Maths NCERT Exemplar Solutions Class 12th Chapter Five

$e^{4x}+4e^{3x}-58e^{2x}+4e^x+1=0$

$\left(e^{2x}+\frac{1}{e^{2x}}\right)+4\left(e^x+\frac{1}{e^x}\right)-58=0$

$\Rightarrow {\left(e^x+\frac{1}{e^x}+2\right)}^2=64$

$e^x$$=$$\frac{6\pm \text{exception 25:}}{2}$$=$$3$$\pm$$2$

A = {1, 2, 3, ….50}

R₁ = (2, 1), (2, 2), (2, 4)…. (2, 32)

(3, 1) (3, 3) (3, 9) (3, 27)

(13, 1) (13, 13)   etc.

R₂ = { (2, 1), (2, 2), (3, 1), (3, 3), (5, 1), (5, 5), (7, 1), (7, 7), (1, 1), (11, 11)…}

$\therefore R_1-R_2$ contains only 9 elements.

 $t_n=\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)},n=1,2,3,....,99$

$=\frac{1}{2}\left(\frac{1}{\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+2\right)\left(n+3\right)}\right)=V_n-V_{n+1}$

$=\frac{1}{12}\left(\frac{1716}{101*17}\right)=\frac{143}{101*17}=\frac{k}{101}\Rightarrow k=\frac{143}{17}$

34 k = 286

 $f\left(\alpha \right)={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{lo{g}_{10}t}{1+t}}dt$

$f\left(\frac{1}{\alpha }\right)={\displaystyle \underset{1}{\overset{\frac{1}{\alpha }}{\int }}\frac{lo{g}_{10}t}{1+t}dt={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{-lo{g}_{10}z}{1+\frac{1}{z}}-\frac{1}{z^2}dz}}$

$={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{lo{g}_{10}z}{z\left(z+1\right)}dz}$

$={\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{lo{g}_{10}t}{t}}dt=\frac{1}{\mathcal{l}n10}{\displaystyle \underset{1}{\overset{\alpha }{\int }}\frac{\mathcal{l}nt}{t}dt=\frac{1}{\mathcal{l}n10}{\left[\frac{{\left(\mathcal{l}nt\right)}^2}{2}\right]}_1^{\alpha }=\frac{{\left(\mathcal{l}n\alpha \right)}^2}{2\mathcal{l}n10}}$

$f\left(e^3\right)+f\left(e^{-3}\right)=\frac{{\left(\mathcal{l}n{e}^3\right)}^2}{2\mathcal{l}n10}=\frac{9}{2\mathcal{l}n10}$

 $z^5+{\left(\overline{z}\right)}^5$

$={\left(2+3i\right)}^5+{\left(2-3i\right)}^5$

$=2\left(32-720+810\right)=244$

Since $\left(\mathrm{3,3}\right)$ lies on $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{9}{a^2}-\frac{9}{b^2}=1$

Now, normal at $\left(\mathrm{3,3}\right)$ is $y-3=-\frac{a^2}{b^2}(x-3)$ ,

which passes through $\left(\mathrm{9,0}\right)\Rightarrow b^2=2a^2$

So,  ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=3$

Also,  ${\mathrm{a}}^2=\frac{9}{2}$

(From (i) and (ii)

Thus,  $\left({\mathrm{a}}^2,{\mathrm{e}}^2\right)=\left(\frac{9}{2},3\right)$

$f\left(x\right)=\stackrel{?}{a}\cdot (\stackrel{?}{b}*\stackrel{?}{c})=\left|\begin{array}{ccc}x& -2& 3\\ -2& x& -1\\ 7& -2& x\end{array}\right|$

$=x^3-27x+26$

$f^{\mathrm{\text{'}}}\left(x\right)=3x^2-27=0\Rightarrow x=\pm 3$ and $f^{\mathrm{\text{'}}\mathrm{\text{'}}}(-3)<0$

$\Rightarrow$ local maxima at $\mathrm{x}={\mathrm{x}}_0=-3$

Thus, $\stackrel{?}{a}=-3\hat{i}-2\hat{j}+3\hat{k},\stackrel{?}{b}=2\hat{i}-3\hat{j}-\hat{k}$ , and $\stackrel{?}{c}=7\hat{i}-2\hat{j}-3\hat{k}$

$\Rightarrow \stackrel{?}{a}\cdot \stackrel{?}{b}+\stackrel{?}{b}\cdot \stackrel{?}{c}+\stackrel{?}{c}\cdot \stackrel{?}{a}=9-5-26=-22$

This is Other Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:\frac{{\left(a^2+1\right)}^2}{2a-i}=x+yi\dots \left(i\right)\\ Takingconjugateonbothsides\\ \frac{{\left(a^2+1\right)}^2}{2a+i}=x-yi\dots \left(ii\right)\\ Multiplyingeqn.\left(i\right)and\left(ii\right)wehave\\ \frac{{\left(a^2+1\right)}^2{\left(a^2+1\right)}^2}{\left(2a-i\right)\left(2a+i\right)}=x^2+y^2\\ \Rightarrow \frac{{\left(a^2+1\right)}^4}{4a^2-i^2}=x^2+y^2\\ \Rightarrow \frac{{\left(a^2+1\right)}^4}{4a^2+1}=x^2+y^2\\ Hence,thevalueof{x}^2+y^2=\frac{{\left(a^2+1\right)}^4}{4a^2+1}.\end{array}$