JEE Advanced 2019 Paper 1 Offline | Differential Equations Question 5 | Mathematics

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

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

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

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

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

JEE Advanced 2016 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

A solution curve of the differential equation

$$\left( {{x^2} + xy + 4x + 2y + 4} \right){{dy} \over {dx}} - {y^2} = 0,$$ $$x>0,$$ passes through the

point $$(1,3)$$. Then the solution curve

intersects $$y=x+2$$ exactly at one point

intersects $$y=x+2$$ exactly at two points

intersects $$y = {\left( {x + 2} \right)^2}$$

does NOT intersect $$\,y = {\left( {x + 3} \right)^2}$$

JEE Advanced 2016 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Let $$f:(0,\infty ) \to R$$ be a differentiable function such that $$f'(x) = 2 - {{f(x)} \over x}$$ for all $$x \in (0,\infty )$$ and $$f(1) \ne 1$$. Then

$$\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = 1$$

$$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = 2$$

$$\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = 0$$

$$\left| {f(x)} \right| \le 2$$ for all $$x \in (0,2)$$

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