Class 11th

5. From the given data we have,

4. Arranging the given data in ascending order we get,

36,42,45,46,46,49,51,53,60,72

As n = 10 (even)

$=\frac{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\text{\hspace{0.17em}thObservation}}^{}+{(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1)thObservation}^{}}{2}$

$\Rightarrow =\frac{{\mathrm{5thobservation}}^{}+{\mathrm{6thobservation}}^{}}{2}$

$=\frac{46+49}{2}$

$=\frac{95}{2}$

= 47.5

$\therefore$ M.D. (M) $=\frac{1}{n}*{\displaystyle \sum _{i=1}^n\left|x_i-M\right|}$

$=\frac{1}{10}*\left(11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5\right)$

$=\frac{70}{10}=7.$

3. Arranging the data in ascending order we get,

10,11,1112,13,13,14,16,16,17,17,18

As n=12, even

So, median is the mean of ${(\frac{M}{2})th}^{}$ and ${(\frac{M}{2}+1)th}^{}$ observation.

$\Rightarrow =\frac{{\mathrm{6thobservation}}^{}+{\mathrm{7thobservation}}^{}}{2}$

$M=\frac{13+14}{2}=\frac{27}{2}=13.5.$

So, deviation of respective observation about the median. $M,\left|x_i-M\right|$ are

Therefore the mean deviation about the mean is

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

$=\frac{1}{12}*(3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5)$

$=\frac{28}{12}=2.33.$

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}$

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

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

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