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Quantitative Aptitude Question on Linear & Quadratic Equations

f(x)=f(x) = x2+2x−15x2−7x−18\frac{x^2+2x−15}{x^2−7x−18} is negative if and only if

A

−5<x<−2-5 < x < -2 or 3<x<9 3 < x < 9

B

−2<x<3-2 < x < 3 or x>9x > 9

C

x<−5x < -5 or 3<x<93 < x < 9

D

x<−5x < -5 or −2<x<3-2 < x < 3

Answer

−5<x<−2-5 < x < -2 or 3<x<9 3 < x < 9

Explanation

Solution

To find when f(x)f(x) is negative, we need to determine the sign of f(x)f(x) in the different intervals given by its zeros. Let's first find the zeros of the numerator and denominator:
For (x2+2x−15):(x+5)(x−3)=0\left( x^2 + 2x - 15 \right): (x + 5)(x - 3) = 0
This gives us zeros at x = -5 and x = 3.

For (x2−7x−18):(x+2)(x−9)=0\left( x^2 - 7x - 18 \right): (x + 2)(x - 9) = 0
This gives us zeros at x = -2 and x = 9.

Now, using these zeros, we get the following intervals:
(-∞, -5), (-5, -2), (-2, 3), (3, 9), and (9, ∞).

Let's test the sign of f(x)f(x) in each of these intervals by picking a test point from each:
1. Test (x=−6):(f(−6))=(−)(−)=( x = -6 ): ( f(-6) ) = \frac{(-)}{ (-)} =positive
2. Test (x=−3):(f(−3))=(+)(−)=\left( x = -3 \right): \left( f(-3) \right) =\frac{ (+) }{ (-) }= negative
3. Test (x=0):(f(0))=(−)(−)=\left( x = 0 \right): \left( f(0) \right) = \frac{(-) }{ (-) }= positive
4. Test(x=6):(f(6))=(+)(−)=\left( x = 6 \right): \left( f(6) \right) =\frac{ (+) }{ (-)} = negative
5. Test (x=10):(f(10))=(+)(+)=( x = 10 ): ( f(10) ) =\frac{ (+) }{ (+)} = positive
From the above evaluations, f(x)f(x) is negative in the intervals: (-5, -2) and (3, 9).

So, the correct option is option(A): −5<x<−2-5 < x < -2 or 3<x<9.3 < x < 9.