1
WB JEE 2024
+1
-0.25

Let $$\mathrm{f}$$ be a differential function with $$\lim _\limits{x \rightarrow \infty} \mathrm{f}(x)=0$$. If $$\mathrm{y}^{\prime}+\mathrm{yf}^{\prime}(x)-\mathrm{f}(x) \mathrm{f}^{\prime}(x)=0$$, $$\lim _\limits{x \rightarrow \infty} y(x)=0$$ then

A
$$\mathrm{y}+1=\mathrm{e}^{\mathrm{f}(x)}+\mathrm{f}(x)$$
B
$$\mathrm{y}+1=\mathrm{e}^{-\mathrm{f}(x)}+\mathrm{f}(x)$$
C
$$\mathrm{y}+2=\mathrm{e}^{-\mathrm{f}(\mathrm{x})}+\mathrm{f}(x)$$
D
$$\mathrm{y}-1=\mathrm{e}^{-\mathrm{f}(x)}+\mathrm{f}(x)$$
2
WB JEE 2024
+1
-0.25

If $$x y^{\prime}+y-e^x=0, y(a)=b$$, then $$\lim _\limits{x \rightarrow 1} y(x)$$ is

A
$$e+2 a b-e^a$$
B
$$e^2+a b-e^{-a}$$
C
$$\mathrm{e}-\mathrm{ab}+\mathrm{e}^{\mathrm{a}}$$
D
$$\mathrm{e}+\mathrm{ab}-\mathrm{e}^{\mathrm{a}},\left(\mathrm{y}^{\prime}=\frac{\mathrm{dy}}{\mathrm{d} x}\right)$$
3
WB JEE 2023
+1
-0.25

If $$y = {x \over {{{\log }_e}|cx|}}$$ is the solution of the differential equation $${{dy} \over {dx}} = {y \over x} + \phi \left( {{x \over y}} \right)$$, then $$\phi \left( {{x \over y}} \right)$$ is given by

A
$${{{y^2}} \over {{x^2}}}$$
B
$$- {{{y^2}} \over {{x^2}}}$$
C
$${{{x^2}} \over {{y^2}}}$$
D
$$- {{{x^2}} \over {{y^2}}}$$
4
WB JEE 2023
+1
-0.25

Given $${{{d^2}y} \over {d{x^2}}} + \cot x{{dy} \over {dx}} + 4y\cos e{c^2}x = 0$$. Changing the independent variable x to z by the substitution $$z = \log \tan {x \over 2}$$, the equation is changed to

A
$${{{d^2}y} \over {d{z^2}}} + {3 \over y} = 0$$
B
$$2{{{d^2}y} \over {d{z^2}}} + {e^y} = 0$$
C
$${{{d^2}y} \over {d{z^2}}} - 4y = 0$$
D
$${{{d^2}y} \over {d{z^2}}} + 4y = 0$$
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