Match List-I with List-II List - I List - II (a) If x=t^2 and y=t^3 t…

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)

f_min = $f'(\frac{-2}{3}) = -2$       (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