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JEE Main 20204 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
4 September
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2020 · 4 September · Morning Shift · Mathematics · Functions

Suppose a differentiable function f (x) satisfies the identity f(x...

Suppose a differentiable function ff\left(x\right)satisfiestheidentitysatisfies the identityf\left(x + y\right) = f\left(x\right) + f\left(y\right) + x{y}^{2} + {x}^{2}y,forallrealfor all realxandandy.If. If\lim_{x \to 0}\dfrac{f\left(x\right)}{x} = 1,thenthen{f}^{'}\left(3\right) is equal to :

Correct answer

0

Step-by-step solution

∵lim⁡x→0f(x)x=1⇒f′(0)=1∵ \lim_{x \to 0}\dfrac{f\left(x\right)}{x} = 1 ⇒ f ' \left(0\right) = 1

f(x+y)=f(x)+f(y)+xy2+x2yf\left(x + y\right) = f\left(x\right) + f\left(y\right) + x{y}^{2} + {x}^{2}y

Differentiate w.r.t. xx keeping yy constant

f′(x+y)=f′(x)+0+y2+2xyf ' \left(x + y\right) = f ' \left(x\right) + 0 + {y}^{2} + 2xy

put y=−xy = - x

f′(0)=f′(x)+x2−2x2f ' \left(0\right) = f ' \left(x\right) + {x}^{2} - 2{x}^{2}

1=f′(x)−x21 = f ' \left(x\right) - {x}^{2}

f′(x)=1+x2f ' \left(x\right) = 1 + {x}^{2}

f′(3)=10f ' \left(3\right) = 10.

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