MCQ 11 Mark
If $\text{f}(\text{x})=\frac{1}{4\text{x}^{2}+2\text{x}+1}$, then its maximum value is :
- ✓
$\frac{4}{3}$
- B
$\frac{2}{3}$
- C
$1$
- D
$\frac{3}{4}$
AnswerCorrect option: A. $\frac{4}{3}$
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 $\frac{1}{4\text{x}^{2}+2\text{x}+1}=\frac{1}{4\Big(\frac{-1}{4}\Big)^{2}+2\Big(\frac{-1}{4}\Big)+1}$
$=\frac{16}{12}=\frac{4}{3}$
View full question & answer→MCQ 21 Mark
If a cone of maximum volume is inscribed in a given sphere, then the ratio of the height of the cone to the diameter of the sphere is :
- A
$\frac{3}{4}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- ✓
$\frac{2}{3}$
AnswerCorrect option: D. $\frac{2}{3}$
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
$h^2 + R^2 - 2hR = R^2 - r^2$
lmplies that $r^2 = 2hr - h^2$
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 $\frac{\text{dV}}{\text{dh}}=0$
View full question & answer→MCQ 31 Mark
The sum of two non-zero number is 8, the minimum value of the sum of the reciprohcle is :
- A
$\frac{1}{4}$
- ✓
$\frac{1}{2}$
- C
$\frac{1}{8}$
- D
AnswerCorrect option: B. $\frac{1}{2}$
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 $=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
View full question & answer→MCQ 41 Mark
if $x$ lies in the interval $[0, 1],$ then the least value of $x^2 + x + 1$ is :
AnswerGiven, $f(x) = x^2 + x + 1$
$\Rightarrow f\ '(x) = 2x + 1$
For a local maxima or a local minima, we must have $f\ '(x) = 0$
$\Rightarrow 2x + 1 = 0$
$\Rightarrow 2x = -1$
$\Rightarrow \frac{-1}{2}\in[0,1]$
At extreme points $f(0) = 0$
$f(1) = 1 +1 +1 =3 > 0$
So, $x = 1$ is a local minima.
View full question & answer→MCQ 51 Mark
The minimum value of $\frac{\text{x}}{\log_{\text{e}}\text{x}}$ is .
- ✓
$\text{e}$
- B
$\frac{1}{\text{e}}$
- C
$1$
- D
AnswerCorrect option: A. $\text{e}$
Given, $\text{f}(\text{x})=\frac{\text{x}}{\log_{\text{e}}\text{x}}$$\Rightarrow \text{f}'(\text{x})=\frac{\log_{e}\text{x}-1}{(\log_{\text{e}}\text{x})^{2}}$
For a local maximum or a local minima, we must have $\text{f}'(\text{x})=0$
$\Rightarrow \frac{\log_{e}\text{x}-1}{(\log_{\text{e}}\text{x})^{2}}=0$
$\Rightarrow \log_{\text{e}}\text{x}-1=0$
$\Rightarrow \log_{\text{e}}\text{x}=1$
$\Rightarrow \text{x}=\text{e}$
Now, $\Rightarrow \text{f}''(\text{x})=\frac{-1}{\text{x}(\log_{\text{e}}\text{x})^{2}}+\frac{2}{\text{x}(\log_{\text{e}}\text{x})^{3}}$
$\Rightarrow \text{f}''(\text{e})=\frac{-1}{\text{e}}+\frac{2}{\text{e}}=\frac{1}{\text{e}}>0$
So, x = e is a local minima.
$\therefore$ minimum value of $\text{f}(\text{x})=\frac{\text{e}}{\log_{\text{e}}\text{e}}=\text{e}$
View full question & answer→MCQ 61 Mark
if $\text{ax}+\frac{\text{b}}{\text{x}}\geq\text{c}$ for all positive x where a, b, > 0, then.
- A
$\text{ab} < \frac{\text{c}^{2}}{4}$
- ✓
$\text{ab} > \frac{\text{c}^{2}}{4}$
- C
$\text{ab} > \frac{\text{c}}{4}$
- D
$\text{None of these}$
AnswerCorrect option: B. $\text{ab} > \frac{\text{c}^{2}}{4}$
$=\text{f}(\text{x})=\text{ax}+\frac{\text{b}}{\text{x}}$$\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.
