Let $f(x) = ax^3 + bx^2 + cx + 41$ be such that $f(1) = 40, f'(1) = 2$ and $f''(1) = 4$. Then $a^2 + b^2 + c^2$ is equal to: Options: (A) 73 (B) 62 (C) 51 (D) 54

### Analyzing the Setup The polynomial is defined as $f(x) = ax^3 + bx^2 + cx + 41$. To analyze its behavior, we determine its derivatives: $$f'(x) = 3ax^2 + 2bx + c$$ $$f''(x) = 6ax + 2b$$ These derivatives serve as our analytical lenses, allowing us to evaluate the slope and curvature of the polynomial at any given point. ### Unlocking the System We utilize the provided keys at $x=1$ to establish a system of linear equations. First, using $f''(1) = 4$: $$6a(1) + 2b = 4 \implies 3a + b = 2$$ Next, using $f'(1) = 2$: $$3a(1)^2 + 2b(1) + c = 2 \implies 3a + 2b + c = 2$$ Finally, using $f(1) = 40$: $$a(1)^3 + b(1)^2 + c(1) + 41 = 40 \implies a + b + c = -1$$ We now have the following system of equations: 1) $3a + b = 2$ 2) $3a + 2b + c = 2$ 3) $a + b + c = -1$ ### The Elegance of Elimination We observe that subtracting Equation (1) from Equation (2) eliminates the $3a$ term: $$(3a + 2b + c) - (3a + b) = 2 - 2 \implies b + c = 0$$ This reveals the relationship $c = -b$. Substituting this into Equation (3): $$a + (b + c) = -1 \implies a + 0 = -1 \implies a = -1$$ ### The Final Reveal With $a = -1$ identified, we substitute this value back into Equation (1): $$3(-1) + b = 2 \implies -3 + b = 2 \implies b = 5$$ Given $b + c = 0$, it follows that $c = -5$. We have successfully determined the coefficients: $a = -1$, $b = 5$, and $c = -5$. The final objective is to calculate the value of $a^2 + b^2 + c^2$: $$(-1)^2 + (5)^2 + (-5)^2 = 1 + 25 + 25 = 51$$ The value of the expression is **51**.

JEE Main 2024 (09 Apr Shift 1)

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Animated Solution for Mathematics - Differentiation: Let $f (x) = a x^{3} + b x^{2} + c x + 41$ be such that $f (1) = 40, f^{'} (1) = 2$ and $f^{''} (1) = 4$ . Then $a^{2} + b^{2} + c^{2}$ is equal to:

Q: Wait, why do we need to find the derivatives first? Can't we just solve for $a, b, c$ using $f(1)$?

Great question! If you only use $f(1)=40$, you get one equation: $a + b + c + 41 = 40$, which simplifies to $a + b + c = -1$. This is one equation with three unknowns. You cannot solve for three variables with only one equation. The derivatives $f'(1)$ and $f''(1)$ provide the two additional, independent equations needed to solve the system uniquely.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. You are dealing with a polynomial

f (x) = a x^{3} + b x^{2} + c x + 41

. Notice that you have three unknown coefficients (

a, b, c

) and three pieces of information (

f (1)

f^{'} (1)

, and

f^{''} (1)

). This is a classic system of linear equations problem. Your goal is to translate these three conditions into three distinct algebraic equations.

Hint 2

Hint 2: The Arsenal. Remember that differentiation is a linear operator. To find

f^{'} (x)

and

f^{''} (x)

, simply apply the power rule to each term of

f (x)

. Once you have the general expressions for the derivatives, substitute

x = 1

into them to create your system of equations.

