Maths M.C.Q (1 Marks)

1. $\frac{4}{3}$

Solution:

Maximum value of  $\frac{1}{4\text{x}^{2}+2\text{x}+1}$ = Minimum value of $4\text{x}^{2}+2\text{x}+1$

Now, $\text{f}(\text{x})=\text4\text{x}^{2}+2\text{x}+1$

lmplies that $\text{f}'(\text{x})=8\text{x}+2$

For a local maxima or a local minima, We must have f'(x) = 0

lmplies that $8\text{x}+2 =0$

lmplies that $8\text{x}=-2$

lmplies that $\text{x}=-14$

Now, $\text{f}''(\text{x})=8$

lmplies that $\text{f}''(\text{1})=8 >0$

Therefore, $\text{x}=\frac{-1}{4}$ is a local minima.

Thus, $\frac{1}{4\text{x}^{2}+2\text{x}+1}$ is maximum at $\text{x}=\frac{-1}{4}$

lmplies that maximum value of $\frac{1}{4\text{x}^{2}+2\text{x}+1}=\frac{1}{4\Big(\frac{-1}{4}\Big)^{2}+2\Big(\frac{-1}{4}\Big)+1}$

$=\frac{16}{12}=\frac{4}{3}$

4. $\frac{2}{3}$

Solution:

Let h, r, V and R be the height, radius of the base, volume of the cone and the radius of the sphere, respectively.

Given, $\text{h}=\text{R}+\sqrt{\text{R}^{2}-\text{r}^{2}}$

lmplies that $\text{h}-\text{R}=\sqrt{\text{R}^{2}-\text{r}^{2}}$

Squaring both side, we get 

h² + R² - 2hR = R² - r²

lmplies that r² = 2hr - h² 

Now, Volume $\frac{1}{3}\pi\text{r}^{2}\text{h}$

lmplies that $\text{V}=\frac{\pi}{3}(2\text{h}^{2}\text{R}-\text{h}^{3})$

lmplies that $\frac{\text{dV}}{\text{dh}}=\frac{\pi}{3}(4\text{h}\text{R}-3\text{h}^{2})$

For maximum or minimum value os V, we must have $\frac{\text{dV}}{\text{dh}}=0$ 

2. $\frac{1}{2}$

Solution:

Let the two non-zero number be x and y. Then,

x + y = 8

⇒ y = 8 - x ...(i)

Now, $\text{f}(\text{x})=\frac{1}{\text{x}}+\frac{1}{\text{y}}$

$\Rightarrow\text{f}(\text{x})=\frac{1}{\text{x}}+\frac{1}{8-\text{x}}$ [from eq.(i)]

$\Rightarrow\text{f}'(\text{x})=\frac{1}{\text{x}}+\frac{1}{8-\text{x}^{2}}$

For a local minima or a local maxima, we must have f'(x) = 0

$\Rightarrow\frac{-1}{\text{x}^{2}}+\frac{1}{8-\text{x}^{2}}=0$

$\Rightarrow \frac{-(8-{\text{x}^{2}})+\text{x}^{2}}{(\text{x})^{2}(8-\text{x})^{2}}=0$

$\Rightarrow -64-\text{x}^{2}+16\text{x}+\text{x}^{2}=0$

$\Rightarrow 16\text{x}-64 =0$

$\Rightarrow \text{x}=4$

$ \text{f}''(\text{x})=\frac{2}{\text{x}^{3}}-\frac{2}{(8-{\text{x}})^{3}}$

$ \Rightarrow\text{f}''(\text{4})=\frac{2}{\text{x}^{3}}-\frac{2}{(8-{\text{4}})^{3}}$

$ \Rightarrow\text{f}''(\text{4})=\frac{2}{\text{4}^{3}}-\frac{2}{(8-{\text{4}})^{3}}$

$ \Rightarrow\text{f}''(\text{4})=\frac{2}{\text{64}}-\frac{2}{64}=0$

$\therefore$ minimum value $=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$ 

3. $1$

Solution:

Given, f(x) =  x² + x + 1

⇒ f'(x) = 2x + 1

For a local maxima or a local minima, we must have f'(x) = 0

⇒ 2x + 1 = 0

⇒ 2x = -1

$\Rightarrow \frac{-1}{2}\in[0,1]$

At extreme points f(0) = 0

f(1) = 1 +1 +1 =3 > 0

So, x = 1 is a local minima.

