JEE Main 2017 (Offline) | Differential Equations Question 133 | Mathematics

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point:

$$\left( {{1 \over 2},{1 \over 2}} \right)$$

$$\left( {{1 \over 2}, - {1 \over 3}} \right)$$

$$\left( {{1 \over 2},{1 \over 3}} \right)$$

$$\left( { - {1 \over 2}, - {1 \over 3}} \right)$$

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

If $$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$$ and y(0) = 1,

then $$y\left( {{\pi \over 2}} \right)$$ is equal to

$$ - {2 \over 3}$$

$$ - {1 \over 3}$$

$${4 \over 3}$$

$${1 \over 3}$$

JEE Main 2016 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

The solution of the differential equation

$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$

where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by :

y = 1 $$-$$ $${x \over {\sec x + \tan x}}$$

y² = 1 + $${x \over {\sec x + \tan x}}$$

y² = 1 $$-$$ $${x \over {\sec x + \tan x}}$$

y = 1 + $${x \over {\sec x + \tan x}}$$

JEE Main 2016 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

If f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x} $$ $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :

$${{13} \over 6}$$

$${{23} \over 18}$$

$${{25} \over 9}$$

$${{31} \over 18}$$

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