Hint 3

Hint 3: The Execution. Start by writing out

f^{'} (x) = 3 a x^{2} + 2 b x + c

and

f^{''} (x) = 6 a x + 2 b

. Now, substitute

x = 1

into these expressions and set them equal to the given values

f^{'} (1) = 2

and

f^{''} (1) = 4

. This will give you two equations. The third equation comes from

f (1) = 40

. Solve this system for

a, b,

and

c

Visualized Solution

Analyzing the Given Polynomial

Given: $f (x) = a x^{3} + b x^{2} + c x + 41$
Conditions: $f (1) = 40$ , $f^{'} (1) = 2$ , $f^{''} (1) = 4$
Goal: Find the value of $a^{2} + b^{2} + c^{2}$

Finding the Derivatives

First derivative: $f^{'} (x) = 3 a x^{2} + 2 b x + c$
Second derivative: $f^{''} (x) = 6 a x + 2 b$

Applying the Second Derivative Condition

Given: $f^{''} (1) = 4$
Substitute $x = 1$ : $6 a (1) + 2 b = 4$
Divide by 2: $\frac{6 a + 2 b}{2} = \frac{4}{2}$
Simplify: $3 a + b = 2$ --- (Equation 1)

Applying the First Derivative Condition

Given: $f^{'} (1) = 2$
Substitute $x = 1$ : $3 a (1)^{2} + 2 b (1) + c = 2$
Simplify: $3 a + 2 b + c = 2$ --- (Equation 2)

Applying the Function Value Condition

Given: $f (1) = 40$
Substitute $x = 1$ : $a (1)^{3} + b (1)^{2} + c (1) + 41 = 40$
Simplify: $a + b + c = - 1$ --- (Equation 3)

Finding a Relation Between $b$ and $c$

Equation 1: $3 a + b = 2$
Equation 2: $3 a + 2 b + c = 2$
Subtract Eq 1 from Eq 2: $(3 a + 2 b + c) - (3 a + b) = 2 - 2$
Result: $b + c = 0$

Solving for $a$

Equation 3: $a + b + c = - 1$
Substitute $b + c = 0$ into Equation 3
$a + 0 = - 1$
Result: $a = - 1$

Solving for $b$ and $c$

Substitute $a = - 1$ into Equation 1: $3 (- 1) + b = 2$
$- 3 + b = 2 ⟹ b = 5$
Substitute $b = 5$ into $b + c = 0$
$5 + c = 0 ⟹ c = - 5$

Final Calculation: $a^{2} + b^{2} + c^{2}$

Values found: $a = - 1$ , $b = 5$ , $c = - 5$
Calculate: $(- 1)^{2} + (5)^{2} + (- 5)^{2}$
$= 1 + 25 + 25$
$= 51$

00:00 / 00:00

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The polynomial is defined as

f (x) = a x^{3} + b x^{2} + c x + 41

. To analyze its behavior, we determine its derivatives:

f^{'} (x) = 3 a x^{2} + 2 b x + c

f^{''} (x) = 6 a x + 2 b

These derivatives serve as our analytical lenses, allowing us to evaluate the slope and curvature of the polynomial at any given point.

Unlocking the System

We utilize the provided keys at

x = 1

to establish a system of linear equations. First, using

f^{''} (1) = 4

6 a (1) + 2 b = 4 ⟹ 3 a + b = 2

Next, using

f^{'} (1) = 2

3 a (1)^{2} + 2 b (1) + c = 2 ⟹ 3 a + 2 b + c = 2

Finally, using

f (1) = 40

a (1)^{3} + b (1)^{2} + c (1) + 41 = 40 ⟹ a + b + c = - 1

We now have the following system of equations:

3 a + b = 2

3 a + 2 b + c = 2

a + b + c = - 1

The Elegance of Elimination

We observe that subtracting Equation (1) from Equation (2) eliminates the

3 a

term:

(3 a + 2 b + c) - (3 a + b) = 2 - 2 ⟹ b + c = 0

This reveals the relationship

c = - b

. Substituting this into Equation (3):

a + (b + c) = - 1 ⟹ a + 0 = - 1 ⟹ a = - 1

The Final Reveal

With

a = - 1

identified, we substitute this value back into Equation (1):

3 (- 1) + b = 2 ⟹ - 3 + b = 2 ⟹ b = 5

Given

b + c = 0

, it follows that

c = - 5

. We have successfully determined the coefficients:

a = - 1

b = 5

, and

c = - 5

The final objective is to calculate the value of

a^{2} + b^{2} + c^{2}

(- 1)^{2} + (5)^{2} + (- 5)^{2} = 1 + 25 + 25 = 51

The value of the expression is 51.

