If the domain of the function f(x) = log_e le | vedclass

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

Question

(B) For the function to be defined,we need: $1$) $\frac{2x+3}{4x^2+x-3} > 0$ $2$) $-1 \leq \frac{2x-1}{x+2} \leq 1$ Step $1$: Solve $\frac{2x+3}{(4x-3

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) For the function to be defined,we need: $1$) $\frac{2x+3}{4x^2+x-3} > 0$ $2$) $-1 \leq \frac{2x-1}{x+2} \leq 1$ Step $1$: Solve $\frac{2x+3}{(4x-3)(x+1)} > 0$. The critical points are $x = -3/2, -1, 3/4$. Using the wavy curve method,the solution is $x \in (-3/2, -1) \cup (3/4, \infty)$. Step $2$: Solve $-1 \leq \frac{2x-1}{x+2} \leq 1$. Part $A$: $\frac{2x-1}{x+2} + 1 \geq 0 \implies \frac{3x+1}{x+2} \geq 0$. Solution: $x \in (-\infty, -2) \cup [-1/3, \infty)$. Part $B$: $\frac{2x-1}{x+2} - 1 \leq 0 \implies \frac{x-3}{x+2} \leq 0$. Solution: $x \in (-2, 3]$. Intersection of Part $A$ and Part $B$: $x \in [-1/3, 3]$. Step $3$: Find the intersection of Step $1$ and Step $2$. $((-3/2, -1) \cup (3/4, \infty)) \cap [-1/3, 3] = (3/4, 3]$. Thus,$\alpha = 3/4$ and $\beta = 3$. Step $4$: Calculate $5\beta - 4\alpha$. $5(3) - 4(3/4) = 15 - 3 = 12$.

If the domain of the function $f(x) = \log_e \left( \frac{2x+3}{4x^2+x-3} \right) + \cos^{-1} \left( \frac{2x-1}{x+2} \right)$ is $(\alpha, \beta]$,then the value of $5\beta - 4\alpha$ is equal to

Similar Questions

If $[x]$ denotes the greatest integer $\leq x$,then the range of the real-valued function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

The domain of the real valued function $f(x) = \log_2 \log_3 \log_5(x^2 - 5x + 11)$ is

The domain of the function defined by $f(x) = \frac{-5}{4x^2+1} + \sqrt{x^2-4}$ is

The function $f(x) = \frac{\sec^{-1}x}{\sqrt{x - [x]}}$,where $[.]$ denotes the greatest integer less than or equal to $x$,is defined for all $x$ belonging to:

The number of integral values of $\lambda$ for which the function $f(x) = \sqrt{\ln(2\lambda \cos x + 5)}$ is defined for all $x \in R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module