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1.Relation and Function
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જો વિધેય $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$ નો પ્રદેશ $[\alpha, \beta) U (\gamma, \delta]$ હોય, તો $|3 \alpha+10(\beta+\gamma)+21 \delta|=..........$
A
$23$
B
$22$
C
$24$
D
$21$
(JEE MAIN-2023)
Solution
$f(x)=\sec ^{-1} \frac{2 x}{5 x+3}$
$\left|\frac{2 x}{5 x+3}\right|$
$\left|\frac{2 x}{5 x+3}\right| \geq 1 \Rightarrow|2 x| \geq|5 x+3|$
$(2 x)^2-(5 x+3)^2 \geq|5 x+3|$
$(7 x+3)(-3 x-3) \geq 0$
$\frac{-\quad-\quad-}{-1} \quad-\frac{3}{7}$
$\text { domain }\left[-1, \frac{-3}{5}\right) \cup\left(\frac{-3}{5}, \frac{-3}{7}\right]$
$\alpha=-1, \beta=\frac{-3}{5}, \gamma=\frac{-3}{5}, \delta=\frac{-3}{7}$
$3 \alpha+10(\beta+\gamma)+21 \delta=-3$
$-3+10\left(\frac{-6}{5}\right)+\left(\frac{-3}{7}\right) 21=-24$
Standard 12
Mathematics
