Maths

6. From the given data we tabulate the following.

We have,

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,

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

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

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