Domain of Inverse Trigonometric Terms
The function is defined as $f(x) = \cos^{-1}\left( \frac{2x - 5}{11 - 3x} \right) + \sin^{-1}(2x^2 - 3x + 1)$. Its domain is the intersection of the domains of the individual inverse trigonometric terms.
Domain for $\cos^{-1}\left( \frac{2x - 5}{11 - 3x} \right)$
The argument of $\cos^{-1}(u)$ must be in the interval $[-1, 1]$. Thus, we require:
$-1 \le \frac{2x - 5}{11 - 3x} \le 1$
This inequality splits into two parts:
- $\frac{2x - 5}{11 - 3x} \le 1 \implies \frac{5x - 16}{11 - 3x} \le 0$. The solution is $x \in (-\infty, 16/5] \cup (11/3, \infty)$.
- $\frac{2x - 5}{11 - 3x} \ge -1 \implies \frac{6 - x}{11 - 3x} \ge 0$. The solution is $x \in (-\infty, 11/3) \cup [6, \infty)$.
The intersection of these solutions gives the domain for the first term:
$D_1 = (-\infty, 16/5] \cup [6, \infty)$
Domain for $\sin^{-1}(2x^2 - 3x + 1)$
The argument of $\sin^{-1}(v)$ must be in the interval $[-1, 1]$. Thus, we require:
$-1 \le 2x^2 - 3x + 1 \le 1$
This inequality splits into two parts:
- $2x^2 - 3x + 1 \le 1 \implies 2x^2 - 3x \le 0$. Factoring gives $x(2x - 3) \le 0$. The solution is $x \in [0, 3/2]$.
- $2x^2 - 3x + 1 \ge -1 \implies 2x^2 - 3x + 2 \ge 0$. The discriminant is $(-3)^2 - 4(2)(2) = 9 - 16 = -7$. Since the leading coefficient is positive and the discriminant is negative, this quadratic is always positive. The solution is $x \in \mathbb{R}$.
The intersection gives the domain for the second term:
$D_2 = [0, 3/2]$
Overall Domain and Calculation
The domain of the function $f(x)$ is the intersection $D_1 \cap D_2$:
$D = ( (-\infty, 16/5] \cup [6, \infty) ) \cap [0, 3/2] = [0, 3/2]$
The problem states the domain is the interval $[\alpha, \beta]$. Comparing this with our result, we have:
$\alpha = 0$
$\beta = 3/2$
We need to compute $\alpha + 2\beta$:
$\alpha + 2\beta = 0 + 2 \times \frac{3}{2} = 0 + 3 = 3$.
The calculated value is 3. The provided correct answer is Option C (1).
Similar Questions
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
- The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
- The number of points of discontinuity of the function $f (x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}], x \in [0,4]$, where $[.]$denotes the greatest integer function, is ________
- The number of functions $f: \{1, 2, 3, 4\} \rightarrow \{a, b, c\}$, which are not onto, is:
- If $y = \tan^{-1}\left(\frac{3\cos x - 4\sin x}{4\cos x + 3\sin x}\right) + 2\tan^{-1}\left(\frac{x}{1 + \sqrt{1 - x^2}}\right)$, then $\frac{dy}{dx}$ at $x = \frac{\sqrt{3}}{2}$ is equal to:
- Let $f(x) = \lim_{\theta \rightarrow 0} \left(\frac{\cos\pi x - x^{\left(\frac{2}{\theta}\right)}\sin(x - 1)}{1 + x^{\left(\frac{2}{\theta}\right)}(x - 1)}\right)$, $x \in \mathbf{R}$. Consider the following two statements :
(I) $f(x)$ is discontinous at $x = 1$.
(II) $f(x)$ is continous at $x = -1$.
Then, - The sum of all the elements in the range of $f(x) = \text{Sgn}(\sin x) + \text{Sgn}(\cos x) + \text{Sgn}(\tan x) + \text{Sgn}(\cot x)$,
$x \neq \frac{n\pi}{2}, n \in \mathbf{Z}$, where $\text{Sgn}(t) = \begin{cases} 1, & \text{if } t > 0 \\ -1, & \text{if } t < 0 \end{cases}$, is : - Let $f: \mathbf{R} \to \mathbf{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbf{R}$ and $f'(a-1) = 0$, where $a$ is a real number. Let $g(x) = f(\tan^2 x - 2\tan x + a)$, $0 < x < \frac{\pi}{2}$.
Consider the following two statements :
(I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$
(II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
Then, - Let $f: \mathbb{R} \rightarrow (0, \infty)$ be a twice differentiable function such that $f(3) = 18, f'(3) = 0$ and $f''(3) = 4$. Then $\lim_{x \to 1} \left( \log_e \left( \frac{f(2+x)}{f(3)} \right)^{\frac{18}{(x-1)^2}} \right)$ is equal to :
Important Questions from Differential Calculus
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that the quadratic equation $f(x)m^2 - 2f'(x)m + f''(x) = 0$ in m, has two equal roots for every $x \in \mathbb{R}$. If $f(0) = 1, f'(0) = 2$, and $(\alpha, \beta)$ is the largest interval in which the function $f(\log_e x - x)$ is increasing, then $\alpha + \beta$ is equal to ____________.
For a real number $y$, consider $[y]$ denotes the greatest integer less than or equal to $y$.
If $f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}$, then- The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[ x^2 - x - \frac{1}{2} \right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is __________.
- The number of points of discontinuity of the function $f (x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}], x \in [0,4]$, where $[.]$denotes the greatest integer function, is ________
