F and g be two positive real valued functions defined on -1-1- such that f-x-1fx and g is an even function with -11gxdx-1 then I-11fxgxdx satisfies

The correct option is A I≥ 1 I=∫_ 1^1f(x)g(x)dx I=∫_ 1^1f( x)g( x)dx ⇒ 2I=∫_ 1^1f(x)g(x)dx+∫_ 1^1f( x)g( x)dx ⇒ 2I=∫_ 1^1g(x)[f(x)+f( x ...

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Byju's Answer

Standard XII

Mathematics

Area between Two Curves

f and g be ...

Question

$f$ and $g$ be two positive real valued functions defined on $[- 1, 1]$ such that $f (- x) = \frac{1}{f (x)}$ and $g$ is an even function with $\int_{- 1}^{1} g (x) d x = 1$ then $I = \int_{- 1}^{1} f (x) g (x) d x$ satisfies

I \geq 1

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I < 1

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\frac{1}{2} < I < 3

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I = 1

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Solution

The correct option is A $I \geq 1$
$I = \int_{- 1}^{1} f (x) g (x) d x$

$I = \int_{- 1}^{1} f (- x) g (- x) d x$

$\Rightarrow 2 I = \int_{- 1}^{1} f (x) g (x) d x + \int_{- 1}^{1} f (- x) g (- x) d x$

$\Rightarrow 2 I = \int_{- 1}^{1} g (x) [f (x) + f (- x)] d x$

$∵ g (x)$ is even.
$\Rightarrow 2 I \geq \int_{- 1}^{1} g (x) d x = 2$

$∴ I \geq 1$

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Similar questions

f

and

g

be two positive real function defined on

[- 1, 1]

such that

f (- x) = \frac{1}{f (x)}

and

g

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\int_{1}^{1} g (x) d x = 1

then

\int_{- 1}^{1} f (x) g (x) d x

satisfies

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f (x) = \sqrt{x + 1}

and

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. Then, describe each of the following functions:
(i) f + g
(ii) g − f
(iii) f g
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(v)

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Q. Let

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f and g be two positive real valued functions defined on [−1,1] such that f(−x)=1f(x) and g is an even function with ∫1−1g(x)dx=1 then I=∫1−1f(x)g(x)dx satisfies

The correct option is A I≥1I=∫1−1f(x)g(x)dxI=∫1−1f(−x)g(−x)dx⇒2I=∫1−1f(x)g(x)dx+∫1−1f(−x)g(−x)dx⇒2I=∫1−1g(x)[f(x)+f(−x)]dx∵g(x) is even.⇒2I≥∫1−1g(x)dx=2∴I≥1

$f$ and $g$ be two positive real valued functions defined on $[- 1, 1]$ such that $f (- x) = \frac{1}{f (x)}$ and $g$ is an even function with $\int_{- 1}^{1} g (x) d x = 1$ then $I = \int_{- 1}^{1} f (x) g (x) d x$ satisfies