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Question: Let f \(\left( \frac{x + y}{2} \right)\)= \(\frac{f(x) + f(y)}{2}\)for real values of x and y. If f...

Let f (x+y2)\left( \frac{x + y}{2} \right)= f(x)+f(y)2\frac{f(x) + f(y)}{2}for real values of x and y.

If f ′(0) exist and equals –1 and f(0) = 1, then f ′(2) is equal to-

A

–1

B

Non-existent

C

12\frac{1}{2}

D

12

Answer

–1

Explanation

Solution

We have,

f (x+y2)\left( \frac{x + y}{2} \right)= f(x)+f(y)2\frac{f(x) + f(y)}{2} for all x, y, ∈ R …(i)

f (x2)\left( \frac{x}{2} \right)= f(1)+12\frac{f(1) + 1}{2} for all x, y ∈ R

⇒\Rightarrowf(x) = 2f (x2)\left( \frac{x}{2} \right)– 1 for all x, y ∈ R .....(ii)

Now, f '(2) = lim⁡h→0f(2+h)−f(2)h\lim_{h \rightarrow 0}\frac{f(2 + h) - f(2)}{h}

⇒\Rightarrowf ′(2) = lim⁡h→0\lim _ { h \rightarrow 0 } f(4+2h2)−f(2)h\frac{f\left( \frac{4 + 2h}{2} \right) - f(2)}{h}

⇒\Rightarrowf ′(2) = lim⁡h→0\lim _ { h \rightarrow 0 } 2f(2)−1+2f(h)−1−2f(2)2h\frac{2f(2) - 1 + 2f(h) - 1 - 2f(2)}{2h}

[Putting x = 4 and 2h respectively in Eq. (ii)]

⇒ f ′(2) = lim⁡h→0\lim _ { h \rightarrow 0 } f(h)−1h\frac{f(h) - 1}{h}

f ′(2) = lim⁡h→0\lim _ { h \rightarrow 0 } f(h)−f(0)h\frac{f(h) - f(0)}{h}= f '(0) = –1