Let f be a functional defined for all x- R- If f is differentiable and fx3-x5 for all x- Rx- 0- then the value of f-27 is-

The correct option is A 15 Given that, f(x^3)=x^5, ∀ xϵ R(x≠ 0) On differentiating w.r.t. x , we get f'(x^3)· 3x^2=5x^4 ⇒ f'(x^3)=5/3x^ ...

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Byju's Answer

Standard XII

Mathematics

Monotonicity in an Interval

Let f be a ...

Question

Let $f$ be a functional defined for all $x ϵ R$ . If $f$ is differentiable and $f (x^{3}) = x^{5}$ for all $x ϵ R (x \neq 0)$ , then the value of $f^{'} (27)$ is.

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Solution

The correct option is A $15$
Given that, $f (x^{3}) = x^{5}, \forall x ϵ R (x \neq 0)$
On differentiating w.r.t. $x$ , we get
$f^{'} (x^{3}) \cdot 3 x^{2} = 5 x^{4}$
$\Rightarrow f^{'} (x^{3}) = \frac{5}{3} x^{2}$
$∴ f^{'} (27) = f^{'} (3^{3}) = \frac{5}{3} (3)^{2} = 15$

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Let f be a functional defined for all xϵR. If f is differentiable and f(x3)=x5 for all xϵR(x≠0), then the value of f′(27) is.

The correct option is A 15Given that, f(x3)=x5,∀xϵR(x≠0)On differentiating w.r.t. x, we getf′(x3)⋅3x2=5x4⇒f′(x3)=53x2∴f′(27)=f′(33)=53(3)2=15

Let $f$ be a functional defined for all $x ϵ R$ . If $f$ is differentiable and $f (x^{3}) = x^{5}$ for all $x ϵ R (x \neq 0)$ , then the value of $f^{'} (27)$ is.

The correct option is A $15$
Given that, $f (x^{3}) = x^{5}, \forall x ϵ R (x \neq 0)$
On differentiating w.r.t. $x$ , we get
$f^{'} (x^{3}) \cdot 3 x^{2} = 5 x^{4}$
$\Rightarrow f^{'} (x^{3}) = \frac{5}{3} x^{2}$
$∴ f^{'} (27) = f^{'} (3^{3}) = \frac{5}{3} (3)^{2} = 15$