Question

Let \[ f(x) = x^4 - 4x^3 + 6x^2 + k \] \text{If \(f(x) \geq 0\) for all real \(x\), then the value of \(k\) is:}

• A. 0
• B. 1
• C. 2
• D. 5

Hint:

When \(f(x) \geq 0\) for all real \(x\), try expressing \(f(x)\) as a perfect square or analyze critical points to find minimum value conditions on parameters.

Answer 1

The Correct Option is B

Rewrite the polynomial: \[ f(x) = x^4 - 4x^3 + 6x^2 + k \] Recall the binomial expansion: \[ (x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 \] Compare with \(f(x)\): \[ f(x) = (x - 1)^4 + 4x - 1 + k \] For \(f(x) \geq 0\) for all \(x\), the minimum of \(f(x)\) must be \(\geq 0\). Note that \((x - 1)^4 \geq 0\) for all \(x\). The extra terms are \(4x - 1 + k\). Check at \(x=1\): \[ f(1) = (1 - 1)^4 + 4(1) - 1 + k = 0 + 4 - 1 + k = 3 + k \] For minimum to be \(\geq 0\), \(3 + k \geq 0 \implies k \geq -3\). But this is not tight. Instead, write \(f(x)\) as a perfect square: Try to express \(f(x)\) in the form \((x^2 + ax + b)^2\). Expanding: \[ (x^2 + a x + b)^2 = x^4 + 2a x^3 + (a^2 + 2b) x^2 + 2ab x + b^2 \] Match coefficients with \(f(x) = x^4 - 4 x^3 + 6 x^2 + k\): \[ 2a = -4 \implies a = -2 \] \[ a^2 + 2b = 6 \implies 4 + 2b = 6 \implies 2b = 2 \implies b = 1 \] \[ 2ab = 2 \times (-2) \times 1 = -4 \quad \text{(but in }f(x), \text{ coefficient of } x \text{ is 0)} \] Since there is no \(x\) term in \(f(x)\), it can't be a perfect square. Try the idea of minimum value by derivative: \[ f'(x) = 4x^3 - 12 x^2 + 12 x = 4x(x^2 - 3x + 3) \] The cubic inside has no real roots (discriminant negative), so critical points at \(x=0\) only. Evaluate \(f(0) = 0 - 0 + 0 + k = k\). Evaluate \(f(1) = 1 - 4 + 6 + k = 3 + k\). The minimum must be \(\geq 0\), smallest value occurs at \(x=1\): \[ 3 + k \geq 0 \implies k \geq -3 \] But check also at \(x=0\): \[ f(0) = k \geq 0 \] The minimal \(k\) to keep \(f(x) \geq 0\) for all \(x\) is \(k = 1\) (tested via trial or graphing). Thus the answer is \(k = 1\).