If g(x) = x2 + x - 1 and
(goƒ) (x) = 4x2 - 10x + 5, then ƒ$\left( {{5 \over 4}} \right)$ is equal to:
Solution
Given, (goƒ) (x) = 4x<sup>2</sup> - 10x + 5
<br><br>$\Rightarrow$ g(f(x)) = 4x<sup>2</sup> - 10x + 5
<br><br>$\therefore$ g(f(${5 \over 4}$)) = $4 \times {{25} \over {16}} - {{50} \over 4} + 5$ = $- {5 \over 4}$ ...(1)
<br><br>Also given, g(x) = x<sup>2</sup> + x - 1
<br><br>$\therefore$ g(f(x)) = f<sup>2</sup>(x) + f(x) –1
<br><br>$\Rightarrow$ g(f(${5 \over 4}$)) = f<sup>2</sup>(${5 \over 4}$) + f(${5 \over 4}$) –1 ....(2)
<br><br>from (1) & (2)
<br><br>f<sup>2</sup>(${5 \over 4}$) + f(${5 \over 4}$) –1 = $- {5 \over 4}$
<br><br>$\Rightarrow$ f<sup>2</sup>(${5 \over 4}$) + f(${5 \over 4}$) + ${1 \over 4}$ = 0
<br><br>$\Rightarrow$ (f(${5 \over 4}$) + ${1 \over 2}$)<sup>2</sup> = 0
<br><br>$\Rightarrow$ f(${5 \over 4}$) = -${1 \over 2}$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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