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Let \( y=y(x) \) be the solutions of the differential equation \( (\tan x)^{\tfrac{1}{2}}\,dy=\left(\sec^3 x-(\tan x)^{\tfrac{3}{2}}\,y\right)dx \),\(0 < x < \frac{\pi}{2},\)\(y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}.\) If \(y\left(\frac{\pi}{3}\right) = \frac{4}{5}\alpha,\) \(\text{then } \alpha^4\) equals_______________

Let \( y = y(x) \) be the solution of the differential equation \( x \, \frac{dy}{dx} - \sin 2y = x^3 (2 - x^3) \cos^2 y, \, x \neq 0 \). If \( y(2) = 0 \), then \( \tan (y(1)) \) is equal to:

Let \( y=y(x) \) be the solution of the differential equation \( \left(x^2-x\sqrt{x^2-1}\right)\,dy+\left(y\left(x-\sqrt{x^2-1}\right)-x\right)\,dx=0,\ x\ge 1 \). If \( y(1)=1 \), then the greatest integer less than \( y(\sqrt{5}) \) is ______.

Let \( y = y(x) \) be the solution of the differential equation \( x \sin\!\left(\frac{y}{x}\right) \, dy = \left( y \sin\!\left(\frac{y}{x}\right) - x \right) dx \), \( y(1) = \frac{\pi}{2} \) and let \( \alpha = \cos\!\left( \frac{y(e^{12})}{e^{12}} \right) \). Then the number of integral value of \( p \), for which the equation \( x^{2} + y^{2} - 2px + 2py + \alpha + 2 = 0 \) represents a circle of radius \( r \leq 6 \), is ______.

If the curve \( y=f(x) \) passes through the point \( (1,e) \) and satisfies the differential equation \( \frac{dy}{y}=(2+\log_e x)dx,\, x>0 \), then \( f(e) \) is equal to:

For the function \( f(x) = e^{\sin|x|} - |x|, \; x \in \mathbb{R} \), consider the following statements:
Statement-I: \( f \) is differentiable for all \( x \in \mathbb{R} \).
Statement-II: \( f \) is increasing in \( \left(-\pi, \frac{\pi}{2}\right) \).
In the light of the above statements, choose the correct answer from the options given below:

Let \( f \) be a twice differentiable non-negative function such that \( (f(x))^{2} = 25 + \int_{0}^{x}\left((f(t))^{2} + (f'(t))^{2}\right)\,dt \). Then the mean of \( f(\log_{e}(1)),\ f(\log_{e}(2)),\ \ldots,\ f(\log_{e}(625)) \) is equal to ______.

Let \( y = y(x) \) be the solution of the differential equation \( \sec x \, \frac{dy}{dx} - 2y = 2 + 3\sin x, \; x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) and \( y(0) = -\frac{7}{4} \). Then \( y\!\left(\frac{\pi}{6}\right) \) is equal to:

Let \( x = x(y) \) be the solution of the differential equation \( 2y^2 \frac{dx}{dy} - 2xy + x^2 = 0, \; y > 1, x(e) = e \). Then \( x(e^2) \) is equal to:

If \( y = y(x) \) satisfies the differential equation \( 16\left(\sqrt{{x}+{9}+\sqrt{x}}\right)\!\left(4+\sqrt{9+\sqrt{x}}\right)\cos y\,dy = (1+2\sin y)\,dx,\ x>0 \) and \( y(256)=\frac{\pi}{2},\ y(49)=\alpha \), then \( 2\sin\alpha \) is equal to:

Let \( y = y(x) \) be the solution curve of the differential equation \( (1 + x^2)\,dy + \left(y - \tan^{-1}x\right)dx = 0,\, y(0) = 1 \). Then the value of \( y(1) \) is:

Let \( y = y(x) \) be the solution of the differential equation:
\( \dfrac{dy}{dx} + \left( \dfrac{6x^2 + (3x^2 - 2x^3 - 4)e^{-2x}}{(x^3 + 2)(2 + e^{-2x})} \right) y = 2 + e^{-2x}, \; x \in (-1, 2), \) satisfying \( y(0) = \dfrac{3}{2} \). If \( y(1) = \alpha(2 + e^{-2}) \), then \( \alpha \) is equal to:

If the solution curve \( y=f(x) \) of the differential equation \( (x^{2}-4)y'-2xy+2x(4-x^{2})^{2}=0,\ x>2 \) passes through the point \( (3,15) \), then the local maximum value of \( f \) is \( \_\_\_\_\_\_\_\_ \).

