3 Marks

Question 13 Marks

Show that $\text{f(x)}=2\text{x}+\cot^{-1}\text{x}+\log\Big(\sqrt{1+\text{x}^1}-\text{x}\Big)$ is increasing in R.

Answer

We have, $\text{f(x)}=2\text{x}+\cot^{-1}\text{x}+\log\Big(\sqrt{1+\text{x}^1}-\text{x}\Big)$
$\therefore\ \text{f}'(\text{x})=2-\frac{1}{1+\text{x}^2}+\frac{1}{\sqrt{1+\text{x}^2}-1}\Big(\frac{\text{x}}{\sqrt{1+\text{x}^2}}-1\Big)$
$=\frac{1}{1+\text{x}^2}+\frac{1}{\sqrt{1+\text{x}^2}-\text{x}}\cdot\frac{\text{x}-\sqrt{1+\text{x}^2}}{\sqrt{1+\text{x}^2}}=2-\frac{1}{1+\text{x}^2}-\frac{1}{\sqrt{1+\text{x}^2}}$
Now $1+\text{x}^2,\sqrt{1+\text{x}^2}\geq1$ for all real x.
$\therefore2-\frac{1}{1+\text{x}^2}-\frac{1}{\sqrt{1+\text{x}^2}}>0$ for all real x,
$\therefore\ \text{f(x)}>0$ for all real x.
Thus f(x) is increasing on R.

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Question 23 Marks

Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3cm and 3.0005cm, respectively.

Answer

Let internal radius = r and external radius = R
$\therefore$ Volume of hollow spherical shell,
$\text{V}=\frac{4}{3}\pi(\text{R}^{3}-\text{r}^3)$
$\text{V}=\frac{4}{3}\pi\big[(3.0005)^3-(3)^3\big]\ \ \dots(\text{i})$
Now, approximate value of (3.0005)³ can be obtained using differentiation
Let $(3.0005)^3=\text{y}+\triangle\text{y}$
And $\text{x}=3,\triangle\text{x}=0.0005$
Also, let y = x³
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=3\text{x}^2$
$\therefore\ \triangle\text{y}=\frac{\text{dy}}{\text{dx}}\times\triangle\text{x}=3\text{x}^3\times0.0005=3\times3^2\times0.0005$
$=27\times0.0005=0.0135$
$\therefore\ (3.0005)^3=\text{y}+\triangle\text{y}=3^3+0.0135=27.0135$
$\therefore\ \text{V}=\frac{4}{3}\pi[27.0135-27000]$
$=\frac{4}{3}\pi[0.0135]=4\pi\times(0.0045)=0.0180\pi\text{cm}^3$

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Question 33 Marks

Prove that $\text{f(x)}=\sin\text{x}+\sqrt{3}\cos\text{x}$ has maximum value at $\text{x}=\frac{\pi}{6}.$

Answer

We have, $\text{f(x)}=\sin\text{x}+\sqrt{3}\cos\text{x}$
$\therefore\ \text{f}'(\text{x})=\cos\text{x}-\sqrt{3}\sin\text{x}$
For $\text{f}'(\text{x)}=0,\cos\text{x}=\sqrt{3}\sin\text{x}$
$\Rightarrow\ \tan\text{x}=\frac{1}{\sqrt{3}}$
$\Rightarrow\ \text{x}=\frac{\pi}{6}$
Differentiating f'(x), we get
$\text{f}''(\text{x})=-\sin\text{x}-\sqrt{3}\cos\text{x}$
$\text{f}''\Big(\frac{\pi}{6}\Big)=-\sin\frac{\pi}{6}-\sqrt{3}\cos\frac{\pi}{6}<0$
$=-\frac{1}{2}-\sqrt{3}.\frac{\sqrt{3}}{2}$
$=-\frac{1}{2}-\frac{3}{2}=-2<0$
Hence, at $\text{x}=\frac{\pi}{6},\text{f(x)}$ has maxima value and $\frac{\pi}{6}$ is the point of local maxima.

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Question 43 Marks

At what point, the slope of the curve y = -x³ + 3x² + 9x - 27 is maximum? Also find the maximum slope.

Answer

We have, y = -x³ + 3x² + 9x - 27
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-3\text{x}^2+6\text{x}+9=$ Slope of the curve
$\Rightarrow\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=-6\text{x}+6=-6(\text{x}-1)$
Find the critical points by equating $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ to 0.
$\therefore\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=0$
$\Rightarrow-6(\text{x}-1)=0$
$\Rightarrow\text{x}=1$
Now, $\frac{\text{d}^3\text{y}}{\text{dx}^3}=-6<0$
So, the maximum slope of given curve is at x = 1
$\therefore\ \Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}=1)}=-3.1^2+6.1+9=12$
Which is maximum slop.
Also, for x = 1, y = -1³ + 3.1² + 9.1 - 27
= -1 + 3 + 39 - 27
= -16
So, the required point is (1, - 16).

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