Question

For K∈R, let 𝑓(𝑥)=𝑥⁴+2𝑥³+𝑘𝑥²−𝑘, X∈R. If 𝑥=\(\frac{3}{2}\) is a point of local minima of 𝑓 and 𝑚 is the global minimum value of 𝑓 then 𝑓(0) − 𝑚 is equal to _______ (in integer).

Answer 1

Correct Answer: 54

Given: 
\(f(x)=x^4+2x^3+kx^2-k,\; x\in\mathbb{R}.\) It is given that \(x=\dfrac{3}{2}\) is a point of local minimum. Let \(m\) be the global minimum value. Find \(f(0)-m\). 

Step 1 — Use the stationary condition at \(x=\tfrac{3}{2}\)
Compute derivative: \[ f'(x)=4x^3+6x^2+2kx. \] Since \(x=\tfrac{3}{2}\) is a stationary point, \(f'\!\big(\tfrac{3}{2}\big)=0\). Thus \[ 4\left(\tfrac{3}{2}\right)^3 + 6\left(\tfrac{3}{2}\right)^2 + 2k\left(\tfrac{3}{2}\right)=0. \] Evaluate powers: \[ \left(\tfrac{3}{2}\right)^2=\tfrac{9}{4},\qquad \left(\tfrac{3}{2}\right)^3=\tfrac{27}{8}. \] So \[ 4\cdot\tfrac{27}{8} + 6\cdot\tfrac{9}{4} + 3k = 0 \quad\Rightarrow\quad \frac{27}{2} + \frac{54}{4} + 3k = 0. \] Simplify \(\frac{54}{4}=\frac{27}{2}\), so \[ \frac{27}{2}+\frac{27}{2}+3k=0 \quad\Rightarrow\quad 27+3k=0. \] Hence \[ 3k=-27 \quad\Rightarrow\quad k=-9. \] 

Step 2 — Substitute \(k=-9\) into \(f(x)\)
\[ f(x)=x^4+2x^3-9x^2+9. \] Step 3 — Find all critical points and identify global minimum
Compute \(f'(x)\) with \(k=-9\): \[ f'(x)=4x^3+6x^2-18x = 2x(2x^2+3x-9). \] Solve \(f'(x)=0\): \[ x=0,\quad 2x^2+3x-9=0. \] Quadratic roots: discriminant \(\Delta=3^2-4\cdot2\cdot(-9)=9+72=81\). So \[ x=\frac{-3\pm9}{4} \implies x=\frac{6}{4}=\tfrac{3}{2}\quad\text{or}\quad x=\frac{-12}{4}=-3. \] Thus critical points: \(x=-3,\,0,\,\tfrac{3}{2}\). Evaluate \(f\) at these points: \[ f\!\left(\tfrac{3}{2}\right) = \left(\tfrac{3}{2}\right)^4 +2\left(\tfrac{3}{2}\right)^3 -9\left(\tfrac{3}{2}\right)^2 +9 = \tfrac{81}{16} + 2\cdot\tfrac{27}{8} -9\cdot\tfrac{9}{4} +9. \] Compute stepwise: \[ \tfrac{81}{16}=5.0625,\quad 2\cdot\tfrac{27}{8}= \tfrac{27}{4}=6.75,\quad -9\cdot\tfrac{9}{4}=-\tfrac{81}{4}=-20.25. \] So \[ f\!\left(\tfrac{3}{2}\right)=5.0625+6.75-20.25+9=0.5625. \] \[ f(0)=0+0+0+9=9. \] \[ f(-3)=(-3)^4+2(-3)^3-9(-3)^2+9 =81-54-81+9=-45. \] So the smallest (global minimum) value is \(m=f(-3)=-45\). 

Step 4 — Compute \(f(0)-m\)
\[ f(0)-m = 9 - (-45) = 54. \] 
Final Answer: \(\boxed{54}\)