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Friday, 9 October

JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 3 April 2025, Shift 1, Maths Q17

Question 17 of 75 in this shift, Maths question 17 of 25, Section A.

Let f(x)={(1+ax)1/x,x<01+b,x=0(x+4)1/2−2(x+c)1/3−2,x>0f(x) = \begin{cases} (1+ax)^{1/x}, & x < 0 \\ 1+b, & x = 0 \\ \dfrac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2}, & x > 0 \end{cases} be continuous at x=0x = 0. Then eabce^a bc is equal to:
  1. (1)48Official answer
  2. (2)64
  3. (3)72
  4. (4)36

Official answer

Option 1

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 8 Apr 2026, Shift 2 · Q19Let f(x)={13,x≤π/2b(1−sin⁡x)(π−2x)2,x>π/2f(x) = \begin{cases} \frac{1}{3}, & x \le \pi/2 \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2}, & x > \pi/2 \end{cases}. If ff is continuous…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.