Consider the series fx=1+3sin2x+5sin4x+.....∞, gx=cosx+cos3x+cos5x+ …..n terms, then

If I 2 n = ∫ 0 π 2 g x d x , then maximum value of I 2 n + 2 − I 2 n i s 1 5 f o r n ∈ 1 , 6 , n ∈ N , lim x → 0 f x cos e c 2 x g 2 x = e 3 n 2 , ∫ f x d x = tan x + 2 tan 3 x 3 + C (where C is constant of integration, x ≠ 2 n + 1 π 2 )

Consider the series fx=1+3sin2x+5sin4x+.....∞, gx=cosx+cos3x+cos5x+ …..n terms, then

f x = sec 2 x + 2 sec 2 x tan 2 x g x = sin 2 n x 2 sin x A) ∫ 1 + 2 tan 2 x sec 2 x d x = tan x + 2 3 tan 3 x + C B) lim n → 0 sec 2 x + 2 sec 2 x tan 2 x − 1 .4 sin 2 .2 n x = e 3 n 2 C) I 2 n + 2 − I 2 n = 1 2 ∫ 0 π 2 sin 2 n + 2 x − sin 2 n x sin x = ∫ 0 π 2 cos 2 n + 1 x d x = 1 2 n + 1 sin 2 n + 1 π 2 = 1 3 . 1 5 . − 1 7 . 1 9 . − 1 11 . 1 13 f o r n ∈ 1 , 6 D) g x = 0 ⇒ x = π 2 sin 8 x = 0 , sin x ≠ 0 ⇒ x = π 8 . π 4 . 3 π 8 . π 2 . 5 π 8 . 3 π 4 . 7 π 8

If I 2 n = ∫ 0 π 2 g x d x , then maximum value of I 2 n + 2 − I 2 n i s 1 5 f o r n ∈ 1 , 6 , n ∈ N

lim x → 0 f x cos e c 2 x g 2 x = e 3 n 2

∫ f x d x = tan x + 2 tan 3 x 3 + C (where C is constant of integration, x ≠ 2 n + 1 π 2 )

If x ∈ 0 , π a n d n = 4 , then equation g(x) = 0 has 9 solutions

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Consider the series $f (x) = 1 + 3 \sin^{2} x + 5 \sin^{4} x + ..... \infty, g (x) = \cos x + \cos 3 x + \cos 5 x +$ …..n terms, then

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If $I_{2 n} = \int_{0}^{\frac{π}{2}} g (x) d x,$ then maximum value of $(I_{2 n + 2} - I_{2 n}) i s \frac{1}{5} f o r n \in [1, 6], n \in N$

$\lim_{x \to 0} {(f (x))}^{\frac{\cos e c^{2} x}{g^{2} (x)}} = e^{\frac{3}{n^{2}}}$

$\int f (x) d x = \tan x + \frac{2 \tan^{3} x}{3} + C$ (where C is constant of integration, $x \neq (2 n + 1) \frac{π}{2}$ )

If $x \in [0, π] a n d n = 4,$ then equation g(x) = 0 has 9 solutions

answer is A, B, C.

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Detailed Solution

$f (x) = \sec^{2} x + 2 \sec^{2} x \tan^{2} x$
$g (x) = \frac{\sin (2 n x)}{2 \sin x}$
A) $\int (1 + 2 \tan^{2} x) \sec^{2} x d x$ $= \tan x + \frac{2}{3} \tan^{3} x + C$
B) $\lim_{n \to 0} \frac{(\sec^{2} x + 2 \sec^{2} x \tan^{2} x - 1) .4}{\sin^{2} .2 n x} = e^{\frac{3}{n^{2}}}$
C) $I_{2 n + 2} - I_{2 n} = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{\sin (2 n + 2) x - \sin 2 n x}{\sin x}$