\(f( x ) = \ * 20 c cos [ x ], & x ≥ 0 \\ | x | + a, &..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep."> \(f( x )... - Solved | MCA Prep" /> \(f( x ) = \ * 20 c cos [ x ], & x ≥ 0 \\ | x | + a, &..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." /> \(f( x )... - Solved | MCA Prep" /> \(f( x ) = \ * 20 c cos [ x ], & x ≥ 0 \\ | x | + a, &..." - Step-by-step solution, answer key, and detailed explanation on MCA Prep." />

Question 25

Mathematics Limit of Functions Hard

Let&nbsp;<span class="math-tex">\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\cos \left[ x \right],}&amp;{x \ge 0}\\ {\left| x \right| + a,}&amp;{x &lt; 0} \end{array}} \right.,\)</span>&nbsp;where [x] denotes the greatest integer&nbsp;&le; x. If f should be continuous at x = 0, then a must be

(A) 0
(B) 1
(C) 2
(D) -1
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Right-hand value: \(f(0) = \cos[0] = \cos 0 = 1\). For small \(x > 0\), \([x] = 0\), so the right limit is also 1.

Left-hand limit: \(\lim_{x \to 0^-}(|x| + a) = a\).

For continuity, \(a = 1\).

Answer: (B) 1

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