જો વિધેય log _eleft(frac{6 x^2+5 x+1}{2 x-1}r | vedclass

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

Question

$\frac{6 x^2+5 x+1}{2 x-1} > 0$

$\frac{(3 x+1)(2 x+1)}{2 x-1} > 0$

$-\frac{1}{2}-\frac{1}{3} \quad \frac{1}{2}$

$x \in\left(\frac{-1}{2

Vedclass · Accepted Answer

$20$

Vedclass · Answer

$\frac{6 x^2+5 x+1}{2 x-1} > 0$

$\frac{(3 x+1)(2 x+1)}{2 x-1} > 0$

$-\frac{1}{2}-\frac{1}{3} \quad \frac{1}{2}$

$x \in\left(\frac{-1}{2}, \frac{-1}{3}ight) \cup\left(\frac{1}{2}, \inftyight).....(A)$

$x \in\left[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}ight] \cup\left(\frac{5}{3}, \inftyight)..........(B)$

$x < \frac{5}{3}........(C)$

$A \cap B \cap C \equiv\left(\frac{-1}{2}, \frac{-1}{3}ight) \cup\left(\frac{1}{2}, \frac{1}{\sqrt{2}}ight]$

$	ext { So } 18\left(\alpha^2+\beta^2+\gamma^2+\delta^2ight)=18\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{4}+\frac{1}{2}ight)$

$=18+2=20$

જો વિધેય $\log _e\left(\frac{6 x^2+5 x+1}{2 x-1}\right)+\cos ^{-1}\left(\frac{2 x^2-3 x+4}{3 x-5}\right)$ નો પ્રદેશ $(\alpha, \beta) \cup(\gamma, \delta]$ હોય, તો $18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right)=......$

Similar Questions

જો $\phi (x) = {a^x}$, તો ${\{ \phi (p)\} ^3} = . . .$

જો $f(x) = \cos (\log x)$, તો $f(x).f(4) - \frac{1}{2}\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right] =$

જો $f(x) = \frac{{\alpha x}}{{x + 1}},x \ne - 1$, તો $\alpha $ ની . . . . કિમત માટે $f(f(x)) = x$ મળે.

અહી $f(x)=\left\{\begin{array}{l} x \sin \left(\frac{1}{x}\right) \text { when } x \neq 0 \\ 1 \text { when } x=0 \end{array}\right\}$ અને $A=\{x \in R: f(x)=1\} $ હોય તો $A$ માં .. . . .

જો $f(x + ay,\;x - ay) = axy$, તો $f(x,\;y) =$