(a) x=t2, y=t3: dxdy=23t, dx2d2y=4t3. At t=1: 43 -- (iii).
(b) f′(x)=2x1, f′′(x)=−4x3/21. f′′(1)=−41 -- (iv).
(c) Min of 9x2+12x+2 at x=−32: value =4−8+2=−2 -- (i).
(d) Point of inflexion of (x−2)4(x+1)3 is x=−1 -- (ii).
CUET UG 2022 — Mathematics Calculus
Match List-I with List-II
| List-I | List-II |
|---|---|
| (a) If x=t2 and y=t3, then dx2d2y at t=1 | (i) −2 |
| (b) If f(x)=x+1, then f′′(1) | (ii) −1 |
| (c) The minimum value of f(x)=9x2+12x+2 is | (iii) 43 |
| (d) The point of inflexion of the function f(x)=(x−2)4(x+1)3 is | (iv) −41 |
Choose the correct answer from the options given below
Held on 30 Aug 2022 · Verified 13 Jul 2026.
(a) - (i), (b) - (iii), (c) - (ii), (d) - (iv)
(a) - (ii), (b) - (iii), (c) - (i), (d) - (iv)
(a) - (iii), (b) - (iv), (c) - (i), (d) - (ii)
(a) - (iv), (b) - (i), (c) - (iii), (d) - (ii)
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