InterviewSolution
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InterviewSolution
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The function y=f(x) is the solution of the differential equation `[dy]/[dx]+[xy]/[x^2-1]=[x^4+2x]/sqrt[1-x^2]` in (-1, 1), satisfying `f(0)=0`. Then `int_[-sqrt3/2]^[sqrt3/2] f(x)dx` is (A) ` pi/3 - sqrt3/2` (B) ` pi/3 - sqrt3/4` (C) ` pi/6 - sqrt3/4` (D) ` pi/6 - sqrt3/2`A. `(pi)/(3)-(sqrt3)/(2)`B. `(pi)/(3)-(sqrt3)/(4)`C. `(pi)/(6)-(sqrt3)/(4)`D. `(pi)/(6)-(sqrt3)/(2)` |
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Answer» Correct Answer - B We have, `(dy)/(dx)+(xy)/(x^(2)-1)=(x^(4)+2x)/(sqrt(1-x^(2)))" ...(i)"` This is a linear differential equation with `"I.F."=e^(int(x)/(x^(2)-1)dx)=e^((1)/(2)log|x^(2)-1|)=e^((1)/(2)log(1-x^(2)))=sqrt(1-x^(2))` Multiplying both sides of (i) by I.F. and integrating with respect to x, we obtain `ysqrt(1-x^(2))=int(x^(4)+2x)dx+C` or, `ysqrt(1-x^(2))=(x^(5))/(5)+x^(2)+C" ...(ii)"` Putting x = 0 and y = f (0) = 0 in (ii), we get C = 0 Putting C = 0 in (ii), we get `ysqrt(1-x^(2))=(x^(5))/(5)+x^(2)` `rArr" "f(x)-(x^(5))/(5sqrt(1-x^(2)))+(x^(2))/(sqrt(1-x^(2)))` `therefore" "int_(-sqrt3//2)^(sqrt3//2)f(x)dx=int_(-sqrt3//2)^(sqrt3//2)(x^(5))/(sqrt(1-x^(2)))dx+int_(-sqrt3//2)^(sqrt3//2)(x^(2))/(sqrt(1-x^(2)))dx` `=2int_(0)^(sqrt3//2)(x^(2))/(sqrt(1-x^(2)))dx` `=2int_(0)^(pi//3)sin^(2)d theta," where x "=sin theta` `=int_(0)^(pi//3)(1-cos2 theta)d theta` `[theta-(1)/(2)sin2 theta]_(0)^(pi//3)=(pi)/(3)-(sqrt3)/(4)` |
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