Maths

$S=2+\frac{6}{7}+\frac{12}{7^2}+\frac{20}{7^3}+\frac{30}{7^4}+........(i)$  

$\frac{1}{7}S=\frac{2}{7}+\frac{6}{7^2}+\frac{12}{7^3}+\frac{20}{7^4}+........(ii)$               

(i) - (ii)

$\frac{6}{7}S=2+\frac{4}{7}=\frac{6}{7^2}+\frac{8}{7^3}+...........(iii)$               

$\frac{6}{7^2}S=\frac{2}{7}+\frac{4}{7^2}+\frac{6}{7^3}+........(iv)$              

(iii) – (iv)

$=2+\left[\frac{1}{1-\frac{1}{7}}\right]=2\left(\frac{7}{6}\right)$

$\therefore 4S=8{\left(\frac{7}{6}\right)}^3={\left(\frac{7}{3}\right)}^3$      

This is an objective Type Questions as classified in NCERT Exemplar

Sol:

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

40% of K.E. = $mc\Delta T+mL$  

$0.4*\frac{1}{2}*m{v}^2=m\left[125\left(327-127\right)+2.5*10^4\right]$                

$\Rightarrow 0.2v^2=\left[250*10^2+25*10^3\right]$                

$v^2=\frac{2*25*10^3}{0.2}$                

v = 5 * 100

v = 500 m/sec

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:

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