Skip to content
Question

Q.Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : The function f(x)=x2−x+1f(x) = x^2 - x + 1 is strictly increasing on (−1,1)(-1, 1). Reason (R) : If f(x)f(x) is continuous on [a,b][a, b] and derivable on (a,b)(a, b), then f(x)f(x) is strictly increasing on [a,b][a, b] if f′(x)>0f'(x) > 0 for all x∈(a,b)x \in (a, b).

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★est
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

f′(x)=2x−1f'(x)=2x-1 is negative on (−1,12)(-1,\tfrac12), so A is false; the monotonicity theorem in R is a true standard result.

ff is strictly increasing on an interval iff f′(x)>0f'(x)>0 throughout it.

  1. For f(x)=x2−x+1f(x)=x^2-x+1, f′(x)=2x−1f'(x)=2x-1.
  2. On (−1,1)(-1,1), take x=0x=0: f′(0)=−1<0f'(0)=-1<0, so ff is actually decreasing near x=0x=0. Hence ff is NOT strictly increasing on the whole interval (−1,1)(-1,1) — Assertion (A) is false. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.