If (a, b) be the orthocentre of the triangle whose vertices are (1,2),(2,3) and (3,1) , and I_1=\int\limits_a^b x \sin (4 x-x^2) d x, I_2=\int\limits_a^b \sin (4 x-x^2) d x , then 36 \fracI_1I_2 is equal to :

Definite Integration PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q35 with Solution

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Correct answer: B

The vertices are: $A(1,2),\quad B(2,3),\quad C(3,1).$

We find two altitudes.

Slope of $BC$ : $m_{BC}=\frac{1-3}{3-2}=-2.$ So the altitude from $A$ has slope $\frac{1}{2}$ . Its equation is $y-2=\frac12(x-1).$
Slope of $AC$ : $m_{AC}=\frac{1-2}{3-1}=-\frac12.$ So the altitude from $B$ has slope $2$ . Its equation is $y-3=2(x-2).$

Now solve these two equations:

From the first, $y=2+\frac{x-1}{2}=\frac{x+3}{2}.$

From the second, $y=3+2x-4=2x-1.$

Equating, $\frac{x+3}{2}=2x-1$ $x+3=4x-2$ $5=3x$ $x=\frac53.$

Then $y=2x-1=\frac{10}{3}-1=\frac73.$

Hence, $a=\frac53,\qquad b=\frac73.$

\qquad I_2=\int_a^b \sin(4x-x^2)\,dx.$$ Substitute $a=\frac53$, $b=\frac73$. Also note: $$4x-x^2=4-(x-2)^2.$$ The limits are symmetric about $x=2$ because $$\frac53=2-\frac13,\qquad \frac73=2+\frac13.$$ --- 3. **Use symmetry** Let $$u=x-2 \Rightarrow x=u+2,\quad dx=du.$$ Then as $x$ goes from $\frac53$ to $\frac73$, $u$ goes from $-\frac13$ to $\frac13$. Also, $$4x-x^2=4-u^2.$$ So, $$I_2=\int_{-1/3}^{1/3} \sin(4-u^2)\,du.$$ For $I_1$, $$I_1=\int_{-1/3}^{1/3} (u+2)\sin(4-u^2)\,du.$$ Thus, $$I_1=\int_{-1/3}^{1/3} u\sin(4-u^2)\,du+2\int_{-1/3}^{1/3}\sin(4-u^2)\,du.$$ Now, $\sin(4-u^2)$ is an even function of $u$, so $$u\sin(4-u^2)$$ is odd. Therefore, $$\int_{-1/3}^{1/3} u\sin(4-u^2)\,du=0.$$ Hence, $$I_1=2I_2.$$ So, $$\frac{I_1}{I_2}=2.$$ --- 4. **Compute the required value** $$36\cdot \frac{I_1}{I_2}=36\cdot 2=72.$$ --- 5. **Compare with options** The correct option is: $$\boxed{72}$$ which is **Option B**. --- 6. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.