JEE Main 2023 — Mathematics Algebra

Given that $f(x)=\mathrm{log}(4{x}^{2}+11x+6)+{\mathrm{sin}}^{-1}(4x+3)+{\mathrm{cos}}^{-1}(\frac{10x+6}{3})$f(x)=log(4x2+11x+6)+sin−1(4x+3)+cos−1(310x+6​)

We know that the domain of $A(x)+B(x)+C(x)$A(x)+B(x)+C(x) is $D(A)\cap D(B)\cap D(C)$D(A)∩D(B)∩D(C).

Let us find the domain of $\mathrm{log}(4{x}^{2}+11x+6)$log(4x2+11x+6).

We know that domain of $\mathrm{log}(f(x))$log(f(x)) is $f(x)>0$f(x)>0.

$$\Rightarrow 4{x}^{2}+11x+6>0$$⇒4x2+11x+6>0

$$\Rightarrow 4{x}^{2}+8x+3x+6>0$$⇒4x2+8x+3x+6>0

$$\Rightarrow (4x+3)(x+2)>0$$⇒(4x+3)(x+2)>0

$$\Rightarrow x\in (-\infty ,-2)\cup (\frac{-3}{4},\infty )$$⇒x∈(−∞,−2)∪(4−3​,∞)

Domain of $\mathrm{log}(4{x}^{2}+11x+9)$log(4x2+11x+9) is $x\in (-\infty ,-2)\cup (\frac{-3}{4},\infty )$x∈(−∞,−2)∪(4−3​,∞).

Let us find the domain of ${\mathrm{sin}}^{-1}(4x+3)$sin−1(4x+3)

$$\Rightarrow -1\leq 4x+3\leq 1$$⇒−1≤4x+3≤1

$$\Rightarrow -1\leq x\leq \frac{-1}{2}$$⇒−1≤x≤2−1​

$$\Rightarrow x\in [-1,\frac{-1}{2}]$$⇒x∈[−1,2−1​]

Let us find the domain of ${\mathrm{cos}}^{-1}(\frac{10x+6}{3})$cos−1(310x+6​).

$$\Rightarrow -1\leq \frac{10x+6}{3}\leq 1$$⇒−1≤310x+6​≤1

$$\Rightarrow -3\leq 10x+6\leq 3$$⇒−3≤10x+6≤3

$\Rightarrow \frac{-9}{10}\leq x\leq \frac{-3}{10}$⇒10−9​≤x≤10−3​.

Now the common domain is $\frac{-3}{4}\leq x\leq \frac{-1}{2}$4−3​≤x≤2−1​.

$$\Rightarrow x\in [\frac{-3}{4},\frac{-1}{2}]$$⇒x∈[4−3​,2−1​]

But given that the domain of $f(x)$f(x) is $x\in [\alpha ,\beta ]$x∈[α,β].

$\Rightarrow \alpha =\frac{-3}{4},\beta =\frac{-1}{2}$⇒α=4−3​,β=2−1​.

$$\Rightarrow |36(\alpha +\beta )|=|36(\frac{-3}{4}+(\frac{-1}{2}))|$$⇒∣36(α+β)∣=∣36(4−3​+(2−1​))∣

$$\Rightarrow |36(\alpha +\beta )|=|36\times \frac{-5}{4}|$$⇒∣36(α+β)∣=∣36×4−5​∣

$$\Rightarrow |36(\alpha +\beta )|=45$$⇒∣36(α+β)∣=45

Therefore, the required answer is $45$45.