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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

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Question : 64 of 160

Marks: +1, -0

If

f(x)=\begin{cases} \sin x, & \text{if } x \leq 0 \\ x^2+a^2, & \text{if } 0 < x < 1 \\ bx+2, & \text{if } 1 \leq x \leq 2 \\ 0, & \text{if } x > 2 \end{cases}

continuous on IR, then

a + b + ab =

-2
0
2
-1

Solution:

The given function is,

f(x)=\begin{cases} \sin x, & \text{if } x \leq 0 \\ x^{2}+a^{2}, & \text{if } 0 < x < 1 \\ b x+2, & \text{if } 1 \leq x \leq 2 \\ 0, & \text{if } x > 2 \end{cases}

The value of a is calculated as,

\lim\limits_{x \to 0^{-}} f(x) = \lim\limits_{x \to 0^{+}} f(x)

\lim\limits_{x \to 0^{-}} \sin x = \lim\limits_{x \to 0^{+}} x^{2} + a^{2}

0 = 0 + a^{2}

a = 0

The value of

b

is calculated as,

\lim\limits_{x \to 2^{-}} f(x) = \lim\limits_{x \to 2^{+}} f(x)

\lim\limits_{x \to 2^{-}} b x + 2 = \lim\limits_{x \to 2^{+}} 0

2b + 2 = 0

b = -1

The value of

a + b + ab

is calculated as,

a + b + ab

= 0 + (-1) + 0

= -1

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