Limits and Derivatives

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=\frac{x^n-a^n}{x-a}\\ f\text{'}\left(x\right)=\frac{\left(x-a\right)\left(n.x^{n-1}\right)-\left(x^n-a^n\right).1}{{\left(x-a\right)}^2}\\ \therefore f\text{'}\left(a\right)=\frac{\left(a-a\right)\left(n.a^{n-1}\right)-\left(a^n-a^n\right).1}{{\left(a-a\right)}^2}\\ So,f\text{'}\left(a\right)=\frac{0}{0}=doesnotexist\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=1+x+\frac{x^2}{2}+\dots +\frac{x^{100}}{100}\\ f\text{'}\left(x\right)=1+\frac{2x}{2}+\dots +\frac{100x^{99}}{100}\\ \therefore f\text{'}\left(1\right)=1+1+1+\dots +1\left(100times\right)=100\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhaty=\frac{sin\left(x+9\right)}{cos x}\\ \therefore \frac{dy}{dx}=\frac{cos x.cos \left(x+9\right)-sin \left(x+9\right)\left(-sin x\right)}{co{s}^{2}x}\\ =\frac{cos xcos \left(x+9\right)+sin xsin \left(x+9\right)}{co{s}^{2}x}\\ =\frac{cos \left(x+9-x\right)}{co{s}^{2}x}=\frac{cos9}{co{s}^{2}x}\\ \therefore {\left(\frac{dy}{dx}\right)}_{atx=0}=\frac{cos9}{co{s}^{2}0}=\frac{cos9}{{\left(1\right)}^2}=cos9.\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhaty=\frac{sin x+cos x}{sin x-cos x}\\ \therefore \frac{dy}{dx}=\frac{\left(sin x-cos x\right)\left(cosx-sin x\right)-\left(sin x+cos x\right)\left(cosx+sin x\right)}{{\left(sin x-cos x\right)}^2}\\ =\frac{-{\left(sin x-cos x\right)}^2-{\left(sin x+cos x\right)}^2}{{\left(sin x-cos x\right)}^2}\\ =\frac{-\left[si{n}^{2}x+co{s}^{2}x-2sinxcosx+si{n}^{2}x+co{s}^{2}x+2sinxcosx\right]}{{\left(sin x-cos x\right)}^2}=\frac{-2}{{\left(sin x-cos x\right)}^2}\\ \therefore {\left(\frac{dy}{dx}\right)}_{atx=0}=\frac{-2}{{\left(sin 0-cos 0\right)}^2}=\frac{-2}{{\left(-1\right)}^2}=-2\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhaty=\frac{1+\frac{1}{x^2}}{1-\frac{1}{x^2}}\\ \Rightarrow y=\frac{x^2+1}{x^2-1}\\ \therefore \frac{dy}{dx}=\frac{\left(x^2-1\right).2x-\left(x^2+1\right).2x}{{\left(x^2-1\right)}^2}\\ =\frac{2x\left(x^2-1-x^2-1\right)}{{\left(x^2-1\right)}^2}=\frac{2x\left(-2\right)}{{\left(x^2-1\right)}^2}=\frac{-4x}{{\left(x^2-1\right)}^2}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=\frac{x-4}{2\sqrt[]{x}}\\ \therefore f\text{'}\left(x\right)=\frac{1}{2}\left[\frac{\sqrt[]{x}.1-\left(x-4\right).\frac{1}{2\sqrt[]{x}}}{x}\right]\\ =\frac{1}{2}\left[\frac{2x-x+4}{2\sqrt[]{x}.x}\right]=\frac{1}{2}\left[\frac{x+4}{2{\left(x\right)}^{3/2}}\right]\\ \therefore f\text{'}\left(x\right)atx=1=\frac{1}{2}\left[\frac{1+4}{2\times 1}\right]=\frac{5}{4}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhaty=\sqrt[]{x}+\frac{1}{\sqrt[]{x}}\\ \frac{dy}{dx}=\frac{1}{2\sqrt[]{x}}-\frac{1}{2x^{3/2}}\\ {\left(\frac{dy}{dx}\right)}_{atx=1}=\frac{1}{2}-\frac{1}{2}=0\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Given,f\left(x\right)=x-\left[x\right]\\ Wehavetofirstcheckdifferentiabilityoff\left(x\right)atx=\frac{1}{2}\\ \therefore Lf\text{'}\left(\frac{1}{2}\right)=LHD=\underset{h\to 0}{lim}\frac{f\left[\frac{1}{2}-h\right]-f\left[\frac{1}{2}\right]}{-h}\\ =\underset{h\to 0}{lim}\frac{\left(\frac{1}{2}-h\right)-\left[\frac{1}{2}-h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{-h}=\underset{h\to 0}{lim}\frac{\frac{1}{2}-h-0-\frac{1}{2}+0}{-h}=\frac{-h}{-h}=1\\ Lf\text{'}\left(\frac{1}{2}\right)=RHD=\underset{h\to 0}{lim}\frac{f\left(\frac{1}{2}+h\right)-f\left(\frac{1}{2}\right)}{h}\\ =\underset{h\to 0}{lim}\frac{\left(\frac{1}{2}+h\right)-\left[\frac{1}{2}+h\right]-\frac{1}{2}+\left[\frac{1}{2}\right]}{h}=\underset{h\to 0}{lim}\frac{\frac{1}{2}+h-1-\frac{1}{2}+1}{h}=\frac{h}{h}=1\\ LHD=RHD\\ \therefore f\text{'}\left(\frac{1}{2}\right)=1\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Given,\underset{x\to 0}{lim}\frac{tan2x-x}{3x-sinx}\\ =\underset{x\to 0}{lim}\frac{x\left[\frac{tan2x}{x}-1\right]}{x\left[3-\frac{sinx}{x}\right]}=\underset{\begin{array}{l}x\to 0\\ \therefore 2x\to 0\end{array}}{lim}\frac{\frac{tan2x}{2x}\times 2-1}{3-\frac{sinx}{x}}\\ =\frac{1.2-1}{3-1}=\frac{2-1}{2}=\frac{1}{2}\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Given,f\left(x\right)=\{\begin{array}{cc}\hfill x^2-1,\hfill & \hfill 0<x<2\hfill \\ \hfill 2x+3,\hfill & \hfill 2\le x<3\hfill \end{array}\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}\left(x^2-1\right)=\underset{h\to 0}{lim}\left[{\left(2-h\right)}^2-1\right]=\underset{h\to 0}{lim}\left(4+h^2-4h-1\right)\\ =\underset{h\to 0}{lim}\left(h^2-4h+3\right)=3\\ and\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}\left(2x+3\right)=\underset{h\to 0}{lim}\left[2\left(2+h\right)+3\right]=7\\ Therefore,thequadraticequationwhoserootsare3and7is\\ x^2-\left(3+7\right)x+3\times 7=0i.e.,x^2-10x+21=0.\\ Hence,thecorrectoptionis\left(d\right).\end{array}$