Ncert Solutions Maths class 12th

$e^{2x}-4=0or6e^{2x}-5e^x+1=0$

$\Rightarrow x=\mathcal{l}n2x=\mathcal{l}n\left(\frac{1}{3}\right),\mathcal{l}n\left(\frac{1}{2}\right)$

= $-\mathcal{l}n3,-\mathcal{l}n2$

p                          q                        &nb

  $\left(1,1\right)\left(1,4\right)\left(4,1\right)\left(2,4\right)\left(4,2\right)\left(3,4\right)\left(4,3\right)\left(4,4\right)$  all have only one image.

(2, 1) (1, 2), (2, 2) each element has 3 choice.

(3, 2) (2, 3) (3, 1) (1, 3) (3, 3) each element has two choices.

$\therefore$ total function = 3 × 3 × 2 × 2 × 2 = 72

Case I

None of the pre image have 3 as image, total functions = 2 × 2 × 1 × 1 × 1 = 4

Case II

None o

cota = 1        & secβ =   $\frac{-5}{3}$

< < $\frac{}{}$       $\frac{3\pi }{2}$  secβ =   $\frac{-5}{3}$

$\alpha =\left(\pi +\frac{\pi }{4}\right)$  cosβ = $\frac{-3}{5}=cos\left(180-53\right)$  

tanb =  $\frac{-4}{3}$  

tan(α + β) =   $\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }$

$\therefore A-\frac{1}{4}\&4thquadrant.$

$=\frac{1-\frac{4}{3}}{1+\frac{4}{3}}=\frac{-1}{7}$

$\therefore -\frac{1}{4}\&4thquadrant.$

    &n

No of one – one functions ® ⁵P₄ = 120

f(a) + 2f(b) – f(c) = f(d)  $\left\{1,2,3,4,5\right\}$                   

2f(b) = f(d) + f(c) – f(a)

So, f(d) + f(c) – f(a) should be even.

Only possibilities of        f(d)    &n

ax + by + cz = d

2a + 3b – 5c = d               2a + 3b – 5c = d …………….(v)

Putting (i) & (iv) in (v) we get a = -9d.

In conditions

$\left(2\widehat{i}+\widehat{j}-5\widehat{k}\right),\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right)=0$

2b = d ………….(i) d > 0

2a + b – 5c = 0           

$a=\left|\frac{4.2-3\left(-1\right)-21}{5}\right|$

$=\left|\frac{8+3-21}{5}\right|=2$

$\therefore L=4a=8$

$\overrightarrow{a}=\alpha \widehat{i}+2\widehat{j}-\widehat{k}\&\overrightarrow{b}=-2\widehat{i}+\alpha \widehat{j}+\widehat{k}$

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt[]{15\left(a^2+4\right)}$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2=?$

$\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|cos\theta$

$cos\theta =\frac{-1}{\left(a^2+5\right)}$ $\therefore \left|sin\theta \right|=\frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}$

$\left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|\stackrel{?}{a}\right|\times \left|\overrightarrow{b}\right|\times \left|sin\theta \right|$

$=\left(a^5+5\right)\times \frac{\sqrt[]{{\left(a^2+5\right)}^2-1}}{\left(a^2+5\right)}=\sqrt[]{15\left(a^2+4\right)}$

${\left(a^2+5\right)}^2-1=15\left(a^2+4\right)$

$\Rightarrow \left(\alpha =\pm 3\right)$

$2{\left|\overrightarrow{a}\right|}^2+\left(\overrightarrow{a}.\overrightarrow{b}\right){\left|\overrightarrow{b}\right|}^2$

= $2\left(a^2+5\right)-\left(a^2+5\right)$

$\left(a^2+5\right)=14$

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

$e=\sqrt[]{1+\frac{b^2}{a^2}}$ $e\text{'}=\sqrt[]{1+\frac{a^2}{b^2}}$

$\mathcal{l}=\frac{2b^2}{a}$ $\mathcal{l}\text{'}=\frac{2a^2}{b}$

$\left(1+\frac{b^2}{a^2}\right)=\frac{11}{14}\times \frac{2b^2}{a}$ $\left(1+\frac{a^2}{b^2}\right)=\frac{11}{8}\times \frac{2a^2}{b}$

$7\times \left\{{\left(\frac{7b}{4}\right)}^2+b^2\right\}=11\times b^2\times \frac{7b}{4}\Rightarrow a\times 77=65$

65b² = 44b³

65 = b × 44

$\therefore 77a+44b=65\times 2=130$