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જો $f(x) = ax^3 + bx^2 + 11x - 6, x \,\in [1, 3]$ એ રોલના પ્રમેયની શરતોનું પાલન કરે અને ${f}'\,\left( {2\, + \,\frac{1}{{\sqrt 3 }}} \right)\, = \,0$ થાય, તો $a$ અને $b$ શોધો.
$1, -6$
$1, 1$
$0, 6$
$6,-6$
Solution
રોલના પ્રમેયનું પાલન થાય છે તેથી $f(1) = f(3)$
$==> a + b + 11 – 6 = 27a + 9b + 33 – 6 $
$==> 13a + 4b + 11 = 0….. (i)$
હવે, $f'(x) = 3ax^2 + 2bx + 11 = 0$
$\text{x}\,\,=\,\,\left( \text{2}\,\,+\,\,\frac{\text{1}}{\sqrt{\text{3}}} \right)$ આગળ
$\Rightarrow \,\text{f }\!\!'\!\!\text{ }\,\left( \text{2}\,\,+\,\,\frac{\text{1}}{\sqrt{\text{3}}} \right)\,\,=0$
$3a\,{{\left( 2\,\,+\,\,\frac{1}{\sqrt{3}} \right)}^{2}}\,+\,\,2b\,\left( 2\,\,+\,\,\frac{1}{\sqrt{3}} \right)\,\,+\,11\,\,=\,\,0$
$\,\,3a\,\left[ 4\,\,+\,\,\frac{1}{3}\,\,+\,\,\frac{4}{\sqrt{3}} \right]\,+\,2b\,\left( 2\,\,+\,\,\frac{1}{\sqrt{3}} \right)\,+\,11\,\,=\,\,0 $
$\,\,13a\,\,+\,\,4b\,\,+\,\,11\,\,+\,\,\frac{12a}{\sqrt{3}}\,\,+\,\,\frac{2b}{\,\sqrt{3}}\,\,=\,\,0\,\,\,$
$\Rightarrow \,\,\frac{12a}{\sqrt{3}}\,\,+\,\,\frac{2b}{\sqrt{3}}\,\,=\,\,0$
$\Rightarrow \,\frac{\text{12a}}{\sqrt{\text{3}}}\,\,=\,\,-\,\,\frac{2b}{\sqrt{3}}\,\,$
$\Rightarrow \,\,b\,\,=\,\,-6a\,\,\,\,\,\,$
$b$ ની કિમત $\text{(1)}$ માં મુક્તા
$13a + 4b + 11 = 0$
$\Rightarrow$ $ 13a + 4b + 11 = 0 $
$\Rightarrow13a – 24a + 11 = 0$
$\Rightarrow a = 1$ & $b = -6$
