Let the function
\(f(x)\)=\(\begin{cases} \frac{\log_e(1+5x) - \log_e(1+\alpha x)}{x}, & \text{if } x ∈ 0 \\ \space 10 & \text;{if } x = 0 \\ \end{cases}\)
be continuous at x = 0. Then α is equal to
\(f(x)\)=\(\begin{cases} \frac{\log_e(1+5x) - \log_e(1+\alpha x)}{x}, & \text{if } x ∈ 0 \\ \space 10 & \text;{if } x = 0 \\ \end{cases}\)
be continuous at x = 0. Then α is equal to
- 10
- -10
- 5
- -5
The Correct Option is D
Solution and Explanation
\(It_{{x \to 0}}\)\(\frac{(1+5x)-In(1+αx)}{x}\)
⇒5–α=10
⇒ α=−5
So, the correct option is (D): -5
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Concepts Used:
Types of Functions
Types of Functions
One to One Function
A function is said to be one to one function when f: A → B is One to One if for each element of A there is a distinct element of B.
Many to One Function
A function which maps two or more elements of A to the same element of set B is said to be many to one function. Two or more elements of A have the same image in B.
Onto Function
If there exists a function for which every element of set B there is (are) pre-image(s) in set A, it is Onto Function.
One – One and Onto Function
A function, f is One – One and Onto or Bijective if the function f is both One to One and Onto function.
Read More: Types of Functions