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#1 Posted 2:30pm 16-03-10 please see( more than one choice)1.. [sqrt]1-x[p]2[/p][/sqrt] + [sqrt]1-y[p]2[/p][/sqrt] = a(x-y) then dy/dx is a) (1-x[p]2[/p])/(1-y[p]2[/p]) b) (1-y[p]2[/p])/ (1-x[p]2[/p]) c) ([sqrt]1-x[p]2[/p])/(1-y[p]2[/p][/sqrt]) d) [sqrt]1-y[p]2[/p])/ (1-x[p]2[/p])[/sqrt] 2. If y= cosec x (cot[p]–1[/p]x) thendy/dx a. x/[sqrt]1-x[p]2[/p][/sqrt] b) -x/[sqrt]1+x[p]2[/p][/sqrt] c) x/[sqrt]1+x[p]2[/p][/sqrt] d) None 3. The tangents to the circle x[p]2[/p] + y[p]2[/p] = 25 which is parallel to line 2x-y+1 = 0 are a) y=2x+4[sqrt]5[/sqrt] b) y=2x-4[sqrt]5[/sqrt] c) y=2x+5[sqrt]5[/sqrt] d) y=2x-5[sqrt]5[/sqrt] 4. ∫1/sin[p]2[/p]x+cos[p]2[/p]xdx = a) -2cot2x+c b) tanx-cotx+c c) 9+(sec[p]2[/p]x)/2 d) 3-tan[p]2[/p]x/2 e) (3-sec[p]2[/p]x)/2 f) (tan[p]2[/p]x/2)-5 5. lim 0 to pi/4∫tan[p]n[/p] xdx , n belongs to I (more than one choice) a) I[ss]n[/ss]+I[ss]n-2[/ss]=1/n-1 b) I[ss]n[/ss]+I[ss]n-2[/ss]=1/n c) I1 = I3 + 2I5 d) None 6. At x=3/2 the value is real for (more tahn one choice) (a) cos[p]–1 [/p]2x (b) cosec[p]–1[/p] x (c) tan[p]–1[/p] x (d) None
Edited on 3:02pm 16-03-10 |
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#2 Posted 2:45pm 16-03-10 Re: please see( more than one choice)For the third one just use the formula for the tanget with slope m y = mx ± a[sqrt]1+ m[p]2[/p][/sqrt] ... |
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#3 Posted 3:11pm 16-03-10 Re: please see( more than one choice)let the eqn be y=2x+c... putting thge eqn..in the eqn of the circle.... we get, 5x[p]2[/p]+4cx+(c[p]2[/p]-25)=0 as tangent touches the circle at 1 pt.... D=0 ie.c=5√5.... hence aans is ..(c)
SUCCESS IS NOT FINAL..FAILURE IS NOT FATAL...ITS THE COURAGE TO CONTINUE THAT COUNTS .... |
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#4 Posted 4:17pm 16-03-10 Re: please see( more than one choice)For the first one substitute x = sinα and y = sinβ [im]http://latex.codecogs.com/gif.latex?\sqrt%20{1-sin^2\alpha}%20+%20\sqrt%20{1-sin^2\beta}%20=%20a(sin\alpha%20-%20sin\beta)[/im] [im]http://latex.codecogs.com/gif.latex?=%3E%20cos\alpha%20+%20cos\beta%20=%202asin(\frac{\alpha-\beta}{2})cos(\frac{\alpha+\beta}{2})[/im] => [im]http://latex.codecogs.com/gif.latex?2cos(\frac{\alpha-\beta}{2})cos(\frac{\alpha+\beta}{2})%20=%202asin(\frac{\alpha-\beta}{2})cos(\frac{\alpha+\beta}{2})[/im] => [im]http://latex.codecogs.com/gif.latex?cot(\frac{\alpha-\beta}{2})%20=%20a[/im] [im]http://latex.codecogs.com/gif.latex?=%3E\frac{\alpha-\beta}{2}%20=%20arccot(a)[/im] [im]http://latex.codecogs.com/gif.latex?=%3E%20arcsin(x)%20-%20arcsin(y)%20=%202arccot(a)[/im] Now differentiate. It's much easier. Differentiation of cot[p]-1[/p]a is zero.
Self belief is such a peculiar thing. As long as you have it, nothing can stop you, you feel invulnerable. But once it's gone, nothing can bring you up from the "I'm-useless" state you end up in. Edited on 4:18pm 16-03-10 |
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#5 Posted 6:01pm 16-03-10 Re: please see( more than one choice)6)b,c see cos[p]-1[/p]3 is not possible.as cos can take values in interval [-1,1]
For those who seek perfection there can be no rest this side of the grave. |
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#6 Posted 6:27pm 16-03-10 Re: please see( more than one choice)6th agreed with b,c! |
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