Ncert Solutions Maths class 11th

5. i. The distance between points (2, 3, 5) and (4, 3, 1) is

4. i. The x-axis and y-axis taken together determine a plane known as XY plane.

ii. The coordinates of points in the XY-plane are of the form (x, y, 0).

iii. Coordinate planes divide the space into 8 (eight) octants.

3. 

(1, 2, 3) lies in octant I.

(4, –2, 3) lies in octant IV.

(4, –2, –5) lies in octant VIII.

(4, 2, –5) lies in octant V.

(–4, 2, –5) lies in octant VI.

(–4, 2, 5) lies in octant II.

(–3, –1, 6) lies in octant III.

(–2, –4, –7) lies in octant VII.

2. When a point lies on XZ-plane, its y-coordinate is zero.

1. When a point lies on x-axis, its y-coordinate and z-coordinate are zero.

(c) Given, f(x)=(x-a)(x-b)

where a and b are constants.

So,

$f^{\prime }(x)=(x-a)\frac{d}{dx}(x-b)+(x-b)\frac{d}{dx}(x-a)$

=(x-a)+(x-b)

= 2x– a- b. $$

(ii) Given f(x)= ${\left(a{x}^2+b\right)}^2.$ where ab are constant

So, $f^{\prime }(x)=\frac{d}{dx}{\left(a{x}^2+b\right)}^2$

$=\frac{d}{dx}\left(a^2{x}^4+b^2+2a{x}^{2}b\right)$

$=\frac{d}{dx}{a}^2{x}^4+\frac{d}{dx}{b}^2+\frac{d}{dx}2a{x}^{2}b$

= $4a^2{x}^3+0+4axb$

= $4ax\left(a{x}^2+b\right).$

(iii) Given, f(x)= $\frac{x-9}{x-b}$ where a and bare constants

So, $f(x)=\frac{(x-b)\frac{d}{dx}(x-a)-(x-a)\frac{d}{dx}(x-b)}{{(x-b)}^2}$

$=\frac{(x-b)-(x-a)}{{(x-b)}^2}$

$=\frac{a-b}{{(x-b)}^2}$

38. Given, $f(x)=x^n+a{x}^{n-1}+a^2{x}^{n-2}+?+a^{n-1}x+a^n.$

We know that,

$\frac{d}{dx}\left(x^x\right)=n{x}^{x-1}$

So,

$f^{\prime }(x)=\frac{d}{dx}{x}^x+\frac{d}{dx}a{x}^{x-1}+\frac{d}{dx}{a}^2{x}^{x-2}+?+\frac{d}{dx}$ $a^{x-1}x+\frac{d}{dx}{a}^x=n{x}^{n-1}+a(n-1)x^{n-2}+a^2(n-2)x^{n-3}+?+a^{n-1}+0.$

$\begin{array}{l}(?\frac{dax}{dx}=a\frac{dx}{dx}\frac{\mathrm{and\; d}a}{dx}=\mathrm{0whereaisconstant}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$

37. 

Given $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+?+\frac{x^2}{2}+x+1$

$\underset{h\to 0}{lim}\frac{{(x+h)}^{100}-x^{100}}{100\cdot h}+\underset{h\to 0}{lim}\frac{{(x+h)}^{99}-x^{99}}{99\cdot h}+\dots +\underset{x\to 0}{lim}\frac{{(x+h)}^2-x^2}{2h}$ $+\underset{x\to 0}{lim}x+\underset{x\to 0}{lim}1$

$=\frac{100\cdot x^{99}}{100}+\frac{99x^{98}}{99}+?+\frac{2x}{2}+0+1$

$=x^{99}+x^{98}+?+x+1$

At x=0,

f(X)  =1.

and at x=1,

$f^{\prime }(1)=1^{99}+1^{98}+\dots +1^2+1+1$

=100 $*$ 1

=100 $*f^{\prime }(0)$

Hence, $f^{\prime }(1)=100f^{\prime }(0)$

41. In the 13 letter word ASSASSINATION there are 3-A, 4-S, 2-I, 2-N, 1-T and 1-O.

Since all the S are to be occurred together we treat them i.e. (SSSS) as single object. This single object together with 13 – 4 = 9 remaining object will account for 10 objects having 3-A, 2-I, 2-N, 1-T and 1-O and can be rearranged in $\frac{10!}{3!2!2!}$

= 10 * 9 * 8 * 7 * 6 * 5

= 151200