Let f( x )= operatornameMin ⋅((1/2)-(3 x 2/4), (5 x 2/4)) for 0 ≤ x ≤ 1. The maximum value of f( x ) is

Question

Let f( x )= operatornameMin ⋅((1/2)-(3 x 2/4), (5 x 2/4)) for 0 ≤ x ≤ 1. The maximum value of f( x ) is (A) 0 (B) -(1/4) (C) (5/16) (D) none

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/eb60678be84ca84359c86d0e6862ac64-.png" /> Sodving, (1/2)-(3 x2/4)=(5 x2/4) ∴ 2-3 x2=5 x2 ⇒ 8 x2=2 ⇒ x=(1/2) text or -(1/2) ∴ f is maximum when x =1 / 2 f max =(5/4) ⋅ (1/4)=(5/16)

Q. Let $f (x) = Min \cdot (\frac{1}{2} - \frac{3 x ^{2}}{4}, \frac{5 x ^{2}}{4})$ for $0 \leq x \leq 1$ . The maximum value of $f (x)$ is

0

$- \frac{1}{4}$

$\frac{5}{16}$

none

Sodving, $\frac{1}{2} - \frac{3 x ^{2}}{4} = \frac{5 x ^{2}}{4}$
$∴ 2 - 3 x^{2} = 5 x^{2} \Rightarrow 8 x^{2} = 2 \Rightarrow x = \frac{1}{2} or - \frac{1}{2}$
$∴ f$ is maximum when $x = 1/2$
$f_{m a x} = \frac{5}{4} \cdot \frac{1}{4} = \frac{5}{16}$

Q. Let f(x)=Min⋅(21​−43x2​,45x2​) for 0≤x≤1. The maximum value of f(x) is

0

−41​

165​

none

Sodving, 21​−43x2​=45x2​ ∴2−3x2=5x2⇒8x2=2⇒x=21​ or −21​ ∴f is maximum when x=1/2 fmax​=45​⋅41​=165​

Q. Let $f (x) = Min \cdot (\frac{1}{2} - \frac{3 x ^{2}}{4}, \frac{5 x ^{2}}{4})$ for $0 \leq x \leq 1$ . The maximum value of $f (x)$ is

$- \frac{1}{4}$

$\frac{5}{16}$

Sodving, $\frac{1}{2} - \frac{3 x ^{2}}{4} = \frac{5 x ^{2}}{4}$
$∴ 2 - 3 x^{2} = 5 x^{2} \Rightarrow 8 x^{2} = 2 \Rightarrow x = \frac{1}{2} or - \frac{1}{2}$
$∴ f$ is maximum when $x = 1/2$
$f_{m a x} = \frac{5}{4} \cdot \frac{1}{4} = \frac{5}{16}$