Consider the function.\(f(x) = \begin{cases} \frac{a(7x - 12 - x^2)}{b(x^2 - 7x + 12)} & , \quad x < 3 \\[8pt] \frac{\sin(x - 3)}{2^{x - \lfloor x \rfloor}} & , \quad x > 3 \\[8pt] b & , \quad x = 3 \end{cases}\)Where \(\lfloor x \rfloor\)denotes the greatest integer less than or equal to \(x\). If \(S\) denotes the set of all ordered pairs \((a, b)\) such that \(f(x)\) is continuous at \(x = 3\), then the number of elements in \(S\) is:
- 2
- Infinitely many
- 4
- 1
The Correct Option is D
Approach Solution - 1
To determine the number of ordered pairs \((a, b)\) such that the function \(f(x)\) is continuous at \(x = 3\), we need to ensure that the left-hand limit, right-hand limit, and the value of the function at \(x = 3\) are equal.
Step 1: Check Continuity at \(x = 3\)
The function is given by:
- \(f(x) = \frac{a(7x - 12 - x^2)}{b(x^2 - 7x + 12)}\) for \(x < 3\)
- \(f(x) = \frac{\sin(x - 3)}{2^{x - \lfloor x \rfloor}}\) for \(x > 3\)
- \(f(x) = b\) for \(x = 3\)
Step 2: Evaluate the Limit as \(x \to 3^-\)
For \(x < 3\), the denominator becomes:
\(x^2 - 7x + 12 = (x - 3)(x - 4)\)
So, the function simplifies to:
\(\frac{a(7x - 12 - x^2)}{b(x - 4)}\)
As \(x \to 3^-\), simplify the numerator:
\(7x - 12 - x^2 = -(x - 3)(x + 4)\)
The expression becomes:
\(\frac{a(-(x-3)(x+4))}{b(x-4)} = \frac{a(-(3-4)(3+4))}{b(3-4)}\)
Thus, the left-hand limit is:
\(L_1 = \frac{-7a}{b}\)
Step 3: Evaluate the Limit as \(x \to 3^+\)
For \(x > 3\), the function becomes:
\(f(x) = \frac{\sin(x-3)}{2^{x - \lfloor x \rfloor}}\)
Since \(\lfloor x \rfloor = 3\), it simplifies to:
\(\frac{\sin(x-3)}{1} = \sin(x-3)\)
The right-hand limit as \(x \to 3^+\) is:
\(L_2 = \sin(0) = 0\)
Step 4: Equate Limits and the Function Value
For continuity, \(L_1 = L_2 = f(3)\):
- \(\frac{-7a}{b} = 0 \Rightarrow a = 0\)
- And, \(f(3) = b = 0\)
Thus, the ordered pair is \((a, b) = (0, 0)\).
Conclusion
There is only one ordered pair \((0, 0)\) where the function \(f(x)\) is continuous at \(x = 3\). Therefore, the number of elements in set \(S\) is:
1
Approach Solution -2
Step 1. Continuity Condition at \( x = 3 \): For \( f(x) \) to be continuous at \( x = 3 \), we must have:
\(f(3^-) = f(3) = f(3^+)\)
Step 2. Calculate \( f(3^-) \): For \( x < 3 \),
\(f(x) = \frac{a(7x - 12 - x^2)}{b|x^2 - 7x + 12|} = \frac{-a(x - 3)(x - 4)}{b(x - 3)(x - 4)} = \frac{-a}{b}\)
So, \(f(3^-) = -\frac{a}{b}\).
Step 3. Calculate \( f(3^+) \): For \( x > 3 \),
\(f(x) = \frac{\sin(x - 3)}{2x - |x|} \Rightarrow \lim_{x \to 3^+} f(x) = 2\)
Step 4. Set Up Continuity Condition: Since \( f(3^-) = f(3) = f(3^+) \),
\(-\frac{a}{b} = 2 \quad \text{and} \quad b = 2 \Rightarrow a = -4\)
Therefore, the only solution is \( (a, b) = (-4, 2) \).
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 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:
- 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:
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,