Let A - B are two points on the curve y -log 1 - 2 x -1-2-log 2-4 x 2-4 x -1 and A is also on the circle whose radius is -10 and centre atO 0-0 - B lies inside the circle such that its abscissa is integer- then A- OA - OB -4B- OA - OB -7C- - AB -2D- - AB -1

The correct options are A OA·OB = 4 B OA·OB = 7 C |AB| = 2 D |AB| =1y = log _1/2(x 12) + log _2 √(4x^2 4x +1) ⇒ y = log _2 (2x 12)+ log _2 √((2x 1)^2) We k ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

| - - \to A B | = 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

| - - \to A B | = 1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A $- - \to O A \cdot - - \to O B = 4$
B $- - \to O A \cdot - - \to O B = 7$
C $| - - \to A B | = 2$
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,
$- - \to O A \cdot - - \to O B = 4 or 7$
$| - - \to A B | = 1 or 2$

Suggest Corrections

Similar questions

Q. 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

Q. Let B and C be two points on a circle with centre A Let P be a point on radius AB such that

B C = A P, A B . B P = {(A P)}^{2} a n d ∠ C A B = θ t h e n θ

is equal to

B

and

C

are points on the circle

x^{2} + y^{2} = a^{2}

A point

A (b, c)

lies on that circle such that

A B = A C = d

. The equation to

B C

\dots \dots

A, B, C

are three points on a circle such that

A B

is the chord and

C P

is the perpendicular to

O P

, where

O

is the centre and

P

is any point on

A B

. The radius

r

of the circle is given by

Q. The line

2 x - y + 1 = 0

is a tangent to the circle at the point (2, 5) and the centre of the circle lies on

x - 2 y = 4

. Then radius of the circle is :

Join BYJU'S Learning Program

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