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Standard XII

Mathematics

Fallacy

p,q,r and s...

Question

p,q,r and $s$ are integers. If the A.M. of the roots of $x^{2} - p x + q^{2} = 0$ and G.M. of the roots of $x^{2} - r x + s^{2} = 0$ are equal, then

q

is an odd integer

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r

is an even integer

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p

is an even integer

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s

is an odd integer

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Solution

The correct option is A $p$ is an even integer
$A . M = \frac{α_{1} + α_{2}}{2} = \frac{p}{2} = \sqrt{γ δ} = \sqrt{s^{2}} = G . M$
$\frac{p}{2} = s \Rightarrow p = 2 s$
As s is an integer , p is an even integer.

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Q. If p < 0 < q < r < s, where p, q, r and s are integers, pq = –1, s – p = 10 and r is an odd integer, then which of the following is largest?

यदि p < 0 < q < r < s, जहाँ p, q, r व s पूर्णांक हैं, pq = –1, s – p = 10 तथा r एक विषम पूर्णांक है, तब निम्नलिखित में से कौनसा विकल्प महत्तम है?

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Check whether the following ststement are true or not :

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(ii) q : If x, y are integers such that xy is even, then at least one of x and y is an even integer.

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p,q,r and s are integers. If the A.M. of the roots of x2−px+q2=0 and G.M. of the roots of x2−rx+s2=0 are equal, then

The correct option is A p is an even integerA.M=α1+α22=p2=√γδ=√s2=G.Mp2=s⇒p=2sAs s is an integer , p is an even integer.

p,q,r and $s$ are integers. If the A.M. of the roots of $x^{2} - p x + q^{2} = 0$ and G.M. of the roots of $x^{2} - r x + s^{2} = 0$ are equal, then

The correct option is A $p$ is an even integer
$A . M = \frac{α_{1} + α_{2}}{2} = \frac{p}{2} = \sqrt{γ δ} = \sqrt{s^{2}} = G . M$
$\frac{p}{2} = s \Rightarrow p = 2 s$
As s is an integer , p is an even integer.