Question

Let \( f:\mathbb{R}\rightarrow \mathbb{R} \) be a function such that \( f(x)=x^{3} + x^{2}f^{\prime}(1) + x f^{\prime\prime}(2) + 6 \) for \( x \in \mathbb{R} \), then \( f(2) \) equals

• A. 30
• B. -4
• C. -2
• D. 8

Hint:

Calculus Tip:When a function's definition includes its own derivatives evaluated at specific points (like $f'(1)$ or $f''(2)$), treat those terms strictly as unknown constants, then differentiate and substitute those specific $x$-values to form a solvable system of linear equations.

Answer 1

The Correct Option is C

Concept: Calculus - Higher Order Derivatives and Function Evaluation.

Step 1: Find the first derivative of the function. Differentiate the given function $f(x)$ with respect to $x$. Remember that $f^{\prime}(1)$ and $f^{\prime\prime}(2)$ are simply constant values. $f^{\prime}(x) = 3x^{2} + 2xf^{\prime}(1) + f^{\prime\prime}(2)$. Let's call this Equation (i).

Step 2: Find the second derivative of the function. Differentiate the first derivative $f^{\prime}(x)$ to get the second derivative. $f^{\prime\prime}(x) = 6x + 2f^{\prime}(1)$. Let's call this Equation (ii).

Step 3: Evaluate the first derivative at $x=1$. Substitute $x=1$ into Equation (i) to create a linear equation involving our unknown constants: $f^{\prime}(1) = 3(1)^{2} + 2(1)f^{\prime}(1) + f^{\prime\prime}(2)$. Rearranging the terms yields $f^{\prime}(1) + f^{\prime\prime}(2) = -3$. Let's call this Equation (iii).

Step 4: Evaluate the second derivative at $x=2$. Substitute $x=2$ into Equation (ii) to create another linear equation: $f^{\prime\prime}(2) = 6(2) + 2f^{\prime}(1)$. This simplifies to $f^{\prime\prime}(2) = 12 + 2f^{\prime}(1)$. Let's call this Equation (iv).

Step 5: Solve the system of equations and calculate $f(2)$. We now have a system of two linear equations, (iii) and (iv). We can substitute the expression for $f^{\prime\prime}(2)$ from (iv) into (iii): $f^{\prime}(1) + (12 + 2f^{\prime}(1)) = -3$. Combining like terms gives $3f^{\prime}(1) + 12 = -3$, which leads to $3f^{\prime}(1) = -15$, so $f^{\prime}(1) = -5$.

Now, substitute this value back into Equation (iii) to find the other constant: $-5 + f^{\prime\prime}(2) = -3$, meaning $f^{\prime\prime}(2) = 2$.

With both constants identified, we can write the complete, exact original function: $f(x) = x^{3} - 5x^{2} + 2x + 6$. Finally, we substitute $x=2$ into this complete function to find the required value: $f(2) = (2)^{3} - 5(2)^{2} + 2(2) + 6 = 8 - 20 + 4 + 6 = 18 - 20 = -2$. $$ \therefore \text{The value of } f(2) \text{ is } -2. $$