Q.$f(x) = x^x$ has a stationary point at:
(A) $x = e$
(B) $x = \dfrac{1}{e}$
(C) $x = 1$
(D) $x = \sqrt{e}$

The key idea is that a turning point (also called a stationary point) occurs where the derivative is zero. Once we find $a$ from $f'(1)=0$, we determine the nature (max or min) using the second derivative.

1. Find the first derivative

   $f(x) = x^4 - 31x^2 + a x + 5$

   Differentiate term by term:

   $f'(x) = 4x^3 - 62x + a$

2. Apply the turning point condition

   At $x=1$, $f'(1)=0$:

   $4(1)^3 - 62(1) + a = 0$

   $4 - 62 + a = 0$

   $-58 + a = 0$

   So $a = 58$.

3. Determine the nature of the turning point

   Compute the second derivative:

   $f''(x) = 12x^2 - 62$

   Evaluate at $x=1$:

   $f''(1) = 12(1)^2 - 62 = 12 - 62 = -50$ …

We are given the curve

The question asks what happens at $x = -\frac{1}{\sqrt{3}}$ and $x = \frac{1}{\sqrt{3}}$: are these both minima, both maxima, or one of each?

Concept and intuition:

To decide whether a critical point is a local maximum or minimum, we use the second derivative test. If the second derivative is positive, the curve is concave up (minimum); if negative, concave down (maximum). Since the function is even (only even powers of $x$), the two points are symmetric, so they will behave identically — meaning both are the same type of extremum. That already hints that options (B) and (C) (which claim they are different) are unlikely.

Step-by-step reasoning:

1. Find the first derivative to locate critical points.

Factor:

Critical points occur when $y' = 0$:

2. Find the second derivative to test each critical point.

3. Evaluate the second derivative at each given point.
   • At $x = \frac{1}{\sqrt{3}}$:

 Positive → local minimum.  

• At $x = -\frac{1}{\sqrt{3}}$:

 Also positive → local minimum. …

Concept & Intuition

When a function involves a square root, the derivative may fail to exist at points where the radicand is zero, especially if the zero occurs at the boundary of the domain. Here, $4ax - x^2 = x(4a - x)$ is zero at $x=0$ and $x=4a$, but we are asked about $x=2a$, which is the vertex of the quadratic inside the root. At $x=2a$, the radicand is $4a(2a) - (2a)^2 = 8a^2 - 4a^2 = 4a^2 > 0$, so the square root is well-defined. However, the derivative involves a term $\frac{1}{\sqrt{4ax - x^2}}$ after differentiation, which blows up when the radicand is zero — but here it isn’t zero. Wait: let’s check carefully. The radicand at $x=2a$ is $4a^2$, not zero. So why might the derivative not exist? The pitfall is that the derivative formula from the product rule gives a finite value, but we must check the limit definition because the function might have a cusp or vertical tangent. Actually, let’s compute properly.

Step-by-step solution

1. Write the function and its domain

   $f(x) = x \sqrt{4ax - x^2}$, with $a>0$. The square root requires $4ax - x^2 \ge 0$, i.e., $x(4a - x) \ge 0$, so $0 \le x \le 4a$. At $x=2a$, we are inside the domain.

2. Differentiate using the product rule

   Let $u = x$, $v = \sqrt{4ax - x^2} = (4ax - x^2)^{1/2}$.

   Then $u' = 1$, and $v' = \frac{1}{2}(4ax - x^2)^{-1/2} \cdot (4a - 2x) = \frac{2a - x}{\sqrt{4ax - x^2}}$.

   So

3. Combine into a single fraction

4. Evaluate at $x = 2a$

   Numerator: $6a(2a) - 2(2a)^2 = 12a^2 - 8a^2 = 4a^2$.

   Denominator: $\sqrt{4a(2a) - (2a)^2} = \sqrt{8a^2 - 4a^2} = \sqrt{4a^2} = 2a$ (since $a>0$).

   So $f'(2a) = \frac{4a^2}{2a} = 2a$, which is finite and positive.

5. But wait — check the limit definition from both sides

   The algebraic simplification assumed the derivative exists. However, note that the denominator $\sqrt{4ax - x^2}$ is zero at the endpoints $x=0$ and $x=4a$, but not at $x=2a$. So the derivative appears to exist and equal $2a$. Yet the problem suggests it might not exist. Let’s re-examine the original function: $f(x) = x \sqrt{4ax - x^2}$. At $x=2a$, the radicand is $4a^2$, fine. But consider the behavior of the derivative formula: the expression $\frac{2a - x}{\sqrt{4ax - x^2}}$ in the product rule is undefined if the denominator is zero — but it isn’t. So why would the derivative not exist?

   Watch out

   A common mistake is to assume that because the derivative formula simplifies nicely, the derivative exists. But we must check the one-sided limits of the difference quotient, especially if the function has a cusp. Here, the function is smooth at $x=2a$ — it’s actually the maximum point of the quadratic inside the root, but the square root is smooth there. Let’s compute the difference quotient directly to be sure.

6. Difference quotient at $x=2a$

   $f(2a) = 2a \cdot \sqrt{4a(2a) - (2a)^2} = 2a \cdot 2a = 4a^2$.

   For $h \neq 0$, …

We are given $f(x) = \mu \sin x + \frac{1}{3} \sin 3x$ and told that its derivative is zero at $x = \frac{\pi}{3}$. This is a straightforward application of differentiation and trigonometric evaluation — but the trap is forgetting the chain rule on $\sin 3x$ or mis-evaluating $\cos \pi$ and $\cos \pi/3$.

Why this approach works:

The derivative of a sum is the sum of derivatives. For $\sin 3x$, the chain rule gives $3\cos 3x$. Then we plug in the specific $x$ and set the expression equal to zero. This yields a simple linear equation in $\mu$.

1. Differentiate $f(x)$ term by term.
   • The derivative of $\mu \sin x$ is $\mu \cos x$.
   • The derivative of $\frac{1}{3} \sin 3x$ is $\frac{1}{3} \cdot 3 \cos 3x = \cos 3x$. So

2. Apply the given condition: $f'\left(\frac{\pi}{3}\right) = 0$. Substitute $x = \frac{\pi}{3}$:

3. Evaluate the trigonometric values.
   • $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.
   • $3 \cdot \frac{\pi}{3} = \pi$, and $\cos(\pi) = -1$. So the equation becomes:

4. Solve for $\mu$. $$$$

Derivative and turning-point conditions

A turning point requires $f'(x)=0$. Applying this at the two given locations:

• At $x=-1$: $-1-2b+a=0\ \Rightarrow\ a-2b=1.$
• At $x=2$: $\frac12+4b+a=0\ \Rightarrow\ a+4b=-\frac12.$

Solve the system

Subtract the first equation from the second: …

$f(x)=-4\sin x+2x$, $f'(x)=-4\cos x+2$. Setting $f'=0$: $\cos x=\tfrac12\Rightarrow x=\tfrac{\pi}{3}\in[0,\pi/2]$.

Evaluate at the critical point and endpoints:

• $f(0)=0$
• $f(\pi/2)=-4(1)+\pi=\pi-4\approx-0.86$ …