JEE Main 2023 — Mathematics Algebra

We need to find the domain of the function ${\mathrm{log}}_{e}(\frac{6{x}^{2}+5x+1}{2x-1})+{\mathrm{cos}}^{-1}(\frac{2{x}^{2}-3x+4}{3x-5})$loge​(2x−16x2+5x+1​)+cos−1(3x−52x2−3x+4​).

$\Rightarrow \frac{6{x}^{2}+5x+1}{2x-1}>0.....(1)$⇒2x−16x2+5x+1​>0.....(1) and

$$\Rightarrow -1\leq \frac{2{x}^{2}-3x+4}{3x-5}\leq 1......(2)$$⇒−1≤3x−52x2−3x+4​≤1......(2)

Now From $(1)$(1),

$$\Rightarrow \frac{6{x}^{2}+5x+1}{2x-1}>0$$⇒2x−16x2+5x+1​>0

$$\Rightarrow \frac{(3x+1)(2x+1)}{(2x-1)}>0$$⇒(2x−1)(3x+1)(2x+1)​>0

Now we get the common region here as

$$\Rightarrow x\in (-\frac{1}{2},-\frac{1}{3})\cup (\frac{1}{2},\infty ).......(a)$$⇒x∈(−21​,−31​)∪(21​,∞).......(a)

From $(2)$(2) we get that

$$\Rightarrow -1\leq \frac{2{x}^{2}-3x+4}{3x-5}\leq 1$$⇒−1≤3x−52x2−3x+4​≤1

$$\Rightarrow \frac{2{x}^{2}-3x+4}{3x-5}\geq -1......(3)$$⇒3x−52x2−3x+4​≥−1......(3)

and $\frac{2{x}^{2}-3x+4}{3x-5}\leq 1.....(4)$3x−52x2−3x+4​≤1.....(4)

From $(3)$(3) we get

$$\Rightarrow \frac{2{x}^{2}-3x+4}{3x-5}+1\geq 0$$⇒3x−52x2−3x+4​+1≥0

$$\Rightarrow x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]\cup (\frac{5}{3},\infty ).....(b)$$⇒x∈[2​−1​,2​1​]∪(35​,∞).....(b)

From $(4)$(4) we get

$$\Rightarrow \frac{2{x}^{2}-3x+4}{3x-5}-1\leq 0$$⇒3x−52x2−3x+4​−1≤0

$$\Rightarrow \frac{2{x}^{2}-6x+9}{3x-5}\leq 0$$⇒3x−52x2−6x+9​≤0

As $D={(-6)}^{2}-4(2)(9)<0\text{and}2>0$D=(−6)2−4(2)(9)<0and2>0, we observe that $2{x}^{2}-6x+9>0\forall x\in \mathbb{R}$2x2−6x+9>0∀x∈R

$$\Rightarrow \frac{1}{3x-5}\leq 0(\because 2{x}^{2}-6x+9>0\forall x\in R)$$⇒3x−51​≤0(∵2x2−6x+9>0∀x∈R)

$$\Rightarrow x\in (-\infty ,\frac{5}{3})......(c)$$⇒x∈(−∞,35​)......(c)

$\Rightarrow$⇒ Intersection of $(a),(b)$(a),(b) and $(c)$(c) gives us

$x\in (-\frac{1}{2},-\frac{1}{3})\cup (\frac{1}{2},\frac{1}{\sqrt{2}})$x∈(−21​,−31​)∪(21​,2​1​).

On comparing this with $(\alpha ,\beta )\cup (\gamma ,\delta )$(α,β)∪(γ,δ),

$$\Rightarrow {\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}+{\delta }^{2}=\frac{10}{9}$$⇒α2+β2+γ2+δ2=910​

$\Rightarrow 18({\alpha }^{2}+{\beta }^{2}+{\gamma }^{2}+{\delta }^{2})=20$⇒18(α2+β2+γ2+δ2)=20.

Hence this is the required answer.