The function f - R - R satisfiesf x 2 f - x - f - x f - x 2 - x - R- given that f 1-1 and f -1-8- thenA- f -1-2B- f -1-4C- f -1-4D- f -1-2

The correct options are A f'(1) = 2 B f”(1) = 4Let f'(1) = a, f”(1) = b Put x = 1 ⇒ f(1)f”(1) = f'(1)f'(1) b = a^2 differentiating again nbsp; f'(x^2) . 2xf”( ...

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Standard XII

Mathematics

Integrating Factor

The function ...

Question

The function $f : R \to R$ satisfies
$f (x^{2}) f^{''} (x) = f^{'} (x) f^{'} (x^{2}) \forall x \in R,$ given that $f (1) = 1$ and $f^{'''} (1) = 8$ , then

f^{'} (1) = 2

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f^{''} (1) = 4

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f^{'} (1) = 4

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f^{''} (1) = 2

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Solution

The correct options are
A $f^{'} (1) = 2$
B $f^{''} (1) = 4$
Let $f^{'} (1) = a, f^{''} (1) = b$
Put $x = 1 \Rightarrow f (1) f^{''} (1) = f^{'} (1) f^{'} (1)$
$b = a^{2}$
differentiating again
$f^{'} (x^{2}) . 2 x f^{''} (x) + f (x^{2}) f^{'''} (x) = f^{'} (x) f^{''} (x^{2}) . 2 x + f^{'} (x^{2}) . f^{''} (x)$
Put $x = 1$
$2 a b + 8 = a b + 2 a b \Rightarrow a b = 8$
So, $a = 2, b = 4$

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The function f:R→R satisfies f(x2)f′′(x)=f′(x) f′(x2) ∀ x ∈ R, given that f(1)=1 and f′′′(1)=8, then

The correct options are A f′(1)=2 B f′′(1)=4Let f′(1)=a,f′′(1)=b Put x=1⇒f(1)f′′(1)=f′(1)f′(1) b=a2 differentiating again f′(x2).2xf′′(x)+f(x2)f′′′(x)=f′(x)f′′(x2).2x+f′(x2).f′′(x) Put x=1 2ab+8=ab+2ab⇒ab=8 So, a=2,b=4

The function $f : R \to R$ satisfies
$f (x^{2}) f^{''} (x) = f^{'} (x) f^{'} (x^{2}) \forall x \in R,$ given that $f (1) = 1$ and $f^{'''} (1) = 8$ , then