A function f(x) is defined by f(x)=x+1,when x less than equal to 1 and f(x)=3-ax^2, when x>1.If f(x) is continuous at point x=1. Then what will be the value of a?

1 Answer
martin23239
Feb 21, 2018

# "The answer is:"\qquad \qquad \qquad \qquad \qquad a \ = \ 1.#

Explanation:

# "Let's look at the value of" \ \ f(x) \ \"at" \ \ x = 1, "and the left- and" #
# "right-sided limits as" \quad x rarr 1. #

# \qquad \qquad \qquad \qquad \qquad f(x) \ "will be continuous at" \ x = 1 \quad hArr \quad #

# f(1), \quad \lim_{x rarr 1^-} f(x), \quad \lim_{x rarr 1^+} f(x) \qquad "all exist, and are all equal" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad"to each other." #

# "By the construction of the function, we have:" #

#"a)" \qquad \qquad f(1)\ = \ 1 + 1 \ = \ 2. #

# :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad f(1) \ = \ 2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (1) #

# "b)" \qquad \qquad \lim_{x rarr 1^-} f(x) \ = \ \lim_{x rarr 1^-} x + 1 \ = \ 1 + 1 \ = \ \ 2. #

# :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \lim_{x rarr 1^-} f(x) \ = \ 2. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2) #

# "c)" \qquad \qquad \lim_{x rarr 1^+} f(x) \ = \ \lim_{x rarr 1^+} 3 - a x^2 \ = \ 3 - ( a cdot 1^2 ) \ = \ 3 - a. #

# :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \lim_{x rarr 1^+} f(x) \ = \ 3 - a. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ (3) #

# "Thus:" \qquad \qquad \qquad f(1), \quad \lim_{x rarr 1^-} f(x), \quad \lim_{x rarr 1^+} f(x) \qquad \qquad "all exist." #

# "So:" \qquad \quad f(x) \ \ "will now be continuous at" \ \x = 1, "precisely when" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad "all these 3 quantities are equal:" #

# \qquad \qquad \qquad \qquad \quad f(1) \ = \ \lim_{x rarr 1^-} f(x) \ = \ \lim_{x rarr 1^+} f(x). #

# "Using our results from eqns. (1), (2), (3) above, this becomes:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad 2 \ = \ 2 \ = \ 3 - a. #

# "This is the same as just:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad 2 \ = \ 3 - a. #

# "Solving:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3 - a \ = \ 2 . #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 3 \ = \ a + 2. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 1\ = \ a #

# :. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ a \ = \ 1. #

#"This is our answer."#

Related questions
Impact of this question
1725 views around the world
You can reuse this answer
Creative Commons License