JEE Advanced 2026 Paper 2 Online | Differential Equations Question 2 | Mathematics

JEE Advanced 2026 Paper 2 Online

MCQ (More than One Correct Answer)

-1

Let $y = f(x)$ be the real valued function defined on the interval $(0, \infty)$, satisfying $y(1) = 0$ and the differential equation

$$ x \frac{dy}{dx} = y - x^3. $$

Then which of the following statements is (are) TRUE?

The function $f$ has a local minimum at $x = \frac{1}{\sqrt{3}}$

The function $f$ has a local maximum at $x = \frac{1}{\sqrt{3}}$

The function $f$ is increasing in the interval $(1, 2)$

If $g(x) = 4x^3 - 5x^2 + \frac{3}{2}x$ for $x > 0$, then the number of elements in the set $$ \{x \in (0, \infty) : f(x) = g(x) \} $$
is $2$

JEE Advanced 2022 Paper 2 Online

MCQ (More than One Correct Answer)

-2

For $x \in \mathbb{R}$, let the function $y(x)$ be the solution of the differential equation

$$ \frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right), \quad y(0)=0 $$

Then, which of the following statements is/are TRUE ?

$y(x)$ is an increasing function

$y(x)$ is a decreasing function

There exists a real number $\beta$ such that the line $y=\beta \quad$ intersects the curve $y=y(x)$ at infinitely many points

$y(x)$ is a periodic function

JEE Advanced 2019 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

Let $$\Gamma $$ denote a curve y = y(x) which is in the first quadrant and let the point (1, 0) lie on it. Let the tangent to I` at a point P intersect the y-axis at Y_P. If PY_P has length 1 for each point P on I`, then which of the following options is/are correct?

$$xy' + \sqrt {1 - {x^2}} = 0$$

$$xy' - \sqrt {1 - {x^2}} = 0$$

$$y = {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $$

$$y = - {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} $$

JEE Advanced 2017 Paper 2 Offline

MCQ (More than One Correct Answer)

-2

If $$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$$, then

$$g'\left( { - {\pi \over 2}} \right) = 0$$

$$g'\left( { - {\pi \over 2}} \right) = - 2\pi $$

$$g'\left( {{\pi \over 2}} \right) = 2\pi $$

$$g'\left( {{\pi \over 2}} \right) = 0$$

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