Question

If the domain of the function

$f(x) = \log_e (4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}(\frac{10x + 6}{3})$ is $(\alpha, \beta]$, then

$36|\alpha + \beta|$ is equal to :

• A. 72 (KTK2)
• B. 54
• C. 45
• D. 63

Answer 1

Answer: (C)

45

Answer 2

To find the domain of the function $f(x) = \log_e (4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}(\frac{10x + 6}{3})$, we need to ensure that the arguments of each component function satisfy their respective domain conditions.

1. Domain of $\log_e (4x^2 + 11x + 6)$: The argument of a logarithm must be strictly positive. $4x^2 + 11x + 6 > 0$. To find the roots of the quadratic $4x^2 + 11x + 6 = 0$, we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-11 \pm \sqrt{11^2 - 4(4)(6)}}{2(4)}$ $x = \frac{-11 \pm \sqrt{121 - 96}}{8}$ $x = \frac{-11 \pm \sqrt{25}}{8}$ $x = \frac{-11 \pm 5}{8}$ The roots are $x_1 = \frac{-11 - 5}{8} = \frac{-16}{8} = -2$ and $x_2 = \frac{-11 + 5}{8} = \frac{-6}{8} = -\frac{3}{4}$. Since the coefficient of $x^2$ is positive (4), the parabola opens upwards, so $4x^2 + 11x + 6 > 0$ when $x < -2$ or $x > -\frac{3}{4}$. Thus, $D_1 = (-\infty, -2) \cup (-\frac{3}{4}, \infty)$.

2. Domain of $\sin^{-1}(4x + 3)$: The argument of $\sin^{-1}(u)$ must be in the interval $[-1, 1]$. $-1 \le 4x + 3 \le 1$. Subtract 3 from all parts: $-1 - 3 \le 4x \le 1 - 3$ $-4 \le 4x \le -2$. Divide by 4: $\frac{-4}{4} \le x \le \frac{-2}{4}$ $-1 \le x \le -\frac{1}{2}$. Thus, $D_2 = [-1, -\frac{1}{2}]$.

3. Domain of $\cos^{-1}(\frac{10x + 6}{3})$: The argument of $\cos^{-1}(u)$ must also be in the interval $[-1, 1]$. $-1 \le \frac{10x + 6}{3} \le 1$. Multiply by 3: $-3 \le 10x + 6 \le 3$. Subtract 6 from all parts: $-3 - 6 \le 10x \le 3 - 6$ $-9 \le 10x \le -3$. Divide by 10: $\frac{-9}{10} \le x \le \frac{-3}{10}$. Thus, $D_3 = [-\frac{9}{10}, -\frac{3}{10}]$.

The domain of $f(x)$ is the intersection of $D_1$, $D_2$, and $D_3$. $D = D_1 \cap D_2 \cap D_3$

Converting fractional endpoints to decimals for easier comparison: $-\frac{3}{4} = -0.75$ $-\frac{1}{2} = -0.5$ $-\frac{9}{10} = -0.9$ $-\frac{3}{10} = -0.3$

The domains are: $D_1 = (-\infty, -2) \cup (-0.75, \infty)$ $D_2 = [-1, -0.5]$ $D_3 = [-0.9, -0.3]$

First, find the intersection of $D_2$ and $D_3$: $D_2 \cap D_3 = [-1, -0.5] \cap [-0.9, -0.3]$ The lower bound of the intersection is $\max(-1, -0.9) = -0.9$. The upper bound of the intersection is $\min(-0.5, -0.3) = -0.5$. So, $D_2 \cap D_3 = [-0.9, -0.5]$.

Now, find the intersection of $D_1$ with $[-0.9, -0.5]$: $D = ((-\infty, -2) \cup (-0.75, \infty)) \cap [-0.9, -0.5]$ The interval $[-0.9, -0.5]$ does not overlap with $(-\infty, -2)$ because $-0.5 > -2$. Therefore, we only need to intersect $[-0.9, -0.5]$ with $(-0.75, \infty)$. The lower bound of this intersection is $\max(-0.9, -0.75) = -0.75$. Since $-0.75$ is excluded in $D_1$, it will be excluded in the intersection. The upper bound of this intersection is $\min(-0.5, \infty) = -0.5$. Since $-0.5$ is included in $[-0.9, -0.5]$, it will be included in the intersection. So, the domain of $f(x)$ is $(-0.75, -0.5]$.

Comparing this with the given form $(\alpha, \beta]$, we have: $\alpha = -0.75 = -\frac{3}{4}$ $\beta = -0.5 = -\frac{1}{2}$

We need to calculate $36|\alpha + \beta|$. $\alpha + \beta = -\frac{3}{4} + \left(-\frac{1}{2}\right) = -\frac{3}{4} - \frac{2}{4} = -\frac{5}{4}$ $|\alpha + \beta| = \left|-\frac{5}{4}\right| = \frac{5}{4}$ $36|\alpha + \beta| = 36 \times \frac{5}{4} = 9 \times 5 = 45$.

The final answer is 45.