JEE Main 2024 — Differentiation Question with Solution

Given the polynomial function:

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

We are provided the following conditions from the problem:

1. $$f(1)=40$$

2. $$f^{\prime }(1)=2$$

3. $$f^{\prime \prime }(1)=4$$

First, calculate $f(1)$:

$$f(1)=a(1{)}^3+b(1{)}^2+c(1)+41=40$$

Simplifying, we get:

$$a+b+c+41=40$$

Therefore:

$$a+b+c=-1$$

Next, calculate the first derivative $f^{\prime }(x)$:

$$f^{\prime }(x)=3a{x}^2+2bx+c$$

Given $f^{\prime }(1)=2$:

$$f^{\prime }(1)=3a(1{)}^2+2b(1)+c=2$$

Simplifying, we get:

$$3a+2b+c=2$$

Next, calculate the second derivative $f^{\prime \prime }(x)$:

$$f^{\prime \prime }(x)=6ax+2b$$

Given $f^{\prime \prime }(1)=4$:

$$f^{\prime \prime }(1)=6a(1)+2b=4$$

Simplifying, we get:

$$6a+2b=4$$

Dividing the entire equation by 2:

$$3a+b=2$$

We now have three equations:

1. $$a+b+c=-1$$

2. $$3a+2b+c=2$$

3. $$3a+b=2$$

To solve for $a$, $b$, and $c$, follow these steps:

First, subtract the third equation from the second equation:

$$(3a+2b+c)-(3a+b)=2-2$$

Which simplifies to:

$$b+c=0$$

So,

$$c=-b$$

Substitute $c=-b$ into the first equation:

$$(a+b-b=-1)$$

Simplifying, we get:

$$a=-1$$

Now substitute $a=-1$ into the third equation:

$$3(-1)+b=2$$

Which simplifies to:

$$-3+b=2$$

Therefore:

$$b=5$$

Next, since $c=-b$:

$$c=-5$$

Finally, we need to find $a^2+b^2+c^2$:

$$a^2+b^2+c^2=(-1{)}^2+5^2+(-5{)}^2$$

Simplifying, we get:

$$1+25+25=51$$

Therefore, the answer is:

Option B: 51