Question

16x⁴-72x²+81>0is true for
true for
a All real values of x
b No real value of x
c al values of x Except two
values

d only for two values of se​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\:f(x) = {16x}^{4} - {72x}^{2} + 81$

$\rm :\longmapsto\:f(x) = {16x}^{4} - {36x}^{2} - 36 {x}^{2} + 81$

$\rm :\longmapsto\:f(x) = {4x}^{2}( {4x}^{2} - 9) - 9( {4x}^{2} - 9)$

$\rm :\longmapsto\:f(x) = ( {4x}^{2} - 9)( {4x}^{2} - 9)$

$\rm :\longmapsto\:f(x) = {\bigg(4 {x}^{2} - 9\bigg) }^{2}$

$\rm :\longmapsto\:f(x) = {\bigg({(2x)}^{2} - {(3)}^{2} \bigg) }^{2}$

$\rm :\longmapsto\:f(x) = {\bigg((2x + 3)(2x - 3)\bigg) }^{2}$

$\rm :\longmapsto\:f(x) = {(2x - 3)}^{2} {(2x + 3)}^{2}$

$\rm :\implies\:f(x) = 0 \: when \: x = \dfrac{3}{2} \: or \: \dfrac{ - 3}{2}$

$\rm :\longmapsto\:f(x) > 0 \: when \: x \ne \: \dfrac{3}{2}, \dfrac{ - 3}{2}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: Option \: (c) \: is \: correct}$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)