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1

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)
If $$y=y(x)$$ and it follows the relation $$x\cos \,y + y\,cos\,x = \pi $$ then $$y''(0)=$$
A
$$1$$
B
$$-1$$
C
$${\pi - 1}$$
D
$$ - \pi $$
2

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)
If $$y=y(x)$$ and $${{2 + \sin x} \over {y + 1}}\left( {{{dy} \over {dx}}} \right) = - \cos x,y\left( 0 \right) = 1,$$
then $$y\left( {{\pi \over 2}} \right)$$ equals
A
$$1/3$$
B
$$2/3$$
C
$$-1/3$$
D
$$1$$
3

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)
If $$y(t)$$ is a solution of $$\left( {1 + t} \right){{dy} \over {dt}} - ty = 1$$ and $$y\left( 0 \right) = - 1,$$ then $$y(1)$$ is equal to
A
$$ - 1/2$$
B
$$e+1/2$$
C
$$e-1/2$$
D
$$ 1/2$$
4

IIT-JEE 2000 Screening

MCQ (Single Correct Answer)
If $${x^2} + {y^2} = 1,$$ then
A
$$yy'' - 2{\left( {y'} \right)^2} + 1 = 0$$
B
$$yy'' + {\left( {y'} \right)^2} + 1 = 0$$
C
$$yy'' + {\left( {y'} \right)^2} - 1 = 0$$
D
$$yy'' + 2{\left( {y'} \right)^2} + 1 = 0$$

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