Class 11th

33. Given, R= { (a, b): a, b $\in$ N and a = b²}

(i) Let a = 2 $\in$ N

Then b = 2² = 4 $\in$ N

but a ≠ b.

Hence the given statement is not true.

(ii) For a=b² the inverse b=a² may not hold true

Example (4,2) $\in$ R, a=4, b=2 and a=b²

but (2,4) $\notin$ R.

Hence, the given statement is not true.

(iii) If (a, b) $\in$ R

a=b²…… (1)

and (b, c) $\in$ R

b=c²……. (2)

so for (1) and (2),

a= (c²)²=c⁴.

is, a ≠c²,

Hence, (a, c) $\notin$ R.

? The given statement is false.

32. Given, f(x) = (ax + b)

= {(1,1),(2,3),(0, – 1),(–1, –3)} .

As (1,1) $\in$ f.

Then, f(1)=1   [? f(x) = y for (x, y)]

a * 1+b=1

a+b=1…… (1)

and (0, – 1) $\in$ f .

Then, f(0)= –1

a* 0+b= –1

b= –1…….(2)

Putting value of (2) in (1) we gets

a – 1=1

a=1+1

a=2

So, (a, b)=(2, –1)

31. Given, f(x) = x+1. and g(x) = 2x – 3.

So, (f +g)(x) = f(x)+g(x) = (x+1)+(2x – 3) = x+1+2x – 3 = 3x – 2

(f – g)(x) = f(x) –g(x) = (x+1)–(2x–3) = x+1 – 2x+3 = 4 – x

$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2x-3} such that x\ne \frac{3}{2}$

68. The given eqⁿ of line is

l1: x + y = 4

Let R divides the line joining two points P (?1,1) and Q (5,7) in ratio k:1. Then,

Co-ordinate of R = ($\left(\frac{5k-1}{k+1},\frac{7k+1}{k+1}\right)$)

As l₁ divides line joining PQ, then R lies on l₁

i e,  $\left(\frac{5k-1}{k+1},\frac{7k+1}{k+1}\right)$=4

5k ?1 + 7k + 1= 4 (k + 1)

12k = 4k + 4

8k = 4

k = $\frac{1}{2}$

The ratio in which x + y = 4 divides line joining (?1,1) ad (5,7) is $\frac{1}{2}$ :1 i.e., 1: 2.

30. Given, f (x)= $\left\{\left(x,\frac{x^2}{1+x^2}\right):x\in R\right\}$

We know that, for x $\in$ R.

So,  x²≥ 0 ⇒ $\frac{x^2}{x^2+1}\ge \frac{0}{x^2+1}\Rightarrow \frac{x^2}{x^2+1}\ge 0\Rightarrow f(x)\ge 0$

and x²+1>x²

⇒ $1>\frac{x^2}{x^2+1}$

⇒1 > f (x).

So, 0 ≤ f (x) < 1

∴ Range of f (x) = [0,1).

67. The given eqⁿ of line is.

l₁ : y = mx + c.

Slope of l₁ = m

Let m? be the slope of line passing through origin (0, 0) and making angle θ with l₁

Thus, (y 0) = m? (x 0)

y = m? x

m? =  $\mathrm{y}$
$\frac{\mathrm{}\mathrm{x}}{}$______ (1)

And tanθ = $\left|\frac{m^{\prime }-m}{1+m^{\prime }·m}\right|$ = $\frac{m^{\prime }-m}{1+m^{\prime }m}$

When, tanθ = $\frac{m^{\prime }-m}{1+m^{\prime }m}.$

tanθ + m? m tanθ = m' - m

m + tanθ = m? - m?m tanθ

m' = $\frac{m+tan\theta }{1-mtan\theta }.$

When tan θ = $-\left(\frac{m^{\prime }-m}{1+m^{\prime }m}\right)$

tan θ + m? m tanθ = -m? + m

m' = $\frac{m-tan\theta }{1+mtan\theta }.$

Hence combining the two we get,

$m^{\prime }=\frac{m\pm tan\theta }{1?mtan\theta }.$

$\Rightarrow \frac{y}{x}=\frac{m\pm tan\theta }{1?mtan\theta }.$ {-: eqⁿ (1) }

66. The given eqn of the line is.

4x + 7y – 3 = 0 _____ (1)

2x – 3y + 1 = 0 _______ (2)

