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# Solution for If 4a + 2b + C = 0, Then the Equation 3ax2 + 2bx + C = 0 Has at Least One Real Root Lying in the Interval (A) (0, 1) (B) (1, 2) (C) (0, 2) (D) None of These - CBSE (Commerce) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying in the interval
(a) (0, 1)
(b) (1, 2)
(c) (0, 2)
(d) none of these

#### Solution

(c) (0, 2)

$Let$

$f\left( x \right) = a x^3 + b x^2 + cx + d . . . . . \left( 1 \right)$

$f\left( 0 \right) = d$

$f\left( 2 \right) = 8a + 4b + 2c + d$

$= 2\left( 4a + 2b + c \right) + d$

$= d \left( \because \left( 4a + 2b + c \right) = 0 \right)$

f is continuous in the closed interval [0, 2] and is derivable in the open interval (0, 2).
Also, f(0) = f(2)
By Rolle's Theorem,

$f'\left( \alpha \right) = 0 \text{ for } 0 < \alpha < 2$

$\text { Now }, f'\left( x \right) = 3a x^2 + 2bx + c$

$\Rightarrow f'\left( \alpha \right) = 3a \alpha^2 + 2b\alpha + c = 0$

$\text { Equation } \left( 1 \right) \text{has atleast one root in the interval } \left( 0, 2 \right) .$
$\text { Thus,} f'\left( x \right) \text{must have root in the interval } \left( 0, 2 \right) .$
Is there an error in this question or solution?

#### APPEARS IN

Solution If 4a + 2b + C = 0, Then the Equation 3ax2 + 2bx + C = 0 Has at Least One Real Root Lying in the Interval (A) (0, 1) (B) (1, 2) (C) (0, 2) (D) None of These Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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