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Correct Answer – 4
Let unknow polynomial be P(x). Let Q(x) and R(x) be the quatient and remainder, respectively, when it is divided by the `(x-3) (x-4)`. Then
`P(x) = (x-3)(x-4) Q(x) + R(x)`
Then we have
`R(x) = ax + b`
` rArr P(x) = (x-3) (x-4)Q(x) + ax+ b`
Given that `P(3) =2 ` and `P(4)=1` . Hence,
`3a + b = 2 and 4a + b = 1`
`rArr a = -1 and b= 5`
`rArr R(x) = 5 -x`
`5-x = x^(2) + ax + 1 `
`rArr x^(2) + (a + 1) x – 4 =0`
Given that roots and reall distinct. Therefore,
`D lt 0 rArr (a+1)^(2) + 16 lt 0`
Which is true for all real a.
Correct Answer – 3
Let unknow polynomial be P(x). Let Q(x) and R(x) be the quatient and remainder, respectively, when it is divided by the `(x-3) (x-4)`. Then
`P(x) = (x-3)(x-4) Q(x) + R(x)`
Then we have
`R(x) = ax + b`
` rArr P(x) = (x-3) (x-4)Q(x) + ax+ b`
Given that `P(3) =2 ` and `P(4)=1` . Hence,
`3a + b = 2 and 4a + b = 1`
`rArr a = -1 and b= 5`
`rArr R(x) = 5 -x`
`- x + 5 = px^(2) + (q- 1)x + 6`
`rArr px^(2) + qx + 1 = 0`
Now, `p gt 0 ` and equation has no distinct real roots or equation has real and equal or imaginary roots. Then,
`px^(2) + qx + 1 ge 0, AA x in R`
`rArr f(3) ge rArr 9p + 3q + 1 ge 0 rArr 3p + q ge – 1//3`
Hence, the least value of `3p + q` is `-1//3`.