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BITSAT 2012 Paper

Section: Mathematics

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Question : 105 of 150

Marks: +1, -0

If the function f(x) =

\begin{cases} 1, & x \le 2 \\ ax+b, & 2 < x < 4 \\ 7, & x \ge 4 \end{cases}

is continuous at x = 2 and 4, then the values of a and b are

3 , 5
3 , - 5
0 , 3
0 , 5

Solution:

Since f (x) is continuous at x = 2

∴ f (2) =

\lim\limits_{x\to 2^{+}}

f (x) ⇒ 1 =

\lim\limits_{x\to 2^{+}}

(ax + b)

∴ 1 = 2a + b ... (1)

Again f(x) is continuous at x = 4,

∴ f (4) =

\lim\limits_{x\to 4^{-}}

f (x) = 7 =

\lim\limits_{x\to 4^{-}}

(ax + b)

∴ 7 = 4a + b ... (2)

Solving (1) and (2), we get a = 3, b = – 5

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