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Question
If Q.No. 14, c =
Options
b
2b
2b2
−2b
Solution
We have to find the value of c
Given `f(x)= ax^3 +bx -c` is divisible by the polynomial `g (x )= x^2 + bx + c`
We must have
`bx - acx +ab^2x + abc -c = 0 ` for all x
` x (b - ac + ab^2) + c (ab-1)= 0..........(1)`
`c (ab-1)=0`
Since `c ` ≠ `0`, so
`ab - 1 = 0`
`ab = 1`
Now in the equation (1) the condition is true for all x. So put x = 1 and also we have ab = 1
Therefore we have
`b - ac + ab^2 = 0`
`b +ab^2 -ac =0`
`b(a +ab)-ac = 0`
Substituting `a= 1/b`and `ab =1` we get,
`b(1 + 1) - 1/b xxc = 0`
`2b -1/b xxc =0`
`- 1/b xxc = -2b`
`cancel(-)1/bxx cancel(-) 2b`
` c = 2b =b/1`
`c = 2b ^2`
Hence, the correct alternative is `(c)`
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