If the domain of the function log _eleft(frac | vedclass

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

Question

(A) For the function to be defined,both parts must be valid.$1$. For $\log_e\left(\frac{6x^2+5x+1}{2x-1}\right)$,we need $\frac{6x^2+5x+1}{2x-1} > 0$.

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) For the function to be defined,both parts must be valid.$1$. For $\log_e\left(\frac{6x^2+5x+1}{2x-1}ight)$,we need $\frac{6x^2+5x+1}{2x-1} > 0$.$\frac{(3x+1)(2x+1)}{2x-1} > 0$.Using the wavy curve method,the critical points are $x = -1/2, -1/3, 1/2$.The inequality holds for $x \in (-1/2, -1/3) \cup (1/2, \infty) \dots (A)$.$2$. For $\cos^{-1}\left(\frac{2x^2-3x+4}{3x-5}ight)$,we need $-1 \le \frac{2x^2-3x+4}{3x-5} \le 1$ and $3x-5 
eq 0$.Solving $\frac{2x^2-3x+4}{3x-5} \le 1 \implies \frac{2x^2-6x+9}{3x-5} \le 0$. Since $2x^2-6x+9$ has $D Solving $\frac{2x^2-3x+4}{3x-5} \ge -1 \implies \frac{2x^2-1}{3x-5} \ge 0$.Critical points are $x = -1/\sqrt{2}, 1/\sqrt{2}, 5/3$.The inequality holds for $x \in [-1/\sqrt{2}, 1/\sqrt{2}] \cup (5/3, \infty) \dots (C)$.Intersection $A \cap B \cap C = (-1/2, -1/3) \cup (1/2, 1/\sqrt{2}]$.Here $\alpha = -1/2, \beta = -1/3, \gamma = 1/2, \delta = 1/\sqrt{2}$.$18(\alpha^2+\beta^2+\gamma^2+\delta^2) = 18(1/4 + 1/9 + 1/4 + 1/2) = 18(1/2 + 1/9 + 1/2) = 18(1 + 1/9) = 18 + 2 = 20$.

If the domain of the function $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)$ is $(\alpha, \beta) \cup(\gamma, \delta]$,then $18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)$ is equal to $....$.

Similar Questions

If the domain of the function $f(x) = \log_{(10x^{2}-17x+7)}(18x^{2}-11x+1)$ is $(-\infty, a) \cup (b, c) \cup (d, \infty) - \{e\}$,then $90(a+b+c+d+e)$ equals:

Range of the function $\frac{1}{2 - \sin 3x}$ is

The domain of the real-valued function $f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}$ is

The range of $f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}}, (a>0)$ is

Let $A = \{x \in R, x \neq 0, -4 \leq x \leq 4\}$ and $f: A \rightarrow R$ be defined by $f(x) = \frac{|x|}{x}$ for $x \in A$. Then,the range of $f$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module