If the domain of the function f(x) = log_e(4x | vedclass

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(B) The domain of $f(x)$ is the intersection of the domains of its three components.$(i)$ For $\log_e(4x^2 + 11x + 6)$,we require $4x^2 + 11x + 6 > 0$

Vedclass · Accepted Answer

$45$

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(B) The domain of $f(x)$ is the intersection of the domains of its three components.$(i)$ For $\log_e(4x^2 + 11x + 6)$,we require $4x^2 + 11x + 6 > 0$.Factoring gives $(4x + 3)(x + 2) > 0$,so $x \in (-\infty, -2) \cup (-\frac{3}{4}, \infty)$.(ii) For $\sin^{-1}(4x + 3)$,we require $-1 \le 4x + 3 \le 1$.$-4 \le 4x \le -2$,which implies $x \in [-1, -\frac{1}{2}]$.(iii) For $\cos^{-1}\left(\frac{10x + 6}{3}ight)$,we require $-1 \le \frac{10x + 6}{3} \le 1$.$-3 \le 10x + 6 \le 3$,so $-9 \le 10x \le -3$,which implies $x \in [-\frac{9}{10}, -\frac{3}{10}]$.Taking the intersection of all three intervals:$x \in ((-\infty, -2) \cup (-\frac{3}{4}, \infty)) \cap [-1, -\frac{1}{2}] \cap [-\frac{9}{10}, -\frac{3}{10}]$.Intersection of $[-1, -\frac{1}{2}]$ and $[-\frac{9}{10}, -\frac{3}{10}]$ is $[-\frac{9}{10}, -\frac{1}{2}]$.Now,intersecting this with $(-\infty, -2) \cup (-\frac{3}{4}, \infty)$:Since $-\frac{9}{10} Thus,$\alpha = -\frac{3}{4}$ and $\beta = -\frac{1}{2}$.Then $|\alpha + \beta| = |-\frac{3}{4} - \frac{1}{2}| = |-\frac{5}{4}| = \frac{5}{4}$.Finally,$36|\alpha + \beta| = 36 	imes \frac{5}{4} = 9 	imes 5 = 45$.

Column $I$	Column $II$
$(A)$ If $-1 < x < 1$,then $f(x)$ satisfies	$(p)$ $0 < f(x) < 1$
$(B)$ If $1 < x < 2$,then $f(x)$ satisfies	$(q)$ $f(x) < 0$
$(C)$ If $3 < x < 5$,then $f(x)$ satisfies	$(r)$ $f(x) > 0$
$(D)$ If $x > 5$,then $f(x)$ satisfies	$(s)$ $f(x) < 1$

If the domain of the function $f(x) = \log_e(4x^2 + 11x + 6) + \sin^{-1}(4x + 3) + \cos^{-1}\left(\frac{10x + 6}{3}\right)$ is $(\alpha, \beta]$,then $36|\alpha + \beta|$ is equal to:

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