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આપલે પૈકી ક્યૂ વિધેય રોલના પ્રમેયનું પાલન કરે છે ?
$f(x) = \left\{ \begin{array}{l} x,\,\,\,\,\,\,\,0 \le x < 1\\ 0,\,\,\,\,\,\,\,x = 0\,\,\,\,\,\, \end{array} \right.on\,\,\left[ {0,1} \right]$
$f(x) = \left\{ \begin{array}{l} \frac{{\sin x}}{x},\,\,\,\,\,\,\, - \pi \le x < 0\\ 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0\,\,\,\,\,\, \end{array} \right.on\,\,\left[ { - \pi ,0} \right]$
$f(x) = \frac{{{x^2} - x - 6}}{{x - 1}}\,\,\,\,\,on\left[ { - 2,3} \right]$
$f(x) = \left\{ \begin{array}{l} \frac{{{x^3} - 2{x^2} + 5x + 6}}{{x - 1}},\,\,\,\,if\,\,x \ne 0\,\,\,\,\,\,\\ - 6,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,if\,\,x = 1\,\,\,\,\,\,\,\,\,\,\,\,\ \end{array} \right.on\left[ {-2,3} \right]$
Solution
$f(x)=a x^{3}+b x^{2}+11 x-6$
satisfies conditions of Rolle's theorem in $[1,3] .$
Therefore,
$f(1)=f(3)$
or $a+b+11-6=27 a+9 b+33-6$
or $13 a+4 b=-11$ …….$(1)$
and $f'(x)=3 a x^{2}+2 b x+11$
or $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 3a{\left( {2 + \frac{1}{{\sqrt 3 }}} \right)^2} + 2b\left( {2 + \frac{1}{{\sqrt 3 }}} \right) + 11 = 0$
or $3 a\left(4+\frac{1}{3}+\frac{4}{\sqrt{3}}\right)+2 b\left(2+\frac{1}{\sqrt{3}}\right)+11=0$ …….$(2)$
From equations $(1)$ and $(2),$ we get $a=1, b=-6.$
