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Find the value of for which \(f(x) = \begin{cases} kx + 5 ,& When \,x\leq2.\\ x-1 ,& When\, x >2 \end{cases}\) is continuous at x = 2. |
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Answer» \(f(x) = \begin{cases} kx + 5 ,& When \,x\leq2.\\ x-1 ,& When\, x >2 \end{cases}\) Left hand limit of function f(x) at = 2 is f(2−) = \(\lim\limits_{x \to2^-}\)f(x) = \(\lim\limits_{x \to2}\)(kx + 5) = 2k + 5. Right hand limit of function f(x) at = 2 is f(2+) = \(\lim\limits_{x \to2^+}f(x)\) = \(\lim\limits_{x \to2}\)(x-1) = 2 - 1 =1. Since, we know that if function f(x) is continuous at = 2, then left hand limit of function f(x) at x = 2 is equals to the right hand limit of function f(x) at = 2 which is also equals to f(2). i.e., f(2 −) = f(2 +) = (2). Therefore, 2k + 5 = 1 ⇒ 2k = −4 ⇒ k = −2. Hence, k = -2 for which the function \(f(x) = \begin{cases} kx + 5 ,& When \,x\leq2.\\ x-1 ,& When\, x >2 \end{cases}\) is continuous at x = 2. |
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