JEE Main 2019 — Math solution

JEE Main2019Math

Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $\in$ R. If f(x) attains maximum value at $α$ and g(x) attains minimum value at $β$ , then

$x \to α β lim \frac{( x - 1 ) ( x ^{2} - 5 x + 6 )}{x ^{2} - 6 x + 8}$ is equal to :

Solution

f(x) = 5 – |x – 2|

Graph of y = f(x)

f(x) is maximum at x = 2

$α$ = 2

g(x) = |x + 1|

Graph of y = g (x)

g(x) is minimum at x = –1

$β = - 1$

$x \to 2 lim \frac{( x - 1 ) ( x - 2 ) ( x - 3 )}{( x - 2 ) ( x - 4 )} = x \to 2 lim \frac{( x - 1 ) ( x - 3 )}{x - 4} = \frac{1}{2}$

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