`underset0overset(pi/3)int tantheta/(sqrt(2ksectheta))d theta=1-1/sqrt2,(kgt0)`, then the value of k is

Options

A. 1
B. `(1)/(2)`
C. 2
D. 4

Answer

C

Explanation

We have , `int_(0)^(pi//3)(tantheta)/(sqrt(2ksectheta))d theta =1-(1)/(sqrt(2)),(kgt0)`
Let `I=int_(0)^(pi//3)(tantheta)/(sqrt(2k sectheta))d theta=(1)/(sqrt(2k))int_(0)^(pi//3)(tantheta)/(sqrt(sectheta))d theta`
`=(1)/(sqrt(2k))int_(0)^(pi//3)((sintheta))/((cos)thetasqrt((1)/(costheta)))d theta=(1)/(sqrt(2k))int_(0)^(pi//3)(sintheta)/(sqrt(cos theta))d theta`
Let `costheta= t rArr- sinthetad theta = dt rArrsinthetad theta=-dt`
for lower limit , `theta = 0rArr t =cos0=1`
for upper limit , `theta=(pi)/(3)rArrt="cos"(pi)/(3)=(1)/(2)`
`rArrI=(1)/(sqrt(2k))int_(1)^(1//2)(dt)/(sqrt(t))=(-1)/(sqrt(2k))int_(1)^(1//2)t^(-(1)/(2)dt)`
`=-(1)/(sqrt(2k))((t1/(2)+1)/(-(1)/(2)+1))^(1/(2))=-(1)/(sqrt(2k))[2sqrt(t)]_(1)^(1/(2))`
`=-(2)/(sqrt(2k))[sqrt((1)/(2))-sqrt(1)]^(1)=(2)/(sqrt(2k))-(1-(1)/(sqrt(2)))` `:' I=1-(1)/(sqrt(2))` (given)
`:.(2)/(sqrt(2k))(1-(1)/(sqrt(2)))=1-(1)/(sqrt(2))rArr(2)/(sqrt(2k))=1`
`rArr2=sqrt(2k)rArr2k=4rArrk=2`