Question

The curve \( ax^3+bx^2+cx+d \) has a point of minima at \( x=1 \), then

• A. \(3a+b<0\)
• B. \(3a+b>0\)
• C. \(3a+b=0\)
• D. \(a+3b>0\)

Hint:

For minima, use the second derivative test: if \(f'(x)=0\) and \(f''(x)>0\), then the point is a local minimum.

Answer 1

The Correct Option is B

Step 1: Write the given curve.
\[ y=ax^3+bx^2+cx+d. \] 

Step 2: Recall condition for minima. 
For a function to have a point of minima at \(x=1\), we need: 
\[ \frac{dy}{dx}=0 \] and 
\[ \frac{d^2y}{dx^2}>0 \] at \(x=1\). 

Step 3: Find first derivative. 
\[ \frac{dy}{dx}=3ax^2+2bx+c. \]

Step 4: Find second derivative. 
\[ \frac{d^2y}{dx^2}=6ax+2b. \]

Step 5: Apply second derivative condition at \(x=1\). 
\[ \left(\frac{d^2y}{dx^2}\right)_{x=1}=6a+2b. \]
For minima: 
\[ 6a+2b>0. \]

Step 6: Simplify the inequality. 
\[ 2(3a+b)>0. \]
Since \(2>0\), we get: 
\[ 3a+b>0. \]

Step 7: Final conclusion. 
Thus, the required condition is: 
\[ \boxed{3a+b>0}. \]