac{1}{50}}=3^{\frac{1}{50}}$

  $3^{\frac{1}{50}}+13^{\frac{1}{50}}>1^{\frac{1}{50}}$             

3 > 1

$3^{\frac{1}{50}}>2$ $2>3^{\frac{1}{50}}>1$

$3<2^{50}$ $\left[3^{\frac{1}{50}}+1\right]=1+1=2$                     

$f\left(x\right)=\mathcal{l}n\left(x^2+1\right)-e^{-x}+1,g\left(x\right)=e^{-x}-2e^x$

for $f\left(g\left(a\right)\right)>f\left(g\left(b\right)\right)g\text{'}\left(x\right)=\overline{e^x}-2e^x<0\forall x\in R$

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

$\{\begin{array}{l}<1forn\ge 0,g\left(x\right)\downarrow \\ \ge 1forn<0\end{array}$

g(a) > g(b)

$\Rightarrow a<b\left(asg\downarrow \right)$

$\begin{array}{l}\Rightarrow {\alpha }^2-5\alpha +6<0\\ \left(\alpha -3\right)\left(\alpha -2\right)<0\end{array}$

$y=2\left|x^2-\frac{3}{2}x-\frac{7}{2}\right|$

$=2\left|{\left(x-\frac{3}{4}\right)}^2-\frac{65}{16}\right|$

$f\left(x\right)=\{\begin{array}{l}\frac{sin\left(x+2\right)}{x+2},x\in \left(-2,-1\right)\\ 0,x\in \left(-1,0\right]\\ 2x,x\in \left(0,1\right)\\ 1,otherwise\end{array}$

$LHD=\underset{h\to 0}{Lt}\frac{f\left(0-h\right)-f\left(0\right)}{-h}=0$

$RHD=\underset{h\to 0}{Lt}\frac{f\left(0+h\right)-f\left(0\right)}{h}=2$

Hence f (x) is not differentiable at x = 1, 0, 1

$\therefore m=2,n=3$

  $\left(1,1\right)\left(1,4\right)\left(4,1\right)\left(2,4\right)\left(4,2\right)\left(3,4\right)\left(4,3\right)\left(4,4\right)$  all have only one image.

(2, 1) (1, 2), (2, 2) each element has 3 choice.

(3, 2) (2, 3) (3, 1) (1, 3) (3, 3) each element has two choices.

$\therefore$ total function = 3 * 3 * 2 * 2 * 2 = 72

Case I

None of the pre image have 3 as image, total functions = 2 * 2 * 1 * 1 * 1 = 4

Case II

None of the pre images have 2 as image then number of function = 2⁵ = 32

Case III

None of the pre image have either 3 or 2 as image

Total function = 1⁵ = 1

$\therefore$ Total number of onto function

= 72 – 4 – 32 + 1 = 37

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$e=\sqrt[]{1+\frac{b^2}{a^2}}$ $e\text{'}=\sqrt[]{1+\frac{a^2}{b^2}}$

$\mathcal{l}=\frac{2b^2}{a}$ $\mathcal{l}\text{'}=\frac{2a^2}{b}$

$\left(1+\frac{b^2}{a^2}\right)=\frac{11}{14}*\frac{2b^2}{a}$ $\left(1+\frac{a^2}{b^2}\right)=\frac{11}{8}*\frac{2a^2}{b}$

$7*\left\{{\left(\frac{7b}{4}\right)}^2+b^2\right\}=11*b^2*\frac{7b}{4}\Rightarrow a*77=65$

65b² = 44b³

65 = b * 44

$\therefore 77a+44b=65*2=130$

f (x) is an even function

$f\left(-\frac{1}{4}\right)=f\left(-\frac{1}{2}\right)=f\left(\frac{1}{2}\right)=f\left(\frac{1}{4}\right)=0$

So, f (x) has at least four roots in (-2, 2)

$g\left(\frac{-3}{4}\right)=g\left(\frac{3}{4}\right)=0$

So, g (x) has at least two roots in (2, 2)

now number of roots of f (x) $\cdot g"\left(x\right)=f\text{'}\left(x\right)\cdot g\text{'}\left(x\right)=0$

It is same as number of roots of $\frac{d}{dx}\left(f\left(x\right)\cdot g\text{'}\left(x\right)\right)=0$ will have atleast 4 roots in (2, 2)

$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$

option (C) is incorrect, there will be minima.