If 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 ______.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Sets Relations and Functions

If domain of the function

lo g_{e} (\frac{6 x ^{2} + 5 x + 1}{2 x - 1}) + cos^{- 1} (\frac{2 x ^{2} - 3 x + 4}{3 x - 5})

(α, β)

\cup (γ, δ]

, then

18 (α^{2} + β^{2} + γ^{2} + δ^{2})

is equal to ______.

Answer: 20

For the logarithm,

\frac{( 2 x + 1 ) ( 3 x + 1 )}{2 x - 1} > 0

. A sign analysis with critical points

- \frac{1}{2}, - \frac{1}{3}, \frac{1}{2}

gives

x \in (- \frac{1}{2}, - \frac{1}{3}) \cup (\frac{1}{2}, \infty) .

For

cos^{- 1}

, require

- 1 \leq \frac{2 x ^{2} - 3 x + 4}{3 x - 5} \leq 1

. The upper bound gives

\frac{2 x ^{2} - 3 x + 4}{3 x - 5} - 1 = \frac{2 x ^{2} - 6 x + 9}{3 x - 5} \leq 0

; since

2 x^{2} - 6 x + 9 > 0

always, this forces

3 x - 5 < 0

, i.e.

x < \frac{5}{3}

. The lower bound gives

\frac{2 x ^{2} - 3 x + 4}{3 x - 5} + 1 = \frac{2 x ^{2} - 1}{3 x - 5} \geq 0

, i.e.

x \in [- \frac{1}{2}, \frac{1}{2}] \cup (\frac{5}{3}, \infty)

. Intersecting the two bounds leaves

[- \frac{1}{2}, \frac{1}{2}]

.

Intersecting with the logarithm's domain,

(- \frac{1}{2}, - \frac{1}{3}) \cup (\frac{1}{2}, \frac{1}{2}],

α = - \frac{1}{2}, β = - \frac{1}{3}, γ = \frac{1}{2}, δ = \frac{1}{2}

, and

18 (\frac{1}{4} + \frac{1}{9} + \frac{1}{4} + \frac{1}{2}) = 18 \cdot \frac{40}{36} = 20.

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Integer answer