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If f(x)=\mathrm{ \frac{4x-x^2}{(x+2)^2}}, the value of x for which \\\mathrm{ f(x)\geq 0}

Option: 1

x \in [0, 4]


Option: 2

x \in [-\infty, -2)\cup [0, 4]


Option: 3

x \in (-2,0)\cup [4,\infty]


Option: 4

x \in R \;\;\;\;\text{except}\;\;\; x=-2


Answers (1)

best_answer

 

 

Wavy Curve Method -

wavy curve method is used to solve the inequality of the type 

\\\mathrm{\frac{f(x)}{g(x)} > 0 \; or\; (<,\leq, \geq)}

We use following steps in wavy curve method to solve a question

1: Factorize numerator and denominator into linear factors.

2: Make coefficients of x positive in all linear factors. This step may require to change sign of “x” in the linear factor by multiplying inequality with -1. Note that this multiplication will change the inequality sign as well. For example, “less than” will become “greater than” etc.

3: Equate each linear factor to zero and find the values of x in each case. The values are called critical points.

4: Identify distinct critical points on the real number line. The “n” numbers of distinct critical points divide real number lines in (n+1) sub-intervals.

5: The sign of rational function in the rightmost interval is positive. Alternate sign in adjoining intervals on the left.

5: If a linear factor is repeated even times, then sign of function will not alternate about the critical point corresponding to linear factor in question.

 

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\\\mathrm{factorising\; the\; polynomial\; we \; have\;\;\frac{(4-x)x}{(x+2)^2}\geq 0} \\\mathrm{making\; x \;variable\; +ve\; we \; have} \\\mathrm{\frac{(x-4)x}{(x+2)^2}\leq0} \\\mathrm{equating \; all \; factor\; to \; 0,\; we \; have} \\\mathrm{x = 0, x=4, x=-2,-2}

Drawing this on the number line we have

Hence, x \in [0, 4]

 

correct option is (a)

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manish painkra

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