Question

Let t = x² + 4x + 9/ X² + 9 X belongs R. If range of values of t is [a, b], then a+b/3 is equal to​

Answer 1

Step-by-step explanation:

We have,

$\tt{t= \dfrac{ {x}^{2} + 4x + 9 }{ {x}^{2} + 9 } }$

$\sf{ \implies \: t( {x}^{2} + 9) = {x}^{2} + 4x + 9 }$

$\sf{ \implies \: t {x}^{2} + 9t = {x}^{2} + 4x + 9 }$

$\sf{ \implies \: (t - 1){x}^{2} - 4x + 9t - 9 = 0 }$

$\sf{ \implies \: (t - 1){x}^{2} - 4x + 9(t - 1)= 0 }$

Since $\sf{x\in \mathbb{R}}$, so discriminant of the above quadratic equation will be non-negative

So,

$\sf{ \implies \: ( - 4)^{2} - 4 \cdot9(t - 1) \cdot(t - 1) \geqslant 0}$

$\sf{ \implies \:16 - 4 \cdot9(t - 1)^{2} \geqslant 0}$

$\sf{ \implies \:4 \{4 - 9(t - 1)^{2} \}\geqslant 0}$

$\sf{ \implies \:4 - 9(t - 1)^{2} \geqslant 0}$

$\sf{ \implies \: 9(t - 1)^{2} - 4 \leqslant 0}$

$\sf{ \implies \: 9t^{2} - 18t + 9- 4 \leqslant 0}$

$\sf{ \implies \: 9t^{2} - 18t + 5 \leqslant 0}$

$\sf{ \implies \: 9t^{2} - 15t - 3t+ 5 \leqslant 0}$

$\sf{ \implies \: 3t(3t - 5) - 1(3t - 5 ) \leqslant 0}$

$\sf{ \implies \: (3t - 1)(3t - 5) \leqslant 0}$

$\sf{ \implies \:3 \bigg(t - \dfrac{1}{3} \bigg) \cdot3 \bigg(t -\dfrac{5}{3} \bigg) \leqslant 0}$

$\sf{ \implies \:9 \bigg(t - \dfrac{1}{3} \bigg) \bigg(t -\dfrac{5}{3} \bigg) \leqslant 0}$

$\sf{ \implies \: \bigg(t - \dfrac{1}{3} \bigg) \bigg(t -\dfrac{5}{3} \bigg) \leqslant 0}$

$\sf{ \implies \: t \in \bigg [ \dfrac{1}{3} , \dfrac{5}{3} \bigg] }$

So,

$\sf{a = \dfrac{1}{3} \: \: \: and \: \: \: b = \dfrac{5}{3} }$

Now,

$\sf{a + b = \dfrac{1}{3} + \dfrac{5}{3} }$

$\sf{ \implies \: a + b = \dfrac{1 + 5}{3} }$

$\sf{ \implies \: a + b = \dfrac{6}{3} }$

$\sf{ \implies \: a + b = 2 }$

$\sf{ \implies \: \dfrac{a + b}{3} = \dfrac{2}{3} }$