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Tricks for II
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Here are few mathematical shorcuts to solve probs asked in IIT-JEE.... 1. (IIT 04) then the interval in which 'a' lies is (a)a<-5 (b)-5<a<2 (c)a>5 (d)2<a<5 Comment: the alternatives do no have a common value. Put a convenient value for a. Let us try : a=0. Then Therefore, (b) is the correct answer!!! 2. (AIEEE04) The domain of the function is (a) [1,2) (b) [2,3) (c) [1,2] (d) [2,3] Comment: Look at the alternatives: Neither (a) nor (c) is the correct ans. bçoz x=1, does not exist. now he prob reduces to whether 3 belongs to domain or not. When x=3, denominator = 0. Hence, (b) is the ans. 3. (IIT 2K) The domain of function y(x) given by is (a) [0,1] (b) (0,1] (c) (d) Comment: When x=1, 2+2^y = 2 is not possible. Hence, (a) and (b) are not the ans. Now, put the value for x betn. (0,1) and check. That is cumbersome. Therefore don't think this kind of technique works for every prob. 4. (IIT 05) In any triangle ABC (with usual notations of the sides a,b,c) (a) b-c/2 sin A = a cos A (b)(b-c) cos A/2 = a sin B-C/2 (c)b+c/2 cos A/2 = a sin B- C/2 (d)(b+c) cos A/2 = 2a sin B+C/2 Comment : Try an isosceles triangle in which b=c, i.e. B = C. Take (b): 0 =0 ; no other alternatives satisfy them. 5. A hyperbola having transverse axis of length , is confocal with the ellipse Then its eqn. is (a) (b) (c) (d)
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Transcript of Tricks for II
28 Oct 2008 23:27:28 IST
Here are few mathematical shorcuts to solve probs asked in IIT-JEE....
1. (IIT 04)
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then the interval in which 'a' lies is(a)a (t+1)^2 - (a-3)t(t+1) + (a-4)t^2 = 0=> (a-5)t - 1 = 0
=> now if a = 6=> t-1 = 0=> x^2 + x = 0=> x(x+1) = 0 and the condition is satisfied ..
if we replace a with 7/8/9 the we'll get 3 other quadratic equations in which the discriminants are < 0 .. hence they dont satisfy the condition..
Complex No. - Assertion Reason
by nikhil_ Tue Mar 08, 2011 8:12 pm
Let z1 and z2 be two complex numbers such that |z1-z2|=|z1+z2|Statement I:If z1, z2 are end point of diameter of circle then it passing through the originStatement II:|arg(z1/z2)| = /2
(A)Both true. II is correct explanation of I.(B)Both true. II is not correct explanation of I.(C)I true, II false.(D)I false, II true.
Both true and II is the correct explanation.
note that the equation of a circle which has z1 and z2 as end points of one of its diameter is |z - (z1 + z2)/2| = |(z1-z2)/2|
z = 0 satisfies this and hence, the circle does pass through the origin.
also, |z|^2 = z * z(bar)put z = z1 - z2 in the above equation to get,z1 * z2(bar) = - z2 * z1(bar)take argument on both the sides
arg(zbar) = - arg(z) and arg(z1 * z2) = arg(z1) + arg(z2) this gives, |arg(z1/z2)| = pi/2
