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Question: If the function\(\left( \frac{x - 2}{2} \right) + \frac{\pi}{6}\), is continuous in the interval [0...

If the function(x−22)+π6\left( \frac{x - 2}{2} \right) + \frac{\pi}{6}, is

continuous in the interval [0, π] then the values of (a, b) are

A

(–1, –1)

B

(0, 0)

C

(–1, 1)

D

(1, –1)

Answer

(0, 0)

Explanation

Solution

Since f is continuous at x=π4x = \frac { \pi } { 4 };

∴f(π4)=fh→0(π4+h)=fh→0(π4−h)\therefore f \left( \frac { \pi } { 4 } \right) = \underset { h \rightarrow 0 } { f } \left( \frac { \pi } { 4 } + h \right) = \underset { h \rightarrow 0 } { f } \left( \frac { \pi } { 4 } - h \right)

⇒ π4(1)+b=(π4−0)+a22sin⁡(π4−0)\frac { \pi } { 4 } ( 1 ) + b = \left( \frac { \pi } { 4 } - 0 \right) + a ^ { 2 } \sqrt { 2 } \sin \left( \frac { \pi } { 4 } - 0 \right)

⇒\Rightarrow π4+b=π4+a22sin⁡π4\frac { \pi } { 4 } + b = \frac { \pi } { 4 } + a ^ { 2 } \sqrt { 2 } \sin \frac { \pi } { 4 }⇒b=a22⋅12⇒b=a2b = a ^ { 2 } \sqrt { 2 } \cdot \frac { 1 } { \sqrt { 2 } } \Rightarrow b = a ^ { 2 }

Also as f is continuous at x=π2x = \frac { \pi } { 2 };

⇒\Rightarrow bsin⁡2π2−acos⁡2π2=lim⁡h→0[(π2−h)cot⁡(π2−h)+b]b \sin 2 \frac { \pi } { 2 } - a \cos 2 \frac { \pi } { 2 } = \lim _ { h \rightarrow 0 } \left[ \left( \frac { \pi } { 2 } - h \right) \cot \left( \frac { \pi } { 2 } - h \right) + b \right]

⇒ b.0−a(−1)=0+b⇒a=bb .0 - a ( - 1 ) = 0 + b \Rightarrow a = b .

Hence (0, 0) satisfy the above relations.