Let $f(x)=5-|x-2|$ and $g(x)=|x+1|$, $x \in \mathbb{R}$. If $f(x)$ attains maximum value at $\alpha$ and $g(x)$ attains minimum value at $\beta$, then $\lim_{x \rightarrow-\alpha \beta} \frac{(x-1)(x^{2}-5 x+6)}{x^{2}-6 x+8}$ is equal to :

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Limits

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

lim_{x \to - α β} \frac{( x - 1 ) ( x ^{2} - 5 x + 6 )}{x ^{2} - 6 x + 8}

is equal to :

Answer: (C)

The peak of

f (x) = 5 - ∣ x - 2∣

is at

α = 2

, and the minimum of

g (x) = ∣ x + 1∣

is at

β = - 1

, so

- α β = 2

and the limit is taken as

x \to 2

. Factor the fraction as

\frac{( x - 1 ) ( x - 2 ) ( x - 3 )}{( x - 2 ) ( x - 4 )}

, cancel the

x - 2

, and put

x = 2

to get

\frac{( 1 ) ( - 1 )}{- 2} = \frac{1}{2}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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