If the function \( f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 \), where \( a>0 \), attains its local maximum and minimum at \( p \) and \( q \), respectively, such that \( p^2 = q \), then \( f(3) \) is equal to:
Show Hint
- 23
- 55
- 10
37
The Correct Option is D
Approach Solution - 1
The problem asks for the value of \( f(3) \) for the function \( f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 \) (with \( a>0 \)), given that its local maximum and minimum occur at points \( p \) and \( q \) respectively, and that these points are related by the condition \( p^2 = q \).
Concept Used:
To find the local maximum and minimum of a differentiable function, we use the following steps:
1. First Derivative Test: The local extrema of a function occur at its critical points, which are the points where the first derivative is zero or undefined. For a polynomial function, we find the roots of \( f'(x) = 0 \).
2. Second Derivative Test: To classify the critical points, we use the second derivative. Let \( c \) be a critical point such that \( f'(c) = 0 \).
- If \( f''(c) < 0 \), then the function has a local maximum at \( x = c \).
- If \( f''(c) > 0 \), then the function has a local minimum at \( x = c \).
Step-by-Step Solution:
The given function is \( f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 \).
First, we find the first derivative of \( f(x) \) to locate the critical points:
\[ f'(x) = \frac{d}{dx}(2x^3 - 9ax^2 + 12a^2x + 1) = 6x^2 - 18ax + 12a^2 \]
To find the critical points, we set \( f'(x) = 0 \):
\[ 6x^2 - 18ax + 12a^2 = 0 \]
We can simplify this quadratic equation by dividing the entire equation by 6:
\[ x^2 - 3ax + 2a^2 = 0 \]
This equation can be factored by finding two numbers that multiply to \( 2a^2 \) and add up to \( -3a \). These numbers are \( -a \) and \( -2a \).
\[ (x - a)(x - 2a) = 0 \]
The critical points are \( x = a \) and \( x = 2a \). These are the values of \( p \) and \( q \).
To determine which point corresponds to the maximum and which to the minimum, we use the second derivative test. We find the second derivative, \( f''(x) \):
\[ f''(x) = \frac{d}{dx}(6x^2 - 18ax + 12a^2) = 12x - 18a \]
Now, we evaluate \( f''(x) \) at each critical point:
At \( x = a \):
\[ f''(a) = 12(a) - 18a = -6a \]
Since it is given that \( a > 0 \), we have \( f''(a) = -6a < 0 \). Therefore, the function has a local maximum at \( x = a \). This means \( p = a \).
At \( x = 2a \):
\[ f''(2a) = 12(2a) - 18a = 24a - 18a = 6a \]
Since \( a > 0 \), we have \( f''(2a) = 6a > 0 \). Therefore, the function has a local minimum at \( x = 2a \). This means \( q = 2a \).
We are given the condition \( p^2 = q \). Substituting our values for \( p \) and \( q \):
\[ (a)^2 = 2a \implies a^2 - 2a = 0 \] \[ a(a - 2) = 0 \]
This gives two possible solutions: \( a = 0 \) or \( a = 2 \). Since the problem states that \( a > 0 \), we must have \( a = 2 \).
Now that we have the value of \( a \), we can write the specific function for \( f(x) \):
\[ f(x) = 2x^3 - 9(2)x^2 + 12(2)^2x + 1 = 2x^3 - 18x^2 + 48x + 1 \]
Final Computation & Result:
We are asked to find the value of \( f(3) \). We substitute \( x = 3 \) into the function:
\[ f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1 \] \[ f(3) = 2(27) - 18(9) + 144 + 1 \] \[ f(3) = 54 - 162 + 144 + 1 \] \[ f(3) = 199 - 162 = 37 \]
Hence, the value of \( f(3) \) is 37.
Approach Solution -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 Differential Calculus Questions
- If $y(\theta) = \frac{2\cos\theta + \cos2\theta}{\cos3\theta + 4\cos2\theta + 5\cos\theta + 2}$, then at $\theta = \frac{\pi}{2}, y'' + y' + y$ is equal to:
Let \( f : \mathbb{R} \to \mathbb{R} \) be a twice differentiable function such that \[ (\sin x \cos y)(f(2x + 2y) - f(2x - 2y)) = (\cos x \sin y)(f(2x + 2y) + f(2x - 2y)), \] for all \( x, y \in \mathbb{R}. \)
If \( f'(0) = \frac{1}{2} \), then the value of \( 24f''\left( \frac{5\pi}{3} \right) \) 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 \(x = \frac{\sqrt{5}}{2}\) is equal to:
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,