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QJEEM23Ma11eb: JEE Main 2023 Mathematics - Class 12 JEEM23Ma11eb Solution: If the domain of the function is , then is equal to _________.

If the domain of the function f(x)=sec⁡−1(2x5x+3)f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right) is [α,β)U(γ,δ][\alpha, \beta) \mathrm{U}(\gamma, \delta] , then ∣3α+10(β+γ)+21δ∣|3 \alpha+10(\beta+\gamma)+21 \delta| is equal to _________.

Step-by-Step Verified Solution

Given that f(x)=sec⁡−1(2x5x+3)f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)

Since, the domain for sec⁡−1x\sec ^{-1} x is ∣x∣≥1|x| \geq 1

Therefore ∣2x5x+3∣≥1\left|\frac{2 x}{5 x+3}\right| \geq 1

amp;2x5x+3≤−1 or 2x5x+3≥1amp;2x+5x+35x+3≤0 or 2x−5x−35x+3≥0amp;7x+35x+3≤0 or −3(x+1)5x+3≥0\begin{aligned}& \frac{2 x}{5 x+3} \leq-1 \text { or } \frac{2 x}{5 x+3} \geq 1 \\\\& \frac{2 x+5 x+3}{5 x+3} \leq 0 \text { or } \frac{2 x-5 x-3}{5 x+3} \geq 0 \\\\& \frac{7 x+3}{5 x+3} \leq 0 \text { or } \frac{-3(x+1)}{5 x+3} \geq 0\end{aligned}

Case I : 7x+3≤07 x+3 \leq 0 and 5 x+3>0

amp;x≤−37 or xgt;−35amp;⇒−35lt;x≤−37\begin{array}{rlrl} & x \leq-\frac{3}{7} \text { or } x > -\frac{3}{5} \\\\& \Rightarrow -\frac{3}{5} < x \leq-\frac{3}{7}\end{array}

Case II : 7x+3≥07 x+3 \geq 0 and 5 x+3<0

x \geq-\frac{3}{7} \text { and } x<-\frac{3}{5}

Which is not possible

Case III : x+1≥0x+1 \geq 0 and 5 x+3<0

amp;x≥−1 and xlt;−35amp;⇒−1≤xlt;−35\begin{aligned}& x \geq-1 \text { and } x<-\frac{3}{5} \\\\& \Rightarrow -1 \leq x<-\frac{3}{5}\end{aligned}

Case IV : x+1≤0x+1 \leq 0 and 5x+3≥05 x+3 \geq 0

x≤−1 and x≥−35x \leq-1 \text { and } x \geq-\frac{3}{5}

Which is not possible

∴\therefore Domain is [−1,−35)∪(−35,−37]\left[-1,-\frac{3}{5}\right) \cup\left(-\frac{3}{5},-\frac{3}{7}\right]

∴α=−1,β=−35,γ=−35,δ=−37\therefore \alpha=-1, \beta=-\frac{3}{5}, \gamma=-\frac{3}{5}, \delta=-\frac{3}{7}

∴\therefore ∣3α+10(β+γ)+21δ∣=∣−3+10(−35−35)+21(−37)∣=24|3 \alpha+10(\beta+\gamma)+21 \delta|=\left|-3+10\left(-\frac{3}{5}-\frac{3}{5}\right)+21\left(-\frac{3}{7}\right)\right|=24

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