IPMAT Indore 2024 - QA (SA) Q9 Explanation

Explanation:

(6) f(x) = |x +|x|| and g(x) = 1/x for x ≠ 0.

Given f(a) + g(f(a)) = 13/6

⇒f(a)+ 1/f(a) =13/6

⇒[f(a)]^2 + 1 = f(a)⋅13/6

⇒6[f(a)]^2 +6 = 13⋅f(a)

⇒ let f(a) = t

∴ 6t^2 −13t+6 = 0.

Solving above quadratic equation, we get, t = f(a) = 3/2 or 2/3

Case I: If f(a) = 3/2

∴ |x+|x|| = 3/2

⇒x+|x| =3/2 or−3/2 ---------eqn. (i)

We know |x| = +x for x > 0 & |x| = −x for x < 0,

Taking x < 0 will not lead to any solution of x. So, taking x > 0, and substituting |x| with +x in eqn.(i),

we get x +x= 2x=3/2 or −3/2

We will get, x = 3/4 (valid) & x = −3/4 (invalid)

Case II: If f(a) = 2/3

∴ |x+|x|| = 2/3

⇒ x+|x|=2/3 or −2/3

When x > 0, |x| = +x,

Substituting in eqn.(i) we get, x + x = 2x = 2/3 or = −2/3

x =2/6 = 1/3 (valid) or −2/6 = −1/3 (invalid)

Now, f(g(a)) = |g(a) +|g(a)|| = |1/x + |1/x ||

For x = 3/4 ⇒ f(g(a)) = |4/3 + |4/3 || = 8/3 and for x = 1/3 ⇒ f (g(a)) = |3 +|3||= 6

So, the maximum value of f(g(a)) = 6. Ans