Method of DifferentiationmediumFree

Method of Differentiation — JEE Maths practice question

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

Question
A function f:R→Rf:\mathbb{R}\to\mathbb{R} satisfies sin⁡xcos⁡y (f(2x+2y)−f(2x−2y))=cos⁡xsin⁡y (f(2x+2y)+f(2x−2y))\sin x\cos y\,(f(2x+2y)-f(2x-2y)) = \cos x\sin y\,(f(2x+2y)+f(2x-2y)). If f′(0)=12f'(0) = \dfrac{1}{2}, then the value of 4f′′(x)+f(x)4f''(x)+f(x) is.
Solution
Answer: 0
Step 1: Simplify the functional equation Rearranging:
f(2x+2y)sin⁡(x−y)=f(2x−2y)sin⁡(x+y)f(2x+2y)\sin(x-y) = f(2x-2y)\sin(x+y)
f(2x+2y)f(2x−2y)=sin⁡(x+y)sin⁡(x−y)\frac{f(2x+2y)}{f(2x-2y)} = \frac{\sin(x+y)}{\sin(x-y)}
 This identifies f(x)=ksin⁡x2f(x) = k\sin\dfrac{x}{2}. Step 2: Apply f′(0)=12f'(0) = \dfrac{1}{2}
f′(x)=k2cos⁡x2  ⟹  f′(0)=k2=12  ⟹  k=1f'(x) = \frac{k}{2}\cos\frac{x}{2} \implies f'(0) = \frac{k}{2} = \frac{1}{2} \implies k = 1
 So f(x)=sin⁡x2f(x) = \sin\dfrac{x}{2}. Step 3: Evaluate 4f′′(x)+f(x)4f''(x)+f(x)
f′′(x)=−14sin⁡x2f''(x) = -\frac{1}{4}\sin\frac{x}{2}
4f′′(x)+f(x)=−sin⁡x2+sin⁡x2=04f''(x)+f(x) = -\sin\frac{x}{2}+\sin\frac{x}{2} = 0
 Answer: 0
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Method of Differentiation · easy
If y= 1+x 2 +x 4 1+x+x 2 and dy dx =ax+b , then the values of a and b are
Method of Differentiation · easy
If f(x) = | x| , then f' ( 3 4 ) is:
Method of Differentiation · easy
If f(x)= x , then f'( x) equals
Method of Differentiation · easy
d dx [ x + x 1 + 2x ] , where 0 < x < 4 , is:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath