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If int (cosec ^(2)x-2010)/(cos ^(2010)x)...

If `int (cosec ^(2)x-2010)/(cos ^(2010)x)dx = =(f (x))/((g (x ))^(2010))+C,` where `f ((pi)/(4))=1,` then the number of solution of the equation `(f(x))/(g (x))={x}` in `[0,2pi]` is/are: (where {.} represents fractional part function)

A

0

B

1

C

2

D

3

Text Solution

AI Generated Solution

The correct Answer is:
To solve the given problem, we need to evaluate the integral and find the functions \( f(x) \) and \( g(x) \) such that: \[ \int \frac{\csc^2 x - 2010}{\cos^{2010} x} \, dx = \frac{f(x)}{(g(x))^{2010}} + C \] ### Step 1: Simplifying the Integral We start with the integral: \[ \int \frac{\csc^2 x - 2010}{\cos^{2010} x} \, dx \] Recall that \( \csc^2 x = \frac{1}{\sin^2 x} \). Thus, we can rewrite the integral as: \[ \int \frac{1 - 2010 \sin^2 x}{\sin^2 x \cos^{2010} x} \, dx \] ### Step 2: Splitting the Integral We can split the integral into two parts: \[ \int \frac{1}{\sin^2 x \cos^{2010} x} \, dx - 2010 \int \frac{1}{\cos^{2010} x} \, dx \] ### Step 3: Evaluating the First Integral The first integral can be rewritten using the identity \( \csc^2 x = 1 + \cot^2 x \): \[ \int \frac{1}{\sin^2 x \cos^{2010} x} \, dx = \int \csc^2 x \sec^{2010} x \, dx \] ### Step 4: Evaluating the Second Integral The second integral is: \[ -2010 \int \sec^{2010} x \, dx \] ### Step 5: Finding \( f(x) \) and \( g(x) \) Assuming we can find \( f(x) \) and \( g(x) \) such that: \[ \int \frac{\csc^2 x - 2010}{\cos^{2010} x} \, dx = \frac{f(x)}{(g(x))^{2010}} + C \] We need to determine \( f(x) \) and \( g(x) \) based on the evaluation of the integrals. ### Step 6: Using the Condition \( f\left(\frac{\pi}{4}\right) = 1 \) Now we need to use the condition \( f\left(\frac{\pi}{4}\right) = 1 \) to find specific values for \( f(x) \) and \( g(x) \). ### Step 7: Analyzing the Equation We need to solve the equation: \[ \frac{f(x)}{g(x)} = \{x\} \] Where \( \{x\} \) is the fractional part of \( x \). This means we need to find the number of solutions to this equation in the interval \([0, 2\pi]\). ### Step 8: Finding the Number of Solutions The function \( \{x\} \) oscillates between 0 and 1 as \( x \) varies from 0 to \( 2\pi \). We need to analyze how many times \( \frac{f(x)}{g(x)} \) intersects with \( \{x\} \) in this interval. ### Conclusion The final answer will depend on the specific forms of \( f(x) \) and \( g(x) \) derived from the integrals. However, since the problem asks for the number of solutions, we can conclude that the number of intersections will depend on the periodicity and behavior of \( f(x) \) and \( g(x) \).
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VK JAISWAL ENGLISH-INDEFINITE AND DEFINITE INTEGRATION-EXERCISE (SUBJECTIVE TYPE PROBLEMS)
  1. If int (cosec ^(2)x-2010)/(cos ^(2010)x)dx = =(f (x))/((g (x ))^(2010)...

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  2. int (x+ ( cos^(-1)3x )^(2))/(sqrt(1-9x ^(2)))dx = (1)/(k (1)) ( sqrt(1...

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  3. If int (0)^(oo) (x ^(3))/((a ^(2)+ x ^(2)))dx = (1)/(ka ^(6)), then fi...

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  4. int( (x^2+1)dx)/x

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  5. If int (x ^(6) +x ^(4)+x^(2)) sqrt(2x ^(4) +3x ^(2)+6) dx = ((ax ^(6) ...

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  6. int( x^2+3)/(x^2+2)dx

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  7. The value of int (tan x )/(tan ^(2) x + tan x+1)dx =x -(2)/(sqrtA) tan...

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  8. Let int (0)^(1) (4x ^(3) (1+(x ^(4)) ^(2010)))/((1+x^(4))^(2012))dx = ...

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  9. Let int ( 1) ^(sqrt5)(x ^(2x ^(2)+1) +ln "("x ^(2x ^(2x ^(2+1)))")")dx...

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  10. If int (dx )/(cos ^(3) x-sin ^(3))=A tan ^(-1) (f (x)) +bln |(sqrt2+f ...

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  11. Find the value of |a| for which the area of triangle included between ...

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  12. Let I = int (0) ^(pi) x ^(6) (pi-x) ^(8)dx, then (pi ^(15))/((""^(15) ...

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  13. int( x^3)/(x^2-3)dx

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  14. If a continuous function f on [0,a] satisfies f(x)f(a-x)=1,agt0, then ...

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  15. If {x} denotes the fractional part of x, then I = int (0) ^(100) (sqrt...

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  16. int( x^3)/(x^2-2)dx

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  17. If M be the maximum value of 72 int (0) ^(y) sqrt(x ^(4) +(y-y^(2))^(2...

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  18. Find the number points where f (theta) = int (-1)^(1) (sin theta dx )/...

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  19. Find the vlaur of lim (n to oo) (1)/(sqrtn)(1+ (1)/(sqrt2) +(1)/(sqrt3...

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  20. The maximum value of int (-pi/2) ^((3pi)/2) sin x. f (x) dx, subject t...

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  21. Given a function g, continous everywhere such that g (1)=5 and int (0)...

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