Q2 of 96 Page 15

Mark the correct alternative in the following:

If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval.


Let f(x) = ax3 + bx2 + cx + d --------- (i)


f(0) = d


f(2) = a(2)3 + b(2)2 + c(2) + d


= 8a + 4b + 2c + d


= 2(4a + 2b + c) + d


4a + 2b + c = 0 {Given}


= 2 (0) + d


= 0 + d


= d


f is continuous in closed interval [0, 2] and f is derivable in the open interval (0, 2).


Also, f(0) = f(2)


As per Rolle’s Theorem,


f’(α) = 0 for 0 < α < 2


f’(x) = 3ax2 + 2bx + c


f’(α) = 3aα2 + 2b(α) + c


3aα2 + 2b(α) + c = 0


Hence equation (i) has at least one root in the interval (0, 2).


Thus, f’(x) must have one root in the interval (0, 2).


Hence, Option (C) is the answer.

More from this chapter

All 96 →
11

Using Lagrange’s mean value theorem, prove that

(b – a ) sec2 a < tan b – tan a < (b – a) sec2 b, where 0 < a < b < π/2.

 1

Mark the correct alternative in the following:

If the polynomial equation a0xn + an–1xn–1 + an–2xn–2 + …. a2x2 + a1x + a0 = 0


n being a positive integer, has two different real roots α and β, then between α and β, the equation n anxn–1 + (n–1) an–1xn–2 + … + a1 = 0 has


 3

Mark the correct alternative in the following:

For the function the value of c for the Lagrange’s mean value theorem is


 4

Mark the correct alternative in the following:

If from Lagrange’s mean value theorem, we have then