JEE MAIN Increasing Decreasing Function Previous Year Questions (PYQs) – Page 3 of 3
Study Stuff for MCA Examinations
Solution
Given quadratic equation has equal roots, thus
$D = 0 \Rightarrow (r(x))^2 = r'(x)\cdot f(x)$
$\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}$
Integrate
$\ln(f'(x)) = \ln(f(x)) + \ln C \Rightarrow f'(x) = C f(x)$
Put $x = 0$
$1 = C \cdot 2 \Rightarrow C = \frac{1}{2}$
Now $2f'(x) = f(x)$
$\Rightarrow \frac{f'(x)}{f(x)} = 2$
Integrate
$\ln(f(x)) = 2x + d$
$\Rightarrow d = 0$
$\Rightarrow \ln(f(x)) = 2x \Rightarrow f(x) = e^{2x}$
Now let $g(x) = f(\ln x - x) = e^{2(\ln x - x)}$
$\therefore g'(x) = 2e^{2(\ln x - x)}\left(\frac{1}{x} - 1\right) \ge 0$
$\Rightarrow \frac{1 - x}{x} \ge 0$
$\Rightarrow x \in (0,1]$
$\Rightarrow \alpha = 0,; \beta = 1$
$\alpha + \beta = 1$
Let $f : R \to R$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in R$ and $f'(a-1) = 0$, where $a$ is real number. Let
$g(x) = f(\tan x - 2\tan x + a), \quad 0 < x < \frac{\pi}{2}$
Consider the following two statements:
(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Then,
Solution
JEE MAIN
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