Practicing Success
Match List-I with List-II
Choose the correct answer from the options given below |
(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) |
(a) - (iii), (b) - (iv), (c) - (i), (d) - (ii) |
x = t2, y = t3 a. $\frac{d x}{d t}=2 t ~~~~\frac{d y}{d t}=3 t^2$ so $\frac{d y}{d x}=\frac{3 t}{2} \Rightarrow \frac{d^2 y}{d x^2}=\frac{3}{2} \frac{d t}{d x} \Rightarrow \frac{d^2 y}{d x^2}=\frac{3}{4} t$ at t = 1 → $\left.\frac{d^2 y}{d x^2}\right]_{t=1}=\frac{3}{4}$ → (iii) b. $f(x)=\sqrt{x}+1$ so $f'(x)=\frac{1}{2} \sqrt{x}$ so $f''(x) = \frac{-1}{4x^{3/2}}$ so $f''(1) = \frac{-1}{4}$ → (iv) c. $f(x) = 9x^2 + 12x + 2$ $f'(x) = 18x + 12$ $\Rightarrow f'(x)=0 \Rightarrow 18 x+12=0$ $x=\frac{-2}{3}$ $f''(x) = 18$ so $f''(\frac{-2}{3}) > 0$ (point of minima) fmin = $f'(\frac{-2}{3}) = -2$ (minimum value) → (i) d. $f(x)= (x-2)^4(x+1)^3$ $f'(x) =4(x-2)^3(x+1)^3+3(x-2)^4(x+1)^2$ $=(x-2)^3(x+1)^2[4 x+4+3 x-6]$ $=(x-2)^3(x+1)^2(7 x-2)$, so $f'(x)=0$ $\Rightarrow x=2, -1, \frac{2}{7}$ $f''(x) = 6(x + 1) (x-2)^4 + 12(x-2)^3 (x+1)^2 +12(x+1)^2 (x-2)^3 + 12(x-2)^2(x+1)^3$ so $f''(-1) = 0$ → (ii) (point of intersection) Option: 3 |