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5, 5r, 5r^2 are sides of a triangle. Whi...

`5, 5r, 5r^2 `are sides of a triangle. Which value of r cannot be possible (a) `3//2` (b) `5//4` (c) `3//4` (d) `7//4`

A

`5/4`

B

`7/4`

C

`3/2`

D

`3/4`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `alpha =5,b=5rand c=5r^(2)` We know that, in a triangle sum of 2 sides is always greater than the traid side.
`thereforea +b gtc, b+cgtalpha and c+agtb`
Now, `a+b gtc`
`thereforea +b gtc, b+cgtalpha and c+agtb`
Now, `a+b gtcimplies5+5r gt5r^(2)-5r-5 lt0`
`impliesr^(2)-r-1lt0`
`impliesr^(2)-r-1lt0`
`implies[r-((1-sqrt5)/(2))][r-((a+sqrt5)/(2))]lt0`
`" ""["because"roots of"ax^(2)+bx+x=0"are given by"x=(-b+-sqrt(b^(2)-4ac))/(2alpha)and r^(2)-r1=0impliesr=(1+-sqrt(1+4))/(2)=(1+-sqrt5)/(2)"]"`
`implies" "r in((1-sqrt5)/(2),(1+sqrt5)/(2))" "(i)`

Similarly, `b+c gt a`
`implies 5r + 5r^(2) gt 5`
`implies r^(2)+r-1gt0`
`[r-((-1-sqrt5)/(2))][r-((-1+sqrt5)/(2))]gt0`
`[becauser^(2)+r-1=0impliesr=(-1+-sqrt(1+4))/(2)=(-1+-sqrt5)/(2)]`
`impliesr in (-oo,(-1-sqrt5)/(2))uu((-1+sqrt5)/(2),oo)" "(ii)`

and `c+a gt b`
`implies5r^(2)+5gt5r`
`implies r^(2)-r+1 gt0`
`impliesr^(2)-2.(1)/(2)r+((1)/(2))^(2)+1-((1)/(2))^(2)gt0`
`implies(r-(1)/(2))^(2)+3/4gt0`
`impliesr inR" "...(iii)`
From Eqs. (i), (ii) and (iii), we get `r in((-1+sqrt5)/(2),(1+sqrt5)/(2))`

and `7/4` is the only value that does not satisfy.
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