Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If 5f(x)+3f ⁣(1x)=x+25f(x)+3f\!\left(\dfrac{1}{x}\right)=x+2 and y=xf(x)y=xf(x), then dydx\dfrac{dy}{dx} at x=1x=1 is
A1414
B7/87/8correct
C11
DNone
Solution
Step 1: Write the given equation:
5f(x)+3f ⁣(1x)=x+2…(1)5f(x)+3f\!\left(\dfrac{1}{x}\right)=x+2\quad\ldots(1)
 Step 2: Replace xx by 1x\dfrac{1}{x}:
5f ⁣(1x)+3f(x)=1x+2…(2)5f\!\left(\dfrac{1}{x}\right)+3f(x)=\dfrac{1}{x}+2\quad\ldots(2)
 Step 3: Compute 5×(1)−3×(2)5\times(1)-3\times(2):
16f(x)=5(x+2)−3 ⁣(1x+2)=5x+10−3x−6=5x−3x+416f(x)=5(x+2)-3\!\left(\dfrac{1}{x}+2\right)=5x+10-\dfrac{3}{x}-6=5x-\dfrac{3}{x}+4
f(x)=116 ⁣(5x−3x+4)f(x)=\dfrac{1}{16}\!\left(5x-\dfrac{3}{x}+4\right)
 Step 4: Compute y=xf(x)y=xf(x):
y=x⋅116 ⁣(5x−3x+4)=116(5x2+4x−3)y=x\cdot\dfrac{1}{16}\!\left(5x-\dfrac{3}{x}+4\right)=\dfrac{1}{16}(5x^{2}+4x-3)
 Step 5: Differentiate and evaluate at x=1x=1:
dydx=10x+416  ⇒  dydx∣x=1=10+416=1416=78\dfrac{dy}{dx}=\dfrac{10x+4}{16}\;\Rightarrow\;\dfrac{dy}{dx}\bigg|_{x=1}=\dfrac{10+4}{16}=\dfrac{14}{16}=\dfrac{7}{8}
 Correct answer: (2)
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