Ncert Solutions Maths class 12th

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

Sol:

$\begin{array}{l}Giventhat:\\ \text{?}\text{?}\text{?}BagI=3redballsandnowhiteball\\ BagII=2redballsand1whiteball\\ BagIII=noredballand3whiteballs\\ Let{E}_1,E_{2}and{E}_{3}betheeventsofBagII,andBagIIIrespectivelyandaballis\\ drawnfromit.\\ \therefore P\left(E_1\right)=\frac{1}{6},P\left(E_2\right)=\frac{2}{6}andP\left(E_3\right)=\frac{3}{6}\\ \left(i\right)LetEbetheeventthatredballisselected\\ \therefore P\left(E\right)=P\left(E_1\right).P\left(E/E_1\right)+P\left(E_2\right).P\left(E/E_2\right)+P\left(E_3\right).P\left(E/E_3\right)\\ =\frac{1}{6}.\frac{3}{3}+\frac{2}{6}.\frac{2}{3}+\frac{3}{6}.0=\frac{3}{18}+\frac{4}{18}=\frac{7}{18}\\ \left(ii\right)LetFbetheeventthatwhiteballisselected\\ \therefore P\left(F\right)=1-P\left(E\right)\left[P\left(E\right)+P\left(F\right)=1\right]\\ =1-\frac{7}{18}=\frac{11}{18}\\ Hence,therequiredprobabilitiesare\frac{7}{18}and\frac{11}{18}.\end{array}$

Consider, $I={\displaystyle {\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)}dx$

$\begin{array}{l}\Rightarrow I={\displaystyle {\int }_0^{1}ta{n}^{-1}\left(\frac{x-\left(1-x\right)}{1+x\left(1-x\right)}\right)}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}x-ta{n}^{-1}\left(1-x\right)\right]}dx-----(1)\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(1-1+x\right)\right]}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}x\right]}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(x\right)\right]}dx-----(2)\end{array}$

Adding (1) and (2), we get

$\begin{array}{l}\Rightarrow 2I={\displaystyle {\int }_0^1\left[ta{n}^{-1}x-ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}x\right]}dx\\ \Rightarrow 2I=0\\ \Rightarrow I=0\end{array}$

Thus, the correct option is B.

Giver, f(a + bx) = f(x). _________ (1)

Let I $={\displaystyle {\int }_a^{b}x}+(x)={\displaystyle {\int }_a^b(}a+b-x)f(a+b-x)dx\_\_\_\_\_\left(2\right)$

$\{?{\displaystyle {\int }_a^{b}f}(x)dx={\displaystyle {\int }_a^{b}f}(a+b-x)dx.$

$={\displaystyle {\int }_a^b(}a+b-x)f(x)dx.$ ___________ (3) {because Equation (1)}

$+={\displaystyle {\int }_a^b[(a+b-x)f(x)+xf(x)]}dx$

$\Rightarrow \mathrm{2I}={\displaystyle {\int }_a^b(}a+b-x+x)\stackrel{}{f}(x)dx$

$\Rightarrow \mathrm{2I}={\displaystyle {\int }_a^b(}a+b)f(x)dx$

$\Rightarrow =\frac{a+b}{2}{\displaystyle {\int }_a^{b}f}(x)dx.$

therefore, Option D is correct.

Let $=\int \frac{cos2xdx}{{(sinx+cosx)}^2}$

$=\int \frac{co{s}^{2}x-si{n}^{2}x}{{(sinx+cosx)}^2}dx\{?cos2x=co{s}^{2}x-si{n}^{2}x$

$=\int \frac{(cosx+sinx)(cosx-sinx)}{{(sinx+cosx)}^2}dx\{?a^2-b^2=(x+b)(x-b)$

$=\int \frac{(cosx-sinx)}{sinx+cosx.}dx$

$=log|sinx+cosx|+c\{{\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=log|f(x)|+c}$

So, option B is correct.

Let $=\int \frac{dx}{e^x+e^{-x}}$

$=\int \frac{dx}{e^x+\frac{1}{e^x.}}=\int \frac{e^{x}dx}{e^x·e^x+1}$

$=\int \frac{e^{x}dx}{e^{2x}+1}$

Putting e^x = t

e^xdx = dt.

