RD Chapter 3 Functions Ex 3.2 Solutions
Question - 1 : - If f (x) = x2 тАУ 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).
Answer - 1 : -
Given:
f(x) = x2┬атАУ 3x + 4.
Let us find x satisfying f (x) = f (2x + 1).
We have,
f (2x + 1) = (2x + 1)2┬атАУ3(2x + 1) + 4
= (2x)┬а2┬а+2(2x) (1) + 12┬атАУ 6x тАУ 3 +4
= 4x2┬а+ 4x + 1 тАУ 6x+ 1
= 4x2┬атАУ 2x + 2
Now, f (x) = f (2x + 1)
x2┬атАУ 3x + 4 = 4x2┬атАУ 2x + 2
4x2┬атАУ 2x + 2 тАУ x2┬а+ 3x тАУ 4 = 0
3x2┬а+ x тАУ 2 = 0
3x2┬а+ 3x тАУ 2x тАУ 2 =0
3x(x + 1) тАУ 2(x + 1) = 0
(x + 1)(3x тАУ 2) = 0
x + 1 = 0 or 3x тАУ 2 = 0
x = тАУ1 or 3x = 2
x = тАУ1 or┬а2/3
тИ┤The values of x are тАУ1 and 2/3.
Question - 2 : - If f (x) = (x тАУ a)2 (x тАУ b)2, find f (a + b).
Answer - 2 : -
F (x) = (x тАУ a)2(x тАУ b)2
Let us find f (a + b).
We have,
f (a + b) = (a + b тАУ a)2┬а(a+ b тАУ b)2
f (a + b) = (b)2┬а(a)2
тИ┤┬аf(a + b) = a2b2
Question - 3 : - If y = f (x) = (ax тАУ b) / (bx тАУ a), show that x = f (y).
Answer - 3 : -
Given:
y = f (x) = (ax тАУ b) / (bx тАУ a) тЗТ f (y) = (ay тАУ b) / (by тАУ a)
Let us prove that x = f (y).
We have,
y = (ax тАУ b) / (bx тАУ a)
By cross-multiplying,
y(bx тАУ a) = ax тАУ b
bxy тАУ ay = ax тАУ b
bxy тАУ ax = ay тАУ b
x(by тАУ a) = ay тАУ b
x = (ay тАУ b) / (by тАУ a) = f (y)
тИ┤┬аx= f (y)
Hence proved.
Question - 4 : - If f (x) = 1 / (1 тАУ x), show that f [f {f (x)}] = x.
Answer - 4 : -
Given:
f (x) = 1 / (1 тАУ x)
Let us prove that f [f {f (x)}] = x.
Firstly, let us solve for f {f (x)}.
f {f (x)} = f {1/(1 тАУ x)}
= 1 / 1 тАУ (1/(1 тАУ x))
= 1 / [(1 тАУ x тАУ 1)/(1 тАУ x)]
= 1 / (-x/(1 тАУ x))
= (1 тАУ x) / -x
= (x тАУ 1) / x
тИ┤ f {f (x)} = (x тАУ 1) / x
Now, we shall solve for f [f {f (x)}]
f [f {f (x)}] = f [(x-1)/x]
= 1 / [1 тАУ (x-1)/x]
= 1 / [(x тАУ (x-1))/x]
= 1 / [(x тАУ x + 1)/x]
= 1 / (1/x)
тИ┤ f [f {f (x)}] = x
Hence proved.
Question - 5 : - If f (x) = (x + 1) / (x тАУ 1), show that f [f (x)] = x.
Answer - 5 : -
Given:
f (x) = (x + 1) / (x тАУ 1)
Let us prove that f [f (x)] = x.
f [f (x)] = f [(x+1)/(x-1)]
= [(x+1)/(x-1) + 1] / [(x+1)/(x-1) тАУ 1]
= [[(x+1) + (x-1)]/(x-1)] / [[(x+1) тАУ (x-1)]/(x-1)]
= [(x+1) + (x-1)] / [(x+1) тАУ (x-1)]
= (x+1+x-1)/(x+1-x+1)
= 2x/2
= x
тИ┤ f [f (x)] = x
Hence proved.
Question - 6 : - If
┬а
Find:
(i) f (1/2)
(ii) f (-2)
(iii) f (1)
(iv) f (тИЪ3)
(v) f (тИЪ-3)
Answer - 6 : -
(i)┬аf(1/2)
When, 0 тЙд x тЙд 1, f(x) = x
тИ┤f (1/2) = ┬╜
(ii)┬аf(-2)
When, x < 0, f(x) = x2
f (тАУ2) = (тАУ2)2
= 4
тИ┤┬аf(тАУ2) = 4
(iii)┬аf(1)
When, x тЙе 1, f (x) = 1/x
f (1) = 1/1
тИ┤┬аf(1)= 1
(iv)┬аf(тИЪ3)
We have тИЪ3 = 1.732 > 1
When, x тЙе 1, f (x) = 1/x
тИ┤f (тИЪ3) = 1/тИЪ3
(v)┬аf(тИЪ-3)
We know┬атИЪ-3┬аis not a real number and the function f(x) isdefined only when x┬атИИ┬аR.
тИ┤┬аf(тИЪ-3)┬аdoes not exist.