Question

for maths genius only

solve question no. 17​

Answer 1

Step-by-step explanation:

We have,

$f(x) = \frac{ {a}^{x} }{ {a}^{x} + \sqrt{x} }$

$f(1 - x) = \frac{ {a}^{1 - x} }{ {a}^{1 - x} + \sqrt{a} }$

$= \frac{ \frac{a}{ {a}^{x} } }{ \frac{a}{ {a}^{x} } + \sqrt{a} }$

$= \frac{a}{a + \sqrt{a} . {a}^{x} }$

$= \frac{ \sqrt{a} }{ \sqrt{a} + {a}^{x} }$

Now,

$f(x) + f(1 - x) = \frac{ {a}^{x} }{ {a}^{x} + \sqrt{a} } + \frac{ \sqrt{a} }{ \sqrt{a} + {a}^{x} }$

$= 1$

Answer 2

Required Answer:-

Given:

$\rm \mapsto f(x) = \dfrac{ {a}^{x}}{ {a}^{x} + \sqrt{a} }$

To find:

• Value of f(x) + f(1 - x)

Solution:

Given that,

$\rm f(x) = \dfrac{ {a}^{x}}{ {a}^{x} + \sqrt{a}}$

So,

$\rm f(1 - x) = \dfrac{ {a}^{1 - x}}{ {a}^{1 - x} + \sqrt{a}}$

$\rm = \dfrac{a}{ {a}^{x} } \div ({a}^{1 - x} + \sqrt{a})$

$\rm = \dfrac{a}{ {a}^{x} } \div \bigg( \dfrac{a}{ {a}^{x} } + \sqrt{a} \bigg)$

$\rm = \dfrac{a}{ {a}^{x} } \div \bigg( \dfrac{a + {a}^{x} \sqrt{a} }{ {a}^{x} } \bigg)$

$\rm = \dfrac{a}{ {a}^{x} } \times \dfrac{ {a}^{x} }{a + {a}^{x} \sqrt{a} }$

$\rm = \dfrac{ a}{a + {a}^{x} \sqrt{a} }$

$\rm = \dfrac{ \sqrt{a} \times \sqrt{a} }{ \sqrt{a} ( \sqrt{a} + {a}^{x} )}$

$\rm = \dfrac{\sqrt{a} }{\sqrt{a} + {a}^{x}}$

So,

$\rm f(x) + f(1 - x)$

$\rm = \dfrac{ {a}^{x} }{ {a}^{x} + \sqrt{a} } + \dfrac{ \sqrt{a} }{ {a}^{x} + \sqrt{a} }$

$\rm = \dfrac{ {a}^{x} + \sqrt{a} }{ {a}^{x} + \sqrt{a} }$

$\rm = 1$

Hence,

$\rm \mapsto f(x) + f(1 - x) = 1$