Let f(x) = begin{cases} frac{ax^{2}+2ax+3}{4x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For $f(x)$ to be continuous at $x = -\frac{3}{2}$, the limit $\lim_{x 	o -\frac{3}{2}} f(x)$ must exist and equal $f(-\frac{3}{2}) = b$.Since the

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) For $f(x)$ to be continuous at $x = -\frac{3}{2}$, the limit $\lim_{x 	o -\frac{3}{2}} f(x)$ must exist and equal $f(-\frac{3}{2}) = b$.Since the denominator $(2x-1)(2x+3)$ approaches $0$ as $x 	o -\frac{3}{2}$, the numerator $ax^2 + 2ax + 3$ must also approach $0$.$a(-\frac{3}{2})^2 + 2a(-\frac{3}{2}) + 3 = 0$ $\Rightarrow \frac{9a}{4} - 3a + 3 = 0$ $\Rightarrow -\frac{3a}{4} = -3$ $\Rightarrow a = 4$.Substituting $a=4$, $f(x) = \frac{4x^2+8x+3}{(2x-1)(2x+3)} = \frac{(2x+1)(2x+3)}{(2x-1)(2x+3)} = \frac{2x+1}{2x-1}$ for $x 
eq -\frac{3}{2}, \frac{1}{2}$.Now, $f(f(x)) = f\left(\frac{2x+1}{2x-1}ight) = \frac{2(\frac{2x+1}{2x-1}) + 1}{2(\frac{2x+1}{2x-1}) - 1} = \frac{4x+2+2x-1}{4x+2-2x+1} = \frac{6x+1}{2x+3}$.Given $f(f(x)) = \frac{7}{5}$, we have $\frac{6x+1}{2x+3} = \frac{7}{5}$.$5(6x+1) = 7(2x+3)$ $\Rightarrow 30x + 5 = 14x + 21$ $\Rightarrow 16x = 16$ $\Rightarrow x = 1$.

Let $f(x) = \begin{cases} \frac{ax^{2}+2ax+3}{4x^{2}+4x-3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases}$ be continuous at $x=-\frac{3}{2}$. If $f(f(x)) = \frac{7}{5}$, then $x$ is equal to:

Similar Questions

Let $f: R \rightarrow R$ be a continuous function satisfying $f(0)=1$ and $f(2x)-f(x)=x$ for all $x \in R$. If $\lim_{n \rightarrow \infty} \{f(x)-f(\frac{x}{2^n})\} = G(x)$,then $\sum_{r=1}^{10} G(r^2)$ is equal to

Let $f(x) = \begin{cases} x^2 - a & x < 3 \\ b\sqrt{x - 2} + a & 3 \leqslant x < 6 \\ 2x + b & x \geqslant 6 \end{cases}$. If $f(x)$ is continuous $\forall x \in R$,then find the value of $\frac{f(1) - f(3)}{4}$.

Is the function defined by $f(x) = |x|$ a continuous function?

If $f(x) = \begin{cases} ax + 7, & \text{if } x < 1 \\ 3x - 1, & \text{if } x = 1 \\ \frac{x + 3}{b}, & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then

Prove that the function $f(x) = x^{n}$ is continuous at $x = n$,where $n$ is a positive integer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module