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ВОРОТА | GATE-CS-2014- (Set-2) | Вопрос 65

Если x вещественное и | x 2 — 2x + 3 | = 11, то возможные значения | — x 3 + x 2 — x | включают
(А) 2, 4
(Б) 2, 14
(С) 4, 52
(D) 14, 52

Ответ: (D)
Объяснение:

Here we use the modulus property, which says:

|x| = x when x >= 0

|x| = -x when x < 0

i.e. range of a modulus function is always positive.

Now, given that |x^2 – 2x + 3| = 11, we can say that

x^2 – 2x + 3 = +11 ----------------(1)

and

x2 – 2x + 3 = -11------------------(2)

Solving 1st equation, we get real roots as 4 and -2.

Solving 2nd eq, we get imaginary roots, hence we ignore them.

Now, for eq |- x^3 + x^2 – x|, we put 4 and -2 in place of x.

putting x = 4, we get |-4^3 + 4^2-4| = |-64+16-4| =  52

putting x = -2 we get |-(-2)^3 + (-2)^2 - (-2)| = 14

so |- x^3 + x^2 – x| has possible values as 52 and 14 

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ВОРОТА | GATE-CS-2014- (Set-2) | Вопрос 65

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