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Question: The largest negative integer for which \[\dfrac{(x-4)(x-2)}{(x-1)(x-5)}>0\] is a). -2 b). -1 ...

The largest negative integer for which (x4)(x2)(x1)(x5)>0\dfrac{(x-4)(x-2)}{(x-1)(x-5)}>0 is
a). -2
b). -1
c). -3
d). -4

Explanation

Solution

Hint: To solve the question, we have to check whether the condition for the given expression is satisfied for the given options or not. Then arrange the largest number among the options which satisfy the given condition.

Complete step by step solution:

The given expression is (x4)(x2)(x1)(x5)\dfrac{(x-4)(x-2)}{(x-1)(x-5)}
Substitute the given options to find the largest negative integer for which the given expression be greater than 0.
Thus, by substituting the option (a) value, x = -2 in the given expression we get

& =\dfrac{(-2-4)(-2-2)}{(-2-1)(-2-5)} \\\ & =\dfrac{-(2+4)\left( -(2+2) \right)}{-(2+1)\left( -(2+5) \right)} \\\ & =\dfrac{(2+4)\left( 2+2 \right)}{(2+1)\left( 2+5 \right)} \\\ & =\dfrac{6\left( 4 \right)}{(3)\left( 7 \right)} \\\ & =\dfrac{2\left( 4 \right)}{\left( 7 \right)} \\\ & =\dfrac{8}{7}=1.143 \\\ \end{aligned}$$ Thus, the value of $$\dfrac{(x-4)(x-2)}{(x-1)(x-5)}$$ for x = -2 is 1.143 By substituting the option (b) value, x = -1 in the given expression we get $$\begin{aligned} & =\dfrac{(-1-4)(-1-2)}{(-1-1)(-1-5)} \\\ & =\dfrac{-(1+4)\left( -(1+2) \right)}{-(1+1)\left( -(1+5) \right)} \\\ & =\dfrac{(1+4)\left( 1+2 \right)}{(1+1)\left( 1+5 \right)} \\\ & =\dfrac{5\left( 3 \right)}{(2)\left( 6 \right)} \\\ & =\dfrac{5}{2\left( 2 \right)} \\\ & =\dfrac{5}{4}=1.25 \\\ \end{aligned}$$ Thus, the value of $$\dfrac{(x-4)(x-2)}{(x-1)(x-5)}$$ for x = -1 is 1.25 By substituting the option (c) value, x = -3 in the given expression we get $$\begin{aligned} & =\dfrac{(-3-4)(-3-2)}{(-3-1)(-3-5)} \\\ & =\dfrac{-(3+4)\left( -(3+2) \right)}{-(3+1)\left( -(3+5) \right)} \\\ & =\dfrac{(3+4)\left( 3+2 \right)}{(3+1)\left( 3+5 \right)} \\\ & =\dfrac{7\left( 5 \right)}{(4)\left( 8 \right)} \\\ & =\dfrac{35}{32}=1.094 \\\ \end{aligned}$$ Thus, the value of $$\dfrac{(x-4)(x-2)}{(x-1)(x-5)}$$ for x = -3 is 1.094 Thus, by substituting the option (d) value, x = -4 in the given expression we get $$\begin{aligned} & =\dfrac{(-4-4)(-4-2)}{(-4-1)(-4-5)} \\\ & =\dfrac{-(4+4)\left( -(4+2) \right)}{-(4+1)\left( -(4+5) \right)} \\\ & =\dfrac{(4+4)\left( 4+2 \right)}{(4+1)\left( 4+5 \right)} \\\ & =\dfrac{8\left( 10 \right)}{(5)\left( 9 \right)} \\\ & =\dfrac{8(2)}{\left( 9 \right)} \\\ & =\dfrac{16}{9}=1.78 \\\ \end{aligned}$$ Thus, the value of $$\dfrac{(x-4)(x-2)}{(x-1)(x-5)}$$ for x = -1 is 1.78 In the given values of options, all values result in positive values when substituted in the given expression. But, among the given options, -1 is greater than the other values. Since -1 > -2 > -3 > -4 Thus, -1 is the largest negative integer for which $$\dfrac{(x-4)(x-2)}{(x-1)(x-5)}>0$$ is Hence, option(b) is the right choice. Note: The possibility of mistake can be using the direct method for solving the given inequation, but the given expression is positive for all negative integers. Thus, solving by that method will not lead to the right answer. The other possibility of mistake can be directly assuming -1 is the right choice because often in this type of question it is better to check all the options to arrive at the right answer with no mistake.