Find:Β f (β-3) Answer: f (β-3) β-3 is not a real numberΒ [Function f(x) is defined only when x β R] β΄ f (β-3) do not exist.
If
Find: (i) f (1) (ii) f (β3) Answers: (i) If x β₯ 1, f (x) = 1/x f (1) = 1/1 β΄Β f(1) = 1 (ii) β3 = 1.732 > 1 If x β₯ 1, f (x) = 1/x β΄ f (β3) = 1/β3...
If
Find: (i) f (1/2) (ii)Β f (-2) Answers: (i) If 0 β€ x β€ 1, f(x) = x β΄ f (1/2) = Β½ (ii) If x < 0, f(x) = x2 f (β2) = (β2)2 β΄Β f (β2) = 4
If f (x) = (x + 1) / (x β 1), show that f [f (x)] = x.
Answer: f [f (x)] = f [(x+1)/(x-1)] f [f (x)] = [(x+1)/(x-1) + 1] / [(x+1)/(x-1) β 1] f [f (x)] = [[(x+1) + (x-1)]/(x-1)] / [[(x+1) β (x-1)]/(x-1)] f [f (x)] = [(x+1) + (x-1)] / [(x+1) β (x-1)] f [f...
If y = f (x) = (ax β b) / (bx β a), show that x = f (y).
Answer: Consider, y = (ax β b) / (bx β a) [Cross multiply] y(bx β a) = ax β b bxy β ay = ax β b bxy β ax = ay β b x(by β a) = ay β b x = (ay β b) / (by β a) = f (y) β΄Β x = f (y) Thus,...
If f (x) = (x β a)2 (x β b)2, find f (a + b).
Answer: f (a + b) = (a + b β a)2Β (a + b β b)2 f (a + b) = (b)2Β (a)2 β΄Β f (a + b) = a2b2
If f (x) = x2 β 3x + 4, then find the values of x satisfying the equation f (x) = f (2x + 1).
Answer: f(x) = x2Β β 3x + 4. Consider, f (x) = f (2x + 1). Let us take, f (2x + 1) = (2x + 1)2Β β 3(2x + 1) + 4 f (2x + 1) = (2x)Β 2Β + 2(2x) (1) + 12Β β 6x β 3 + 4 f (2x + 1) = 4x2Β + 4x + 1 β 6x + 1 f...
For the given system of equation show graphically that equation has infinitely many solution: $aβ2b+11=0$ $3a+6b+33=0$
Given, $aβ2b+11=0$β¦β¦. (i) $3aβ6b+33=0$β¦β¦. (ii) From equation (i), β $b=(a+11)/2$ When $a=-1$, we get $b=(-1+11)/2=5$. When $a=-3$, we get $b=(-3+11)/2=4$. Thus, we have the following table giving...