1. $\text{e}$

Solution:

Given, $\text{f}(\text{x})=\frac{\text{x}}{\log_{\text{e}}\text{x}}$

$\Rightarrow \text{f}'(\text{x})=\frac{\log_{e}\text{x}-1}{(\log_{\text{e}}\text{x})^{2}}$

For a local maximum or a local minima, we must have $\text{f}'(\text{x})=0$

$\Rightarrow \frac{\log_{e}\text{x}-1}{(\log_{\text{e}}\text{x})^{2}}=0$

$\Rightarrow \log_{\text{e}}\text{x}-1=0$

$\Rightarrow \log_{\text{e}}\text{x}=1$

$\Rightarrow \text{x}=\text{e}$

Now, $\Rightarrow \text{f}''(\text{x})=\frac{-1}{\text{x}(\log_{\text{e}}\text{x})^{2}}+\frac{2}{\text{x}(\log_{\text{e}}\text{x})^{3}}$

$\Rightarrow \text{f}''(\text{e})=\frac{-1}{\text{e}}+\frac{2}{\text{e}}=\frac{1}{\text{e}}>0$

So, x = e is a local minima.

$\therefore$ minimum value of $\text{f}(\text{x})=\frac{\text{e}}{\log_{\text{e}}\text{e}}=\text{e}$ 

2. $\text{ab} > \frac{\text{c}^{2}}{4}$

Solution:

$=\text{f}(\text{x})=\text{ax}+\frac{\text{b}}{\text{x}}$

$\text{f}(\text{x})=0$

$\Rightarrow \text{a}-\frac{\text{b}}{\text{x}^{2}}=0$

$\Rightarrow\text{x}=\pm\sqrt{\frac{\text{b}}{\text{a}}}$

$\text{f}''(\text{x})=\frac{2\text{b}}{\text{x}^{3}}$

$\text{f}''\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=\frac{2\text{b}}{\Big(\sqrt{\frac{\text{b}}{\text{c}}}\Big)^{3}}>0$

$\Rightarrow\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}$ has a minima.

$\text{f}''\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=2\sqrt{\text{ab}}\geq\text{c}$

$\frac{\text{c}}{2}\le\sqrt{\text{ab}}$

$\Rightarrow\frac{\text{c}}{4}\le{\text{ab}}$

3. $\frac{\pi}{6}$

Solution:

$\text{f}(\text{x})=\sin +\sqrt{3}\cos\text{x}$

$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sqrt{3}\sin\text{x}$

For maxima or maxima,

f'(x) = 0

$\cos\text{x}-\sqrt{3}\sin\text{x}=0$

$\Rightarrow\ \tan\text{x}=\frac{1}{\sqrt{3}}\Rightarrow\text{x}=\frac{\pi}{6}$

$\text{f}''\Big(\frac{\pi}{6}\Big)=-\sin\frac{\pi}{6}-\sqrt{3}\cos\frac{\pi}{6}=\frac{-1-\sqrt{3}}{2}<0$

function has local maima at $\text{x}=\frac{\pi}{6}$

1. 75

Sloution:

$\text{f}(\text{x})=\text{x}^{2}+\frac{250}{\text{x}}$

$\text{f}'(\text{x})=2\text{x}-\frac{250}{\text{x}^{2}}$

For the local minima a or maxima. We must have f'(x) = 0

 $=2\text{x}-\frac{250}{\text{x}^{2}}=0$

⇒ x = 5

$=2\text{x}-\frac{250}{\text{x}^{2}}=0$

$\text{f}''(\text{x})=2+\frac{500}{\text{x}^{3}}$

$\text{f}''(\text{x})=2+\frac{500}{125}>0$

function has minima at x = 5

f(5) = 75.

4. maximum value < minimum value.

