If the domain of the function \[ f(x) = \cos^{-1} \left( \frac{2 - |x|}{4} \right) + \left( \log_e (3 - x) \right)^{-1} \] is \([-\alpha, \beta) - \{ \gamma \}\), then \( \alpha + \beta + \gamma \) is equal to:
- 12
- 9
- 11
- 8
The Correct Option is C
Approach Solution - 1
To find the domain of the function \(f(x) = \cos^{-1} \left( \frac{2 - |x|}{4} \right) + \left( \log_e (3 - x) \right)^{-1}\), we need to consider the domain restrictions of both parts separately.
Domain of \(\cos^{-1} \left( \frac{2 - |x|}{4} \right)\):
The inverse cosine function, \(\cos^{-1}(y)\), is defined for \(-1 \leq y \leq 1\). Thus, we need:
\(-1 \leq \frac{2 - |x|}{4} \leq 1\)
Solving the inequality:
- \(\frac{2 - |x|}{4} \leq 1\):
- \(2 - |x| \leq 4\)
- \(-|x| \leq 2\)
- \(-1 \leq \frac{2 - |x|}{4}\):
- \(-4 \leq 2 - |x|\)
- \(-6 \leq x \leq 6\)
Domain of \(\left( \log_e(3 - x) \right)^{-1}\):
The natural logarithm, \(\log_e(3-x)\), is defined and positive if \(3 - x > 1\), i.e., \(x < 3\).
But since \(\left(\log_e(3-x)\right)^{-1}\) should be defined, \(\log_e(3-x) \neq 0\).
This implies \(3 - x \neq 1\), hence, \(x \neq 2\).
Combining all conditions:
\(-6 \leq x \leq 6\), \(x < 3\), and \(x \neq 2\).
This means the domain is \([-6, 3) \setminus \{2\}\), which translates to \([-6, 2) \cup (2, 3)\).
Thus, we have:
- \(\alpha = 6\) (since \([-\alpha, \beta)\) implies \([-6, 3)\))
- \(\beta = 3\)
- \(\gamma = 2\)
Therefore, \(\alpha + \beta + \gamma = 6 + 3 + 2 = 11\).
The correct answer is 11.
Approach Solution -2
To find the domain of \( f(x) \), analyze each component individually.
For \( \cos^{-1} \left( \frac{2 - |x|}{4} \right) \) to be defined, \( -1 \leq \frac{2 - |x|}{4} \leq 1 \). Solving these inequalities:
\[ -1 \leq \frac{2 - |x|}{4} \leq 1 \]
leads to \( |x| \leq 6 \), so \( x \in [-6, 6] \).
For \( (\log_e (3 - x))^{-1} \) to be defined, \( \log_e (3 - x) \neq 0 \) and \( 3 - x > 0 \).
- \( 3 - x > 0 \Rightarrow x < 3 \).
- \( \log_e (3 - x) \neq 0 \Rightarrow x \neq 2 \) (since \( \log_e (3 - x) = 0 \) when \( x = 2 \)).
Combining these conditions, we have:
\[ x \in [-6, 3) - \{2\} \]
Thus, the domain is \( [-\alpha, \beta) - \{\gamma\} \) where \( \alpha = 6 \), \( \beta = 3 \), and \( \gamma = 2 \).
\( \alpha + \beta + \gamma = 11 \)
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Functions Questions
The domain of \(y= cos^{-1}|\frac{2-|x|}{4}| log(3 - x)^{-1}\) is [α, β) - {y} then the value of α+β-y =?
- If the domain of the function \(f(x)\) = \(\frac{\sqrt{x^2-25}}{(4-x^2)}+ \log(x^2+2x-15)\) is \((-∞, α) ∪ (β, ∞)\), then \(α^2+β^2\) is equal to \(b\)
- If the function \( f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 2 \\ ax^2 + 2b, & |x| < 2 \end{cases} \) is differentiable on \( \mathbb{R} \), then \( 48 (a + b) \) is equal to \(\_\_\_\_\_\_\_\_\_\).
- Consider the function \( f : (0, 2) \to \mathbb{R} \) defined by \[ f(x) = \frac{x}{2} + \frac{2}{x} \] and the function \( g(x) \) defined by \[ g(x) = \begin{cases} \min\{f(t)\}, & 0 < t \leq x \text{ and } 0 < x \leq 1 \\ \frac{3}{2} + x, & 1 < x < 2 \end{cases} \] Then:
- Let \( f : R \setminus \left\{ -\frac{1}{2} \right\} \to R \) and \( g : R \setminus \left\{ -\frac{5}{2} \right\} \to R \) be defined as \( f(x) = \frac{2x + 3}{2x + 1} \) and \( g(x) = \frac{|x| + 1}{2x + 5} \). Then the domain of the function \( f(g(x)) \) is:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,