Ncert Solutions Maths class 11th

2. Mean of the given observation is.

$\overline{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}$

$=\frac{500}{10}=50.$

So,

Therefore, the required mean deviation about the mean is

$\text{\hspace{0.17em}M.D.}(\overline{x})=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left|x_i-\overline{x}\right|}$

$=\frac{12+20+2+10+8+5+13+4+4+6}{10}$

$=\frac{84}{10}$

= 8.4

1. Mean of the given observation is.

$\overline{x}=\frac{4+7+8+9+10+12+13+17}{8}$

$=\frac{80}{8}=10.$

Deviation of the respective observation about the mean $\overline{x}$ i.e.,  $x_i-\overline{x}$ are 4–10,7–10,8–10,9–10,10–10,12–10,13–10,17–10

=6, -3, -2, -1,0,2,3,7

The absolute value of the deviation i.e.,  $\left|x_i-\overline{x}\right|$ are 6,3,2,1,0,2,3,7.

Therefore, the required mean deviation about the mean is

$\mathrm{M.D}=(\overline{x})=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left|x_i-\overline{x}\right|}=\frac{6+3+2+1+0+2+3+7}{8}$

$=\frac{24}{8}$

= 3.

67. Given, f (x) = (ax² + sin x) (p +q cos x).

So, f? (x) = (ax² + sin x) $\frac{d}{dx}$ $(p+qcosx)+(p+qcosx)\frac{d}{dx}(a{x}^2+sinx)$

$=\left(a{x}^2+sinx\right)\left(0+q\frac{d}{dx}cosx\right)+(p+qcosx)\left(a\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}sinx\right)$

$=\left(a{x}^2+sinx\right)(q(-sinx)+(p+qcosx)(a·2x+cosx)$

= q sin x (ax² + sin x) + (p + q cos x) (2ax + cos x)

66. Given, f (x) = (x² + 1) cos x

f? (x) = (x² + 1) $\frac{d}{dx}cosx+cosx\frac{d}{dx}\left(x^2+1\right)$

$=\left(x^2+1\right)(-sinx)+cosx(2x+0)\left[?\frac{d}{dx}cosx=-sinx\right]$

= x² sin x sin x + 2x cos x.

65. Given, f (x) = x⁴. (5 sin x 3 cos x)

$f^{\prime }(x)=x^4\frac{d}{dx}(5sinx-3cosx)+(5sinx-3cosx)\frac{d{x}^4}{dx}.$

$=x^4\left[5\frac{d}{dx}sinx-3·\frac{dcosx}{dx}\right]+[5sinx-3cosx]·4x^3$

As $\frac{d}{dx}sinx=cosx$

and $\frac{d}{dx}cosx=-sinx$

Thus,

$f^{\prime }(x)=x^4[5cosx+3sinx]+[5sinx-3cosx]·4x^3$

$=x^3[5cosx+3xsinx+20sinx-12cosx]$

$=x^3[5xcosx+3xsinx+20sinx-12cosx].$

64.  Given, f (x) = $\frac{sin(x+a)}{cosx}$

$f^{\prime }(x)=\frac{cosx\frac{d}{dx}sin(x+a)-sin(x+a)\frac{d}{dx}cosx}{co{s}^{2}x}$

Let g?(x) = sin (x + a)

So, g?(x) = $\underset{}{}$$\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

= cos (x + a)

And P(x) = cos x

So, P?(x) = $\underset{h\to 0}{lim}\frac{p(x+h)-p(x)}{h}$

Thus, f?(x) = $\frac{}{}$$\frac{cosx·cos(x+a)-sin(x+a)(-sinx)}{co{s}^{2}x}$

$=\frac{cosx·cos(x+a)+sin(x+a)sinx}{co{s}^{2}x}$

$=\frac{cos(x+a-x)}{co{s}^{2}x}$

$=\frac{cosa}{co{s}^{2}x}$

63. Given, f (x) = $\frac{a+bsinx}{c+dcosx}$

f?(x) = $\frac{(c+dcosx)\frac{d}{dx}(bsinx)-(a+bsinx)\frac{d}{dx}(c+dcosx)}{{(c+dcosx)}^2}$

