Continuity and Differentiability
Madhya Pradesh Board · Class 12 · Mathematics
Flashcards for Continuity and Differentiability — Madhya Pradesh Board Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.
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See them allजांचें कि क्या f(x) = 2x + 3, x = 1 पर सतत है।
Answer
चरण 1: f(1) = 2(1) + 3 = 5 ज्ञात कीजिए। चरण 2: x -> 1 पर सीमा ज्ञात कीजिए: lim x->1 (2x + 3) = 2(1) + 3 = 5. चरण 3: सीमा और मान की तुलना कीजिए। चूँकि lim x->1 f(x) = f(1) = 5 है, इसलिए फलन x = 1 पर सत…
Check whether f(x) = 2x + 3 is continuous at x = 1.
Answer
Step 1: Find f(1) = 2(1) + 3 = 5. Step 2: Find the limit as x -> 1: lim x->1 (2x + 3) = 2(1) + 3 = 5. Step 3: Compare limit and value. Since lim x->1 f(x) = f(1) = 5, the function is continuous at x =…
Check whether f(x) = x^2 is continuous at x = 0.
Answer
Step 1: Compute f(0) = 0^2 = 0. Step 2: Find lim x->0 x^2 = 0. Step 3: Since lim x->0 f(x) = f(0) = 0, the function is continuous at x = 0.
जांचें कि क्या f(x) = x^2, x = 0 पर सतत है।
Answer
चरण 1: f(0) = 0^2 = 0 निकालिए। चरण 2: lim x->0 x^2 = 0 ज्ञात कीजिए। चरण 3: चूँकि lim x->0 f(x) = f(0) = 0 है, इसलिए फलन x = 0 पर सतत है।…
f(x) = x^3 + 3 for x ≠ 0 और f(0) = 1 के लिए असातत्य का बिंदु ज्ञात कीजिए।
Answer
चरण 1: x = 0 पर, f(0) = 1 है। चरण 2: x ≠ 0 के लिए, x^3 + 3 वाले व्यंजक का उपयोग कीजिए। चरण 3: lim x->0 (x^3 + 3) = 3. चरण 4: सीमा और मान की तुलना कीजिए: 3 ≠ 1. अतः, f, x = 0 पर असतत है। यह असातत्य का …
Find the point of discontinuity of f(x) = x^3 + 3 for x ≠ 0 and f(0) = 1.
Answer
Step 1: At x = 0, f(0) = 1. Step 2: For x ≠ 0, use the expression x^3 + 3. Step 3: lim x->0 (x^3 + 3) = 3. Step 4: Compare limit and value: 3 ≠ 1. Therefore, f is discontinuous at x = 0. This is the o…
Is f(x) = |x| continuous at x = 0? Check using limits.
Answer
Step 1: f(0) = 0. Step 2: Left-hand limit at 0: as x -> 0-, |x| = -x, so the limit is 0. Step 3: Right-hand limit at 0: as x -> 0+, |x| = x, so the limit is 0. Step 4: Since left limit = right limit =…
क्या f(x) = |x|, x = 0 पर सतत है? सीमाओं का उपयोग करके जाँचें।
Answer
चरण 1: f(0) = 0. चरण 2: 0 पर बाईं ओर की सीमा: जब x -> 0-, |x| = -x, इसलिए सीमा 0 है। चरण 3: 0 पर दाईं ओर की सीमा: जब x -> 0+, |x| = x, इसलिए सीमा 0 है। चरण 4: चूँकि बाईं सीमा = दाईं सीमा = f(0) = 0 है…
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