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

Answer: (D)

72

Answer 2

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.