Let \( y=y(x) \) be a differentiable function in the interval \( (0,\infty) \) such that \( y(1)=2 \), and \( \lim_{t\to x} \left(\dfrac{t^{2}y(x)-x^{2}y(t)}{x-t}\right)=3 \) for each \( x>0 \). Then \( 2y(2) \) is equal to:

If the roots of the quadratic equation \( ax^2 + bx + 6 = 0 \) are equal, then minimum value of \( (6a - b + 2) \) is: (\( a, b \in \mathbb{R} \))

Let \( f:\mathbb{R}\to \mathbb{R} \) be a twice differentiable function such that the quadratic equation \( f(x)m^{2}-2f'(x)m+f''(x)=0 \) in \( m \) has two equal roots for every \( x\in\mathbb{R} \). If \( f(0)=1 \), \( f'(0)=2 \), and \( (\alpha,\beta] \) is the largest interval in which the function \( f\left(\log_{e}x - x\right) \) is increasing, then \( \alpha+\beta \) is equal to ________.

Let \( y=y(x) \) be the solution curve of the differential equation \( (1+\sin x)\frac{dy}{dx}+(y+1)\cos x=0, \; y(0)=0 \). If the curve \( y=y(x) \) passes through the point \( \left(\alpha,-\frac{1}{2}\right) \), then a value of \( \alpha \) is:

Let \( y=y(x) \) be the solution of the differential equation \( \dfrac{dy}{dx}=(1+x+x^{2})(1-y+y^{2}) \), \( y(0)=\dfrac{1}{2} \). Then \( (2y(1)-1) \) is equal to:

Let \( y=y(x) \) be the solution of the differential equation \( x\sqrt{1-x^{2}}\,dy+(y\sqrt{1-x^{2}}-x\cos^{-1}x)\,dx=0,\ x\in(0,1) \). If \( \displaystyle \lim_{x\to 1^{-}}y(x)=1 \). Then \( y\!\left(\dfrac{1}{2}\right) \) equals:

Let the solution curve of the differential equation \( x\,dy-y\,dx=\sqrt{x^{2}+y^{2}}\,dx ,\ x>0 ,\ y(1)=0 \), be \( y=y(x) \). Then \( y(3) \) is equal to:

Let \( y = y(x) \) be the solution of the differential equation \( x^4 \, dy + \left(4x^3 y + 2 \sin x\right) \, dx = 0 \), \( x > 0 \), \( y\!\left(\frac{\pi}{2}\right) = 0 \). Then \( \pi^4 \, y\!\left(\frac{\pi}{3}\right) \) is equal to:

Let \( f:\mathbb{R}\rightarrow \mathbb{R} \) be such that \( f(xy)=f(x)f(y) \), for all \( x,y\in \mathbb{R} \) and \( f(0)\neq 0 \). Let \( g:[1,\infty)\rightarrow \mathbb{R} \) be a differentiable function such that
\( x^{2}g(x)=\displaystyle\int_{1}^{x}\big(t^{2}f(t)-t\,g(t)\big)\,dt \)
Then \( g(2) \) is equal to:

Let \( f:[1,\infty)\to \)(\(\mathbb{R}\)) be a differentiable function. If \( 6\int_{1}^{x} f(t)\,dt=3x f(x)+x^{3}-4 \) for all \( x\ge 1 \), then the value of \( f(2)-f(3) \) is:

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JEE Main 2026 23rd Jan Evening Shift

If the solution curve \( y=f(x) \) of the differential equation \( (x^{2}-4)y'-2xy+2x(4-x^{2})^{2}=0,\ x>2 \) passes through the point \( (3,15) \), then the local maximum value of \( f \) is \( \_\_\_\_\_\_\_\_ \).