\(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." />
Let <span class="math-tex">\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\cos \left[ x \right],}&{x \ge 0}\\ {\left| x \right| + a,}&{x < 0} \end{array}} \right.,\)</span> where [x] denotes the greatest integer ≤ x. If f should be continuous at x = 0, then a must be
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
Reinforce your conceptual grasp with related questions from MAH MCA CET 2019:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →