If fx-y-fx-kxy-2y2 - x-y - and f1-2-f2-6- then which of the following is-are true -

F(x+y)=f(x)+kxy 2y^2 Put, x=y=1 ⇒ nbsp;f(2)=f(1)+k 2 ⇒ 6=2+k 2 ⇒ k=6 ∴ f(x+y)=f(x)+6xy 2y^2 Now, f'(x) =lim_h → 0 f(x+h) f(x)h =lim_h → 0 f(x)+6xh 2h^2 f(x)h ...

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Byju's Answer

Standard XII

Mathematics

First Principle of Differentiation

If fx+y=fx+kx...

Question

If $f (x + y) = f (x) + k x y - 2 y^{2} \forall x, y \in R$ and $f (1) = 2, f (2) = 6,$ then which of the following is/are true :

k = 6

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f^{'} (3) = 18

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f^{''} (100) = 600

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f^{'''} (100) = 0

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Solution

The correct option is D $f^{'''} (100) = 0$
$f (x + y) = f (x) + k x y - 2 y^{2}$
Put, $x = y = 1$
$\Rightarrow f (2) = f (1) + k - 2$
$\Rightarrow 6 = 2 + k - 2 \Rightarrow k = 6$
$∴ f (x + y) = f (x) + 6 x y - 2 y^{2}$
Now,
$f^{'} (x) = lim h \to 0 \frac{f (x + h) - f (x)}{h}$
$= lim h \to 0 \frac{f (x) + 6 x h - 2 h^{2} - f (x)}{h}$
$\Rightarrow f^{'} (x) = lim h \to 0 (6 x - 2 h) = 6 x$
So, $f^{'} (x) = 6 x \Rightarrow f^{'} (3) = 18$
$\Rightarrow f^{''} (x) = 6 \Rightarrow f^{''} (100) = 6$
$\Rightarrow f^{'''} (x) = 0 \Rightarrow f^{'''} (100) = 0$

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