6u^(2)+17u+12=

Let f be a composite function.of x defined by f(u)=(1)/(u^(3)-6u^(2)+11u-6) where u(x)=(1)/(x) .Then the number of points x where f is discontinuous is

The binding energy of ._(17)^(35 )Cl nncleus. Take atomic mass of ._6^(12) C as 12.000 an u Take R_(0) =1.2 xx 10-^(15) m .

Calculate the the binding energy per nucleon for a _6C^(12) nucleus. Nuclear mass of _6C^(12) = 12.000000 a.m.u.,mass of hydrogen nucleus =1.007825 a.m.u. and mass of neutron = 1.008665 a.m.u.

If u_(0) = 8 , u_(1) = 3 , u_(2) = 12 , u_(3) = 51 , then the value of Delta^(3) u_(0) is

The following integral int_(pi/4)^(pi/2)(2cos e cx)^(17)dx is equal to (a)int_0^("log"(1+sqrt(2)))2(e^u+e^(-u))^(16)d u (b)int_0^("log"(1+sqrt(2)))2(e^u+e^(-u))^(17)d u (c)int_0^("log"(1+sqrt(2)))2(e^u-e^(-u))^(17)d u (d)int_0^("log"(1+sqrt(2)))2(e^u-e^(-u))^(16)d u

Calculate The binding energy per nucleon of ._(6)^(12)C nucleus. Nuclear mass of ._(6)^(12)C=12.000000 u , mass of proton = 1.007825 u and mass of neutron = 1.008665 u.

Calculate the binding energy for nucleon of . _6^(12) C nucleus, if mass of proton m_p = 10078 u , mass of neutron m_n = 1.0087 u , mass of C_12, m_C =12.0000 u , and 1 u =931 .4 MeV .

If u=3x^(12) and v=x^(6) then (du)/(dv) is

If u=3x^(12) and v=x^(6) then (du)/(dv) is