MCQ

If Q.No. 14, *c* =

#### Options

*b*2

*b*2

*b*^{2}−2

*b*

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#### 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)`

Concept: Concept of Polynomials

Is there an error in this question or solution?

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