$\therefore =\int \frac{dt}{t^2+1.}$

= tan^{- 1} t + c

= tan^{- 1} (e^x) + c

therefore, Option A is correct.

Let $I={\displaystyle {\int }_0^1{e}^{2-3x}dx}$

We know that,

${\displaystyle {\int }_a^{b}f\left(x\right)}dx=\left(b-a\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(a\right)+f\left(a+h\right)+...+f\left(a\left(n-1\right)h\right)\right]$

Where, $h=\frac{b-a}{n}$

Here, $a=0,b=1$ and $f\left(x\right)e^{2-3x}$

$\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$

$\therefore {\displaystyle {\int }_0^1{e}^{2-3x}dx}=\left(1-0\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(0\right)+f\left(0+h\right)+...+f\left(0+\left(n-1\right)h\right)\right]$

$\begin{array}{l}=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2+e^{2-3x}+...+e^{2-3\left(n-1\right)h}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{1+e^{-3h}+e^{-6h}+e^{-9h}+...+e^{-3\left(n-1\right)h}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-{\left(e^{-3h}\right)}^n}{1-\left(e^{-3h}\right)}\right\}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-e^{\frac{-3}{n}n}}{1-e^{\frac{-3}{n}}}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\end{array}$

$\begin{array}{l}=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\left(-\frac{1}{3}\right)\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]=\frac{e^2\left(e^{-3}-1\right)}{3}\underset{n\to \infty }{lim}\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]\\ =\frac{-e^2\left(e^{-3}-1\right)}{3}(1)\\ =\frac{-e^{-1}+e^2}{3}\\ =\frac{1}{3}\left(e^2-\frac{1}{e}\right)\end{array}$

Kindly go through the solution

Let I ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}·}2ta{n}^{3}xdx$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}ta{n}^2}xtanxdx$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left(se{c}^{2}x-1\right)}tanxdx\left\{?se{c}^{2}x=ta{n}^{2}x+1\right\}$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}se{c}^2}xtanxdx-2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}tan}xdx$

$=\mathrm{2I}{}_1+2[log]{(cosx)}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

$=\mathrm{2I}{}_1+2\left[log\left(cos\frac{\pi }{4}\right)-log(cos0)\right]$

$=\mathrm{2I}{}_1+2(lo\mathrm{g1/\surd 2}$
$\frac{\mathrm{}}{}-log1)$

$=\mathrm{2I}{}_1+2(log·2^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0)$

$I=\mathrm{2I}{}_1+2*\left(\frac{-1}{2}\right)log2=\mathrm{2I}{}_1-log2.\_\_\_\_(1).$

Where I${}_1={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}se{c}^2}xtanxdx$

Let tan x = t =>sec²xdx = dt

When, x = 0, t = tan 0. = 0

$x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.,t=tan\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.=1$

${}_1={\displaystyle {\int }_0^{1}t}dt={\left[\frac{t^2}{2}\right]}_0^1=\frac{1}{2}-0=\frac{1}{2}$

So, Equation (1) becomes,

$=2*\frac{1}{2}-log2$

=1 - log 2.

LHS = I$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}si{n}^3}xdx.\{sin\mathrm{3A}=3si\mathrm{nA}-4si{n}^{\mathrm{3\; A}}$

$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{4}}(3sinx-sin3x)dx.si{n}^{\mathrm{3A}}=\frac{1}{4}(3si\mathrm{nA}-sin\mathrm{3A})$

$=\frac{1}{4}\left[3{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}sin}xdx-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$q$}\right.}sin}3xdx\right]$

$=\frac{1}{4}\left\{3{[-cosx]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left[\frac{-cos3x}{3}\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}$

$=\frac{3}{4}(-cos\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+cos0)-\frac{1}{12}\left(-cos\frac{3\pi }{2}+cos3*0\right)$

$=\frac{3}{4}(-0+1)-\frac{1}{12}(-0+1)$

$=\frac{3}{4}-\frac{1}{12}=\frac{9-1}{12}=\frac{8}{12}=\frac{2}{3}$