Let k 1 , k 2 be any two integers given in the relation ∣ ( n − a ) ! − t ∣ + ∣ t − ( n − b ) ! ∣ + ∣ a + b − k 1 n − k 2 ∣ ≤ 0 ∀ n such that a < b ≤ n and a , b , n , t ∈ N .
Let P , Q be any two points on the curve
y = lo g 2 1 ( x + 2 k 2 ) + lo g 2 4 x 2 + 4 k 2 x + ( k 1 + k 2 ) .
Also P lies on the circle x 2 + y 2 = k 1 3 − 2 k 2 and Q lies inside the given circle such that its abscissa is a non-zero integer.
O is the centre of the circle.
1) Find the minimum possible value of O P ⋅ O Q .
2) The maximum length of PQ
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@Nishant Rai The solution has made so many assumptions!! If Q has it's abscissa as integer, then why not on the negative X − a x i s , namely ( − 2 , 1 ) and ( − 1 , 1 ) , as they are integers as well. That will give the value of O P . O Q as negative. Please check this.
Secondly, in your question, what do you mean by length of O P . O Q . That is the scalar product and how do you define the length of a scalar. It would be good if you could change that.
Thirdly, due to the anomaly mentioned in the first point, the value of length of P Q also changes. It comes out to be 5 as maximum, i.e. between P = ( 3 , 1 ) and Q = ( − 2 , 1 ) .
Thanks! :)
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No, the x coordinate of P and Q must be x > 1 / 2 , so that the l o g functions are well-defined. Notice that the question simply uses l o g and not ∣ l o g ∣ .
However I agree on the 'length' aspect. He should change that.
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Can you please try the problem named 'spicy 1'?It is a tough question on definite integrals. If you are able to solve it please give me an outline of the procedures.Please do reply here.
Oh ok! Thanks! I kinda solved it by taking Mod in my copy!
Which is this book @Nishant Rai
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