Ncert Solutions Maths class 12th

2. Given, f (x) = 2x2 1

At x = 3

Lim f (x) = $\frac{dy}{dx}=\frac{-\left(3x^2+2xy+y^2\right)}{\left(x^2+2xy+3y^2\right).}$ 2 (3)2 1 = 18 1 = 17.

So, f is continuous at x = 3.

We know,

$\overrightarrow{a}.\overrightarrow{a}=0$ and $\overrightarrow{a}.\overrightarrow{b}=0$

Now,

$\overrightarrow{a}.\overrightarrow{a}=0\Rightarrow {\left|\overrightarrow{a}\right|}^2\Rightarrow \left|\overrightarrow{a}\right|=0$

$\therefore$ $\overrightarrow{a}$ is a zero vector.

Thus, vector $\overrightarrow{b}$ satisfying $\overrightarrow{a}.\overrightarrow{b}=0$ can be any vector.

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

$\begin{array}{l}=\left|\overrightarrow{a}\right|\overrightarrow{b}.\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{a}\right|\overrightarrow{b}.\left|\overrightarrow{b}\right|\overrightarrow{a}+\left|\overrightarrow{b}\right|\overrightarrow{a}.\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}.\left|\overrightarrow{b}\right|\overrightarrow{a}\\ ={\left|\overrightarrow{a}\right|}^2\overrightarrow{b}.\overrightarrow{b}-{\left|\overrightarrow{b}\right|}^2\overrightarrow{a}.\overrightarrow{a}\\ ={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{b}\right|}^2{\left|\overrightarrow{a}\right|}^2\\ =0\end{array}$

$\therefore$ Therefore, $\left|\overrightarrow{a}\right|\overrightarrow{b}+\left|\overrightarrow{b}\right|\overrightarrow{a}$ and $\left|\overrightarrow{a}\right|\overrightarrow{b}-\left|\overrightarrow{b}\right|\overrightarrow{a}$ are perpendicular.

Given,

$\begin{array}{l}\overrightarrow{a}=2\widehat{i}+2\widehat{j}+3\widehat{k}\\ \overrightarrow{b}=-\widehat{i}+2\widehat{j}+\widehat{k}\\ \overrightarrow{c}=3\widehat{i}+\widehat{j}\end{array}$

Now,

$\begin{array}{l}\overrightarrow{a}+\lambda \overrightarrow{b}=\left(2\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\lambda \left(-\widehat{i}+2\widehat{j}+\widehat{k}\right)\\ =\left(2\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\left(-\lambda \widehat{i}+2\lambda \widehat{j}+\lambda \widehat{k}\right)\\ =\left(2-\lambda \right)\widehat{i}+\left(2+2\lambda \right)\widehat{j}+\left(3+\lambda \right)\widehat{k}\end{array}$

If $\left(\overrightarrow{a}+\lambda \overrightarrow{b}\right)$ is perpendicular to $\overrightarrow{c}$ , then $\left(\overrightarrow{a}+\lambda \overrightarrow{b}\right).\overrightarrow{c}=0$

$\begin{array}{l}=\left[\left(2-\lambda \right)\widehat{i}+\left(2+2\lambda \right)\widehat{j}+\left(3+\lambda \right)\widehat{k}\right].\left(3\widehat{i}+\widehat{j}\right)\\ =3\left(2-\lambda \right)+1\left(2+2\lambda \right)+0\left(3+\lambda \right)\\ =6-3\lambda +2+2\lambda +0\\ =8-\lambda \\ \Rightarrow \lambda =8\end{array}$

Therefore, the required value of $\lambda$ is 8.

$\begin{array}{l}\left(\overrightarrow{x}-\overrightarrow{a}\right).\left(\overrightarrow{x}-\overrightarrow{a}\right)=12\\ \overrightarrow{x}.\overrightarrow{x}+\overrightarrow{x}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{x}-\overrightarrow{a}.\overrightarrow{a}=12\\ {\left|\overrightarrow{x}\right|}^2-{\left|\overrightarrow{a}\right|}^2=12\\ {\left|\overrightarrow{x}\right|}^2-1=12\left[\left|\overrightarrow{a}\right|=1asa\overrightarrow{a}isunitvector\right]\\ {\left|\overrightarrow{x}\right|}^2=12+1=13\\ \therefore \left|\overrightarrow{x}\right|=\surd 13\end{array}$

Let $\theta$ be the angle between the vectors $\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ .

