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Listen if x,y,z are positive : then let the power of 2 in x,y,z be respectively . so, by the problem: a+b+c=1. Number of non-negative integral solution 0f this = similarly, let the powers of 5 in x,y,z be respectively. so, no of solution of this = = so, total number of solutions = rate if u like it! !
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The function is not exactly symmetric with respect to p,q. So, your assumption is not right, there is l;oss of generality.
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See , when we generally count 3 digit number we put the restriction that 0 cannot be at the first place, if it is allowed to go to all the places we get the 2 digits and the single digits number aswell. so, total number of numbers with five at unit place (1digit +2 digit +3digit ) = 10 * 10 = 100 similarly there are 100 each for five at tens place and thousand's place respectively. So, total number of ways = 100+100+100 = 300
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I did 2,3 and 6.Actually my solution to the inequality is similar to bhatt sir's solution though there is a better solution as pointed out by anant sir.I thought i had done the 5th one but there is unfortunately a small flow in my solution.I don't think this is enough to get through in class 12.question 3 is quite easy just on quadratic equations. 2 and 6 has already been solved
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So, that we can start a discussion :
Find all real numbers such that:
Let be an acute angled triangle and let be its ortho centre.Let denote the largest altitude of the triangle .Prove that:
 Let be positive real numbers such that .Prove that: . | |
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