JEE MAIN - Mathematics (2019 - 12th April Evening Slot - No. 9)
Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $$ \in $$ R. If f(x) attains maximum value at $$\alpha $$ and g(x) attains
minimum value at $$\beta $$, then
$$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$$ is equal to :
$${1 \over 2}$$
$$-{1 \over 2}$$
$${3 \over 2}$$
$$-{3 \over 2}$$
Explanation
From f(x) = 5 - | x - 2 |
maximum value of f(x) is at x = 2
From g(x) = | x + 1 |
minimum value of g(x) is at x = -1
$$ \therefore $$ $$\alpha \beta $$ = - 2
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}$$
$$ \Rightarrow $$ $${{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}$$
maximum value of f(x) is at x = 2
From g(x) = | x + 1 |
minimum value of g(x) is at x = -1
$$ \therefore $$ $$\alpha \beta $$ = - 2
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}$$
$$ \Rightarrow $$ $${{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}$$
Comments (0)
