Let f(x) = 5- |x-2| and g(x) = |x + 1|, x ∈ R. If f(x) attains maximum value at α and g(x) attains minimum value at β, then displaystyle limx→-αβ ((x-1)(x2 -5x+6)/x2 - 6x + 8) is equal to :

Question

Let f(x) = 5- |x-2| and g(x) = |x + 1|, x ∈ R. If f(x) attains maximum value at α and g(x) attains minimum value at β, then displaystyle limx→-αβ ((x-1)(x2 -5x+6)/x2 - 6x + 8) is equal to : (A) 1/2 (B) -3/2 (C) 3/2 (D) -1/2

Tardigrade Admin · Accepted Answer

Maxima of f(x) occured at x = 2 i.e. α = 2 Minima of g(x) occured at x = -1 i.e. β = -1 ∴ limx → 2 ((x-1)(x-2)(x-3)/(x-2)(x-4)) = (1/2)

Q. 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 :

1/2

-3/2

3/2

-1/2

Maxima of $f (x)$ occured at $x = 2$ i.e. $α$ = $2$
Minima of g(x) occured at $x = - 1$ i.e. $β$ = $- 1$
$∴ lim_{x \to 2} \frac{( x - 1 ) ( x - 2 ) ( x - 3 )}{( x - 2 ) ( x - 4 )} = \frac{1}{2}$

Q. Let f(x)=5−∣x−2∣ and g(x)=∣x+1∣,x∈ R. If f(x) attains maximum value at α and g(x) attains minimum value at β, then x→−αβlim​x2−6x+8(x−1)(x2−5x+6)​ is equal to :

1/2

-3/2

3/2

-1/2

Maxima of f(x) occured at x=2 i.e. α = 2 Minima of g(x) occured at x=−1 i.e. β = −1 ∴limx→2​(x−2)(x−4)(x−1)(x−2)(x−3)​=21​

Q. 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 :

Maxima of $f (x)$ occured at $x = 2$ i.e. $α$ = $2$
Minima of g(x) occured at $x = - 1$ i.e. $β$ = $- 1$
$∴ lim_{x \to 2} \frac{( x - 1 ) ( x - 2 ) ( x - 3 )}{( x - 2 ) ( x - 4 )} = \frac{1}{2}$