Question
Let x = f(t) and y = g(t) be parametric forms with t as a parameter, then
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\times\frac{\text{dt}}{\text{dx}}=\frac{\text{g}'(\text{t})}{\text{f}'(\text{t})},$ where $\text{f}'(\text{t})\neq0.$
On the basis of above information, answer the following questions.
  1. The derivative of $\text{f}(\tan\text{x})\text{w.r.t.}\text{ g}(\sec\text{x})\text{ at}\text{ x}=\frac{\pi}{4},$ where f'(1) = 2 and $\text{g}'(\sqrt{2})=4,$ is:
  1. $\frac{1}{\sqrt{2}}$
  2. ${\sqrt{2}}$
  3. 1
  4. 0
  1. The derivative of $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$ is:
  1. -1
  2. 1
  3. 2
  4. 4
  1. The derivative of $\text{e}^{\text{x}^3}$ with respect to log x is:
  1. $\text{e}^{\text{x}^3}$
  2. $3\text{x}^22\text{e}^{\text{x}^3}$
  3. $3\text{x}^3\text{e}^{\text{x}^3}$
  4. $3\text{x}^2\text{e}^{\text{x}^3}+3\text{x}$
  1. The derivative of $\cos^{-1}(2\text{x}^2-1)\text{w.r.t.}\cos^{-1}\text{x}$ is:
  1. $2$
  2. $\frac{-1}{2\sqrt{1-\text{x}^2}}$
  3. $\frac{2}{\text{x}}$
  4. $1-\text{x}^2$
  1. If $\text{y}=\frac{1}{4}\mu^4$ and $\mu=\frac{2}{3}\text{x}^3+5,$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$
  2. $\frac{2}{7}\text{x}^2(2\text{x}^3+15)^3$
  3. $\frac{2}{27}\text{x}(2\text{x}^3+5)^3$
  4. $\frac{2}{7}(2\text{x}^3+15)^3$

Answer

  1. (a) $\frac{1}{\sqrt{2}}$
Solution:
Now, $\frac{\text{df}(\tan\text{x})}{\text{dg}(\sec\text{x})}=\frac{\text{f}'(\tan \text{x})\sec^2\text{x}}{\text{g}'(\sec\text{x})\sec\text{x}\tan \text{x}}$
$=\frac{\text{f}'(\tan \text{x})\sec\text{x}}{\text{g}'(\sec\text{x})\tan \text{x}}$
$\therefore\Big[\frac{\text{df}(\tan\text{x})}{\text{dg}(\sec\text{x})}\Big]_{\text{x}=\frac{\pi}{4}}=\frac{\text{f}'(1)\sqrt{2}}{\text{g}'(\sqrt{2})\cdot1}=\frac{2\sqrt{2}}{4\cdot1}=\frac{1}{\sqrt{2}}$
  1. (b) 1
  1. (c) $3\text{x}^3\text{e}^{\text{x}^3}$
Solution:
Let $\text{y}=\text{e}^{\text{x}^3},\text{z}=\log\text{x}$
Differentiating w.r.t. x, we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^3}(3\text{x}^2)=3\text{x}^2\text{e}^{\text{x}^3}$ and $\therefore\frac{\text{dy}}{\text{dz}}=\frac{\frac{\text{dy}}{\text{dx}}}{\frac{\text{dz}}{\text{dx}}}=\frac{3\text{x}^2\text{e}^{\text{x}^3}}{\Big(\frac{1}{\text{x}}\Big)}=3\text{x}^3\text{e}^{\text{x}^3}$
  1. (a) $2$
Solution:
Let $\text{y}=\cos^{-1}(2\text{x}^2-1)=2\cos^{-1}\text{x}$
Differentiating w.r.t. $\cos^{-1}\text{x},$ we get
$\frac{\text{dy}}{\text{d}(\cos^{-1}\text{x})}=\frac{2\text{d}(\cos^{-1}\text{x})}{\text{d}(\cos^{-1}\text{x})}=2$
  1. (a) $\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$
Solution:
We have, $\text{y}=\frac{1}{4}\text{u}^4\Rightarrow\frac{\text{dy}}{\text{du}}=\frac{1}{4}\cdot4\text{u}^3=\text{u}^3$
and $\text{u}=\frac{2}{3}\text{x}^3+5\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{3}\cdot3\text{x}^2=2\text{x}^2$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{du}}\cdot\frac{\text{du}}{\text{dx}}=\text{u}^3\cdot2\text{x}^2=\Big(\frac{2}{3}\text{x}^3+5\Big)^3(2\text{x})^2$
$=\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$

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Based on the above information, answer the following questions.
  1. Distance between House (A) and ATM (B) is:
  1. $3\text{ units}$
  2. $3\sqrt{2}\text{ units}$
  3. $\sqrt{2}\text{ units}$
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Based on the above information, answer the following questions.
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  2. $5:30$ p.m.
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Based on the above information, answer the following questions.
  1. By Newton's law of cooling, $\frac{\text{dT}}{\text{dt}}$ is proportional to:
  1. $T - 60$
  2. $T - 50$
  3. $T - 70$
  4. $T - 98.6$
  1. When $t = 0$, then body temperature is equal to:
  1. $50^\circ F$
  2. $60^\circ F$
  3. $70^\circ F$
  4. $98.6^\circ F$
  1. When $t = 2$, then body temperature is equal to:
  1. $50^\circ F$
  2. $60^\circ F$
  3. $70^\circ F$
  4. $98.6^\circ F$
  1. The value of T at any time t is:
  1. $50+20\Big(\frac{1}{2}\Big)^\text{t}$
  2. $50+20\Big(\frac{1}{2}\Big)^\text{t-1}$
  3. $50+20\Big(\frac{1}{2}\Big)^\frac{\text{t}}{2}$
  4. None of these
  1. If it is given that $\log_\text{e} (2.43) = 0.88789$ and $\log_\text{e} (0.5) = -0.69315,$ then the time at which the murder occur is:
  1. $7:30$ p.m.
  2. $5:30$ p.m.
  3. $6:00$ p.m.
  4. $5:00$ p.m.
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Based on the above information, answer the following questions.
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  3. $\begin{bmatrix}3\\-3\end{bmatrix}$
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  1. The value of $|U|$ is:
  1. $2$
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Based on the above information, answer the following questions
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    2. $\text{y}=\frac{3}{2}(\text{x}-1)$
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    4. $\text{y}=\frac{1}{2}(\text{x}-1)$
  2. Equation of line BC is.
    1. $\text{y}=\frac{1}{2}\text{x}-\frac{7}{2}$
    2. $\text{y}=\frac{3}{2}\text{x}-\frac{7}{2}$
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  3. Area of region ABCD is.
  1. 2 sq. units
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(i) If $\mathrm{r} \mathrm{cm}$ be the radius and $\mathrm{h} \mathrm{cm}$ be the height of the cylindrical tin can, then express the surface area as a function of radius (r)

(ii) Find the radius of the can that will minimize the cost of tin used for making can?

(iii) Find the height that will minimize the cost of tin used for making can ?

OR

Find the minimum cost of material used to manufacture the tin can.