Ncert Solutions Maths class 12th

35. Let f (x) = cos (sin x).

f' (x) $\frac{d}{dx}$ cos (sin x)

= - sin (sin x) $\frac{d}{dx}$ sin x

= - sin (sin x) cos x.

$\begin{array}{l}\overrightarrow{r}=\left(1-t\right)\widehat{i}+\left(t-2\right)\widehat{j}+\left(3-2t\right)\widehat{k}\\ \overrightarrow{r}=\widehat{i}-t\widehat{i}+t\widehat{j}-2\widehat{j}+3\widehat{k}-2t\widehat{k}\\ \overrightarrow{r}=\widehat{i}-2\widehat{j}+3\widehat{k}+t\left(-\widehat{i}+\widehat{j}-2\widehat{k}\right)-----(1)\end{array}$

$\begin{array}{l}\Rightarrow \overrightarrow{r}=\left(s+1\right)\widehat{i}+\left(2s-1\right)\widehat{j}-\left(2s+1\right)\widehat{k}\\ \overrightarrow{r}=s\widehat{i}+\widehat{i}+2s\widehat{j}-\widehat{j}-2s\widehat{k}-\widehat{k}\\ \overrightarrow{r}=\widehat{i}-\widehat{j}-\widehat{k}+s\left(\widehat{i}+2\widehat{j}-2\widehat{k}\right)-----(2)\end{array}$

Here,   

Hence, the shortage distance between the line is given by

$d=\left|\frac{\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right).\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}\right|=\frac{8}{}$

$\begin{array}{l}\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\lambda \left(\widehat{i}-3\widehat{j}+2\widehat{k}\right)and-----(1)\\ \overrightarrow{r}=4\widehat{i}+5\widehat{j}+6\widehat{k}+\mu \left(2\widehat{i}+3\widehat{j}+\widehat{k}\right)-----(2)\end{array}$

Here, comparing (1) and (2) with $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ , we have

$\begin{array}{l}{a}_1=\widehat{i}+2\widehat{j}+3\widehat{k},b_1=\widehat{i}-3\widehat{j}+2\widehat{k}\\ a_2=4\widehat{i}+5\widehat{j}+6\widehat{k},b_2=2\widehat{i}+3\widehat{j}+\widehat{k}\end{array}$

Therefore,

              $\begin{array}{l}{\overrightarrow{a}}_2-{\overrightarrow{a}}_1=3\widehat{i}+3\widehat{j}+3\widehat{k}\\ {\overrightarrow{b}}_1*{\overrightarrow{b}}_2=\\ =\widehat{i}\left(-3-6\right)-\widehat{j}\left(1-4\right)+\widehat{k}\left(3+6\right)\\ =-9\widehat{i}+3\widehat{j}+9\widehat{k}\\ \left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|==\\ =\\ =3\end{array}$

34. Let f (x) = sin (x2 + 5)

Differentiating w. r t. x we get,

f'¢ (x) = $\frac{d}{dx}$sin (x² + 5)

= cos (x² + 5)  $\frac{d}{dx}$ = cos (x² + 5) 

= cos (x² + 5) [2x].

= 2x cos (x² + 5).

$\begin{array}{l}\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\\ \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\end{array}$

Shortest distance between two lines is given by,

$\begin{array}{l}{x}_1=-1,y_1=-1,z_1=-1\\ a_1=7,b_1=-6,c_1=1\\ x_2=3,y_2=5,z_2=7\\ a_2=1,b_2=-2,c_2=1\end{array}$

Then,

$\begin{array}{l}=4\left(-6+2\right)-6\left(7-1\right)+8\left(-14+6\right)\\ =-16-36-64\\ =-116\end{array}$

$\begin{array}{l}\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}+\widehat{k}\right)+\lambda \left(\widehat{i}-\widehat{j}+\widehat{k}\right)and-----(1)\\ \overrightarrow{r}=2\widehat{i}-\widehat{j}-\widehat{k}+\mu \left(2\widehat{i}+\widehat{j}+2\widehat{k}\right)-----(2)\end{array}$

Solution. Comparing (1) and (2) with $\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$ respectively.

