First slide

Functions (XII)

Question

Consider a differentiable $f : R \to R$ for which f(1) = 2 and $f (x + y) = 2^{x} f (y) + 4^{y} f (x) \forall x, y \in R .$

Moderate

Question

The value of f(4) is

A

160
B

240
C

200
D

none of these

Solution

$\begin{array}{l} f (x + y) = 2^{x} f (y) + 4^{y} f (x) \\ Interchanging x and y, we get \\ \Rightarrow f (x + y) = 2^{y} f (x) + 4^{x} f (y) \\ \Rightarrow \frac{f (x)}{4^{x} - 2^{x}} = \frac{f (y)}{4^{y} - 2^{y}} = k \\ \Rightarrow f (x) = k (4^{x} - 2^{x}) \\ f (1) = 2 \\ \Rightarrow k = 1 . \\ Hence, f (x) = 4^{x} - 2^{x} \\ f (4) = 4^{4} - 2^{4} = 240 \end{array}$

Question

The minimum value of f(x) is

A

1
B

$- \frac{1}{2}$
C

$- \frac{1}{4}$
D

none of these

Solution

$\begin{array}{l} f (x + y) = 2^{x} f (y) + 4^{y} f (x) \\ Interchanging x and y, we get \\ \Rightarrow f (x + y) = 2^{y} f (x) + 4^{x} f (y) \\ \Rightarrow \frac{f (x)}{4^{x} - 2^{x}} = \frac{f (y)}{4^{y} - 2^{y}} = k \\ \Rightarrow f (x) = k (4^{x} - 2^{x}) \\ f (1) = 2 \\ \Rightarrow k = 1 . \\ Hence, f (x) = 4^{x} - 2^{x} \\ = {(2^{x})}^{2} - 2^{x} = {(2^{x} - 1 / 2)}^{2} - 1 / 4 \\ Thus f (x) has least value - 1 / 4 \end{array}$

Question

The number of solutions of f(x) = 2 is

A

0
B

1
C

2
D

infinite

Solution

$\begin{array}{l} f (x + y) = 2^{x} f (y) + 4^{y} f (x) \\ Interchanging x and y, we get \\ \Rightarrow f (x + y) = 2^{y} f (x) + 4^{x} f (y) \\ \Rightarrow \frac{f (x)}{4^{x} - 2^{x}} = \frac{f (y)}{4^{y} - 2^{y}} = k \\ \Rightarrow f (x) = k (4^{x} - 2^{x}) \\ f (1) = 2 \\ \Rightarrow k = 1 . \\ Hence, f (x) = 4^{x} - 2^{x} \\ Also 4^{x} - 2^{x} = 2 \\ \Rightarrow {(2^{x})}^{2} - 2^{x} - 2 = 0 \\ \Rightarrow (2^{x} - 2) (2^{x} + 1) = 0 \\ \Rightarrow 2^{x} - 2 = 0 \\ \Rightarrow x = 1 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App