[ 0,dfrac (pi )(2)] D. [ -sqrt (dfrac {pi )(2)},0]

sqrt (2)pi -|||-C. dfrac (sqrt {2pi )}(4)-|||-D. dfrac (sqrt {2pi )}(8)

dfrac (2pi )(sqrt {3)} D. dfrac (pi )(6)

dfrac (2pi )(sqrt {3)}-|||-D. dfrac (pi )(6)

dfrac (1)(2)pi .-|||-C.π-|||-D. dfrac (5)(4)pi .-|||-0

dfrac ({mu )_(0)I}(pi R)(dfrac (1)(2)+dfrac (pi )(6))-|||-dfrac ({mu )_(0)I}(pi

3.要使函数φ(x )= ,dfrac {pi )(2)] (B)[π,2π] (C) [ 0,dfrac (pi )(2)] (D) [ dfrac

dfrac (qQ)(4pi {varepsilon )_(0)l} C. dfrac (qQ)(2pi {varepsilon )_(0)l} D. -dfr

dfrac (pi )(4) C. dfrac (pi )(2) D.π

(B) (mu ,dfrac (1)(sqrt {2pi )})-|||-(C) (mu ,dfrac (1)(2)), (D)(0,σ).

dfrac (1)(2)pi (R)^2E-|||-D. sqrt (2)(R)^2E-|||-E. dfrac (pi {R)^2E}(sqrt {2)}