Let $f(x)=a x^3+b x^2+c x+41$ be such that $f(1)=40, f^{\prime}(1)=2$ and $f^{\prime \prime}(1)=4$. Then $a^2+b^2+c^2$ is equal to:

<p>Given the polynomial function:</p> <p>$f(x) = ax^3 + bx^2 + cx + 41$</p> <p>We are provided the following conditions from the problem:</p> <p>1. $f(1) = 40$</p> <p>2. $f^{\prime}(1) = 2$</p> <p>3. $f^{\prime \prime}(1) = 4$</p> <p>First, calculate $f(1)$:</p> <p>$f(1) = a(1)^3 + b(1)^2 + c(1) + 41 = 40$</p> <p>Simplifying, we get:</p> <p>$a + b + c + 41 = 40$</p> <p>Therefore:</p> <p>$a + b + c = -1$</p> <p>Next, calculate the first derivative $f^{\prime}(x)$:</p> <p>$f^{\prime}(x) = 3ax^2 + 2bx + c$</p> <p>Given $f^{\prime}(1) = 2$:</p> <p>$f^{\prime}(1) = 3a(1)^2 + 2b(1) + c = 2$</p> <p>Simplifying, we get:</p> <p>$3a + 2b + c = 2$</p> <p>Next, calculate the second derivative $f^{\prime \prime}(x)$:</p> <p>$f^{\prime \prime}(x) = 6ax + 2b$</p> <p>Given $f^{\prime \prime}(1) = 4$:</p> <p>$f^{\prime \prime}(1) = 6a(1) + 2b = 4$</p> <p>Simplifying, we get:</p> <p>$6a + 2b = 4$</p> <p>Dividing the entire equation by 2:</p> <p>$3a + b = 2$</p> <p>We now have three equations:</p> <p>1. $a + b + c = -1$</p> <p>2. $3a + 2b + c = 2$</p> <p>3. $3a + b = 2$</p> <p>To solve for $a$, $b$, and $c$, follow these steps:</p> <p>First, subtract the third equation from the second equation:</p> <p>$(3a + 2b + c) - (3a + b) = 2 - 2$</p> <p>Which simplifies to:</p> <p>$b + c = 0$</p> <p>So,</p> <p>$c = -b$</p> <p>Substitute $c = -b$ into the first equation:</p> <p>$(a + b - b = -1)$</p> <p>Simplifying, we get:</p> <p>$a = -1$</p> <p>Now substitute $a = -1$ into the third equation:</p> <p>$3(-1) + b = 2$</p> <p>Which simplifies to:</p> <p>$-3 + b = 2$</p> <p>Therefore:</p> <p>$b = 5$</p> <p>Next, since $c = -b$:</p> <p>$c = -5$</p> <p>Finally, we need to find $a^2 + b^2 + c^2$:</p> <p>$a^2 + b^2 + c^2 = (-1)^2 + 5^2 + (-5)^2$</p> <p>Simplifying, we get:</p> <p>$1 + 25 + 25 = 51$</p> <p>Therefore, the answer is:</p> <p><strong>Option B: 51</strong></p>