Continuity and Differentiability

Note : n should be given as a natural number:

$f\left(x\right)=\{\begin{array}{l}\begin{array}{ccc}\hfill \frac{-sin\left(x-1\right)}{x-1}\hfill & \hfill ,\hfill & \hfill x<-1\hfill \\ \hfill -\left(sin2+1\right)\hfill & \hfill ,\hfill & \hfill x=-1\hfill \\ \hfill cos2\pi x\hfill & \hfill ,\hfill & \hfill -1<x<1\hfill \end{array}\\ 1,x=1\end{array}$

$\frac{-sin\left(x-1\right)}{x-1},x>1$

f (x) is discontinuous at x = 1 and x = 1

Since,  ${\mathrm{l}\mathrm{i}\mathrm{m}}_{x\to 0}?\frac{f\left(x\right)}{x}$ exist $\Rightarrow f\left(0\right)=0$

Now,  $f^{\mathrm{\text{'}}}\left(x\right)={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)+x{h}^2+x^{2}h}{h}($ take $y=h)$

$={\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to 0}?\frac{f\left(h\right)}{h}+{\mathrm{l}\mathrm{i}\mathrm{m}}_{h\to .0}?\left(xh\right)+x^2$

$\Rightarrow {\mathrm{f}}^{\mathrm{\text{'}}}\left(\mathrm{x}\right)=1+0+{\mathrm{x}}^2$

$\Rightarrow f^{\mathrm{\text{'}}}\left(3\right)=10$

$e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$

$\left(e^{2x}+\frac{1}{e^{2x}}\right)+4\left(e^x+\frac{1}{e^x}\right)-58=0$

$\Rightarrow {\left(e^x+\frac{1}{e^x}+2\right)}^2=64$

$e^x$$=$$\frac{6\pm \text{exception 25:}}{2}$$=$$3$$\pm$$2$

A = {1, 2, 3, ….50}

R₁ = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R₂ = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

$\therefore R_1-R_2$ contains only 9 elements.

Since $\left(\mathrm{3,3}\right)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{9}{a^2}-\frac{9}{b^2}=1$

Now, normal at $\left(\mathrm{3,3}\right)$ is $y-3=-\frac{a^2}{b^2}(x-3)$ ,

which passes through $\left(\mathrm{9,0}\right)\Rightarrow b^2=2a^2$

So,  ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=3$

Also,  ${\mathrm{a}}^2=\frac{9}{2}$

(From (i) and (ii)

Thus,  $\left({\mathrm{a}}^2,{\mathrm{e}}^2\right)=\left(\frac{9}{2},3\right)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ab)\frac{2b^2}{a}=10\Rightarrow b^2=5a$

Now,  $?\left(t\right)=\frac{5}{12}+t-t^2=\frac{8}{12}-{\left(t-\frac{1}{2}\right)}^2$
$?(\mathrm{t}{)}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\Rightarrow {\mathrm{e}}^2=1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{4}{9}$

$\Rightarrow {\mathrm{a}}^2=81\mathrm{}$  (From (i) and (ii)

So,  $a^2+b^2=81+45=126$

$f\left(x\right)={\displaystyle {\int }_0^{x^2}\frac{t^2-5t+4}{2+e^t}dt}$

$f\text{'}\left(x\right)=2x\left(\frac{x^4-5x^2+4}{2+e^{x^2}}\right)=0$

$x=0,or\left(x^2-4\right)\left(x^2-1\right)=0$

$x=0,x=\pm 2,\pm 1$

Now,

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

Changes sign from positive to negative at x = 1, 1 So, number of local maximum points = 2

Changes sign from negative to positive at

x = 2, 0, 2 So, number of local minimum points = 3

$\therefore m=2,n=3$

This is a True or False Type questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}False.\\ Letustakeanexample:f\left(x\right)=sinxandg\left(x\right)=cotx\\ \therefore f\left(x\right).g\left(x\right)=sinx.cotx=sinx.\frac{cosx}{sinx}=cosxwhichiscontinuousatx=0butcotxisnot\\ continuousatx=0.\end{array}$

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $True.$

This is a True or False Type questions as classified in NCERT Exemplar

Sol: $True.Weknowthatthesumanddifferenceoftwoormorefunctionsisalwayscontinuous.$