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Question: If \(f(x) = \left\\{ \begin{gathered} mx - 1,x \leqslant 5 \\\ 3x - 5,x > 5 \\\ \end{gath...

If f(x) = \left\\{ \begin{gathered} mx - 1,x \leqslant 5 \\\ 3x - 5,x > 5 \\\ \end{gathered} \right. is continuous then the value of m is :
A.115\dfrac{{11}}{5}
B.511\dfrac{5}{{11}}
C.53\dfrac{5}{3}
D.35\dfrac{3}{5}

Explanation

Solution

We are given that the function f(x) is continuous at 5 and we know that any function is continuous at x = a then limxaf(x)=limxa+f(x)\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) and applying this we can get the value of m

Complete step-by-step answer:
The given function is f(x) = \left\\{ \begin{gathered} mx - 1,x \leqslant 5 \\\ 3x - 5,x > 5 \\\ \end{gathered} \right.
And we are given that f(x) is continuous at x = 5
Whenever a function is continuous at x = a then
limxaf(x)=limxa+f(x)\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)
Same way, applying this in our given function
Since the value of f(x) when x is less than 5 is mx – 1
We get limx5f(x)=m(5)1=5m1\mathop {\lim }\limits_{x \to {5^ - }} f(x) = m(5) - 1 = 5m - 1 ……….(1)
Since the value of f(x) when x is more than 5 is 3x - 5
We get limx5+f(x)=3(5)5=155=10\mathop {\lim }\limits_{x \to {5^ + }} f(x) = 3(5) - 5 = 15 - 5 = 10 ……….(2)
Equating (1) and (2)
5m1=10 5m=10+1 m=115  \Rightarrow 5m - 1 = 10 \\\ \Rightarrow 5m = 10 + 1 \\\ \Rightarrow m = \dfrac{{11}}{5} \\\
The correct option is a.

Note: A function is said to be continuous if a small change in the input only causes a small change in the output.
A function is continuous when its graph is a single unbroken curve