Let fx=sin2x, 0,x≤π6ax+b, π6<x<1, If f(x) and f(x) are continuous, then

Clearly, f (x) is continuous for all .x except possibly at x=π6For f(x) to be continuous at x=π6 we must havelimx→π6- f(x)=limx→π6+ f(x) or lim x→π6 sin 2x =limx→π6 ax+b or sin π3=π6a+b or 32=π6a+b (1)For f (x) to be differentiable at x=π6. we must have (L.H.D. at x=π6) = : (R.H.D. at x=π6)or lim x→π6 2 cos 2x =limx→π6 a or 2 cos π3=a or a=1 Putting a =1 in equation (1), we get b =36-π6

Let fx=sin2x, 0,x≤π6ax+b, π6<x<1, If f(x) and f(x) are continuous, then

Clearly, f (x) is continuous for all .x except possibly at x=π6For f(x) to be continuous at x=π6 we must havelimx→π6- f(x)=limx→π6+ f(x) or lim x→π6 sin 2x =limx→π6 ax+b or sin π3=π6a+b or 32=π6a+b (1)For f (x) to be differentiable at x=π6. we must have (L.H.D. at x=π6) = : (R.H.D. at x=π6)or lim x→π6 2 cos 2x =limx→π6 a or 2 cos π3=a or a=1 Putting a =1 in equation (1), we get b =36-π6

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$Let f (x) = \{\begin{cases} \sin 2 x, 0, x \leq \frac{π}{6} \\ ax + b, \frac{π}{6} < x < 1 \end{cases}$ , If f(x) and f(x) are continuous, then

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$a = 1, b = \frac{1}{\sqrt{2}} + \frac{π}{6}$

$a = \frac{1}{\sqrt{2}}, b = \frac{1}{\sqrt{2}}$

$a = 1, b = \frac{\sqrt{3}}{2} - \frac{π}{6}$

none of these

answer is C.

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Detailed Solution

Clearly, f (x) is continuous for all .x except possibly at $x = \frac{π}{6}$

For f(x) to be continuous at $x = \frac{π}{6}$ we must have

$\lim_{x \to \frac{π}{6^{-}}} f (x) = \lim_{x \to \frac{π}{6^{+}}} f (x) or \lim_{x \to \frac{π}{6}} \sin 2 x = \lim_{x \to \frac{π}{6}} ax + b or \sin (\frac{π}{3}) = (\frac{π}{6}) a + b or \frac{\sqrt{3}}{2} = \frac{π}{6} a + b (1)$

For f (x) to be differentiable at $x = \frac{π}{6}$ . we must have (L.H.D. at $x = \frac{π}{6}$ ) = : (R.H.D. at $x = \frac{π}{6}$ )

$or \lim_{x \to \frac{π}{6}} 2 \cos 2 x = \lim_{x \to \frac{π}{6}} a or 2 \cos \frac{π}{3} = a or a = 1 Putting a = 1 in equation (1), we get b = (\frac{\sqrt{3}}{6}) - \frac{π}{6}$