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  • Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 11 sub question (i)

Need solution for RD Sharma math class 12 chapter Mean value theorem exercise 14.1 question 11 sub question (i)

Answers (1)

Answer:

               b=-5, c=8

Hint:

               f(x)=x^{3}+b x^{2}+c x, x \in[1,2]

Given:

               f(x)=x^{3}+b x^{2}+c x, x \in[1,2], x=\frac{4}{3}

Explanation:

It is given that the Rolle’s Theorem holds for the function

               f(x)=x^{3}+b x^{2}+c x, x \in[1,2]

At the point,\begin{array}{l} x=\frac{4}{3} \end{array}

We need to find the value of b and c

               f(x)=x^{3}+b x^{2}+c x

Since it satisfies the Rolle’s Theorem, we have

               \begin{aligned} & f(1)=f(2) \\\\ \Rightarrow &(1)^{3}+b(1)^{2}+c(1)=(2)^{3}+b(2)^{2}+c(2) \\\\ \Rightarrow \quad & 1+b+c=8+4 b+2 c \end{aligned}

               \Rightarrow \quad 3 b+c=-7                                    … (i)

                                                                                                                                

Differentiating the given function, we have    

 \begin{array}{l} f^{\prime}(x)=3 x^{2}+2 b x+c \ \ \ \ \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]\\\\ \Rightarrow f^{\prime}\left(\frac{4}{3}\right)=3\left(\frac{4}{3}\right)^{2}+2 b\left(\frac{4}{3}\right)+c \\\\ \Rightarrow \quad 0=\frac{16}{3}+\frac{8 b}{3}+c \end{array}          … (ii)

Solving the equation (i) and (ii), we have

b=-5, c=8

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