Question

For the function \(f(x) = x^4 - x^2\), which of the following statements is/are TRUE?

• A. The function is symmetric about \(x = 0\).
• B. The minimum value of the function is \(-0.5\).
• C. The function has two minima.
• D. The function is an odd function.

Hint:

- Even functions satisfy \(f(-x)=f(x)\), odd functions satisfy \(f(-x)=-f(x)\). - Use first and second derivative tests to identify local maxima/minima. - Always substitute back into the original function to check exact minimum/maximum values.

Answer 1

The Correct Option is A, C

Step 1: Check symmetry. We test \(f(-x)\): \[ f(-x) = (-x)^4 - (-x)^2 = x^4 - x^2 = f(x). \] Thus, \(f(x)\) is an even function, symmetric about \(x=0\). Hence (A) is true. Step 2: Find critical points. Differentiate: \[ f'(x) = 4x^3 - 2x = 2x(2x^2 - 1). \] So critical points are \(x=0\) and \(x=\pm \tfrac{1}{\sqrt{2}}\). Step 3: Classify critical points. Second derivative: \[ f''(x) = 12x^2 - 2. \] - At \(x=0\): \(f''(0) = -2<0 \Rightarrow\) local maximum. - At \(x=\pm \tfrac{1}{\sqrt{2}}\): \(f''(\pm \tfrac{1}{\sqrt{2}}) = 12\cdot \tfrac{1}{2} - 2 = 6 - 2 = 4>0\). Hence local minima. Step 4: Minimum value. \[ f\left(\pm \tfrac{1}{\sqrt{2}}\right) = \left(\tfrac{1}{\sqrt{2}}\right)^4 - \left(\tfrac{1}{\sqrt{2}}\right)^2 = \tfrac{1}{4} - \tfrac{1}{2} = -\tfrac{1}{4} = -0.25 \] So the actual minimum value is \(-0.25\), not \(-0.5\). Thus, statement (B) is False. Step 5: Check other statements. - (C) Two minima exist at \(x=\pm \tfrac{1}{\sqrt{2}}\). This is True.
- (D) Since the function is even, not odd, this is False. Final Answer: (A) and (C)