Question 79Asked in BPSC TRE 1.0 (2023) Mathematics Class 11-12

If the function f(x) = √(1 − √(1 − x²)), then f(x) isयदि फलन f(x) = √(1 − √(1 − x²)), तो f(x) है

यदि फलन f(x) = √(1 − √(1 − x²)), तो f(x) है
  1. Acontinuous in [−1, 1] and differentiable in (−1, 1)[−1, 1] में संतत तथा (−1, 1) में अवकलनीय
  2. Bcontinuous in [−1, 1] and differentiable in (−1, 0) ∪ (0, 1)[−1, 1] में संतत तथा (−1, 0) ∪ (0, 1) में अवकलनीय
  3. Ccontinuous and differentiable in [−1, 1][−1, 1] में संतत एवं अवकलनीय
  4. DMore than one of the aboveउपर्युक्त में से एक से अधिक
  5. ENone of the aboveउपर्युक्त में से कोई नहीं
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Correct answer: (B) continuous in [−1, 1] and differentiable in (−1, 0) ∪ (0, 1)[−1, 1] में संतत तथा (−1, 0) ∪ (0, 1) में अवकलनीय

Explanation

f is defined and continuous on [−1, 1]. Near x = 0, 1 − √(1 − x²) ≈ x²/2, so f(x) ≈ |x|/√2, which has a corner at 0. Away from 0 inside (−1, 1) it is differentiable, so f is differentiable on (−1, 0) ∪ (0, 1).f, [−1, 1] पर परिभाषित और संतत है। x = 0 के निकट 1 − √(1 − x²) ≈ x²/2, अतः f(x) ≈ |x|/√2, जिसका 0 पर कोना है। (−1, 1) में 0 को छोड़कर यह अवकलनीय है, इसलिए f, (−1, 0) ∪ (0, 1) में अवकलनीय है।

From BPSC TRE 1.0 (2023) Mathematics Class 11-12 — Previous Year Paper

This is question 79 of BPSC TRE 1.0 (2023) Mathematics Class 11-12 — Previous Year Paper. See all questions of this paper with answers, or attempt the paper as a timed online test to get your score and all-India rank.