सही विकल्प है (B) 1
Given \(f(x) = \begin{cases} |x| + 3, &\text{if } x \le -3\\-2x, & \text{if }-3< x<3\\6x +2, & \text{if } x \ge 3\end{cases}\)
When x = -3
If f(x) is continuous for x = -3 then
\(\lim\limits_{x\to 3^-}f(x) = \lim\limits_{x \to -3^+} f(x) = f(-3)\)
Finding L.H.L.
\(\lim\limits_{x\to 3^-} |x| + 3\)
\(= \lim\limits_{h\to 0} |-3-h| +3\)
Putting h = 0 then we get,
= |-3 - 0| + 3
= |-3| + 3
= 6
Finding R.H.L.
\(\lim\limits_{x \to -3^+} -2x = \lim\limits_{h \to 0} - 2(-3 +h)\)
\(= \lim\limits_{h\to 0} 6 -2h\)
Putting h = 0 then we get,
= 6 - 2 \(\times\) 0
= 6
Find f(x) at x = -3
f(-3) = |-3| + 3 = 6
Hence, \(\lim\limits_{x\to 3^-}f(x) = \lim\limits_{x \to -3^+} f(x) = f(-3)\)
Therefore, the function f(x) is continuous at x = -3.
When x = 3
If f(x) is continuous for x = 3 then
\(\lim\limits_{x\to 3^-}f(x) = \lim\limits_{x \to -3^+} f(x) = f(3)\)
Finding L.H.L.
\(\lim\limits_{x\to 3^-}-2x = \lim\limits_{h \to 0} - 2(-3 -h)\)
\(= \lim\limits_{h\to 0} -6 +2h\)
Putting h=0 then we get,
= -6 + 2 \(\times\) 0
= -6
Finding R.H.L.
\(\lim\limits_{x \to -3^+} 6x+ 2 = \lim\limits_{h \to 0} 6(3 + h ) +2\)
\(= \lim\limits_{h\to 0} 18 + 6h + 2\)
Putting h = 0 then we get,
= 18 + 6 \(\times\) 0 + 2
= 20
Find f(x) at x = 3
f(3) = 6x + 2 at x = 3
f(3) = 6 \(\times\) 3 + 2 = 20
Hence, \(\lim\limits_{x\to 3^-}f(x) \ne \lim\limits_{x \to -3^+} f(x) = f(-3)\)
Therefore, the function
f(x) is discontinuous at x = 3
When x < -3,
For x < -3, f(x) = |x| + 3
Since the function f(x) = |x| + 3 is a modulus function so it is continuous.
∴ f(x) is continuous for x < -3
When x > 3,
For x > 3, f(x) = 6x + 2
Since the function f(x) = 6x + 2 is a polynomial so it is continuous.
∴ f(x) is continuous for x > 3
When -3 < x < 3
For -3 < x < 3, f(x) = -2x
Since the function f(x) = -2x is a polynomial so it is continuous.
∴ f(x) is continuous for -3 < x < 3
∴ \(f(x) = \begin{cases} |x| + 3, &\text{if } x \le -3\\-2x, & \text{if }-3< x<3\\6x +2, & \text{if } x \ge 3\end{cases}\) discontinuous at x = 3 only.
