Question

If \[ y=x^3-ax^2+48x+7 \] is an increasing function for all real values of \(x\), then \(a\) lies in the interval

• A. \((-14,14)\)
• B. \((-12,12)\)
• C. \((-16,16)\)
• D. \((-21,-21)\)

Hint:

For a polynomial function to be increasing for all real \(x\), check whether its derivative is non-negative for all real \(x\). For a quadratic derivative \(Ax^2+Bx+C\), with \(A\gt 0\), use the condition \(D\leq 0\).

Answer 1

The Correct Option is B

Step 1: Use the condition for increasing function.
For a function to be increasing for all real values of \(x\), we must have \[ \frac{dy}{dx}\geq 0 \] for all real \(x\).
Given, \[ y=x^3-ax^2+48x+7 \] Differentiate with respect to \(x\): \[ \frac{dy}{dx}=3x^2-2ax+48 \]

Step 2: Apply non-negative quadratic condition.
We need \[ 3x^2-2ax+48\geq 0 \] for all real \(x\).
Since the coefficient of \(x^2\) is positive, this quadratic will be non-negative for all real \(x\) if its discriminant is less than or equal to zero.
So, \[ D\leq 0 \]

Step 3: Find the discriminant.
For \[ 3x^2-2ax+48, \] we have \[ A=3,\qquad B=-2a,\qquad C=48 \] Thus, \[ D=B^2-4AC \] \[ D=(-2a)^2-4(3)(48) \] \[ D=4a^2-576 \] Now, \[ 4a^2-576\leq 0 \] \[ 4a^2\leq 576 \] \[ a^2\leq 144 \] \[ -12\leq a\leq 12 \]

Step 4: Select the correct interval.
The value of \(a\) lies in \[ [-12,12] \] From the given options, the corresponding interval is \[ (-12,12) \]

Step 5: Final conclusion.
Therefore, \[ \boxed{(-12,12)} \]