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Maths-

General

Easy

Question

Let $P open parentheses ct subscript 1 comma straight c over straight t subscript 1 close parentheses$ and $Q open parentheses ct subscript 2 comma straight c over straight t subscript 2 close parentheses$ be the points of a rectangular hyperbola $x y equals c squared$ which subtends at another point R ' of the given hyperbola, if ' s ' be the mid polint of PQ and ' $straight O to the power of straight prime$ centre of the hyperbola then area of the $straight triangle ROS$ is equal to

$\frac{c^{2} (1 + {(t_{1} t_{2})}^{2}) |t_{1} + t_{2}|}{4 |t_{1} t_{2}| \sqrt{|t_{1} t_{2}|}}$
$\frac{c^{2} |t_{1} + t_{2}|}{4 |t_{1} t_{2}| \sqrt{|t_{1} t_{2}|}}$
$\frac{c^{2} |1 + t_{1} t_{2}| |t_{1} + t_{2}|}{2 |t_{1} t_{2}| \sqrt{|t_{1} t_{2}|}}$
none of these

The correct answer is: $fraction numerator straight c squared open parentheses 1 plus open parentheses straight t subscript 1 straight t subscript 2 close parentheses squared close parentheses open vertical bar straight t subscript 1 plus straight t subscript 2 close vertical bar over denominator 4 open vertical bar straight t subscript 1 straight t subscript 2 close vertical bar square root of open vertical bar straight t subscript 1 straight t subscript 2 close vertical bar end root end fraction$

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Statement 2: In a triangle the longest sides is less than the sum of the other two sides.

Statement 1: The number of triangles that can be formed with sides of length a,b,c where a,b,c are integers such that $a less or equal than b less or equal than c equals 15$ is 64 and
Statement 2: In a triangle the longest sides is less than the sum of the other two sides.

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Number of positive integers that are divisors of atleast one of the numbers $10 to the power of 10 end exponent semicolon 15 to the power of 7 end exponent semicolon 18 to the power of 11 end exponent$ is

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What is the final product(B)of this sequence?

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