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
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."