CBSE Class 12 Sample Paper for 2025 Boards
Question 2
Question 3
Question 4
Question 5 Important
Question 6
Question 7
Question 8 Important
Question 9 Important
Question 10 Important
Question 11
Question 12 Important
Question 13 Important
Question 14
Question 15 Important
Question 16 Important
Question 17
Question 18
Question 19 [Assertion Reasoning] Important
Question 20 [Assertion Reasoning] Important
Question 21 Important
Question 22
Question 23 (A)
Question 23 (B)
Question 24 (A)
Question 24 (B) Important
Question 25 Important
Question 26 Important
Question 27 Important
Question 28 (A)
Question 28 (B)
Question 29 (A) Important
Question 29 (B)
Question 30 Important
Question 31 (A) Important
Question 31 (B)
Question 32 Important
Question 33 Important
Question 34 (A)
Question 34 (B)
Question 35 (A)
Question 35 (B) Important
Question 36 (i) [Case Based]
Question 36 (ii)
Question 36 (iii) (A) Important You are here
Question 36 (iii) (B) Important
Question 37 (i) [Case Based]
Question 37 (ii) Important
Question 37 (iii) (A) Important
Question 37 (iii) (B)
Question 38 (i) [Case Based] Important
Question 38 (ii) Important
CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Oct. 3, 2024 by Teachoo
Question 36 (iii) (A) For what value of 𝒙, the volume of each container is maximum?Now, V = 2(2𝑥^3−65𝑥^2+500𝑥) And, 𝐝𝐕/𝒅𝒙= 4(𝑥−5)(3𝑥−50) Putting 𝐝𝐕/𝒅𝒙= 0 4(𝑥−5)(3𝑥−50)=0 So, x = 5 and x = 𝟓𝟎/𝟑 If 𝒙 = 𝟓𝟎/𝟑 Breadth of box = 25 – 2𝑥 = 25 – 2(𝟓𝟎/𝟑) = 25 – 33.3 = –8.3 Since, breadth cannot be negative, ∴ x = 𝟓𝟎/𝟑 is not possible Hence, 𝒙 = 5 only Finding V’’(𝒙) V’(𝑥)=" 4" [𝟑𝒙^𝟐−𝟔𝟓𝒙+𝟐𝟓𝟎] V’’(𝑥)=4[6𝑥−65] Putting 𝒙=𝟓 V’’(𝟓)=4(6(5)−65)= 4(30−65)= 4(−35)= –140 V’’(𝒙)<𝟎 when 𝑥=5 Thus, V(𝑥) is maximum at 𝑥=5 ∴ Square of side 5 cm is cut off from each Corner