Question:

If $f: R \rightarrow R$ is a differentiable function such that $f'(x)>2 f(x)$ for all $x \in R$, and $f(0)=1$, then

Updated On: Jun 14, 2022
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The Correct Option is C

Solution and Explanation

$f'(x)-2 f(x)>0$
$\Rightarrow \frac{d}{d x}\left(f(x) \cdot e^{-2 x}\right)>0$
$\Rightarrow f(x) \cdot e^{-2 x}$ is increasing function
$\Rightarrow f(x) \cdot e^{-2 x}>f(0)=1 \,\,\,\,\forall x>0$
$f(x)>e^{2 x} \forall x>0$
$\Rightarrow f'(x)>2 f(x)>2 e^{2 x}>0 \forall x>0$
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Questions Asked in JEE Advanced exam

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Concepts Used:

Application of Derivatives

Various Applications of Derivatives-

Rate of Change of Quantities:

If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)

This is also known to be as the Average Rate of Change.

Increasing and Decreasing Function:

Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).

  • If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
  • Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
  • The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.

Read More: Application of Derivatives