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In a A B C ,ifA B=x , B C=x+1,/C=pi/3 ,...

In a ` A B C ,ifA B=x , B C=x+1,/_C=pi/3` , then the least integer value of `x` is 6 (b) 7 (c) 8 (d) none of these

A

6

B

7

C

8

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B
Using cosine rule, we get
`x^(2) = (x + 1)^(2) + b^(2) -2 (x + 1) b cos.(pi)/(3)`
`rArr 0 = 2x + 1 + b^(2) - (x +1) b`
`rArr b^(2) -(x +1) b + 2x + 1 = 0`
Since b is real, we have
`rArr (x + 1)^(2) - 4 (2x +1) ge 0`
`rarr x^(2) -6x - 3 ge 0`
`rArr x ge 3 + sqrt12`
The least integral value of x is 7
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