If g(x)=x2+x−1 and (gof)(x)=4x2−10x+5, then f(54) is equal to

g(x)=x2+x−1 and gof(x)=4x2−10x+5∴ g(f(x))=4x2−10x+5put x=54⇒ g(f(54))=4(54)2−10(54)+5⇒ f2(54)+f(54)−1=254−504+=25−50+204=−54⇒ f2(54)+f(54)−1+54=0⇒ f2(54)+f(54)+14=0⇒ (f(54)+12)2=0⇒ f(54)=−12

If g(x)=x2+x−1 and (gof)(x)=4x2−10x+5, then f(54) is equal to

g(x)=x2+x−1 and gof(x)=4x2−10x+5∴ g(f(x))=4x2−10x+5put x=54⇒ g(f(54))=4(54)2−10(54)+5⇒ f2(54)+f(54)−1=254−504+=25−50+204=−54⇒ f2(54)+f(54)−1+54=0⇒ f2(54)+f(54)+14=0⇒ (f(54)+12)2=0⇒ f(54)=−12

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$I f g (x) = x^{2} + x - 1 a n d (g o f) (x) = 4 x^{2} - 10 x + 5, t h e n f (\frac{5}{4}) i s e q u a l t o$

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$- \frac{3}{2}$

$\frac{1}{2}$

$- \frac{1}{2}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} g (x) = x^{2} + x - 1 a n d g o f (x) = 4 x^{2} - 10 x + 5 \\ ∴ g (f (x)) = 4 x^{2} - 10 x + 5 \\ p u t x = \frac{5}{4} \\ \Rightarrow g (f (\frac{5}{4})) = 4 {(\frac{5}{4})}^{2} - 10 (\frac{5}{4}) + 5 \\ \Rightarrow f^{2} (\frac{5}{4}) + f (\frac{5}{4}) - 1 = \frac{25}{4} - \frac{50}{4} + \\ = \frac{25 - 50 + 20}{4} \\ = \frac{- 5}{4} \end{array}$

$\begin{array}{l} \Rightarrow f^{2} (\frac{5}{4}) + f (\frac{5}{4}) - 1 + \frac{5}{4} = 0 \\ \Rightarrow f^{2} (\frac{5}{4}) + f (\frac{5}{4}) + \frac{1}{4} = 0 \\ \Rightarrow {(f (\frac{5}{4}) + \frac{1}{2})}^{2} = 0 \\ \Rightarrow f (\frac{5}{4}) = \frac{- 1}{2} \end{array}$