Test Index

Manipal Entrance Test 2013 Math Paper

© examsnet.com

Question : 45 of 80

Marks: +1, -0

\text{Let } f(xy)=f(x)\cdot f(y)

x, y \in R.

f'(1)=2

f(4)=4,

f'(4)

equal to

4
1
$\frac{1}{2}$
8

Solution:

We have f(xy) = f{x) f(y) for all x,y ∈ R

Putting x = y = 1, we get

f(1)=f(1)f(1)

⇒

f(1)[1-f(1)]=0

⇒

f(1)=1\;\;\;\;[\because f(1)\neq 0]

.....(i)

f'(1)=2

⇒

\lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}=2

⇒

f(1)\lim\limits_{h\to 0}\frac{f(h)-1}{h}=2

⇒

\lim\limits_{h\to 0}\frac{f(1)f(h)-f(1)}{h}=2\;\;\;\;[\text{ using }f(1)=1]

⇒

\lim\limits_{h\to 0}\frac{f(h)-1}{h}=2

Now

f'(4)=\lim\limits_{h\to 0}\frac{f(4+h)-f(4)}{h}

=\lim\limits_{h\to 0}\frac{f(4)\cdot f(h)-f(4)}{h}

=\left(\lim\limits_{h\to 0}\frac{f(h)-1}{h}\right)\cdot f(4)=2f(4)

{from (i)}

=2\times 4=8

© examsnet.com

Go to Question:

Prev Question

Next Question