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JEE Mathematics Solved Question Paper 2017

Multiple Choice Questions

21.

The general solution of the equation $\frac{dy}{dx} = \frac{y^{2} - x}{2 y (x + 1)}$ is

y² = (1 + x)log(1 + x) - c
$y^{2} = (1 + x) \log (\frac{c}{(1 - x)}) - 1$
$y^{2} = (1 - x) \log (\frac{c}{(1 - x)}) - 1$
$y^{2} = (1 + x) \log (\frac{c}{(1 + x)}) - 1$

22.

The general solution of the differential equation $\frac{dy}{dx} + \sin (\frac{x + y}{2}) = \sin (\frac{x - y}{2})$ is

$\log_{e} |\tan (\frac{y}{2})| = - 2 \sin (\frac{x}{2}) + C$
$\log_{e} |\tan (\frac{y}{4})| = 2 \sin (\frac{x}{2}) + C$
$\log_{e} (\tan (\frac{y}{2})) = - 2 \sin (\frac{x}{2}) + C$
$\log_{e} (\tan (\frac{y}{4})) = - 2 \sin (\frac{x}{2}) + C$

23.

If a, b, c are three vectors such that [a b c] = 5, then the value of [a x b, b x c, c x a] is

24.

The point of inflection of the function y = $\int_{0}^{x} (t^{2} - 3 t + 2) dt$ is

$(\frac{3}{2}, \frac{3}{4})$
$(- \frac{3}{2}, - \frac{3}{4})$
$(- \frac{1}{2}, - \frac{3}{2})$
$(\frac{1}{2}, \frac{3}{2})$

25.

The value of integral $\int \frac{dx}{x \sqrt{x^{2} - a^{2}}}$

$c - \frac{1}{a} \sin^{- 1} (\frac{a}{|x|})$
$c - \frac{1}{a} \cos^{- 1} (\frac{a}{|x|})$
$\sin^{- 1} (\frac{a}{|x|}) + c$
$c + \frac{1}{a} \sin^{- 1} (\frac{a}{|x|})$

26.
The function y specified implicitly by the relation $\int_{0}^{y} e^{t} dt + \int_{0}^{x} \cos (t) dt$ = 0 satisfies the differential equation

$e^{2 y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$

$e^{y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (2 x)$

$e^{y} (2 \frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$

$e^{y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$

$e^{y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$

$Given, \int_{0}^{y} e^{t} dt + \int_{0}^{x} \cos (t) dt \Rightarrow {[e^{t}]}_{0}^{y} + {[\sin (x)]}_{0}^{x} = 0 \Rightarrow e^{y} + \sin (x) = 0 By differentiating w . r . t .' x', we get e^{y} \frac{dy}{dx} + \cos (x) = 0 Again, by differentiating w . r . t .' x', we get e^{y} \frac{dy}{dx} . \frac{dy}{dx} + e^{y} \frac{d^{2} y}{{dx}^{2}} - \sin (x) = 0 \Rightarrow e^{y} (\frac{d^{2} y}{{dx}^{2}} + {(\frac{dy}{dx})}^{2}) = \sin (x)$