$\text{f}''\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=2\sqrt{\text{ab}}\geq\text{c}$
$\frac{\text{c}}{2}\le\sqrt{\text{ab}}$
$\Rightarrow\frac{\text{c}}{4}\le{\text{ab}}$
View full question & answer→MCQ 71 Mark
$\text{f}(\text{x})=\sin +\sqrt{3}\cos\text{x}$ is maximum when x =
- A
$\frac{\pi}{3}$
- B
$\frac{\pi}{4}$
- ✓
$\frac{\pi}{6}$
- D
$0$
AnswerCorrect option: C. $\frac{\pi}{6}$
$\text{f}(\text{x})=\sin +\sqrt{3}\cos\text{x}$
$\Rightarrow\text{f}'(\text{x})=\cos\text{x}-\sqrt{3}\sin\text{x}$
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 $\text{x}=\frac{\pi}{6}$
View full question & answer→MCQ 81 Mark
The minimum value of $(\text{x}^{2}+\frac{250}{\text{x}})$ is:
Answer$\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.
View full question & answer→MCQ 91 Mark
The function $f(x) = 2x^3- 15x^2 + 36x + 4$ is maximum at $x =$
AnswerGiven, $f(x)=2 x^3-15 x^2+36 x+4$
Implies that $f^{\prime}(x)=6 x^2-30 x+36$
For a local maxima or a local minima, we must have $f^{\prime}(x)=0$
Implies that $6 x^2-30 x+36=0$
Implies that $x^2-5 x+6=0$
$(x-2)(x-3)=0$
Implies that $x=2,3$
Now, $\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}-30$
Implies that $\mathrm{f}^{\prime \prime}(2)=24-30=6<0$
Therefore, $x=1$ is a local maxima.
Also, $f^{\prime \prime}(3)=36-30=6 > 0$
Therefore, $x=2$ is a local maxima.
View full question & answer→MCQ 101 Mark
For the function $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}$
- A
x = 1 is a point of maximum.
- B
x = -1 is a point of minimum.
- C
maximum value > minimum value.
- ✓
maximum value < minimum value.
AnswerCorrect option: D. maximum value < minimum value.
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.
View full question & answer→MCQ 111 Mark
The minimum value of $f(x) = x^4 - x^2 - 2x + 6$ is.
AnswerGiven $,f(x)=x^4-x^2-2 x+6$
$\Rightarrow f^{\prime}(x)=4 x^3-2 x-2$
$\Rightarrow f^{\prime}(x)=(x-1)\left(4 x^2+4 x+2\right)$
For a local maxima or a local minima, we must have $f^{\prime}(x)=0$
$\Rightarrow(x-1)\left(4 x^2+4 x+2\right)=0$
$\Rightarrow(x-1)=0$
$\Rightarrow x=1$
Now, $\mathrm{f}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}^2-2$
$\Rightarrow f^{\prime \prime}(x)=12-2=10>0$
So, $x=1$ is a local minima.
The local minimum value is given by
$f(1)=1-1-2+6=4$
View full question & answer→MCQ 121 Mark
The minimum value of $\text{x}\log_{\text{e}}\text{x}$ is equal to :
- A
$\text{e}$
- B
$\frac{1}{\text{e}}$
- ✓
$\frac{-1}{\text{e}}$
- D
$\text{2}{\text{e}}$
AnswerCorrect option: C. $\frac{-1}{\text{e}}$
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}}$
View full question & answer→MCQ 131 Mark
Let $f(x) = 2x^3 - 3x^2 - 12x + 5$ on $[-2, 4].$ The relative maximum occurs at $x =$
AnswerGiven, $f(x) = 2x^3 - 3x^2 - 12x + 5$
$\Rightarrow f'(x) = 6x^2 - 6x - 12$
For a local maxima or a local minima, we must have $f'(x) = 0$
$\Rightarrow 6x^2 - 6x - 12 = 0$
$\Rightarrow x^2- x - 2 = 0$
$\Rightarrow (x - 2)(x + 1) = 0$
$\Rightarrow x = 2, -1$
Now, $f''(x) = 12x - 6$
$\Rightarrow f''(-1) = -12 - 6 = 18 > 0$
So, $x = 1$ is a local maxima.
Also, $f''(2) = 24 - 6 = 18 > 0$
So, $x = 2$ is a local minima.
View full question & answer→MCQ 141 Mark
If $\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}, \text{x}>0$, then its greatest value is:
AnswerGiven, $\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.
View full question & answer→MCQ 151 Mark
If x + y = 8, then the maximum value of xy is:
AnswerGiven, x + y =8lmplies that y = 8 - x ..(i)
Let f(x) = x(8 - x) [From eq.(i)]
lmplies that f'(x) = 8 - 2x
For a local maxima or a local mimima, we must have f'(x) = 0
lmplies that 8 - 2x = 0
lmplies that 8 = 2x
lmplies that x = 4
lmplies that y = 8 - 4 = 4 [From eq.(i)]
Now, f''(x) = -2
lmplies that f''(4) = -2 < 0
Therefore, x = 4 is a local maxima.
