Let gi-8-3-8-i-1-2 and f-8-3-8- be functions such that g1x-1-g2x-4x- and fx-sin2x- for all x-8-3-8-If Si-83-8fx- gixdx-i-1-2- then the value of 16S1- is

S_1=∫_π/8^3π/8f(x)· g_1(x)dx Given : g_1(x)=1 ⇒ S_1=∫_π/8^3π/8sin^2xdx ⇒ S_1=12∫_π/8^3π/8(1 cos 2x)dx S_1=12[π4 (sin 2x2)_π/8^3π/8] S_1=π8 ∴16S_1π=2 ...

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Standard XII

Mathematics

Standard Formulae - 2

Let gi:[π8,3π...

Question

Let $g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2$ and $f : [\frac{π}{8}, \frac{3 π}{8}] \to R$ be functions such that $g_{1} (x) = 1, g_{2} (x) = | 4 x - π |$ and $f (x) = {sin}^{2} x,$ for all $x \in [\frac{π}{8}, \frac{3 π}{8}] .$
If $S_{i} = 3 π / 8 \int π / 8 f (x) \cdot g_{i} (x) d x, i = 1, 2,$ then the value of $\frac{16 S_{1}}{π}$ is

3

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\frac{1}{2}

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Solution

The correct option is C $2$
$S_{1} = 3 π / 8 \int π / 8 f (x) \cdot g_{1} (x) d x$
Given : $g_{1} (x) = 1$
$\Rightarrow S_{1} = 3 π / 8 \int π / 8 {sin}^{2} x d x$
$\Rightarrow S_{1} = \frac{1}{2} 3 π / 8 \int π / 8 (1 - cos 2 x) d x$
$S_{1} = \frac{1}{2} [\frac{π}{4} - {(\frac{sin 2 x}{2})}_{π / 8}^{3 π / 8}]$
$S_{1} = \frac{π}{8}$
$∴ \frac{16 S_{1}}{π} = 2$

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Similar questions

Q. Let

g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2

and

f : [\frac{π}{8}, \frac{3 π}{8}] \to R

be functions such that

g_{1} (x) = 1, g_{2} (x) = | 4 x - π |

and

f (x) = {sin}^{2} x,

for all

x \in [\frac{π}{8}, \frac{3 π}{8}] .

Define

S_{i} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} f (x) \cdot g_{i} (x) d x, i = 1, 2

The value of

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Q. Let

g_{i} : [\frac{π}{8}, \frac{3 π}{8}] \to R, i = 1, 2

and

f : [\frac{π}{8}, \frac{3 π}{8}] \to R

be functions such that

g_{1} (x) = 1, g_{2} (x) = | 4 x - π |

and

f (x) = {sin}^{2} x,

for all

x \in [\frac{π}{8}, \frac{3 π}{8}] .

Define

S_{i} = \int_{\frac{π}{8}}^{\frac{3 π}{8}} f (x) \cdot g_{i} (x) d x, i = 1, 2

The value of

\frac{16 S_{1}}{π}

Q. If

a = {sin}^{4} (\frac{3 π}{2} - α) + {sin}^{4} (3 π + α)

and

b = {sin}^{6} (\frac{π}{2} + α) + {sin}^{6} (5 π - α)

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Let

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - tan x}{4 x - π}, & x \neq \frac{π}{4} λ, & x = \frac{π}{4} \end{matrix}

. and

x ϵ [0, \frac{π}{2})

f (x)

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λ

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{3} x}{3 {cos}^{2} x}, i f x < \frac{π}{2} a, i f x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}}, i f x > \frac{π}{2} \end{matrix}

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f (x)

is continuous at

x = \frac{π}{2}

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a

and

b

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Standard Formulae - 2

Standard XII Mathematics

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Let gi:[π8,3π8]→R,i=1,2 and f:[π8,3π8]→R be functions such that g1(x)=1,g2(x)=|4x−π| and f(x)=sin2x, for all x∈[π8,3π8]. If Si=3π/8∫π/8f(x)⋅gi(x)dx,i=1,2, then the value of 16S1π is

The correct option is C 2S1=3π/8∫π/8f(x)⋅g1(x)dx Given : g1(x)=1 ⇒S1=3π/8∫π/8sin2xdx ⇒S1=123π/8∫π/8(1−cos2x)dx S1=12[π4−(sin2x2)3π/8π/8] S1=π8 ∴16S1π=2