We get,

$\begin{array}{l}{a}_1=\widehat{i}+2\widehat{j}+\widehat{k},b_1=\widehat{i}-\widehat{j}+\widehat{k}\\ a_2=2\widehat{i}-\widehat{j}-\widehat{k},b_2=2\widehat{i}+\widehat{j}+2\widehat{k}\end{array}$

Therefore,

$\begin{array}{l}{\overrightarrow{a}}_2-{\overrightarrow{a}}_1=\widehat{i}-3\widehat{j}-2\widehat{k}\\ {\overrightarrow{b}}_1*{\overrightarrow{b}}_2=\left(\widehat{i}-\widehat{j}+\widehat{k}\right)*\left(2\widehat{i}+\widehat{j}+2\widehat{k}\right)\end{array}$

$\begin{array}{l}=\left(2-1\right)\widehat{i}-\left(2-2\right)\widehat{j}+\left(1+2\right)\widehat{k}\\ =-3\widehat{i}+3\widehat{k}\\ \left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|====3\\ \left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right).\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)=\left(-3\widehat{i}+3\widehat{k}\right)\left(\widehat{i}-3\widehat{j}-2\widehat{k}\right)\\ =-3-6\\ =-9\end{array}$

Hence, the shortest distance between the given line is given by

$\begin{array}{l}d=\left|\frac{\left({\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right).\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)}{\left|{\overrightarrow{b}}_1*{\overrightarrow{b}}_2\right|}\right|\\ =\left|\frac{-9}{3}\right|\\ =\frac{3}{}=\frac{3}{2}\end{array}$

$\begin{array}{l}\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\\ \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\end{array}$

Direction ratios of given lines are (7, -5,1) and (1,2,3).

i.e.,  $\begin{array}{l}{a}_1=7,b_1=-5,c_1=1\\ a_2=1,b_2=2,c_2=3\end{array}$

Now,

$\begin{array}{l}=a_1{a}_2+b_1{b}_2+c_1{c}_2\\ =7*1+\left(-5\right)*2+1*3\\ =7-10+3\\ =10-10\\ =0\end{array}$

$\therefore$ These two lines are perpendicular to each other.

33. Let g (x) = $x$ is continuous being a modules f x and h (x) = $x$ is also continuous being a modules $\forall x\in ?$

Then, f (x) = g (x) h (x).is also continuous for all x. E. R.

Hence, there is no point of discontinuous for f (x).

$\frac{1-x}{3}=\frac{7y-14}{2\rho }=\frac{z-3}{2}and\frac{7-7x}{3\rho }=\frac{y-5}{1}=\frac{6-z}{5}$

The standard form of a pair of Cartesian lines is;

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}and\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}-----(1)$

So,

$\begin{array}{l}\frac{-\left(x-1\right)}{3}=\frac{7\left(y-5\right)}{2\rho }=\frac{z-3}{2}and\frac{-7\left(x-1\right)}{3\rho }=\frac{y-5}{1}=\frac{-\left(z-6\right)}{5}\\ \frac{x-1}{-3}=\frac{y-2}{2\rho /7}=\frac{z-3}{2}and\frac{x-1}{-3\rho /7}=\frac{y-5}{1}=\frac{z-6}{-5}-----(2)\end{array}$

Comparing (1) and (2) we get

$\begin{array}{l}{a}_1=-3,b_1=\frac{2\rho }{7},c_1=2\\ a_2=\frac{-3\rho }{7},b_2=1,c_2=-5\end{array}$

Now, both the lines are at right angles

So, $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\begin{array}{l}\Rightarrow \left(-3\right)*\frac{\left(-3\rho \right)}{7}+\frac{2\rho }{7}*1+2*\left(-5\right)=0\\ \Rightarrow \frac{9\rho }{7}+\frac{2\rho }{7}+\left(-10\right)=0\\ \Rightarrow \frac{9\rho +2\rho }{7}=10\\ \Rightarrow 11\rho =70\\ \rho =\frac{70}{11}\end{array}$

$\therefore$ The value of $\rho$ is $\frac{70}{11}$

(i) Let  ${\overrightarrow{b}}_1$  and  ${\overrightarrow{b}}_2$  be the vectors parallel to the pair of lines,  $\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\&\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$ , respectively.

$\begin{array}{l}{\overrightarrow{b}}_1=2\widehat{i}+5\widehat{j}-3\widehat{k}\&{\overrightarrow{b}}_2=-\widehat{i}+8\widehat{j}+4\widehat{k}\\ \therefore \left|{\overrightarrow{b}}_1\right|==\\ \left|{\overrightarrow{b}}_2\right|===9\\ {\overrightarrow{b}}_1.{\overrightarrow{b}}_2=\left(2\widehat{i}+5\widehat{j}-3\widehat{k}\right).\left(-\widehat{i}+8\widehat{j}+4\widehat{k}\right)\\ =2(-1)+5*8+(-3).4\\ =-2+40-12\\ =26\end{array}$