If 2a+3b+6c=0\left ( a,b,c\: \in R\right )

then the quadratic equation ax^{2}+bx+c= 0 has

  • Option 1)

    At least one in \left ( 0.1 \right )

     

  • Option 2)

    At least one root in  \left [ 2,3 \right ]

  • Option 3)

    At least one root in \left [ 4,5 \right ]

  • Option 4)

    none of these

 

Answers (1)

As we learnt in 

Rolle's Theorems -

Let f(x) be a function of x subject to the following conditions.

1.  f(x) is continuous function of    x:x\epsilon [a,b]

2.  f'(x) is exists for every point :  x\epsilon [a,b]

3.  f(a)=f(b)\:\:\:then\:\:f'(c)=0\:\:such \:that\:\:a<c<b.

-

 

 ax^{2}+bx^{2}+c=0

\\ Let, \: {f}'(x)= ax^{2}+bx+c \\ \\ \therefore f(x)=\frac{ax^{3}}{3}+\frac{bx^{2}}{2}+cx

f(0)=0+0+0=0

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

        =\frac{2a+3b+6c}{6} \: \: \: \therefore \frac{0}{6}=0

 \\So \: that \: f(0)=f(1) \\ \\ It \: means \: it \: follows \: Rolle's \: Theorem

So at least one root between (0,1)


Option 1)

At least one in \left ( 0.1 \right )

 

This option is correct

Option 2)

At least one root in  \left [ 2,3 \right ]

This option is incorrect

Option 3)

At least one root in \left [ 4,5 \right ]

This option is incorrect

Option 4)

none of these

This option is incorrect

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