If $a x^{2} + b c + c = 0, a, b, c \in R$ then find the condition that this equation would have atleast one root in $(0, 1)$ .

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Solution

Let $f^{'} (x) = a x^{2} + b x + c$

$\Rightarrow f (x) = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + c x + d$

$f (0) = d$

$f (1) = \frac{2 a + 3 b + 6 c}{6} + d = d$

Hence all the conditions of Rolle's theorem are satisfied in $[0, 1]$

So $f^{'} (x) = 0$ for atleast one value in (0,1)

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If ax2+bc+c=0,a,b,c∈R then find the condition that this equation would have atleast one root in (0,1).

Let f′(x)=ax2+bx+c⇒f(x)=ax33+bx22+cx+df(0)=df(1)=2a+3b+6c6+d=dHence all the conditions of Rolle's theorem are satisfied in [0,1]So f′(x)=0 for atleast one value in (0,1)

If $a x^{2} + b c + c = 0, a, b, c \in R$ then find the condition that this equation would have atleast one root in $(0, 1)$ .

Let $f^{'} (x) = a x^{2} + b x + c$

$\Rightarrow f (x) = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + c x + d$

$f (0) = d$

$f (1) = \frac{2 a + 3 b + 6 c}{6} + d = d$

Hence all the conditions of Rolle's theorem are satisfied in $[0, 1]$

So $f^{'} (x) = 0$ for atleast one value in (0,1)