Domain of cos ^{-1}left[log _5left(x^2+7 x+15 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The function $f(x) = \cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]$ is defined when the argument of $\cos ^{-1}$ lies in the interval $[-1

Vedclass · Accepted Answer

$[-5,-2]$

Vedclass · Answer

(D) The function $f(x) = \cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]$ is defined when the argument of $\cos ^{-1}$ lies in the interval $[-1, 1]$.So,$-1 \leq \log _5\left(x^2+7 x+15\right) \leq 1$.Applying the base $5$ exponential function,we get $5^{-1} \leq x^2+7 x+15 \leq 5^1$,which is $\frac{1}{5} \leq x^2+7 x+15 \leq 5$.First,consider $x^2+7 x+15 \leq 5$,which implies $x^2+7 x+10 \leq 0$.Factoring the quadratic,we get $(x+5)(x+2) \leq 0$,which gives $x \in [-5, -2]$.Next,consider $x^2+7 x+15 \geq \frac{1}{5}$,which implies $x^2+7 x + 14.8 \geq 0$.The discriminant of $x^2+7 x + 14.8$ is $D = 7^2 - 4(1)(14.8) = 49 - 59.2 = -10.2 Since the leading coefficient is positive and $D Thus,the domain is the intersection of $x \in [-5, -2]$ and $x \in \mathbb{R}$,which is $x \in [-5, -2]$.

Domain of $\cos ^{-1}\left[\log _5\left(x^2+7 x+15\right)\right]$ is

Similar Questions

The range of $\tan^{-1} x$ is

The domain of the function $f(x) = \frac{\sin^{-1}(x - 3)}{\sqrt{9 - x^2}}$ is

The range of $\operatorname{Sin}^{-1} x + \operatorname{Cos}^{-1} x + \operatorname{Tan}^{-1} x$ is

Domain of function $f(x) = \sin^{-1}(5x)$ is

The range of $f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module