Solution:

Given, $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$

lmplies that $\text{f}''(\text{x})=\text{x}-\frac{1}{\text{x}}$

For a local maxima or a local minima, we must have f'(x) = 0

lmplies that $1-\frac{1}{\text{x}^{2}}=0$

lmplies that $\text{x}^{2}-1=0$

lmplies that $\text{x}^{2}=0$

lmplies that $\text{x}=\pm1$

Now, $\text{f}'(\text{x})=\frac{2}{\text{x}^{3}}$

lmplie that $\text{f}'(\text{1})=\frac{2}{\text{1}}=2<0$

Threrefore, x = 1 is a local minima.

Also, f''(1) - 2 < 0

Threrefore, x = -1 is a local maxima.

The local minimum value is given by

f(1) = 2

The local maximum value is given by

f(-1) = -2

$\therefore$ maximum value < minimum value.

3. $\frac{-1}{\text{e}}$

Solution :

Here, $\text{f}(\text{x})=\text{x}\log_{\text{e}}\text{x}$

lmplies that $\text{f}'(\text{x})=\log_{\text{e}}\text{x}+1$

For a local maxima or a local minima, we must have f'(x) = 0

lmplies that $\log_{\text{e}}\text{x}+1=0$ 

lmplies that $\log_\text{e}\text{x}=-1$

lmplies that $\text{x}=\text{e}^{-1}$

Now, $\text{f}''(\text{x})=\frac{1}{\text{x}}$

lmplies that $\text{f}''(\text{e}^{-1})=\text{e}>0$

Threrfore, $\text{f}(\text{e}^{-1})$ is a local minima.

Hence, the minimum value of $\text{f}(\text{x})=\text{f}(\text{e}^{-1})$

lmplies that $\text{e}^{-1}\log_\text{e}(\text{e}^{-1})=-\text{e}^{-1}=\frac{-1}{\text{e}}$

4. none of these.

Solution:

Given, $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$

$\Rightarrow \text{f}'(\text{x})=1-\frac{1}{\text{x}^{2}}$

For a local maxima or a local minima, we must have f'(x) = 0

$\Rightarrow 1-\frac{1}{\text{x}^{2}}=0$

$\Rightarrow \text{x}^{2}-1=0$

$\Rightarrow \text{x}^{2}=1$

$\Rightarrow \text{x}=\pm1$

$\Rightarrow \text{x}=1$

Now, $\text{f}''(\text{x})=\frac{2}{\text{x}^{3}}$

$\text{f}''(1)=2>0$

So, x = 1 is a local minima.

3. $\frac{1}{6}$

Solution :

$\text{f}(\text{x})=\frac{\text{x}}{4+\text{x}+\text{x}^{2}}$

$\Rightarrow \text{f}'(\text{x})=\frac{4-\text{x}^{2}}{(4+\text{x}+\text{x}^{2})^{2}}$

For a local maxima or minima, f'(x) = 0

$\frac{4-\text{x}^{2}}{(4+\text{x}+\text{x}^{2})^{2}}=0$

$\Rightarrow\text{x}=\pm2\notin [-1,1] $

$\text{f}(1)=\frac{1}{6}>0$

$\text{f}(-1)=\frac{-1}{4}<0$

$\Rightarrow \frac{1}{6}$ is the maximum value.

3. $3$

Solution:

Given, $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$

lmplies that f(x) = (x - 1)² + (x - 2)² + (x - 3)² + (x - 4)² + x - 5²

lmplies that f'(x) = 2(x - 1 + x - 2 + x - 3 + x - 4 + x - 5)

lmplies that f'(x) = 2(5x - 15)

For a local maxima and a local minima, we must have f'(x) = 0

limplies that 2(5x - 15) = 0

limplies that 5x - 15 = 0

limplies that x = 3

Now, f''(x) = 10

f''(x) = 10 > 0

Therefore, x = 3 is a local minima.

