If function f(x) = a{x^3} + b{x^2} + 11x - 6 satisfies conditions of Rolle's theorem for interva

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5. Continuity and Differentiation

medium

If the function $f(x) = a{x^3} + b{x^2} + 11x - 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3$] and $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$, then the values of $a$ and $b$ are respectively

$1, -6$

$-2, 1$

$-1$, $\frac{1}{2}$

$-1, 6$

Solution

(a) Given $f(x) = a{x^3} + b{x^2} + 11x – 6$
From $f(1) = f(3)$
$a + b + 11 – 6 = 27a + 9b + 33 – 6$
$13a + 4b = – 11$…..$(i)$
and $f'(x) = 3a{x^2} + 2bx + 11$
$\because$$f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$
$\therefore 3a\left( {\frac{{13}}{3} + \frac{4}{{\sqrt 3 }}} \right) + 2b\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = – 11$…..$(ii)$
From $(i)$ and $(ii),$
$13a + 4b = 3\left( {\frac{{13}}{3} + \frac{4}{{\sqrt 3 }}} \right){\rm{ }}a + 2b\left( {2 + \frac{1}{{\sqrt 3 }}} \right)$
$b = – 6a$ ……$(iii)$
From $(i)$ and $(iii),$ $a = 1$ and $b = – 6$.

Standard 12

Mathematics

If function $f(x) = x(x + 3) e^{-x/2} ;$ satisfies the rolle's theorem in the interval $[-3, 0],$ then find $C$

normal

Functions $f(x)$ and $g(x)$ are such that $f(x) + \int\limits_0^x {g(t)dt = 2\,\sin \,x\, – \,\frac{\pi }{2}} $ and $f'(x).g (x) = cos^2\,x$ , then number of solution $(s)$ of equation $f(x) + g(x) = 0$ in $(0,3 \pi$) is-

normal

If $f: \mathrm{R} \rightarrow \mathrm{R}$ is a twice differentiable function such that $f^{\prime \prime}(x)>0$ for all $x \in \mathrm{R}$, and $f\left(\frac{1}{2}\right)=\frac{1}{2}, f(1)=1$, then

normal

(IIT-2017)

Let $f (x)$ and $g (x)$ are two function which are defined and differentiable for all $x \ge x_0$. If $f (x_0) = g (x_0)$ and $f ' (x) > g ' (x)$ for all $x > x_0$ then

normal

Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
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\end{array}} \right\}$, Rolle’s theorem is applicable to $ f $ for $x \in [0,1]$, if $\alpha = $

hard

(IIT-2004)

If the function $f(x) = a{x^3} + b{x^2} + 11x - 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3$] and $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$, then the values of $a$ and $b$ are respectively

Solution

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Let $f(x) = \left\{ {\begin{array}{*{20}{c}}
{{x^2}\ln x,\,x > 0} \\
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\end{array}} \right\}$, Rolle’s theorem is applicable to $ f $ for $x \in [0,1]$, if $\alpha = $