Let A- B are two points on the curve y - log 1-2x - 12 - log 2 -4x2 -4x -1 and A is also on the circle whose radius is -10 and centre at O0-0- B lies inside the circle such that its abscissa is integer- then

Y = log _1/2(x 12) + log _2 √(4x^2 4x +1) ⇒ y = log _2 (2x 12)+ log _2 √((2x 1)^2) We know that, 2x 1 0 ⇒ x 1 2 ⇒ y = log _2 (2x 1) +1 + log _2 (2x 1) ...

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Standard XIII

Mathematics

Dot Product of Two Vectors

Let A, B are ...

Question

Let $A, B$ are two points on the curve $y = {log}_{1 / 2} (x - \frac{1}{2}) + {log}_{2} \sqrt{4 x^{2} - 4 x + 1}$ and $A$ is also on the circle whose radius is $\sqrt{10}$ and centre at $O (0, 0)$ . $B$ lies inside the circle such that its abscissa is integer, then

- - \to O A \cdot - - \to O B = 4

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- - \to O A \cdot - - \to O B = 7

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| - - \to A B | = 2

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| - - \to A B | = 1

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Solution

The correct option is D $| - - \to A B | = 1$
$y = {log}_{1 / 2} (x - \frac{1}{2}) + {log}_{2} \sqrt{4 x^{2} - 4 x + 1} \Rightarrow y = - {log}_{2} (\frac{2 x - 1}{2}) + {log}_{2} \sqrt{(2 x - 1)^{2}}$
We know that,
$2 x - 1 > 0 \Rightarrow x > \frac{1}{2}$
$\Rightarrow y = - {log}_{2} (2 x - 1) + 1 + {log}_{2} (2 x - 1) \Rightarrow y = 1 \dots (1)$
The equation of the circle is,
$x^{2} + y^{2} = 10 \dots (2)$
Using $(1)$ and $(2)$ ,
$x^{2} = 9 \Rightarrow x = 3 [∵ x > \frac{1}{2}]$

Therefore,
$A = (3, 1)$ and $B = (1, 1) or (2, 1)$
Now,
$- - \to O A = 3^i +^j - - \to O B =^i +^j or 2^i +^j - - \to A B = -^i or - 2^i$

Therefore,
$- - \to O A \cdot - - \to O B = 4 or 7$
$| - - \to A B | = 1 or 2$

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Let A,B are two points on the curve y=log1/2(x−12)+log2√4x2−4x+1 and A is also on the circle whose radius is √10 and centre at O(0,0). B lies inside the circle such that its abscissa is integer, then

Let $A, B$ are two points on the curve $y = {log}_{1 / 2} (x - \frac{1}{2}) + {log}_{2} \sqrt{4 x^{2} - 4 x + 1}$ and $A$ is also on the circle whose radius is $\sqrt{10}$ and centre at $O (0, 0)$ . $B$ lies inside the circle such that its abscissa is integer, then