Let f(x) is a continuous function which takes positive values for x (x>0), and satisfy ∫x0f(t)dt=x√f(x) with f(1)=12. Then the value of f(√2+1) equals
A
1
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B
√2−1
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C
14
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D
1√2−1
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Solution
The correct option is C14 We have ∫x0f(t)dt=x√f(x) ....(1)
Differentiating both the sides of equation (1) w.r.t. x, we get f(x)=xf′x2√f(x)+√f(x);Letf(x)=y2⇒f′(x)=2ydydxy2=x.2y.dydx.12y+y⇒y2=x.dydx+y⇒y2−y=x.dydx∫dyy(y−1)=∫dxx⇒∫y−(y−1)y(y−1)dy=∫dxx;ln(y−1)y=lncx⇒(y−1)y=cx ⇒1−1y=cx⇒1y=1−cx⇒y=11−cx∴√f(x)=11−cx ....(1)
If x=1,f(1)=12 (given) 1−c=√2;c=1−√2√f(x)=11+(√2−1)x⇒f(x)=1[1+(√2−1)x]2⇒f(√2+1)=14