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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
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
∴ 6t2 − 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|= 32 or − 32 ........ (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