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I solved the last 2.Can anyone please help me with the first.If anyone wants the solutions to the last 2, I can post.
But here is the logic anyways:
2) Assume a+x = y and replace all x with y-a and square.Assume quadratic in a and solve for a.You will then get quadratic in y by squaring.
3) Squaring pr = 2(q+s) and then applying AM GM for q^2 and s^2 , we get p^2r^2 >= 16qs .
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My approach in 1 and 3 is the use of factorization but i cant seem to make factors such that I can use the given conditions.
And I have no idea about the second question.And I dont know how we can use calculus in the questions.
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1) If ax^2 + (b-c)x + a-b-c =0 has unequal real roots for all c  R then prove that either b<a<0 or b>a>0.
2) Prove that for -1/4 <a <0 , the equation [2 ] (a + [2 ] (a+x) )= x has exactly two real solutions of which one is 1 + [ 2] (4a +1) /2
3) Show that if p,q,r and s are real no.s and pr=2(q+s) then at least one of the equations x^2 +px +q =0 , x^2 +rx +s =0 as real roots.
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Wasnt the piston supposed to be fixed?
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^^
I agree completely.It will be a cycloid if seen from the ground frame only but if you are talking about radius of curvature its always performing circular motion about the instantaneous axis of rotation and I dont have any doubt that it should be 2r.
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I dont think derivation is necessary and it is not taught.
Chord of contact is given by : xh/a^2 + yk/b^2 =1 where h,k is the point from where the two tangents are drawn.
NOTE: It is the same as a tangent drawn at the point h,k of an ellipse.
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sin (pi +x) = -sinx
Now reduce equation and see.
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Arey normal force (in the second case) on the lower block is higher so friction will also be higher.But the question is still very vague.Plz type proper questions,.
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Without loss of generality  we can say this SHM is the spring block SHM.
And we know that total energy of a particle executing SHM is 1/2mw^2a^2
Total mechanical energy = KE + PE
So 1/2 mw^2a^2 = 1/2 w^2mx^2 + 1/2mu^2
and 1/2 mw^2a^2 = 1/2 w^2my^2 + 1/2 mv^2 ( because k=w^2m)
Dividing both equations:
w^2(y^2-x^2) = (u^2-v^2)
w^2 = (u^2 - v^2)/(y^2-x^2)
T= 2pi/w = 2 [(y2-x2)/(u2-v2)]1/2
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and Olympiad problems. 
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I suppose you are talking about magnitudes.
|A|^2 = |A|^2 + |B|^2 +2|A||B|cos@
and |B|^2 = |A|^2 + |B|^2 +2|A||B|cos@
=> |A| = |B|
So |A|^2 = 2|A|^2 + 2|A|^2cos@
-1/2 = cos@
@ = 2pi/3
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I was comparing between Sn1 and Sn2.A stronger base favours Sn2 than Sn1 as there is no opportunity for a carbocation to be formed.
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He's being sarcastic man.This question is very complicated,look at the answer yourself. Its not hard but lengthy. Assume the big block to be at rest and apply pseudo forces.
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When concentration of base is very high and/or a very strong base is used, reaction proceeds by Sn2.
Plus the criteria written in the above posts.
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Absolutely, z and w lie at the opposite ends of a line with mid point as origin.So their locus is the circle with centre at origin and radius z or w.
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