2. $(1, 2)$

Solution:

Let the required point be (x, y). Then,

y² = 4x

$\Rightarrow\text{x}=\frac{\text{y}^{2}}{4} ...(\text{i})$

Now, $\text{d}=\sqrt{(\text{x}-2)^{2}+(\text{y}-1)^{2}}$

Squaring both sides, we get

$\Rightarrow\text{d}^{2}=(\text{x}-2)^{2}+(\text{y}-1)^{2}$

$\Rightarrow\text{d}^{2}=\Big(\frac{\text{y}^{2}}{4}-2\Big)^{2}+(\text{y}-1)^{2}$

$\Rightarrow\text{d}^{2}=\frac{\text{y}^{2}}{16}+4-\text{y}^{2}+\text{y}^{2}+1-2\text{y}$ [From eq.(i)]

Now, $\text{Z}=\text{d}^{2}=\frac{\text{y}^{2}}{16}+4-\text{y}^{2}+\text{y}^{2}+1-2\text{y}$

$\Rightarrow \frac{\text{dZ}}{\text{dy}}=\frac{\text{y}^{2}}{4}-2\text{y}+2\text{y}-2$

$\Rightarrow \frac{\text{dZ}}{\text{dy}}=\frac{\text{y}^{2}}{4}-2$

$\Rightarrow \frac{\text{y}^{2}}{4}-2=0$

$\Rightarrow\text{y}^{3}=8$

$\Rightarrow\text{y}=2$

Substituting the value of y in (i), we get x = 1

Now, $\frac{\text{d}^{2}\text{Z}}{\text{dy}^{2}}=\frac{3\text{y}^{2}}{4}$

$\Rightarrow\frac{\text{d}^{2}\text{Z}}{\text{dy}^{2}}=\frac{3\text{(2)}^{2}}{4}=3>0$

So, the nearest point is (1, 2).

3. $\frac{1}{6}$

Solution:

$\text{f}(\text{x})=\frac{\text{x}}{4-\text{x}+\text{x}^{2}}$

$\Rightarrow\text{f}'(\text{x})=\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}$

To find minimum or maximum value f'(x) = 0

$\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}=0$

4 - x² = 0

$\text{x}=\pm2\notin[-1, 1]$

$\text{f}(-1)=\frac{-1}{6}$ and $\text{f}(1)=\frac{1}{4}$

Hence, maximum value is $\frac{1}{4}$.

NOTE: options in the book are not matching with the solution.

Above solution is based on the question given in the book.

4. none of these.

Solution:

Given, $\text{f}(\text{x})=2\sin3\text{x}+3 \cos3\text{x}$

$\Rightarrow \text{f}'(\text{x})=6 \cos3\text{x}-9\cos3\text{x}$

to find maxima or minima f'(x) = 0

$6 \cos3\text{x}-9\cos3\text{x}=0$

$\Rightarrow \tan3\text{x}=\frac{2}{3}$

$\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(3\times\frac{5\pi}{6}\Big)$

$\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(\frac{5\pi}{2}\Big)$

$\Rightarrow\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(2\pi+\frac{\pi}{2}\Big)$

$\Rightarrow\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(\frac{\pi}{2}\Big)$ which is not defined.

Hence, $\text{x}=\frac{5\pi}{6}$ is not a critical point.

1. $\text{e}^\frac{1}{\text{e}}$

Solution:

We have $\text{f}(\text{x})=\text{x}^\frac{1}{\text{x}}$

Taking log on both side, we get 

$\log\text{f}(\text{x})=\frac{1}{\text{x}}\ \log\text{x}$

Differentitating W.r.t. x, we get

$\frac{1}{\text{f}(\text{x})}\text{f}'(\text{x})=\frac{-1}{\text{x}^{2}}\ \log\text{x}+\frac{1}{\text{x}^{2}}$

$\Rightarrow\text{f}'(\text{x})=\text{f}'(\text{x})\frac{1}{\text{x}^{2}}\ (1-\log\text{x})$

$\Rightarrow\text{f}'(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big) ...(\text{i})$

$\Rightarrow\text{f}'(\text{x})=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x})$

For a local maximum or a local minima, we must have f'(x) = 0

$=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x}) =0$

$\Rightarrow \log\text{x}=1$

$\therefore \text{x}=\text{e}$

Now, $\text{f}''(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$

$=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$

At x = e

$\text{f}''(\text{e})=\text{e}^\frac{1}{\text{e}}\Big(\frac{1}{\text{e}^{2}}-\frac{1}{\text{e}^{2}}\log\text{e}\Big)^{2}+\text{e}^\frac{1}{\text{e}}\Big(\frac{-3}{\text{e}^{3}}+\frac{2}{\text{e}^{3}}\log\text{e}\Big)$

$=-\text{e}^\frac{1}{\text{e}}(\frac{1}{\text{e}^{3}})<0$

So, x = e is a point of local maximum.

