If g(x) = x ^ 2 + x - 1 and (gof)(x) = 4 x ^ 2 - 10x + 5, then f((5)/(4)) is equal to

Correct answer: B. - (1)/(2) g(f(x)) = 4 ((5)/(4)) ^ 2 - 10 . (5)/(4) + 5 = - (5)/(4) g(f((5)/(4))) = f ^ 2 ((5)/(4)) + f((5)/(4)) - 1 - (5)/(4) = f ^ 2 ((5)/(4)) + f((5)/(4)) - 1 f ^ 2 ((5)/(4)) + f((5)/(4)) + (1)/(4) = 0 (f((5)/(4)) + (1)/(2)) ^ 2 = 0 f((5)/(4)) = - (1)/(2)

JEE Main 20207 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 7 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2020 · 7 January · Morning Shift · Mathematics · Functions

If g(x) = x ^ 2 + x - 1 and (gof)(x) = 4 x ^ 2 - 10x + 5, then...

g\left(x\right) = {x}^{2} + x - 1 and \left(gof\right)\left(x\right) = 4{x}^{2} - 10x + 5,

then

f\left(\dfrac{5}{4}\right)

is equal to

Options

A $\dfrac{3}{2}$
B $- \dfrac{1}{2}$
C $\dfrac{1}{2}$
D $- \dfrac{3}{2}$

Correct answer

B. $- \dfrac{1}{2}$

Step-by-step solution

g\left(f\left(x\right)\right) = 4{\left(\dfrac{5}{4}\right)}^{2} - 10 . \dfrac{5}{4} + 5 = - \dfrac{5}{4}

g\left(f\left(\dfrac{5}{4}\right)\right) = {f}^{2}\left(\dfrac{5}{4}\right) + f\left(\dfrac{5}{4}\right) - 1

- \dfrac{5}{4} = {f}^{2}\left(\dfrac{5}{4}\right) + f\left(\dfrac{5}{4}\right) - 1

{f}^{2}\left(\dfrac{5}{4}\right) + f\left(\dfrac{5}{4}\right) + \dfrac{1}{4} = 0

{\left(f\left(\dfrac{5}{4}\right) + \dfrac{1}{2}\right)}^{2} = 0

f\left(\dfrac{5}{4}\right) = - \dfrac{1}{2}

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