Let f ( x ) + 2 f ( 1 x ) = x 2 + 5 and 2 g ( x ) – 3 g ( 1 2 ) = x , x 0. If = 1 2 f ( x ) d x , and =1 2 g ( x ) d x , then the value of 9 is : [2025]

Question

Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]

Smart Achievers · Accepted Answer

(2)

We have, $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ ... (i)

$\Rightarrow f (\frac{1}{x}) + 2 f (x) = \frac{1}{x^{2}} + 5$ ... (ii)

On solving (i) and (ii), we get

$f (x) = \frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}$

So, $α = \int_{1}^{2} f (x) d x = \int_{1}^{2} (\frac{2}{3 x^{2}} - \frac{x^{2}}{3} + \frac{5}{3}) d x$

$= {(- \frac{2}{3 x} - \frac{x^{3}}{9} + \frac{5 x}{3})}_{1}^{2} = \frac{6}{3} - \frac{7}{9} = \frac{11}{9}$

Also, we have $2 g (x) - 3 g (\frac{1}{2}) = x$

$\Rightarrow 2 g (\frac{1}{2}) - 3 g (\frac{1}{2}) = \frac{1}{2} \Rightarrow g (\frac{1}{2}) = - \frac{1}{2}$

so, $g (x) = \frac{x}{2} - \frac{3}{4}$

$β = \int_{1}^{2} g (x) d x = \int_{1}^{2} (\frac{x}{2} - \frac{3}{4}) d x$

$= {(\frac{x^{2}}{4} - \frac{3 x}{4})}_{1}^{2} = \frac{3}{4} - \frac{3}{4} = 0$

$∴ 9 α + β = 9 (\frac{11}{9}) + 0 = 11$ .

Solution

Q.
Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]

1 0

2 11

3 10

4 1

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Solution

Q. Let f(x)+2f(1x)=x2+5 and 2g(x)–3g(12)=x, x > 0. If α=∫12f(x)dx, and β=∫12g(x)dx, then the value of 9α+β is : [2025]

1 0

2 11

3 10

4 1

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Q.
Let $f (x) + 2 f (\frac{1}{x}) = x^{2} + 5$ and $2 g (x) - 3 g (\frac{1}{2}) = x$ , x > 0. If $α = \int_{1}^{2} f (x) d x$ , and $β = \int_{1}^{2} g (x) d x$ , then the value of $9 α + β$ is : [2025]