JEE Main 2019 (Online) 11th January Morning Slot | Differential Equations Question 115 | Mathematics

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JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

If xlog_e(log_ex) $$-$$ x² + y² = 4(y > 0), then $${{dy} \over {dx}}$$ at x = e is equal to :

$${{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}$$

$${{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}$$

$${{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}$$

$${e \over {\sqrt {4 + {e^2}} }}$$

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

The curve amongst the family of curves represented by the differential equation, (x² – y²)dx + 2xy dy = 0 which passes through (1, 1) is

a circle with centre on the y-axis

an ellipse with major axis along the y-axis

a circle with centre on the x-axis

a hyperbola with transverse axis along the x-axis

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

Let f be a differentiable function such that f '(x) = 7 - $${3 \over 4}{{f\left( x \right)} \over x},$$ (x > 0) and f(1) $$ \ne $$ 4. Then $$\mathop {\lim }\limits_{x \to 0'} \,$$ xf$$\left( {{1 \over x}} \right)$$

does not exist

exists and equals $${4 \over 7}$$

exists and equals 4

exists and equals 0

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

A helicopter is flying along the curve given by y – x^3/2 = 7, (x $$ \ge $$ 0). A soldier positioned at the point $$\left( {{1 \over 2},7} \right)$$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is -

$${1 \over 6}\sqrt {{7 \over 3}} $$

$${{\sqrt 5 } \over 6}$$

$${1 \over 2}$$

$${1 \over 3}$$$$\sqrt {{7 \over 3}} $$

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