JEE Mathematics 2023 Question with Solution

Given: $f(x)=\log_e(4x^2+11x+6)+\sin^{-1}(4x+3)+\cos^{-1}\!\left(\frac{10x+6}{3}\right)$

Find: $36|\alpha+\beta|$ where the domain is $(\alpha,\beta)$.

For the logarithmic term, the argument must be positive:

$$4x^2+11x+6>0$$ $$(4x+3)(x+2)>0$$

So,

$$x<-2 \quad \text{or} \quad x>-\frac{3}{4}$$

For the inverse sine term, its argument must lie in $[-1,1]$:

$$-1\le 4x+3 \le 1$$ $$-4 \le 4x \le -2$$ $$-1 \le x \le -\frac{1}{2}$$

For the inverse cosine term, its argument must also lie in $[-1,1]$:

$$-1 \le \frac{10x+6}{3} \le 1$$ $$-3 \le 10x+6 \le 3$$ $$-9 \le 10x \le -3$$ $$-\frac{9}{10} \le x \le -\frac{3}{10}$$

Now intersect the interval conditions from the inverse trigonometric terms:

$$-1 \le x \le -\frac{1}{2}$$

and

$$-\frac{9}{10} \le x \le -\frac{3}{10}$$

So the common interval is

$$-\frac{9}{10} \le x \le -\frac{1}{2}$$

Now intersect this with the logarithmic condition $x>-\frac{3}{4}$. Hence,

$$-\frac{3}{4} < x \le -\frac{1}{2}$$

Thus, taking the interval as $(\alpha,\beta)$,

$$(\alpha,\beta)=\left(-\frac{3}{4},-\frac{1}{2}\right)$$

Therefore,

$$\alpha+\beta=-\frac{3}{4}-\frac{1}{2}=-\frac{5}{4}$$ $$|\alpha+\beta|=\frac{5}{4}$$ $$36|\alpha+\beta|=36\times \frac{5}{4}=45$$

Therefore, the correct option is A.

Given: The domain comes from satisfying all three terms simultaneously.

Find: The value of $36|\alpha+\beta|$.

Use the standard domain rules directly:

• For $\log_e(4x^2+11x+6)$, require $4x^2+11x+6>0$.
• For $\sin^{-1}(4x+3)$, require $-1\le 4x+3\le 1$.
• For $\cos^{-1}\!\left(\frac{10x+6}{3}\right)$, require $-1\le \frac{10x+6}{3}\le 1$.

The two inverse trigonometric conditions give bounded intervals, whose intersection is

$$-\frac{9}{10} \le x \le -\frac{1}{2}$$

The logarithmic condition keeps only the part with

$$x>-\frac{3}{4}$$

So the effective domain is

$$-\frac{3}{4}$$