Continuity and Differentiability

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lety=\frac{8^x}{x^8}\\ Takinglogonbothsides,weget,logy=log\frac{8^x}{x^8}\\ \Rightarrow logy=log{8}^x-log{x}^8\Rightarrow logy=xlog8-8logx\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log8.1-\frac{8}{x}\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=y\left[log8-\frac{8}{x}\right]\\ Hence,\frac{dy}{dx}=\frac{8^x}{x^8}\left[log8-\frac{8}{x}\right]\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Lety=2^{co{s}^{2}x}\\ Takinglogonbothsides,weget\\ logy=log{2}^{co{s}^{2}x}\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2.\frac{d}{dx}co{s}^{2}x\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2.\left[2cosx.\frac{d}{dx}cosx\right]\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2.\left[2cosx\left(-sinx\right)\right]\\ \Rightarrow \frac{1}{y}.\frac{dy}{dx}=log2\left(-sin2x\right)\\ \Rightarrow \frac{dy}{dx}=-y.log2sin2x\\ Hence,\frac{dy}{dx}=-2^{co{s}^{2}x}\left(log2sin2x\right)\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf:R\to Rsatisfiestheequationf\left(x+y\right)=f\left(x\right).f\left(y\right)\forall x,y\in R,f\left(x\right)\ne 0.\\ Letustakeanypointx=0atwhichthefunctionf\left(x\right)isdifferentiable.\\ \therefore f\text{'}\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ 2=\underset{h\to 0}{lim}\frac{f\left(0\right).f\left(h\right)-f\left(0\right)}{h}\left[?f\left(0\right)=f\left(h\right)\right]\dots \left(1\right)\\ 2=\underset{h\to 0}{lim}\frac{f\left(0\right).\left[f\left(h\right)-1\right]}{h}\\ Now,f\text{'}\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{lim}\frac{f\left(x\right).f\left(h\right)-f\left(x\right)}{h}\left[?f\left(x+y\right)=f\left(x\right).f\left(y\right)\right]\\ =\underset{h\to 0}{lim}\frac{f\left(x\right).\left[f\left(h\right)-1\right]}{h}=2f\left(x\right)\left[Fromeq\left(1\right)\right]\\ Hence,f\text{'}\left(x\right)=2f\left(x\right)\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Wehavef\left(x\right)=\left|x-5\right|\\ \Rightarrow f\left(x\right)=\{\begin{array}{l}-\left(x-5\right)ifx-5<0orx<5\\ x-5ifx-5>0orx>5\end{array}\\ Forcontinuityatx=5\\ L.H.L.\underset{h\to 5^{-}}{lim}f\left(x\right)=-\left(x-5\right)\\ =\underset{h\to 0}{lim}-\left(5-h-5\right)=\underset{h\to 0}{lim}h=0\\ R.H.L.\underset{h\to 5^{+}}{lim}f\left(x\right)=x-5\\ =\underset{h\to 0}{lim}\left(5+h-5\right)=\underset{h\to 0}{lim}h=0\\ L.H.L.=R.H.L.\\ So,f\left(x\right)iscontinuousatx=5.\\ Now,fordifferentiability\\ Lf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(5-h\right)-f\left(5\right)}{-h}=\underset{h\to 0}{lim}\frac{-\left(5-h-5\right)-\left(5-5\right)}{-h}\\ =\underset{h\to 0}{lim}\frac{h}{-h}=\frac{h}{-h}=1\\ Rf\text{'}\left(5\right)=\underset{h\to 0}{lim}\frac{f\left(5+h\right)-f\left(5\right)}{h}=\underset{h\to 0}{lim}\frac{\left(5+h-5\right)-\left(5-5\right)}{h}\\ =\underset{h\to 0}{lim}\frac{h-0}{h}=\frac{h}{h}=1\\ So,Lf\text{'}\left(5\right)\ne Rf\text{'}\left(5\right)\\ Hence,f\left(x\right)isnotdifferentiableatx=5.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}f\left(x\right)isdifferentiableatx=2if\\ Lf\text{'}\left(2\right)=Rf\text{'}\left(2\right)\\ \therefore Lf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(2-h\right)-f\left(2\right)}{-h}=\underset{h\to 0}{lim}\frac{\left(1+2-h\right)-\left(1+2\right)}{-h}\\ =\underset{h\to 0}{lim}\frac{3-h-3}{-h}=\frac{-h}{-h}=1\\ Rf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{lim}\frac{\left[5-\left(2+h\right)\right]-\left(1+2\right)}{h}\\ =\underset{h\to 0}{lim}\frac{3-h-3}{h}=\frac{-h}{h}=-1\\ So,Lf\text{'}\left(c\right)\ne Rf\text{'}\left(c\right)\\ Hence,f\left(x\right)isnotdifferentiableatx=2.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=\{\begin{array}{l}{x}^{2}sin\frac{1}{x},ifx\ne 0\\ 0,ifx=0\end{array}atx=0.\\ Fordifferentiabilityweknowthat\\ Lf\text{'}\left(c\right)=Rf\text{'}\left(c\right)\\ Lf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(0-h\right)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{{\left(0-h\right)}^{2}sin\frac{1}{\left(0-h\right)}-0}{-h}=\frac{h^{2}sin\left(\frac{1}{-h}\right)}{-h}\\ =\underset{h\to 0}{lim}h.sin\left(\frac{1}{h}\right)=0*\left[-1\le sin\left(\frac{1}{h}\right)\le 1\right]=0\\ Rf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{{\left(0+h\right)}^{2}sin\frac{1}{\left(0+h\right)}-0}{h}=\underset{h\to 0}{lim}\frac{h^{2}sin\left(\frac{1}{h}\right)}{h}\\ =\underset{h\to 0}{lim}h.sin\left(\frac{1}{h}\right)=0*\left[-1\le sin\left(\frac{1}{h}\right)\le 1\right]=0\\ So,Lf\text{'}\left(c\right)=Rf\text{'}\left(c\right)=0\\ Hence,f\left(x\right)isdifferentiableatx=0.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthatafunctionfisdifferentiableatapoint\text{'}a\text{'}initsdomainif\\ Lf\text{'}\left(c\right)=Rf\text{'}\left(c\right)\\ whereLf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(a-h\right)-f\left(a\right)}{-h}andRf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ Here,f\left(x\right)=\{\begin{array}{l}x\left[x\right],if0\le x\le 2\\ \left(x-1\right)x,if2\le x\le 2\end{array}atx=2.\\ Lf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(2-h\right)-f\left(2\right)}{-h}=\underset{h\to 0}{lim}\frac{\left(2-h\right)\left[2-h\right]-\left(2-1\right)2}{-h}\\ =\underset{h\to 0}{lim}\frac{\left(2-h\right).1-2}{-h}\left[?\left[2-h\right]=1\right]\\ =\underset{h\to 0}{lim}\frac{2-h-2}{-h}=1\\ Rf\text{'}\left(c\right)=\underset{h\to 0}{lim}\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{lim}\frac{\left(2+h-1\right)\left(2+h\right)-\left(2-1\right).2}{h}\\ =\underset{h\to 0}{lim}\frac{\left(1+h\right)\left(2+h\right)-2}{h}=\underset{h\to 0}{lim}\frac{2+h+2h+h^2-2}{h}\\ =\underset{h\to 0}{lim}\frac{3h+h^2}{h}\underset{h\to 0}{=lim}\frac{h\left(3+h\right)}{h}=3\\ Lf\text{'}\left(c\right)\ne Rf\text{'}\left(c\right)\\ Hence,f\left(x\right)isnotdifferentiableatx=2.\end{array}$