Let f(x) = x³ + x² f'(1) + 2x f''(2) + f'''(3), x ∈ R. Then the value of f'(5) is:

Q: Wait, why can't I just differentiate the entire equation $f(x) = x^3 + x^2 f'(1) + \dots$ with respect to $x$ immediately?

You can, but you will end up with a recursive mess. By treating the derivatives as constants, you decouple the problem into a simple system of linear equations. It is the difference between trying to solve a puzzle by brute force and finding the key that unlocks the door.

Q: How do I know $f'''(3)$ is just a constant?

Because $f(x)$ is a cubic polynomial. The third derivative of any cubic polynomial is a constant (specifically $3! \times$ the leading coefficient). It does not matter what $x$ you plug in; the result is fixed. In this case, it is $6$.

Q: I keep getting confused with the signs when solving the system of equations. Any tips?

Always write your equations in standard form $Ax + By = C$ before solving. Double-check your arithmetic at each step—JEE Advanced is as much about precision as it is about concepts. If you find a negative sign, treat it like a landmine; step carefully!

Question

Let $f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3), x \in \mathbb{R}$. Then the value of $f'(5)$ is:

Options:

(A) $\frac{117}{5}$

(B) $\frac{657}{5}$

(C) $\frac{2}{5}$

(D) $\frac{62}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a massive, locked gate. It looks imposing, covered in complex symbols and intimidating derivatives. This is how many students feel when they first encounter a functional equation like:

$$f(x) = x^3 + x^2 f'(1) + 2x f''(2) + f'''(3)$$

It looks like a differential equation, but it is a paper tiger. The key to this problem is the realization that $f'(1)$, $f''(2)$, and $f'''(3)$ are not functions; they are fixed constants waiting to be discovered.

### The Transformation

Let us strip away the fear. We define the constants as follows:

$$a = f'(1), \quad b = f''(2), \quad c = f'''(3)$$

Suddenly, the equation becomes a simple, elegant cubic polynomial:

$$f(x) = x^3 + ax^2 + 2bx + c$$

This is the moment the problem shifts from a nightmare to a manageable algebraic exercise. Our mission is now clear: find the values of $a$, $b$, and $c$.

### The Derivative Chain

To find these constants, we must use the information embedded in the derivatives. We differentiate our polynomial $f(x)$ step by step:

$$f'(x) = 3x^2 + 2ax + 2b$$

$$f''(x) = 6x + 2a$$

$$f'''(x) = 6$$

Notice how the function simplifies with each step. Since $c = f'''(3)$ and $f'''(x) = 6$ for all $x$, we immediately know that **$c = 6$**.

### The System of Equations

Now we use our definitions of $a$ and $b$. We know $a = f'(1)$, so we substitute $x=1$ into our expression for $f'(x)$:

$$3(1)^2 + 2a(1) + 2b = a \quad \Rightarrow \quad a + 2b = -3$$

Next, we know $b = f''(2)$, so we substitute $x=2$ into our expression for $f''(x)$:

$$6(2) + 2a = b \quad \Rightarrow \quad 2a - b = -12$$

We now have a system of two linear equations with two variables. From the second equation, we get $b = 2a + 12$. Substituting this into the first equation yields:

$$a + 2(2a + 12) = -3$$

$$5a + 24 = -3 \quad \Rightarrow \quad 5a = -27 \quad \Rightarrow \quad a = -\frac{27}{5}$$

With $a$ in hand, finding $b$ is trivial:

$$b = 2\left(-\frac{27}{5}\right) + 12 = -\frac{54}{5} + \frac{60}{5} = \frac{6}{5}$$

### The Final Victory

We have conquered the constants: $a = -\frac{27}{5}$, $b = \frac{6}{5}$, and $c = 6$. The question asks for $f'(5)$. We return to our derivative expression:

$$f'(5) = 3(5)^2 + 2a(5) + 2b$$

Plugging in our values:

$$f'(5) = 75 + 10\left(-\frac{27}{5}\right) + 2\left(\frac{6}{5}\right)$$

$$f'(5) = 75 - 54 + \frac{12}{5} = 21 + \frac{12}{5}$$

$$f'(5) = \frac{105 + 12}{5} = \frac{117}{5}$$

The final answer is **$\frac{117}{5}$**. You have navigated the complexity, identified the constants, and solved the system. This is the essence of JEE Advanced mathematics: seeing through the noise to find the simple, beautiful truth underneath.

Animated Solution for Mathematics - Differentiation: Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3), x \in R$ . Then the value of $f^{'} (5)$ is:

Select Answer:

Visualized Solution

Identifying the Functional Constants

Rewriting the Function $f (x)$

Finding the First Derivative $f^{'} (x)$

Finding the Second and Third Derivatives

Determining the Constant $c$

Setting up Equation for $a$

Setting up Equation for $b$

Solving for $a$

Calculating the Value of $b$

Setting up the Final Calculation

Final Result

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The Transformation

The Derivative Chain

The System of Equations

The Final Victory

Similar Questions

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3),x∈R. Then the value of f′(5) is:

Select Answer:

Visualized Solution

Identifying the Functional Constants

Rewriting the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Finding the First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Finding the Second and Third Derivatives

Determining the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Setting up Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Setting up Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculating the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b