Question

The maximum value of f(x) = -3 $x^2$ + 5x + 2 ∀ x $ϵ$ [0, 2] is

• A. $\frac{150}{37}$
• B. $\frac{210}{37}$
• C. $\frac{147}{36}$
• D. $\frac{150}{37}$

Answer 1

The correct option is C

$\frac{147}{36}$

f(x) = -3 $x^2$+ 5x + 2

Here a < 0 ⇒ graph is inverted cup shaped & maximum value will occur at x = $\frac{-b}{2a}$ = $\frac{5}{6}$

As 0 <( $\frac{5}{6}$) < 2 ⇒ No other value of x in [0, 2] will have f(x) > f( $\frac{5}{6}$)

So, f( $\frac{5}{6}$) is maximum value

f( $\frac{5}{6}$) = -3x ${\left(\frac{5}{6}\right)}^2$ +5 $\frac{5}{6}$ +2 = $\frac{147}{36}$