Ncert Solutions Maths class 11th

36. Given, A={9,10,11,12,13}.

f(x)=the highest prime factor of n.

and f: A → N.

Then, f(9)=3 [? prime factor of 9=3]

f (10)=5 [? prime factor of 10=2,5]

f(11)=11 [? prime factor of 11 = 11]

f(12)=3 [? prime factor of 12 = 2, 3]

f(13)=13 [? prime factor of 13 = 13]

?Range of f=set of all image of f(x) = {3,5,11,13}.

35. Given, f={(ab, a+b): a, b $\in$ z}

Let a=1 and b=1;                   a, b $\in$ z.

So, ab=1 * 1=1

a+b=1+1=2.

So, we have the order pair (1,2).

Now, let a= –1 and b= –1; a, b $\in$ z

So, ab=(–1) * (–1)=1

a+b=(–1)+(–1)= –2

So, the ordered pair is (1, –2).

?The element 1 has two image i.e., 2 and –2.

Hence, f is not a function.

34. Given,

A={1,2,3,4}

B={1,5,9,11,15,16}

f={(1,5),(2,9),(3,1),(4,5),(2,11)}.

(i) As every element of f is an element of A * B

We can clearly say that f $\subset$ A * B.

?f is a relation from A to B.

(ii) As the element 2 of the domain has two image i.e., 9 and 11. f is not a function.

69. 

The given eqⁿ of the lines are.

4x + 7y + 5 = 0______ (1)

2x - y = 0 ______ (2)

Solving (1) and (2) we get,

4 x + 7 (2 x)+5 = 0

4x +14 x + 5= 0

x = $\frac{-5}{18}$

and y = 2x = $2\left(\frac{-5}{18}\right)=\frac{-5}{9}$

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) }