Question

If \( f(x) = \int_0^x \left[ (a+1)(t+1)^2 - (a-1)(t^2 + t + 1) \right] dt \), then a possible positive value of \( a \), for which \( f'(x) = 0 \) has equal roots, is:

• A. 1
• B. -1
• C. 7
• D. 0

Hint:

For integrals with variable limits, use the fundamental theorem of calculus to differentiate. Ensure the discriminant of the quadratic equation is zero for equal roots.

Answer 1

The Correct Option is A

We are given the function: \[ f(x) = \int_0^x \left[ (a+1)(t+1)^2 - (a-1)(t^2 + t + 1) \right] dt \] To find \( f'(x) \), we differentiate the given integral with respect to \( x \). By the fundamental theorem of calculus: \[ f'(x) = (a+1)(x+1)^2 - (a-1)(x^2 + x + 1) \] For \( f'(x) = 0 \) to have equal roots, the discriminant of the quadratic equation must be zero. After simplifying and solving for \( a \), we find that the possible positive value of \( a \) is \( 1 \). Thus, the correct answer is option (1), \( a = 1 \).

Answer 2

Step 1: Understand the given function
\[ f(x) = \int_0^x \left[ (a+1)(t+1)^2 - (a-1)(t^2 + t + 1) \right] dt \]

Step 2: Differentiate \( f(x) \) using the Fundamental Theorem of Calculus
\[ f'(x) = (a+1)(x+1)^2 - (a-1)(x^2 + x + 1) \]

Step 3: Write \( f'(x) = 0 \) and simplify
\[ (a+1)(x+1)^2 - (a-1)(x^2 + x + 1) = 0 \]
Expand each term:
\[ (a+1)(x^2 + 2x + 1) - (a-1)(x^2 + x + 1) = 0 \]
\[ (a+1)x^2 + 2(a+1)x + (a+1) - (a-1)x^2 - (a-1)x - (a-1) = 0 \]

Step 4: Group like terms
\[ [(a+1) - (a-1)] x^2 + [2(a+1) - (a-1)] x + [(a+1) - (a-1)] = 0 \]
Calculate the coefficients:
\[ (2) x^2 + (2a + 2 + 1 - a) x + (2) = 0 \]
Simplify further:
\[ 2 x^2 + (a + 3) x + 2 = 0 \]

Step 5: Condition for equal roots
For the quadratic equation \( 2x^2 + (a+3)x + 2 = 0 \) to have equal roots, its discriminant must be zero:
\[ D = (a+3)^2 - 4 \times 2 \times 2 = 0 \]
\[ (a+3)^2 = 16 \]
\[ a + 3 = \pm 4 \]

Step 6: Find values of \( a \)
\[ a + 3 = 4 \implies a = 1 \]
\[ a + 3 = -4 \implies a = -7 \]

Step 7: Choose the positive value of \( a \)
Since the problem asks for a possible positive value, the answer is:
\[ a = 1 \]

Final Answer:
\[ \boxed{1} \]