Quadratic and Trigonometry for JEE #2
Let x and y be the roots of the equation a sin 2 θ + b sin θ + c = 0 such that sin x + 2 sin y = 1 then find the value of ( 2 0 1 5 ) ∗ ( a 2 + 2 b 2 + 3 a b + a c )
Given that a,b,c are real numbers such that a = 0 .
The answer is 0.
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4 solutions
Simple and elegant solution. I solved it by bashing though. :3 :P
wow didn't think it could be done this way. Awesome solution!!
SinX+SinY×2=1 from here we can say that x=1 & y=0 or x=0 & y=.5 Then the equation must be Sinx(Sinx -1)=0
Let x=0 and y=pi/6 .....now put the values in given quadratic to get relations between a,b and c.....put the values in the given query to get the ans as 0
Sin x + 2 sin y = 1 => either sinx = 0 and sin y = 0.5 or sinx = -1 and sin y = 1. Case 1 : a= 1, b= 0 and c= -1 Case 2 : a= 2, b = -1 and c= 0. In both cases, answer is zero.
As quadratic equation is in terms of variable a,b , c and theta, this represents a family of quadratic equations. The question asked is to evaluate 2015 * [ a^2 + 2b^2+3ab+ac], and if really depends upon the values of theta chosen, then solution will differ for each value chosen for theta and then answer to the problem would be " can't be determined". I checked my hypothesis with 2 samples of value chosen, if it gives different answer to any other sample values considered, then question can be contested and there can't be any unique solution.
Why are those the only possible cases?
Why can't we have sin x = 0 . 4 , sin y = 0 . 3 ?
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U r right.... it can have many solutions... but it gave the answer zero for two cases I considered. So without loss of generality, it will give same answer for all cases. Chk 4 0.4 & 0.3 too.
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That is not how "without loss of generality" works. What you have done is prove the scenario for 2-3 specific cases only, and not the general case.
See Tijmen's solution for complete details.
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@Calvin Lin – I appreciate your effort in explaining the right way to attempt this problem. I will put again my explanation. As quadratic equation is in terms of variable a,b , c and theta, this represents a family of quadratic equations. The question asked is to evaluate 2015 * [ a^2 + 2b^2+3ab+ac], and if really depends upon the values of theta chosen, then solution will differ for each value chosen for theta and then answer to the problem would be " can't be determined". I checked my hypothesis with 2 samples of value chosen, if it gives different answer to any other sample values considered, then question can be contested and there can't be any unique solution. Please reply if I am incorrect.
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@Shekhar Prasad – Yes, you are correct in saying "If this question has a answer that is independent of the values a, b, c, then the answer must be 0". This is an important "test-taking skill" that can help you save time when working on such problems where you do not need to further explain your answer. E.g. MCQ in JEE.
However, for the solutions, you should also explain the complete logic and reasoning behind the question, so that others who are interested in approaching the problem in it's entity can benefit from the solution that you have written.
For example, a "solution" of the form "I tried 0 and you said I was correct, hence 0 is the answer" is not helpful to others, and I often delete them.
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@Calvin Lin – I agree with u on this " For example, a "solution" of the form "I tried 0 and you said I was correct, hence 0 is the answer" is not helpful to others, and I often delete them."
People should use this trick with discernment. Its a good tool but one can go wrong if not applied properly. It was a nice interaction. Thank You.
The solutions are given by sin θ = 2 a − b ± b 2 − 4 a c , giving us:
sin x + 2 sin y = 2 a − b ∓ b 2 − 4 a c + 2 2 a − b ± b 2 − 4 a c = 1
− 3 b ± b 2 − 4 a c = 2 a
b 2 − 4 a c = ( 2 a + 3 b ) 2 = 4 a 2 + 1 2 a b + 9 b 2
0 = a 2 + 3 a b + 2 b 2 + a c
Therefore the solution is 2 0 1 5 × 0 = 0 .