I n -01 x n tan -1 xdx - If a n I n -2- b n I n - c n - n - N - n - 1 thenA- C 1- c 2- c 3- - - are in H-P-B- a 1- a 2- a 3- - - are in A-P-C- b 1- b 2- b 3- are in G-P-D- a 1- a 2- a 3- - - are in H-P-

I_n = ( x^n+1/n+1 tan^ 1x )^1_0 ∫_0^1x^n+1/n+1.1/1+x^2dx (n+1)I_n = π/4 ∫_0^1x^n+1/1+x^2 dx (n+3)I_n+2 = π/4 ∫_0^1x^n+3/1+x^2dx ∴ (n+1) I_n + (n+3)I_n+ ...

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Byju's Answer

Standard XII

Mathematics

Integration by Parts

I n =∫01 x n ...

Question

$I_{n} = \int_{0}^{1} x^{n} {tan}^{- 1} x d x . If a_{n} I_{n + 2} + b_{n} I_{n} = c_{n} \forall n ϵ N, n \geq 1$ then

a_{1}, a_{2}, a_{3}

.....are in A.P.

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b_{1}, b_{2}, b_{3}

.....are in G.P.

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c_{1}, c_{2}, c_{3}

.....are in H.P.

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a_{1}, a_{2}, a_{3}

.....are in H.P.

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Solution

The correct option is A $a_{1}, a_{2}, a_{3}$ .....are in A.P.
$I_{n} = {(\frac{x^{n + 1}}{n + 1} t a n^{- 1} x)}_{0}^{- 1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} . \frac{1}{1 + x^{2}} d x (n + 1) I_{n} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 1}}{1 + x^{2}} d x (n + 3) I_{n + 2} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 3}}{1 + x^{2}} d x ∴ (n + 1) I_{n} + (n + 3) I_{n + 2} = \frac{π}{2} - \frac{1}{n + 2} ∴ a_{n} = (n + 3) \Rightarrow a_{1}, a_{2}, a_{3}$ .....are in A.P.
$b_{n} = (n + 1) \Rightarrow b_{1}, b_{2}$ ..... are in A.P.
$c_{n} = \frac{π}{2} - \frac{1}{n + 2}$ not in any progression.

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Similar questions

Q. If

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

and

(c_{1}, c_{2}, c_{3})

are linearly independent and

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

then

Q. If

a_{1}, a_{2}, a_{3}, . . . (a_{i} \neq a_{j} \neq 0)

and

b_{1}, b_{2}, b_{3}, . . . (b_{i} \neq b_{j} \neq 0)

are two different A.P., then which of the following is/are always correct?

Q. If each pair of equations

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

and

a_{3} x^{2} + b_{3} x + c_{3} = 0

has a common root, then show that
(i)

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

(ii)

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

If A1, B1, C1 … are the cofactors of the elements a1, b1, c1... of the matrix

$⎡ ⎢ ⎣ \begin{matrix} a 1 & b 1 & c 1 a 2 & b 2 & c 2 a 3 & b 3 & c 3 \end{matrix} ⎤ ⎥ ⎦$ then $∣ ∣ ∣ \begin{matrix} B 2 & C 2 B 3 & C 3 \end{matrix} ∣ ∣ ∣ =$

Q. Suppose four distinct positive number

a_{1}, a_{2}, a_{3}, a_{4}

are in G.P. Let

b_{1} = a_{1}, b_{2} = b_{1} + a_{2}, a_{3} = b_{2} + a_{3}

and

b_{4} = b_{3} + a_{4}

Statement 1 - The numbers

b_{1}, b_{2}, b_{3}, b_{4}

are neither in A.P. nor in G.P.
Statement 2 - The numaber

b_{1}, b_{2}, b_{3}, b_{4}

are in H.P.

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In=∫10xntan−1xdx. If anIn+2+bnIn=cn ∀ nϵN,n≥1 then

The correct option is A a1,a2,a3.....are in A.P.In=(xn+1n+1tan−1x)10−∫10xn+1n+1.11+x2dx(n+1)In=π4−∫10xn+11+x2dx(n+3)In+2=π4−∫10xn+31+x2dx∴(n+1)In+(n+3)In+2=π2−1n+2∴an=(n+3)⇒a1,a2,a3.....are in A.P. bn=(n+1)⇒b1,b2..... are in A.P. cn=π2−1n+2 not in any progression.

$I_{n} = \int_{0}^{1} x^{n} {tan}^{- 1} x d x . If a_{n} I_{n + 2} + b_{n} I_{n} = c_{n} \forall n ϵ N, n \geq 1$ then