$f^{\prime }(x)=\frac{(c+dcosx)·b·\frac{d}{dx}(sinx)-(a+bsinx)·d·\frac{d}{dx}(cosx)}{{(c+dcosx)}^2}\_\_\_\_\_(1)$

{Copy (A)}

So, g?(x) = $\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}[cos(x+h)-cosx]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[-2·sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h-x}{2}\right)\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[-2sin\left(\frac{2x+h}{2}\right)sin\left(\frac{h}{2}\right)\right]$

$=-sin\left(\frac{2x+0}{2}\right)*1$

= sin x ______ (2)

And p?(x) = $\underset{h\to 0}{lim}\frac{p(x+h)-p(x)}{h}$

$=\underset{h\to 0}{lim}\frac{sin(x+h)-sinx}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\cdot 2cos\left(\frac{2x+h}{2}\right)sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

$=\underset{h\to 0}{lim}cos\left(\frac{2x+h}{2}\right)*\underset{h\to 0}{lim}\frac{sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$

= cos x _____ (3)

So, put (2) and (3) in (1) we get,

$f^{\prime }(x)=\frac{(c+dcosx)(b·cosx)-(a+bsinx)(-d·sinx)}{{(c+dcosx)}^2}$

$=\frac{beccosx+bdco{s}^{2}x+adsinx+bdsi{n}^{2}x}{{(c+dcosx)}^2}$

$=\frac{bccosx+adsinx+bd\left(co{s}^{2}x+si{n}^{2}x\right)}{{(c+dcosx)}^2}$

$=\frac{bccosx+adsinx+bd}{{(c+dcosx)}^2}$

62. Given, f (x) =sinⁿx

By chain rule,

f? (x) = n (sin x)^n-1 $\frac{d}{dh}$ sin x

Let (gx) = sinx

So, g? (x) $\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{sin(x+h)-sinx}{h}$

$=\underset{h\to 0}{lim}\frac{2}{h}cos\left(\frac{2a+h}{2}\right)sin\left(\frac{h}{2}\right)$

$=\underset{h\to 0}{lim}cos\left(\frac{2x+h}{2}\right)*\underset{h\to 0}{lim}\frac{sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

= $cos\frac{(2x+0)}{2}*1$

= cos x.

So, f? (x) = n (sin x)^n-1 cos x.

61. Given, f (x) = $\frac{secx-1}{secx+1}$

$=\frac{\frac{1}{cosx}-1}{\frac{1}{cosx}+1}$

$=\frac{1-cosx}{1+cosx}$

So, f?(x) = $\frac{(1+cosx)\frac{d}{dx}(1-cosx)-(1-cosx)\frac{d}{dx}(1+cosx)}{{(1+cosx)}^2}$

$=\frac{(1+cosx)(-1)\frac{d}{dx}(-cosx)-(1-cosx)\frac{d}{dx}(cosx)}{{(1+cosx)}^2}$

Let g(x) = cos x.

So, g?(x) $=\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{cos(x+h)-cosx}{h}$

$=\underset{h\to 0}{lim}-\frac{2}{h}sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h-x}{2}\right)$

$=\underset{h\to 0}{lim}\frac{-2}{h}sin\left(\frac{2x+h}{2}\right)sin\left(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

$=\underset{h\to 0}{lim}-sin\left(\frac{2x+h}{2}\right)*\underset{h\to 0}{lim}\frac{sin\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=-sin\left(\frac{2x+0}{2}\right)*1$

= -sin x.

So, f?(x) $\frac{(1+cosx)(-1)(-sinx)-(1-cosx)(-sinx)}{{(1+cosx)}^2}$

$=\frac{sinx+cosxsinx+sinx-sinxcosx}{{(1+cosx)}^2}$

$=\frac{2sinx}{{(1+cosx)}^2}$

$=\frac{2sinx}{{\left(1+\frac{1}{secx}\right)}^2}=\frac{2sinx*se{c}^{2}x}{{(secx+1)}^2}=\frac{2secx·sinx}{{(sinx+1)}^2·cosx}.$

$=\frac{2secx·tanx}{{(sinx+1)}^2}$