It is given that $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|,\overrightarrow{a}.\overrightarrow{b}=\frac{1}{2}and\theta =60^{?}----(1)$

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

$\begin{array}{l}\therefore \frac{1}{2}=\left|\overrightarrow{a}\right|\left|\overrightarrow{a}\right|cos60^{?}(using(1))\\ \frac{1}{2}={\left|\overrightarrow{a}\right|}^2\times \frac{1}{2}\\ {\left|\overrightarrow{a}\right|}^2=1\\ \left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1\end{array}$

$\therefore$ Magnitude of two vector=1

1. Given, f (x) = 5x 3

At x = 0,  $\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}$ 5x 3 = 5 0 3 = 3.

So f is continuous at x = 1.

At x = 3,  $\pi +h$ 5x 3 = 5 ( 3) 3 = 15 3

= 18.

So f is continuous at x = 3.

At x = 5,  $\forall x\in ?$ .5x 3 = 5.5 3 = 25 3 = 22.

So, f is continuous at x = 5.

$\left(3\overrightarrow{a}-5\overrightarrow{b}\right).\left(2\overrightarrow{a}+7\overrightarrow{b}\right).$

$\begin{array}{l}=3\overrightarrow{a}.2\overrightarrow{a}+3\overrightarrow{a}.7\overrightarrow{b}-5\overrightarrow{b}.2\overrightarrow{a}-5\overrightarrow{b}.7\overrightarrow{b}\\ =6\overrightarrow{a}.\overrightarrow{a}+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{a}.\overrightarrow{b}-35\overrightarrow{b}.\overrightarrow{b}\\ =6{\left|\overrightarrow{a}\right|}^2+21\overrightarrow{a}.\overrightarrow{b}-10\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2\\ =6{\left|\overrightarrow{a}\right|}^2+11\overrightarrow{a}.\overrightarrow{b}-35{\left|\overrightarrow{b}\right|}^2\end{array}$

$\left|\overrightarrow{a}\right|$ and $\left|\overrightarrow{b}\right|$ ,if $\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8$ and $\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|$

$\begin{array}{l}\left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8\\ and\left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|\\ \left(\overrightarrow{a}+\overrightarrow{b}\right).\left(\overrightarrow{a}-\overrightarrow{b}\right)=8\\ \overrightarrow{a}.\overrightarrow{a}-\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}-\overrightarrow{b}.\overrightarrow{b}=8\\ {\left|\overrightarrow{a}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8\\ {\left(8\left|\overrightarrow{b}\right|\right)}^2-{\left|\overrightarrow{b}\right|}^2=8\\ 64{\left|\overrightarrow{b}\right|}^2-{\left|\overrightarrow{b}\right|}^2=8\\ 63{\left|\overrightarrow{b}\right|}^2=8\\ \left|\overrightarrow{b}\right|=\surd 8/\surd 63\left(\therefore magnitudeofavectorisnon-negative\right)\\ \left|\overrightarrow{b}\right|=\frac{\mathrm{2\surd 2}}{\mathrm{3\surd 7}}\\ And\\ \left|\overrightarrow{a}\right|=8\left|\overrightarrow{b}\right|=\frac{2\times \mathrm{2\surd 2}}{\mathrm{3\surd 7}}=\frac{1\mathrm{6\surd 2}}{\mathrm{3\surd 7}}\end{array}$

Here, each of the given three vector is a unit vector.

$\begin{array}{l}\overrightarrow{a}.\overrightarrow{b}=\frac{2}{7}\times \frac{3}{7}+\frac{3}{7}\times \left(\frac{-6}{7}\right)+\frac{6}{7}\times \frac{2}{7}\\ =\frac{6}{49}+\left(\frac{-18}{49}\right)+\frac{12}{49}=\frac{6-18+12}{49}=0\\ \overrightarrow{b}.\overrightarrow{c}=\frac{3}{7}\times \frac{6}{7}+\left(\frac{-6}{7}\right)\times \frac{2}{7}+\frac{2}{7}\times \left(-\frac{3}{7}\right)\\ =\frac{18}{49}-\frac{12}{49}+\left(\frac{-6}{49}\right)=\frac{18-12-6}{49}=0\\ \overrightarrow{c}.\overrightarrow{a}=\frac{6}{7}\times \frac{2}{7}+\frac{2}{7}\times \frac{3}{7}+\left(-\frac{3}{7}\right)\times \frac{6}{7}\\ =\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0\end{array}$

Therefore, the given three vectors are mutually perpendicular to each other.