Answer the following by appropriately matching the lists based on the information in Column I and Column IIColumn IColumn IIa.y=f(x) is given by x=t5−5t3−20t+7 and y=4t3−3t2−18t+3. Then −5×dydx at t=1p. 0b. Let P(x) be a polynomial of degree 4, with P(2)=−1,P′(2)=0,P′′(2)=2,P′′′(2)=−12 and Piv(2)=24, then P′′(3) is q. −2c.y=1x, then dy√1+y4dx√1+x4r. 2d.f(2x+3y5)=2f(x)+3f(y)5 and f′(0)=p and f(0)=q. Then ,f′′(0) is s. −1

Answer the following by appropriately matching the lists based on the

1.	Answer the following by appropriately matching the lists based on the information in Column I and Column IIColumn IColumn IIa.y=f(x) is given by x=t5−5t3−20t+7 and y=4t3−3t2−18t+3. Then −5×dydx at t=1p. 0b. Let P(x) be a polynomial of degree 4, with P(2)=−1,P′(2)=0,P′′(2)=2,P′′′(2)=−12 and Piv(2)=24, then P′′(3) is q. −2c.y=1x, then dy√1+y4dx√1+x4r. 2d.f(2x+3y5)=2f(x)+3f(y)5 and f′(0)=p and f(0)=q. Then ,f′′(0) is s. −1
Answer» Answer the following by appropriately matching the lists based on the information in Column I and Column II $\begin{matrix} C o l u m n I & C o l u m n I I a . y = f (x) is given by x = t^{5} - 5 t^{3} - 20 t + 7 and y = 4 t^{3} - 3 t^{2} - 18 t + 3. Then - 5 \times \frac{d y}{d x} at t = 1 & p. 0 b . Let P (x) be a polynomial of degree 4, with P (2) = - 1, P^{'} (2) = 0, P^{''} (2) = 2, P^{'''} (2) = - 12 and P^{i v} (2) = 24, then P^{''} (3) is & q. - 2 c . y = \frac{1}{x}, then \frac{\frac{d y}{\sqrt{1 + y^{4}}}}{\frac{d x}{\sqrt{1 + x^{4}}}} & r . 2 d . f (\frac{2 x + 3 y}{5}) = \frac{2 f (x) + 3 f (y)}{5} and f^{'} (0) = p and f (0) = q . Then, f^{''} (0) is & s. - 1 \end{matrix}$