JEE Main 2026 — Mathematics Calculus

Given the differential equation:

$$x\sin\left(\dfrac{y}{x}\right)dy = \left(y\sin\left(\dfrac{y}{x}\right) - x\right)dx$$xsin(xy​)dy=(ysin(xy​)−x)dx

Rearranging the terms, we get:

$$x\sin\left(\dfrac{y}{x}\right)dy - y\sin\left(\dfrac{y}{x}\right)dx = -xdx$$xsin(xy​)dy−ysin(xy​)dx=−xdx

Dividing both sides by $x^2$x2:

$$\sin\left(\dfrac{y}{x}\right)\left(\dfrac{xdy - ydx}{x^2}\right) = -\dfrac{dx}{x}$$sin(xy​)(x2xdy−ydx​)=−xdx​

$$\sin\left(\dfrac{y}{x}\right)d\left(\dfrac{y}{x}\right) = -\dfrac{dx}{x}$$sin(xy​)d(xy​)=−xdx​

Integrating both sides:

$$-\cos\left(\dfrac{y}{x}\right) = -\ln|x| - C$$−cos(xy​)=−ln∣x∣−C

$$\cos\left(\dfrac{y}{x}\right) = \ln|x| + C$$cos(xy​)=ln∣x∣+C

Using the initial condition $y(1) = \dfrac{\pi}{2}$y(1)=2π​:

$$\cos\left(\dfrac{\pi}{2}\right) = \ln(1) + C \Rightarrow C = 0$$cos(2π​)=ln(1)+C⇒C=0

Thus, the solution to the differential equation is:

$$\cos\left(\dfrac{y}{x}\right) = \ln|x|$$cos(xy​)=ln∣x∣

We need to find $\alpha = \cos\left(\dfrac{y(e^{12})}{e^{12}}\right)$α=cos(e12y(e12)​). Substituting $x = e^{12}$x=e12:

$$\alpha = \ln(e^{12}) = 12$$α=ln(e12)=12

The given equation of the circle is:

$$x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$$x2+y2−2px+2py+α+2=0

Substituting $\alpha = 12$α=12:

$$x^2 + y^2 - 2px + 2py + 14 = 0$$x2+y2−2px+2py+14=0

The radius $r$r of the circle is given by:

$$r = \sqrt{(-p)^2 + p^2 - 14} = \sqrt{2p^2 - 14}$$r=(−p)2+p2−14​=2p2−14​

For the equation to represent a real circle, $r > 0$r>0:

$$2p^2 - 14 > 0 \Rightarrow p^2 > 7$$2p2−14>0⇒p2>7

We are given that $r \leq 6$r≤6:

$$\sqrt{2p^2 - 14} \leq 6$$2p2−14​≤6

$$2p^2 - 14 \leq 36$$2p2−14≤36

$$2p^2 \leq 50 \Rightarrow p^2 \leq 25$$2p2≤50⇒p2≤25

Combining the inequalities:

$$7 < p^2 \leq 25$$7<p2≤25

Since $p$p is an integer, the possible values for $p^2$p2 are $9, 16, 25$9,16,25.

This gives $p \in \{-5, -4, -3, 3, 4, 5\}$p∈{−5,−4,−3,3,4,5}.

Therefore, there are $6$6 integral values of $p$p.

Answer: $6$6