Question:
The value of \(a\) for which the function
\[
f(x)=a\sin x+\frac{1}{3}\sin 3x
\]
has an extremum value at \(x=\dfrac{\pi}{3}\) is:
The value of \(a\) for which the function
\[
f(x)=a\sin x+\frac{1}{3}\sin 3x
\]
has an extremum value at \(x=\dfrac{\pi}{3}\) is:
Show Hint
For extremum at a point \(x=c\), first use
\[
f'(c)=0
\]
Then verify using
\[
f''(c)\neq 0
\]
to confirm that an extremum exists.
Updated On: Jun 25, 2026
- \(1\)
- \(-1\)
- \(0\)
- \(2\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Condition for extremum.
For a function to have an extremum at \[ x=\frac{\pi}{3}, \] the first derivative must be zero at that point.
So, \[ f'\left(\frac{\pi}{3}\right)=0 \]
Step 2: Differentiate the given function.
Given, \[ f(x)=a\sin x+\frac{1}{3}\sin 3x \] Differentiating with respect to \(x\), \[ f'(x)=a\cos x+\frac{1}{3}\cdot 3\cos 3x \] \[ f'(x)=a\cos x+\cos 3x \]
Step 3: Substitute \(x=\dfrac{\pi}{3}\).
Now, \[ f'\left(\frac{\pi}{3}\right) = a\cos\frac{\pi}{3} + \cos\pi \] Using \[ \cos\frac{\pi}{3}=\frac{1}{2} \] and \[ \cos\pi=-1, \] we get \[ a\left(\frac{1}{2}\right)-1=0 \] \[ \frac{a}{2}-1=0 \] \[ \frac{a}{2}=1 \] \[ a=2 \]
Step 4: Verify the extremum condition.
The second derivative is \[ f''(x)=-a\sin x-3\sin 3x \] At \[ x=\frac{\pi}{3} \] and \[ a=2, \] we get \[ f''\left(\frac{\pi}{3}\right) = -2\sin\frac{\pi}{3}-3\sin\pi \] \[ = -2\cdot \frac{\sqrt{3}}{2}-3\cdot 0 \] \[ =-\sqrt{3} \] Since \[ f''\left(\frac{\pi}{3}\right)\neq 0, \] the function has an extremum at \[ x=\frac{\pi}{3} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{2} \]
For a function to have an extremum at \[ x=\frac{\pi}{3}, \] the first derivative must be zero at that point.
So, \[ f'\left(\frac{\pi}{3}\right)=0 \]
Step 2: Differentiate the given function.
Given, \[ f(x)=a\sin x+\frac{1}{3}\sin 3x \] Differentiating with respect to \(x\), \[ f'(x)=a\cos x+\frac{1}{3}\cdot 3\cos 3x \] \[ f'(x)=a\cos x+\cos 3x \]
Step 3: Substitute \(x=\dfrac{\pi}{3}\).
Now, \[ f'\left(\frac{\pi}{3}\right) = a\cos\frac{\pi}{3} + \cos\pi \] Using \[ \cos\frac{\pi}{3}=\frac{1}{2} \] and \[ \cos\pi=-1, \] we get \[ a\left(\frac{1}{2}\right)-1=0 \] \[ \frac{a}{2}-1=0 \] \[ \frac{a}{2}=1 \] \[ a=2 \]
Step 4: Verify the extremum condition.
The second derivative is \[ f''(x)=-a\sin x-3\sin 3x \] At \[ x=\frac{\pi}{3} \] and \[ a=2, \] we get \[ f''\left(\frac{\pi}{3}\right) = -2\sin\frac{\pi}{3}-3\sin\pi \] \[ = -2\cdot \frac{\sqrt{3}}{2}-3\cdot 0 \] \[ =-\sqrt{3} \] Since \[ f''\left(\frac{\pi}{3}\right)\neq 0, \] the function has an extremum at \[ x=\frac{\pi}{3} \]
Step 5: Final conclusion.
Therefore, \[ \boxed{2} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Maxima and Minima Questions
- If \( x \) is real and \( \alpha, \beta \) are maximum and minimum values of \( \frac{x^2 - x + 1}{x^2 + x + 1} \) respectively, then \( \alpha + \beta = \):
- The sum of the global minimum and global maximum values of the function \[ f(x) = \frac{4}{3}x^3 - 4x \quad \text{in } [0, 2] \text{ is} \]
- The maximum value of \( x^4 y^4 \) when \( a^2 x^4 + b^2 y^4 = c^6 \) is:
- If \( A = \left\{ \frac{x}{9x^2 + 20} \right\} \) and \( f: A \to \mathbb{R} \) is defined by \( f(x) = 2x^3 - 15x^2 + 36x - 48 \), then the maximum value of \( f(x) \) is
- If \( f(x) = 2x^3 - 3x^2 + 4 \), then the minimum value of \( f(x) \) in the interval \( [0, 3] \) is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions