If f(x) = x3-4x+p, and f(0) and f(1) are of
opposite signs,then which of the following is necessarily true?
(1)-1 < p < 2 (2)
0 < p < 3 (3) -2 < p
< 1 (4) -3 < p < 0
Solution follows here:
This is a simple problem and can be noted that the concepts
of quadratic equation are not required to solve this one.
f(x) = x3-4x+p,
f(0) = p f(1)
= 1-4+p = p-3
Given that f(0) and f(1) are of opposite signs =>
f(0)*f(1) < 0
=> p(p-3) < 0, by Theory of Inequality, it can be
concluded that 0 < p < 3
Theory of Inequality fundas:
If (x-a)*(x-b) ≤ 0, then a ≤ x ≤ b
And if (x-a)*(x-b) ≥ 0, then
(i)x ≤ a and x ≥ b if a < b
(ii)x ≤ b and x ≥ a if a > b
(iii) x can be any real number if a=b
Answer can also be arrived by randomly checking for some
integer values of x like -1, 1, 2, 4 etc., and substituting in the answer
options one by one.
Answer (2)
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