Let f , g : ( 0 , ) R be two functions defined by f ( x ) = - x x ( | t | - t 2 ) e - t 2 ? d t and g ( x ) = 0 x 2 t 1 / 2 e - t ? d t . Then, the value of 9 ( f ( log e 9 ) + g ( log e 9 ) ) is equal to [2024]

Question

Let $f, g : (0, \infty) \to R$ be two functions defined by $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t and g (x) = \int_{0}^{x^{2}} t^{1 / 2} e^{- t} d t .$ Then, the value of $9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}))$ is equal to [2024]

Smart Achievers · Accepted Answer

(1)

We have, $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t$

$\Rightarrow f (x) = 2 \int_{0}^{x} (t - t^{2}) e^{- t^{2}} d t$

$= 2 [\int_{0}^{x} t e^{- t^{2}} d t - \int_{0}^{x} t^{2} e^{- t^{2}} d t] = [1 - e^{- x^{2}} - 2 \int_{0}^{x} t^{2} e^{- t^{2}} d t]$

Let $t^{2} = p \Rightarrow 2 t \cdot d t = d p$

$\Rightarrow d t = \frac{d p}{2 \sqrt{p}}$

$∴ f (x) = 1 - e^{- x^{2}} - 2 \int_{0}^{x^{2}} \frac{p \cdot e^{- p}}{2 \sqrt{p}} d p$

$= 1 - e^{- x^{2}} - \int_{0}^{x^{2}} \sqrt{p} e^{- p} d p = 1 - e^{- x^{2}} - g (x)$

$f (x) + g (x) = 1 - e^{- x^{2}}$

Now, $f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}) = 1 - e^{- \log_{e} 9} = 1 - \frac{1}{9} = \frac{8}{9}$

$∴ 9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9})) = 8$

Solution

Q.
Let $f, g : (0, \infty) \to R$ be two functions defined by $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t and g (x) = \int_{0}^{x^{2}} t^{1 / 2} e^{- t} d t .$ Then, the value of $9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}))$ is equal to [2024]

1 8

2 10

3 9

4 6

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Solution

Q. Let f,g:(0,∞)→R be two functions defined by f(x)=∫-xx(|t|-t2)e-t2 dt and g(x)=∫0x2t1/2e-t dt. Then, the value of 9(f(loge9)+g(loge9)) is equal to [2024]

1 8

2 10

3 9

4 6

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Q.
Let $f, g : (0, \infty) \to R$ be two functions defined by $f (x) = \int_{- x}^{x} (| t | - t^{2}) e^{- t^{2}} d t and g (x) = \int_{0}^{x^{2}} t^{1 / 2} e^{- t} d t .$ Then, the value of $9 (f (\sqrt{\log_{e} 9}) + g (\sqrt{\log_{e} 9}))$ is equal to [2024]