First slide

Introduction to limits

Question

Let $f (x)$ be a polynomial satisfying $(f (α))^{2} + {(f^{'} (α))}^{2} = 0 .$ Then, $lim_{x \to α} \frac{f (x)}{f^{'} (x)} [\frac{f^{'} (x)}{f (x)}]$ is equal to (Here $[\cdot]$ denotes the greatest integer function)

Moderate

A

0
B

1
C

$- 1$
D

$f (α) f^{'} (α)$

Solution

It is given that the polynomial $f (x)$ satisfies the relation $(f (α))^{2} + {(f^{'} (α))}^{2} = 0 .$
$∴ f (α) = 0 = f^{'} (α)$
$\Rightarrow x = a$ is a root of $f (x)$ and $f' (x)$
$\Rightarrow {(x - a)}^{2}$ is a factor of $f (x)$
Let $f (x) = (x - α)^{2} ϕ (x) .$ Then,
$\begin{aligned} f^{'} (x) = 2 (x - α) ϕ (x) + (x - α)^{2} ϕ^{'} (x) \\ ∴ \frac{f (x)}{f^{'} (x)} = \frac{(x - α) ϕ (x)}{2 ϕ (x) + (x - α) ϕ^{'} (x)} \end{aligned}$
Now,
$lim_{x \to α} \frac{f (x)}{f^{'} (x)} [\frac{f^{'} (x)}{f (x)}]$
$= lim_{x \to α} \frac{f (x)}{f^{'} (x)} (\frac{f^{'} (x)}{f (x)} - \{\frac{f^{'} (x)}{f (x)}\}),$ where ${\cdot}$ denotes the
fractional part function
$\begin{array}{l} = lim_{x \to α} \frac{f (x)}{f^{'} (x)} \times \frac{f^{'} (x)}{f (x)} - lim_{x \to α} \frac{f (x)}{f^{'} (x)} \{\frac{f^{'} (x)}{f (x)}\} \\ = 1 - lim_{x \to α} \frac{x - α}{2 ϕ (x) + (x - α) ϕ (x)} \{\frac{2 ϕ (x) + (x - α) ϕ^{'} (x)}{(x - α) ϕ (x)}\} \\ = 1 - 0 = 1 . \end{array}$

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