Hence, the local maximum value is given by
f(4) = 4 × 4 = 16.
View full question & answer→MCQ 161 Mark
The maximum value of $\text{f}(\text{x})=\frac{\text{x}}{4+\text{x}+\text{x}^{2}}$ on [-1, 1] is :
- A
$-\frac{1}{4}$
- B
$-\frac{1}{3}$
- ✓
$\frac{1}{6}$
- D
$\frac{1}{5}$
AnswerCorrect option: C. $\frac{1}{6}$
$\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.
View full question & answer→MCQ 171 Mark
The function $\text{f}(\text{x})=\sum_{\text{r}=1}^5(\text{x}-\text{r})^{2}$ assume minimum value at $x =$
- A
$5$
- B
$\frac{5}{2}$
- ✓
$3$
- D
$2$
AnswerGiven, $\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 - 5^2$
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.
View full question & answer→MCQ 181 Mark
The least and greatest value of $f(x) = x^3 - 6x^2+ 9x$ in $[0, 6],$ are.
- ✓
$3 ,4$
- B
$0, 4$
- C
$0, 3$
- D
$3, 6$
AnswerCorrect option: A. $3 ,4$
$f(x)=x^3-6 x^2+9 x$
$\Rightarrow f^{\prime}(x)=3 x^2-12 x+9$
To find minimum or maximum value we have $\mathrm{f}^{\prime}(x)=0$
$3 x^2-12 x+9=0$
$\Rightarrow x^4-4 x+3=0$
$\Rightarrow x=1$ or $ x=3$
$f(0)=0$
$f(6)=54$
$f(1)=4$
$f(3)=0$
Least value and greatest value of the function are $0$ and $54$ respectively.
Note $=$ options are not matching with the solution.
Above solution is according to the question given in the book.
View full question & answer→MCQ 191 Mark
The least value of the function $f(x) = x^3 - 18x^2 + 96x$ in the interval $[0, 9]$ is:
AnswerGiven, $f(x) = x^3 - 18x^2+ 96x$
lmplies that $f'(x) = 3x^2 - 36x + 96$
For a local maxima or a local minima, we must have $f'(x) = 0$
lmplies that $3\times ^2 - 36x + 96$
lmplies that $x^2 - 12x + 32 = 0$
lmplies that $(x - 4)(x - 8) = 0$
lmplies that $x = 4, 8$
Therefore, $f(8) = (8)^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128$
$f(4) = (4)^3- 18(4)^2 + 96(4) = 64 - 288 + 384 = 160$
$f(0) = (0)^3 - 18(0)^2+ 96(0) = 0$
$f(9) = (9)^3 - 18(9)^2 + 96(9) = 729 - 1458 + 864 = 135$
Hence, $0$ is the minimum value in the range $0, 9.$
View full question & answer→MCQ 201 Mark
if x lies in the interval $[0, 1]$, then the least value of $x^2 + x + 1$ is :
AnswerGiven, $f(x) = x^2 + x + 1$
$\Rightarrow f'(x) = 2x + 1$
For a local maxima or a local minima, we muat have $f'(x) = 0$
$\Rightarrow 2x + 1 = 0$
$\Rightarrow 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.
View full question & answer→MCQ 211 Mark
The number which exceeds its square by the greatest possible quantity is,
- ✓
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
$\frac{3}{4}$
- D
AnswerCorrect option: A. $\frac{1}{2}$
Let the required number be $x.$
Then, $f\ '(x) = x - x^2$
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}$.
View full question & answer→MCQ 221 Mark
$\text{f}(\text{x})=1+2\sin\text{x}+3\cos^{2}\text{x}, 0<\text{x}<\frac{2\pi}{3}$ is :
- ✓
Minimum at $\text{x}=\frac{\pi}{2}$
- B
Maximum at $\text{x}=\sin^{-1}(\frac{1}{\sqrt{3}})$
- C
Minimum at $\text{x}=\frac{\pi}{6}$
- D
Maximum at $\sin^{-1}(\frac{1}{6})$
AnswerCorrect option: A. Minimum at $\text{x}=\frac{\pi}{2}$
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.
