Question

If \[f(x) = \{ \]\[
  mx + 1,x \leqslant \dfrac{\pi }{2} \\
  \sin x + n,x > \dfrac{\pi }{2} \\
\] is
Continuous at \[x = \dfrac{\pi }{2}\] then
(A).\[m = \dfrac{\pi }{2} = n\]
(B).\[n = \dfrac{\pi }{2}m\]
(C).\[n = \dfrac{\pi }{2}m\]
(D).\[m = n\dfrac{\pi }{2} + 1\]

Answer 1

Hint: Continuity is the operation of something over time. Here we have to check the conditions given for \[f(x)\] at \[x = \dfrac{\pi }{2}\]. While solving the problem we take the left-hand side (LHS) and the right-hand side (RHS) and then compare both to obtain the answer. A left-hand limit means the limit of a function as it approaches from the left-hand side.
On the other hand, A right-hand limit means the limit of a function as it approaches from the right-hand side. When getting the limit of a function as it approaches a number, the idea is to check the behaviour of the function as it approaches the number. We substitute values as close as possible to the number being approached.
The closest number is the number being approached itself. Hence, one usually just substitutes the number being approached to get the limit. However, we cannot do this if the resulting value is undefined. But we can still check its behaviour as it approaches from one side.

Complete step by step solution:
The conditions are given as:
\[
mx + 1,x \leqslant \dfrac{\pi }{2} \\
\sin x + n,x > \dfrac{\pi }{2} \\
\]
In the question since \[f(x)\] is continuous at \[x = \dfrac{\pi }{2}\]
Therefore,
Step 1: We will find the limit at \[x = \dfrac{\pi }{2}\]
\[\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} f(x) = f(\dfrac{\pi }{2}) = m\dfrac{\pi }{2} + 1\]
We know that the left limit and right limit at a point are equal. So we will calculate both the left and right limit at \[x = \dfrac{\pi }{2}\].
Since, \[\mathop {\lim }\limits_{x \to \dfrac{\pi }{2}} f(x) = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x)\mathop { = \lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x)\]
Step 2: Calculate LHS
L.H.S \[ = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} f(x)\]
\[ = \mathop {\lim }\limits_{x \to {{\dfrac{\pi }{2}}^ - }} (mx + 1) = m\dfrac{\pi }{2} + 1\]
Step 3: Calculate RHS
R.H.S. \[\mathop { = \lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} f(x)\mathop { = \lim }\limits_{x \to {{\dfrac{\pi }{2}}^ + }} (\sin x + n)\]
\[
 = \sin \dfrac{\pi }{2} + n \\
 = 1 + n \\
\]
Step 4: Compare LHS and RHS
L.H.S. = R.H.S
\[
\Rightarrow m\dfrac{\pi }{2} + 1 = 1 + n \\
\therefore n = m\dfrac{\pi }{2} \\
\]

Note: Limits and derivatives are the most important concept in higher secondary and should be paid a lot of time to study. It is important to understand all the underlying concepts. Pay additional time to trigonometry and understanding graphs.