Let fx- x22 x-6 tan x-2 x tan 2 x cos 2 x d x and fx passes through - 0 then number of solutions of the equation fx-x- 31-

F(x) = ∫ [x^2 (2x cos^2 x 2x sin^2 x) + 6 x^2 tan x cos^2 x]dx f(x) = ∫ 2x^3 cos 2x dx + ∫ 3x^2 sin 2x dx f(x) = 2x^3 (sin 2x/2) ∫ 6x^2 (sin 2x/2) dx + ∫ ...

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Byju's Answer

Standard XII

Mathematics

General Solution of a Differential Equation

Let fx=∫ x22 ...

Question

$Let f (x) = \int x^{2} (2 x + 6 tan x - 2 x {tan}^{2} x) {cos}^{2} x d x$ and $f (x)$ passes through $(π, 0)$ then number of solutions of the equation $f (x) = x^{3} in x ϵ [0, 2 π] is$

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Solution

$f (x) = \int [x^{2} (2 x c o s^{2} x - 2 x s i n^{2} x) + 6 x^{2} t a n x c o s^{2} x] d x f (x) = \int 2 x^{3} c o s 2 x d x + \int 3 x^{2} s i n 2 x d x f (x) = 2 x^{3} (\frac{s i n 2 x}{2}) - \int 6 x^{2} (\frac{s i n 2 x}{2}) d x + \int 3 x^{2} s i n 2 x d x f (x) = x^{3} s i n 2 x + c f (x) = 0 \Rightarrow c = 0 f (x) = x^{3} s i n 2 x$
Now, $f (x) = x^{3} \Rightarrow x = 0 & s i n 2 x = 1$
$x = (4 n + 1) \frac{π}{4}$
So, total 3 solutions

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Let f(x)=∫x2(2x+6tanx−2xtan2x)cos2x dx and f(x) passes through (π,0) then number of solutions of the equation f(x)=x3 in x ϵ [0,2π] is

f(x)=∫[x2(2x cos2x−2x sin2x)+6 x2 tan x cos2x]dxf(x)=∫2x3cos 2x dx+∫3x2sin 2x dxf(x)=2x3(sin 2x2)−∫6x2(sin 2x2) dx+∫3x2sin 2x dxf(x)=x3sin 2x+cf(x)=0⇒c=0f(x)=x3sin 2x Now, f(x)=x3⇒x=0 & sin 2x=1 x=(4n+1)π4 So, total 3 solutions

$Let f (x) = \int x^{2} (2 x + 6 tan x - 2 x {tan}^{2} x) {cos}^{2} x d x$ and $f (x)$ passes through $(π, 0)$ then number of solutions of the equation $f (x) = x^{3} in x ϵ [0, 2 π] is$