View full question & answer→MCQ 231 Mark
The point on the curve $y^2 = 4x$ which is nearest to, the point $(2, 1)$ is :
- A
$1,2\sqrt{2}$
- ✓
$(1, 2)$
- C
$(1, -2)$
- D
$(-2, 1)$
AnswerCorrect option: B. $(1, 2)$
Let the required point be $(x, y). $ Then,
$y^2= 4x$
$\Rightarrow\text{x}=\frac{\text{y}^{2}}{4} ...(\text{i})$
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 $(1, 2).$
View full question & answer→MCQ 241 Mark
The maximum value of $\text{f}(\text{x})=\frac{\text{x}}{4-\text{x}+\text{x}^{2}}$ on $[-1, 1]$ is :
- A
$-\frac{1}{4}$
- B
$-\frac{1}{3}$
- ✓
$\frac{1}{4}$
- D
$\frac{1}{5}$
AnswerCorrect option: C. $\frac{1}{4}$
$\text{f}(\text{x})=\frac{\text{x}}{4-\text{x}+\text{x}^{2}}$
$\Rightarrow\text{f}\ '(\text{x})=\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}$
To find minimum or maximum value $f\ '(x) = 0$
$\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}=0$
$4 - x^2 = 0$
$\text{x}=\pm2\notin[-1, 1]$
$\text{f}(-1)=\frac{-1}{6}$ 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.
View full question & answer→MCQ 251 Mark
At $\text{x}=\frac{5\pi}{6}, $ $\text{f}(\text{x})=2\sin3\text{x}+3 \cos3\text{x}$ is:
AnswerGiven, $\text{f}(\text{x})=2\sin3\text{x}+3 \cos3\text{x}$
$\Rightarrow \text{f}'(\text{x})=6 \cos3\text{x}-9\cos3\text{x}$
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.
View full question & answer→MCQ 261 Mark
The maximum value of $\text{x}^\frac{1}{\text{x}}, \text{x}>0 $ is.
AnswerCorrect option: A. $\text{e}^\frac{1}{\text{e}}$
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
$\text{f}(\text{e})=\text{e}^\frac{1}{\text{e}}$
View full question & answer→MCQ 271 Mark
The minimum of the function $f(x) = 2x^3 - 21x^2 + 36x - 20$ is :
AnswerCorrect option: A. $-128$
$f(x)=2 x^3-21 x^2+36 x-20$
$\Rightarrow f^{\prime}(x)=6 x^2-42 x+36$
For local maxima or minima
$6 x^2-42 x+36=0$
$x^2-7 x+36=0$
$\Rightarrow x=1 $ or $x=6$
$f^{\prime \prime}(x)=12 x-42$
$\Rightarrow f^{\prime \prime}(1)=-30<0$
Also, $f^{\prime \prime}(6)=30>0$
function has minima at $x=6$
$\Rightarrow f(6)=-128$
View full question & answer→MCQ 281 Mark
Let $f(x) = x^3 + 3x^2 - 9x + 2.$ Then, $f(x)$ has,
AnswerCorrect option: B. a minimum at $x = 1$
$f(x) = x^3+ 3x^2- 9x + 2$
$\Rightarrow f'(x) = 3x^2+ 6x - 9$
local minima or maxima must have $f'(x) = 0$
$x^3+ 3x^2- 9x + 2 = 0$
$x^2+ 2x - 3 = 0$
$\Rightarrow (x + 3)(x + 1) = 0$
$\Rightarrow x = -3$ or $x = 1$
$f''(x) = 6x + 6$
$\Rightarrow f''(-3) = -12 < 0$
At, $x = -3$ local maxima.
$f''(1) = 12 > 0$
At, $x = 1$ local minima.
View full question & answer→MCQ 291 Mark
Let x, y be two variables and x > 0, xy = 1, then minimum value of x + y is :
- A
$1$
- ✓
$2$
- C
$2\frac{1}{2}$
- D
$3\frac{1}{3}$
AnswerGiven, xy = 1To 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.
View full question & answer→MCQ 301 Mark
Let $f(x) = (x - a)^2+ (x - b)^2 + (x - c)^2$. Then, $f(x)$ has a minimum at $x =$
- ✓
$\frac{\text{a}+\text{b}+\text{c}}{3}$
- B
$\sqrt[3]{\text{a}\text{b}\text{c}}$
- C
$\frac{3}{\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}}$
- D
AnswerCorrect option: A. $\frac{\text{a}+\text{b}+\text{c}}{3}$
$f(x) = (x - a)^2+ (x - b)^2 + (x - c)^2$
$\Rightarrow 2(x - a)^+ 2(x - b) + 2(x - c)$
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 $\text{x}=\frac{\text{a}+\text{b}+\text{c}}{3}$.
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