Let f(x) be defined byf(x)=sin⁡ 2x if 0<x≤π/6ax+b if π/6<x≤1The values of a and b such that f and f' are continuous, are

Let f(x) be defined byf(x)=sin⁡ 2x if 0

A
$a = 1, b = 1 / \sqrt{2} + π / 6$
B
$a = 1 / \sqrt{2}, b = 1 / \sqrt{2}$
C
$a = 1, b = \sqrt{3} / 2 - π / 6$
D
none of these

Solution:

For $f$ to be continuous $f (π / 6) = lim_{x \to π / 6 +} f (x)$

$\begin{array}{l} \sin \frac{π}{3} = a \frac{π}{6} + b i.e. \frac{\sqrt{3}}{2} = a \frac{π}{6} + b \\ f^{'} (π / 6 +) = lim_{h \to 0 +} \frac{f (π / 6 + h) - f (π / 6)}{h} \\ = lim_{h \to 0 +} \frac{a (π / 6 + h) + b - \sin π / 3}{h} \\ = lim_{h \to 0 +} \frac{a (π / 6 + h) + b - (a π / 6 + b)}{h} \\ = a . \\ f^{'} (π / 6 -) = lim_{h \to 0 -} \frac{f (π / 6 + h) - f (π / 6)}{h} \\ = lim_{h \to 0 -} \frac{\sin 2 (π / 6 + h) - \sin π / 3}{h} \\ = 2 \cos 2 \frac{π}{6} = 2 \frac{1}{2} = 1, \\ so a = 1, b = \frac{\sqrt{3}}{2} - \frac{π}{6} . For these values of a and b, \\ f^{'} (x) = \{\begin{array}{cc} 2 \cos 2 x & 0 < x \leq π / 6 \\ 1 & π / 6 < x \leq 1 \end{array} which is continuous. \end{array}$

Let $f (x)$ be defined by
$f (x) = \{\begin{cases} \sin 2 x if 0 < x \leq π / 6 \\ a x + b if π / 6 < x \leq 1 \end{cases}$ The values of $a$ and $b$ such that $f$ and $f'$ are continuous, are

Solution:

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