Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Q1 denotes the first Quartile which is a statistical term to represent value below which 25% of data in dataset falls when arranged in ascending order. Q1 divides dataset into four equal parts which can easily analyze spread and data distribution. It is the median of all the observations of the lower half of the data set. However, it is important to arrange the data in increasing order for accurate results. 

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^n-1}{x}\\ =\underset{x\to 0}{lim}\frac{{\left(1+x\right)}^n-{\left(1\right)}^n}{\left(1+x\right)-\left(1\right)}=\underset{1+x\to 1}{lim}\frac{{\left(1+x\right)}^n-{\left(1\right)}^n}{\left(1+x\right)-\left(1\right)}=n{\left(1\right)}^{n-1}=n\left[?\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n.a^{n-1}\right]\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Given\underset{x\to 3^{+}}{lim}\frac{x}{\left[x\right]}\\ =\underset{h\to 0}{lim}\frac{\left(3+h\right)}{\left[3+h\right]}=1\\ Hence,thevalueofthefilleris1.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhaty=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ \frac{dy}{dx}=0+\frac{1}{1!}+\frac{2x}{2!}+\frac{3x^2}{3!}+\dots \\ =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots =y\\ Hence,thevalueofthefillerisy.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{x\to 0}{lim}\left(sinmxcot\frac{x}{\sqrt[]{3}}\right)=2\\ \Rightarrow \underset{\begin{array}{l}x\to 0\\ \therefore mx\to 0\end{array}}{lim}\frac{sinmx}{mx}*mx\underset{x\to 0}{lim}\left(cot\frac{x}{\sqrt[]{3}}\right)=2\\ \Rightarrow 1*mx\underset{x\to 0}{lim}\frac{1}{tan\frac{x}{\sqrt[]{3}}}=2\\ \Rightarrow \underset{x\to 0}{lim}mx*\frac{\frac{x}{\sqrt[]{3}}}{\frac{x}{\sqrt[]{3}}.tan\frac{x}{\sqrt[]{3}}}=2\\ \Rightarrow \frac{mx}{\frac{x}{\sqrt[]{3}}}*\left(1\right)=2\Rightarrow \sqrt[]{3}m=2\Rightarrow m=\frac{2}{\sqrt[]{3}}=\frac{2\sqrt[]{3}}{3}.\\ Hence,thevalueofthefilleris\frac{2\sqrt[]{3}}{3}.\end{array}$

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenf\left(x\right)=\underset{x\to \pi }{lim}\frac{-tan\left(\pi -x\right)}{x-\pi }\\ =\underset{\pi -x\to 0}{lim}\frac{-tan\left(\pi -x\right)}{-\left(\pi -x\right)}=1\\ Hence,thevalueofthefilleris1.\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=1-x+x^2-x^3+\dots -x^{99}+x^{100}\\ \therefore f\text{'}\left(x\right)=-1+2x-3x^2+\dots -99x^{98}+100x^{99}\\ So,f\text{'}\left(1\right)=-1+2-3+\dots -99+100\\ =\left(-1-3-5\dots -99\right)+\left(2+4+6+\dots +100\right)\\ =\frac{50}{2}\left[2*-1+\left(50-1\right)\left(-2\right)\right]+\frac{50}{2}\left[2*2+\left(50-1\right)\left(2\right)\right]\\ =25\left[-2-98\right]+25\left[4+98\right]=25*-100+25*102\\ =25\left[-100+102\right]=25*2=50.\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatf\left(x\right)=x^{100}+x^{99}+\dots +x+1\\ \therefore f\text{'}\left(x\right)=100x^{99}+99x^{98}+\dots +1\\ So,f\text{'}\left(1\right)=100+99+98+\dots +1\\ =\frac{100}{2}\left[2*100+\left(100-1\right)\left(-1\right)\right]=50\left[200-99\right]=50*101=5050.\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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