If gx - x2 - x - 1 and g - fx - 4x2 - 10x - 5- then f54 is equal to

G(x) = x^2 + x 1 (g ∘ f )(x) = 4x^2 10x + 5 ⇒ g(f(x)) = 4x^2 10x + 5 ⇒ f^2 (x) + f(x) 1 = 4x^2 10x + 5 nbsp; Putting x = 54 and f (54) = t t^2 + t + ...

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Byju's Answer

Standard XII

Mathematics

Constant Function

If gx = x2 + ...

Question

If $g (x) = x^{2} + x - 1$ and $(g \circ f) (x) = 4 x^{2} - 10 x + 5,$ then $f (\frac{5}{4})$ is equal to

- \frac{3}{2}

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- \frac{1}{2}

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\frac{1}{2}

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\frac{3}{2}

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Solution

The correct option is B $- \frac{1}{2}$
$g (x) = x^{2} + x - 1$
$(g \circ f) (x) = 4 x^{2} - 10 x + 5$
$\Rightarrow g (f (x)) = 4 x^{2} - 10 x + 5$
$\Rightarrow f^{2} (x) + f (x) - 1 = 4 x^{2} - 10 x + 5$
Putting $x = \frac{5}{4}$ and $f (\frac{5}{4}) = t$
$t^{2} + t + \frac{1}{4} = 0$
$\Rightarrow t = - \frac{1}{2}$ or $f (\frac{5}{4}) = - \frac{1}{2}$

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Similar questions

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f^{''} (x) = - f (x)

and

g (x) = f^{'} (x) .

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and

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f " (x) = - f (x)

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and given that

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F (x) = {(f (\frac{x}{2}))}^{2} + {(g (\frac{x}{2}))}^{2}, F (5) = 5

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

The correct option is B −12g(x)=x2+x−1 (g∘f)(x)=4x2−10x+5 ⇒g(f(x))=4x2−10x+5 ⇒f2(x)+f(x)−1=4x2−10x+5 Putting x=54 and f(54)=t t2+t+14=0 ⇒t=−12 or f(54)=−12

If $g (x) = x^{2} + x - 1$ and $(g \circ f) (x) = 4 x^{2} - 10 x + 5,$ then $f (\frac{5}{4})$ is equal to