I n -0-2cos n x cos n x d x - n - N then - I 2001- I 2002 can be the eccentricity ofA- CircleB- HyperbolaC- EllipseD- Parabola

The correct option is B HyperbolaI_n+1 = ∫_0^π/2cos^n+1 x cos(n+1) xdx = ∫_0^π/2cos^n+1x (cos nx cos x sin nx sin x)dx ∴ I_n+1 = I_n I_n+1 ⇒ 2 ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

I n =∫0π/2cos...

Question

$I_{n} = \int_{0}^{\frac{π}{2}} {cos}^{n} x cos (n x) d x, n ϵ N$ then $\sqrt{I_{2001} : I_{2002}}$ can be the eccentricity of

Parabola

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Ellipse

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Circle

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Hyperbola

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Solution

The correct option is D Hyperbola
$I_{n + 1} = \int_{0}^{\frac{π}{2}} {cos}^{n + 1} x cos (n + 1) x d x = \int_{0}^{\frac{π}{2}} {cos}^{n + 1} x (cos n x cos x - sin n x sin x) d x ∴ I_{n + 1} = I_{n} - I_{n + 1} \Rightarrow 2 I_{n + 1} = I_{n} ∴ I_{n} : I_{n + 1} = 2$

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In=∫π20cosnxcos(nx)dx,nϵN then √I2001:I2002 can be the eccentricity of

The correct option is D HyperbolaIn+1=∫π20cosn+1xcos(n+1)xdx=∫π20cosn+1x(cosnxcosx−sinnxsinx)dx∴In+1=In−In+1 ⇒2In+1=In∴In:In+1=2

$I_{n} = \int_{0}^{\frac{π}{2}} {cos}^{n} x cos (n x) d x, n ϵ N$ then $\sqrt{I_{2001} : I_{2002}}$ can be the eccentricity of

The correct option is D Hyperbola
$I_{n + 1} = \int_{0}^{\frac{π}{2}} {cos}^{n + 1} x cos (n + 1) x d x = \int_{0}^{\frac{π}{2}} {cos}^{n + 1} x (cos n x cos x - sin n x sin x) d x ∴ I_{n + 1} = I_{n} - I_{n + 1} \Rightarrow 2 I_{n + 1} = I_{n} ∴ I_{n} : I_{n + 1} = 2$