Question

If \( (a, b) \) be the orthocenter of the triangle whose vertices are \( (1, 2) \), \( (2, 3) \), and \( (3, 1) \), and \( I_1 = \int_a^b x \sin(4x - x^2) dx \), \( I_2 = \int_a^b \sin(4x - x^2) dx \), then \( 36 \frac{I_1}{I_2} \) is equal to:

• A. 80
• B. 88
• C. 66
• D. 72

Answer 1

The Correct Option is D

Given the triangle with vertices \( A(1, 2) \), \( B(2, 3) \), and \( C(3, 1) \), we proceed to find the orthocentre \( (a, b) \) which lies on the line \( x + y = 4 \).

Step 1. Equation of Line \( CE \)

The line passing through point \( C(3, 1) \) with slope \( -1 \) is given by:

\(y - 1 = -1(x - 3) \Rightarrow y = -x + 4\)

The equation of the line \( x + y = 4 \) holds for the orthocentre \( (a, b) \). Therefore:

\(a + b = 4\)

Step 2. Evaluation of the Integral \( I_1 \)

Consider the integral:

\(I_1 = \int_a^b x \sin(x(4 - x)) \, dx \quad \text{...(i)}\)

Step 3. Using the King’s Rule

By applying the King’s property of definite integrals, we have:

\(I_1 = \int_a^b (4 - x) \sin(x(4 - x)) \, dx \quad \text{...(ii)}\)

Step 4. Combining the Results

Adding equations (i) and (ii), we obtain:

\(I_1 + I_1 = \int_a^b (x + (4 - x)) \sin(x(4 - x)) \, dx\)
Simplifying:

\(2I_1 = \int_a^b 4 \sin(x(4 - x)) \, dx\)

Therefore:

\(I_1 = 2 \int_a^b \sin(x(4 - x)) \, dx\)

Step 5. Ratio of Integrals

From the problem statement, we have:

\(\frac{I_1}{I_2} = 2\)

Calculating:

\(36 \times \frac{I_1}{I_2} = 36 \times 2 = 72\)
Hence, the value of \(36 \frac{I_1}{I_2}\) is 72.