Solving (1) and (2) using eqn (1) 2 x eqn (2) we get,

(4x + 7y – 3) 2 [ (2x – 3y + 1)] = 0

4x + 7y – 3 – 4x + 6y – 2 = 0

13y = 5

y = $\frac{5}{13}$

And 2x – 3 $\left(\frac{5}{13}\right)$ + 1 = 0

2x = $\frac{15}{13}$ – 1 = $\frac{2}{13}$

$\Rightarrow x=\frac{1}{13}$

Point of intersection of (1) and (2) is $\left(\frac{1}{13},\frac{5}{13}\right)$

Since, the line passing through $\left(\frac{1}{13},\frac{5}{13}\right)$ has equal intercept say c then it is of the form

$\frac{x}{c}+\frac{y}{c}=1.$

x + y = c

$\Rightarrow \frac{1}{13}+\frac{5}{13}=c$

c = $\frac{6}{13}$

the read eqn of line is x + y = $\frac{6}{13}$

13x + 13y – 6 = 0

65. x – 2y = 3

y = $\frac{\mathrm{x}}{2}$ - $\frac{3}{2}$______ (1)

Slope of line (1) is $\frac{1}{2}\cdot$

Let the line through P (3, 2) have slope m

Then, angle between the line = $\left|\frac{m-\frac{1}{2}}{1+m\frac{1}{2}}\right|$

$\Rightarrow tan45^{°}=\left|\frac{2m-1}{2+m}\right|$

$\Rightarrow 1=\left|\frac{2m-1}{2+m}\right|$

$\Rightarrow \frac{2m-1}{2+m}=\pm 1.$

When,  $\frac{2m-1}{2+m}=1$ =>2m – 1 = 2 + m=> m = 3.

The eqn of line through (3, 2) is

y – 2 = 3 (x – 3) 3x – y – 7 = 0.

When $\frac{2m-1}{2+m}$ = – 1=> 2m – 1 = – 2 – m =>3m = – 1 m = $-\frac{1}{3}$

The equation of line through (3,2) is,

y – 2 = $-\frac{1}{3}$ (x – 3) => 3y – 6 = – X + 3

x + 3y – 9 = 0

64. The given eqⁿ of the three lines are

y = m₁ x + c₁ ______ (1)

y = m₂ x + c₂ ______ (2)

y = m₃ x + c₃ ______ (3)

The point of intersection of (2) and (3) is given by.

y - y = (m₂x + c₂) - (m₃ x + c₃)

(m₂ - m₃) x = c₃ - c₂

$\Rightarrow x=\frac{c_3-c_2}{m_2-m_3}.$

Hence, y = $\frac{m_2\left(c_3-c_2\right)}{\left(m_2-m_3\right)}+c_2$

$=\frac{m_2\left(c_3-c_2\right)+c_2\left(m_2-m_3\right)}{m_2-m_3.}$

$=\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}.$

$ie,\left(\frac{c_3-c_2}{m_2-m_3},\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}\right)$

As the three lines are concurrent, the point of intersection of (2) and (3) lies on line (1) also

i e, $\frac{m_2{c}_3-m_3{c}_2}{m_2-m_3}=m_1\left(\frac{c_3-c_2}{m_2-m_3}\right)+c_1$

$\Rightarrow \frac{-m_1\left(c_2-c_3\right)+c_1\left(m_2-m_3\right)}{m_2-m_3}=\frac{-m_2{c}_3-m_3{c}_2}{m_2-m_3}.$

m₁ (c₂ - c₃) - c₁ (m₂ - m₃) + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) - m₂ c₁ + m₃ c₁ + m₂ c₃ - m₃ c₂ = 0

m₁ (c₂ - c₃) + m₂ (c₃ - c₁) + m₃ (c₁ - c₂) = 0

63. The given eqⁿ of the lines are.

3x + y - 2 = 0 _____ (1)

Px + 2y - 3 = 0 ______ (2)

2x - y - 3 = 0 _____ (3)

Point of intersection of (1) and (3) is given by,

(3x + y - 2) + (2x - y - 3) = 0

=> 5x - 5 = 0

=> x = $\frac{5}{5}$

=> x = 1

So, y = 2 - 3x = 2 -3 (1) = 2 - 3 = 1.

i e, (x, y) = (1, -1).

As the three lines interests at a single point, (1, -1) should line on line (2)

i e, P * 1 + 2 * (-1)- 3 = 0

P - 2 - 3 = 0

P = 5