Q: If 3(a + 2c) = 4(b + 3d) ≠ 0 then the equation ax^{3} + bx^{2} + cx + d = 0 will have

(A) no real solution

(B) at least one real root in (–1, 0)

(C) at least one real root in (0, 1)

(D) none of these

Sol: Let $\large f(x) = \frac{a x^4}{4} + \frac{b x^3}{3}+ \frac{c x^2}{2}+ dx $ ; which is continuous and differentiable.

f(0) = 0 ,

$\large f(-1) = \frac{a}{4} – \frac{b}{3} + \frac{c}{2} – d$

$\large = \frac{1}{4}(a+2c)-\frac{1}{3}(b+3d) = 0 $

so, according to Rolle’s theorem, there exist atleast one root of f'(x)= 0 in (–1, 0).