Thus, the maximum value is given by 

$\text{f}(\text{e})=\text{e}^\frac{1}{\text{e}}$

1. $\frac{1}{2}$

Solution:

Let the required number be x.

Then, f'(x) = x - x²

lmplies that f(x) = 1 - 2x = 0

For a local maxima or a local minima, we must have f'(x) = 0

lmplies that 2x = 1

lmplies that $\text{x}=\frac{1}{2}$

Now, f''(x) = -2 < 0

Therefore, $\text{x}=\frac{1}{2}$ is a local maxima.

Hence, the required number is $\frac{1}{2}$.

1. Minimum at $\text{x}=\frac{\pi}{2}$

Solution :

Given, $\text{f}(\text{x})=1+2\sin\text{x}+3\cos^{2}\text{x}$

$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}-6\cos\text{x}\sin\text{x}$

$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}(1-3\sin\text{x})$

For a local maxima or a local minima.

We must have f'(x) = 0

$\Rightarrow2\cos\text{x}(1-3\sin\text{x})=0$

$\Rightarrow2\cos\text{x}=0$ or $(1-3\sin\text{x})=0$

$\Rightarrow \cos \text{x}=0$ or $\sin\text{x}=\frac{1}{3}$

$\Rightarrow \text{x}=\frac{\pi}{2}$ or $\text{x}=\sin^{-1}(\frac{1}{3})$

Now, $\text{f}''(\text{x})=-2\sin\text{x}-6\cos2\text{x}$

$\Rightarrow\text{f}''(\frac{\pi}{2})=-2\sin\frac{\pi}{2}-6\cos(2\times\frac{\pi}{2})$

$=-2+6=4>0$

So, $ \text{x}=\frac{\pi}{2}$ is a local minima.

Also, $\text{f}''(\sin^{-1}\big(\frac{1}{3}\big))=-2\sin(\sin^{-1}\big(\frac{1}{3}\big))-6\cos(\sin^{-1}\big(\frac{1}{3}\big))$

$=\frac{-2}{3}-6\times\frac{2\sqrt{2}}{3}$

$=-\Big(\frac{2}{3}+4\sqrt{2}\Big)<0$

So, $\text{x}=\sin^{-1}(\frac{1}{3})$ is a local maxima.

2. $2$

Solution:

Given, xy = 1

To find minimum value of x + y

$\Rightarrow \text{y}=\frac{1}{\text{x}}$ 

$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$

$\Rightarrow\text{f}'(\text{x})=\text{1}-\frac{1}{\text{x}^{2}}$

To find local maxima or minima. We have f'(x) = 0

$\Rightarrow \text{x}=\pm1\Rightarrow \text{y} =\pm1$

But given that x > 0 ⇒ x = 1, y = 1

$\text{f}''(\text{x})=\frac{2}{\text{x}^{3}}$

f''(x) = 2 > 0

function has minima at x = 1

f(1) = 2.

1. $\frac{\text{a}+\text{b}+\text{c}}{3}$

Solution:

f(x) = (x - a)² + (x - b)² + (x - c)²

⇒ 2(x - a) + 2(x - b) + 2(x - c)

to find minima or maxima f'(x) = 0

2(x - a) + 2(x - b)² + 2(x - c) = 0

^{$\Rightarrow \text{x}=\frac{\text{a}+\text{b}+\text{c}}{3}$}
f''(x) = 6 > 0

function has minima at $\text{x}=\frac{\text{a}+\text{b}+\text{c}}{3}$.