If the roots of x² - 2kx + k² - 4 = 0 are real and exactly one lies in (-2,3), then the integral value of k is
Correct: D
Let f(x)=x²-2kx+k²-4. For exactly one root in (-2,3), f(-2)f(3)<0. f(-2)=4+4k+k²-4=k(k+4), f(3)=9-6k+k²-4=k²-6k+5=(k-1)(k-5). Thus k(k+4)(k-1)(k-5)<0. Sign-chart gives k∈(-4,0)∪(1,5). Integral values: -3,-2,-1,2,3,4. Only 2 appears in choices.