Question

# It is given that for the function  f(x) = x3 - 6x2 + ax + b on [1, 3], Rolle's theorem holds with c = $2+\frac{1}{\sqrt{3}}.$ If f(1) = f(3) = 0, then a =_______, b =________.

Solution

## The given function is f(x) = x3 − 6x2 + ax + b. It is given that Rolle's theorem holds for f(x) defined on [1, 3] with $c=2+\frac{1}{\sqrt{3}}$. $f\left(1\right)=f\left(3\right)=0$     (Given) $\therefore f\left(1\right)=0$ $⇒1-6+a+b=0$ $⇒a+b=5$        .....(1) Also, $f\left(3\right)=0$ $⇒27-54+3a+b=0$ $⇒3a+b=27$     .....(2) Solving (1) and (2), we get a = 11 and b = −6 It can be verified that for a = 11 and b = −6, $f\text{'}\left(2+\frac{1}{\sqrt{3}}\right)=0$. Thus, the values of a and b are 11 and −6, respectively. It is given that for the function  f(x) = x3 − 6x2 + ax + b on [1, 3], Rolle's theorem holds with c = $2+\frac{1}{\sqrt{3}}.$ If f(1) = f(3) = 0, then a = ___11___, b =___−6___.MathematicsRD Sharma XII Vol 1 (2019)All

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