Common Traps & FAQs

Wait, why do we need to find the derivatives first? Can't we just solve for

a, b, c

using

f (1)

Great question! If you only use

f (1) = 40

, you get one equation:

a + b + c + 41 = 40

, which simplifies to

a + b + c = - 1

. This is one equation with three unknowns. You cannot solve for three variables with only one equation. The derivatives

f^{'} (1)

and

f^{''} (1)

provide the two additional, independent equations needed to solve the system uniquely.

Is there a faster way to solve this system than substitution?

You could use matrix methods like Cramer's Rule or Gaussian elimination, but for a

3 \times 3

system like this, simple substitution and elimination are often the fastest and least error-prone. The structure of these equations (where

3 a

appears in both derivative equations) makes elimination particularly elegant.

Why did we subtract Equation 1 from Equation 2?

We subtracted them because both equations contained the term

3 a

. By subtracting, we eliminated the

a

variable entirely, which allowed us to find a direct relationship between

b

and

c

. This is a powerful technique in solving systems of equations—always look for terms that can be cancelled out to simplify your life!

Animated Solution for Mathematics - Differentiation: Let $f (x) = a x^{3} + b x^{2} + c x + 41$ be such that $f (1) = 40, f^{'} (1) = 2$ and $f^{''} (1) = 4$ . Then $a^{2} + b^{2} + c^{2}$ is equal to:

Select Answer:

Visualized Solution

Analyzing the Given Polynomial

Finding the Derivatives

Applying the Second Derivative Condition

Applying the First Derivative Condition

Applying the Function Value Condition

Finding a Relation Between $b$ and $c$

Solving for $a$

Solving for $b$ and $c$

Final Calculation: $a^{2} + b^{2} + c^{2}$

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Unlocking the System

The Elegance of Elimination

The Final Reveal

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)=ax3+bx2+cx+41 be such that f(1)=40,f′(1)=2 and f′′(1)=4. Then a2+b2+c2 is equal to:

Select Answer:

Visualized Solution

Analyzing the Given Polynomial

Finding the Derivatives

Applying the Second Derivative Condition

Applying the First Derivative Condition

Applying the Function Value Condition

Finding a Relation Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

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

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2+b2+c2

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

Unlocking the System

The Elegance of Elimination

The Final Reveal

Similar Questions

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Animated Solution for Mathematics - Differentiation: Let $f (x) = a x^{3} + b x^{2} + c x + 41$ be such that $f (1) = 40, f^{'} (1) = 2$ and $f^{''} (1) = 4$ . Then $a^{2} + b^{2} + c^{2}$ is equal to:

Finding a Relation Between $b$ and $c$

Solving for $a$

Solving for $b$ and $c$

Final Calculation: $a^{2} + b^{2} + c^{2}$

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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:

If $f (x) = x^{3} - x^{2} f^{'} (1) + x f^{''} (2) - f^{'''} (3), x \in R$ , then

Let $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f^{'} (10)$ is equal to

There exist a function $f (x)$ , satisfying $f (0) = 1, f^{'} (0) = - 1, f (x) > 0$ for all $x$ , and

Let $f$ be a twice differentiable function such that $f^{''} (x) = - f (x)$ , and $f^{'} (x) = g (x), h (x) = [f (x)]^{2} + [g (x)]^{2}$ . Find $h (10)$ if $h (5) = 11$ .

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If $y (θ) = \frac{2 c o s θ + c o s 2 θ}{c o s 3 θ + 4 c o s 2 θ + 5 c o s θ + 2}$ , then at $θ = \frac{π}{2}$ , $y^{''} + y^{'} + y$ is equal to :