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Discuss the continuity of the function f, where f is defined by

f(x) = ${- 2; if x \leq - 1 2 x; if - 1 < x \leq 1 2; if x > 1$

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An Intiative by Sri Chaitanya

The given function is continuous at x = -1 and x = 1.

The given function is continuous at x = -1.

The given function is continuous at x = 1.

The given function is not continuous at x = -1 and x = 1.

answer is A.

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Detailed Solution

For a function to be continuous, the functions should have the same value at the left-hand limit, right-hand limit as well as on the limits.
Here in the question, we need to determine the continuity at the point x = 1 and x = -1 as the function given is associated with the same.
Now, the value of the function at x = -1 is defined as f(x) = −2 so, the value of the function at x = -1 is given by
f (−1) = −2
The value of the left-hand limit of the function at x = -1 is defined as f(

x^{-}

) = −2 so, the value of the function at x = −1 – h where h is infinitesimally small and is tending to zero is given by
f(

x^{-}

) =

\lim_{x \to {- 1}^{-}} (- 2)

= (−2)
= −2 −−−−−−−−(i)
The value of the right-hand limit of the function at x = -1 is defined as f(

x^{+}

) = 2x so, the value of the function at x = −1 + h where h is infinitesimally small and is tending to zero is given by
f(

x^{+}

) =

\lim_{x \to {- 1}^{+}} 2 x

\lim_{h \to 0} (- 1 + h)

\lim_{h \to 0} (- 2 + 2 h)

= −2 −−−−−−−−(ii)
From the equations (i) and (ii) we can see that the value of the functions at the left-hand limit is equal to the value of the function at the right-hand limits, so the given function is not continuous at x = -1.
Now, the value of the function at x=1 is defined as f(x) = 2x so, the value of the function at x = 1 is given by
f(1) = 2x
= 2
The value of the left-hand limit of the function at x = 1 is defined as f(

x^{-}

) = 2x so, the value of the function at x = 1 – h where h is infinitesimally small and is tending to zero is given by f(

x^{-}

) =

\lim_{x \to 1^{-}} 2 x

\lim_{h \to 0} (2 (1 - h)

\lim_{h \to 0} (2 - 2 h)

= 2 −−−−−−−−(iii)
The value of the right-hand limit of the function at x = 1 is defined as f(

x^{+}

) = 2 so, the value of the function at x = 1 + h where h is infinitesimally small and is tending to zero is given by f(

x^{+}

) =

\lim_{x \to 1^{+}} (2)

= 2 −−−−−−−−(iv)
From the equations (iii) and (iv) we can see that the value of the functions at the left-hand limit is equal to the value of the function at the right-hand limits, so the given function is not continuous at x = 1.
Hence, the given function is continuous at x = -1 and x = 1.
So, option (1) is correct for this question.

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