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Bunuel -

is f(x) = ((x^3)^3)*(3*(x^5))

or

f(x) = (x^3)^3*(3*(x^5))
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Bunuel -

is f(x) = ((x^3)^3)*(3*(x^5))

or

f(x) = (x^3)^3*(3*(x^5))

It's as written \((x^3){^3}*(x^5)\).
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If x = 0 then the value of f(x) = 0 which is not < 0. For other values of x, you need to know the sign. So A is not the answer. B gives the value of x = -2 but it doesnt tell you what f(x) is, so you need both 1) and 2) to get the result. You dont need to calculate the actual value. The powers are enough to conclude that f(x) where x = -2 is less than 0.

So the answer is (C).
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If x = 0 then the value of f(x) = 0 which is not < 0. For other values of x, you need to know the sign. So A is not the answer. B gives the value of x = -2 but it doesnt tell you what f(x) is, so you need both 1) and 2) to get the result. You dont need to calculate the actual value. The powers are enough to conclude that f(x) where x = -2 is less than 0.

So the answer is (C).

hi...
(x^3)^3*x^5=x^9*x^5=x^14...
so for x=0, the value is 0..
any value other than 0, ans will be >0, as the power is an even no
therefore ans is A
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Bunuel
Is f(x) < 0?

(1) f(x) = (x^3)^3*(x^5)
(2) x = -2

Kudos for a correct solution.

The correct answer is C. Note that x^14 is not necessarily positive, it's non-negative, meaning that for x=0, x^14 = 0.
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Bunuel
Bunuel
Is f(x) < 0?

(1) f(x) = (x^3)^3*(x^5)
(2) x = -2

Kudos for a correct solution.

The correct answer is C. Note that x^14 is not necessarily positive, it's non-negative, meaning that for x=0, x^14 = 0.


hi the statement 1 gives us the reqd ans , which is NO...
the min value is 0 in all cases which is not <0.. so A should be the ans
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Bunuel
Bunuel
Is f(x) < 0?

(1) f(x) = (x^3)^3*(x^5)
(2) x = -2

Kudos for a correct solution.

The correct answer is C. Note that x^14 is not necessarily positive, it's non-negative, meaning that for x=0, x^14 = 0.


hi the statement 1 gives us the reqd ans , which is NO...
the min value is 0 in all cases which is not <0.. so A should be the ans

Sorry, there was a typo in the stem. The question reads is f(x) > 0, not f(x) < 0. Edited.
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