Question 11 Mark
Find the distance between the planes $\overrightarrow{\text{r}}.(2\hat{i} - 3\hat{j} +6\hat{k}) - 4 = 0 \text{ and}\overrightarrow{\text{r}}.(6\hat{i} - 9\hat{j} +18\hat{k}) + 30 = 0.$
AnswerWriting or using, that given planes are parallel
$d = \frac{|\ 4 + 10\ |}{\sqrt{\ 4\ +\ 9 \ + \ 36}} = 2 \text{ units}$
View full question & answer→Question 21 Mark
$\text{If A} = \begin{bmatrix} 2 & 3 \\ 5 & -2 \\ \end{bmatrix}, \text{then write A}^{-1}. $
Answer$|\text{A}| = -19$
$\text{A}^{-1} = -\frac{1}{19}\begin{bmatrix} -2 & -5 \\ -3 & 2 \\ \end{bmatrix}$
View full question & answer→Question 31 Mark
If $\overrightarrow{\text{a}} = 2\hat{\text{i}} + \hat{\text{j}} + 3\hat{\text{k}}$ and $ \overrightarrow{\text{b}} = 3\hat{\text{i}}+ 5\hat{\text{j}} - 2\hat{\text{k}},$ then find $|\overrightarrow{\text{a}}\times|\overrightarrow{\text{b}|}.$
Answer$\overrightarrow{\text{a}}\times\overrightarrow{\text{b}}= -17\hat{\text{i}} + 13\hat{\text{j}} + 7\hat{\text{k}}, \overrightarrow{|\text{a}}\times\overrightarrow{\text{b}|} = \sqrt{507}$
View full question & answer→Question 41 Mark
$\text{If A} = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \\ \end{bmatrix}. \text{find } \alpha \text{ satisfying } 0 < \alpha < \frac{\pi}{2} \text{when A+ A}^{\text{T}} = \sqrt{2}\text{ I}_{2} : $ where $\text{A}^{\text{T}}$ is transpose of $\text{A}$
Answer$\text{Finding A}^{\text{T}}=\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \\ \end{bmatrix} $
$\text{Getting }\alpha = \frac{\pi}{4} \text{or }45^{0}$
View full question & answer→Question 51 Mark
If A is a matrix $3\times3$ If A is a and $\text{|3A| = K|A|,}$ then write the value of k.
View full question & answer→Question 61 Mark
Find the sum of the degree and the order for the following differential equation:
$\frac{\text{d}}{\text{dx}}\Bigg[\bigg(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\bigg)^{4}\Bigg] = 0$
Answer$\text{order 3 , or degree 1}$
$\therefore\text{Degree + order = 4}$
View full question & answer→Question 71 Mark
$\text{If}\overrightarrow{\text{a}} = \hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}},$ $\overrightarrow{\text{b}} =2\hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}$ and $ \overrightarrow{\text{c}} = 5\hat{\text{i}} + 4\hat{\text{j}} - 3\hat{\text{k}},$ then find the value of $(\overrightarrow{\text{a}}+\overrightarrow{\text{b}}).\overrightarrow{\text{c.}}$
Answer$\overrightarrow{\text{a}}+\overrightarrow{\text{b}} = 3\hat{\text{i}} + 3\hat{\text{j}}$
$(\overrightarrow{\text{a}}+\overrightarrow{\text{b}}).\overrightarrow{\text{c}} = 3$
View full question & answer→Question 81 Mark
$\text{If A} = \begin{bmatrix} 2 & 4 \\ 3 & 2 \\ \end{bmatrix} \text{and B} = \begin{bmatrix} -2 & 5 \\ 3 & 4 \\ \end{bmatrix}, $ $\text{then find (3A – B).}$
Answer$3 \begin{bmatrix} 2 & 4 \\ 3 & 2 \\ \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \\ \end{bmatrix} $
$ = \begin{bmatrix} 8 & 7 \\ 6 & 2 \\ \end{bmatrix}$
View full question & answer→Question 91 Mark
Write the number of all possible matrices of order $2\times3$ with each entry 1 or 2.
Answer$2^{6} \text{ or } 64$
View full question & answer→Question 101 Mark
Write the sum of the order and degree of the following differential equation:
$\frac{\text{d}}{\text{dx}} \Bigg\{\bigg(\frac{\text{dy}^{3} }{\text{dx}}\bigg)\Bigg\} = 0$
Answer$\text{order 2, degree 1}$
$\text{sum = 3}$
View full question & answer→Question 111 Mark
Write the integrating factor of the following differential equation:
$(1 + \text{y}^{2}) + (2\text{xy} -\cot\text{y})\frac{\text{dy}}{\text{dx}} = 0$
Answer$\frac{\text{dx}}{\text{dy}} + \frac{2\text{y}}{1 + \text{y}^{2}} . \text{x} = \cot\text{y}$
Integrating factor $= e^{\log(1 + \text{y}^{2})} \text{or} (1 + \text{y}^{2})$
View full question & answer→Question 121 Mark
if A is a $3\times3$ matrix and $\text{|3A| = K|A|,}$ then write the value of k.
View full question & answer→Question 131 Mark
Find the value of $\text{a + b}$ if the points $\text{(2, a, 3), (3, – 5, b) and (– 1, 11, 9) }$are collinear
Answer$\text{d.r's of}\ \overrightarrow{\text{AB}} : \text{1, -5 -a, b - 3; d.r's of } \overrightarrow{\text { BC}}\text{ are - 4, 16, 9-b or d.'s of }\overrightarrow{\text{ AC}}:-3, 11 -\text{a}, 6 $ $\text{getting a = -1, b = 1, a + b = 0}$
View full question & answer→Question 141 Mark
Find the vector and cartesian equations of a line through the point (1, -1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, -1, 0) and (1, 2, -1), (2, 1, 1).
AnswerEquation of line joining (4, 3, 2) and (1, – 1, 0) is
$\frac{\text{x - 4}}{-3} = \frac{\text{y - 3}}{-4} = \frac{\text{z - 2}}{-2}$
Equation of line joining (1, 2, – 1) and (2, 1, 1) is
$\frac{\text{x - 1}}{1} = \frac{\text{y - 2}}{-1} = \frac{\text{z + 1}}{2}$
Let equation of the required line be
$\frac{\text{x - 1}}{\text{a}} = \frac{\text{y + 1 }}{\text{b}} = \frac{\text{z - 1}}{\text{c}} = \lambda\dots\dots\dots\dots\dots\dots\dots\dots\text{(i)}$
According to the question $\text{3a + 4b + 2c = 0}$
$\text{a – b + 2c = 0}$
Solving $\frac{\text{a}}{10} = \frac{\text{b}}{-4} = \frac{\text{c}}{-7} = \mu$
$\Rightarrow \text{a} = 10\mu, \text{b} = -4\mu, \text{c} = -7\mu$
$\text{(i)} \Rightarrow$ Equation of the line is
$\frac{\text{x - 1}}{10} = \frac{\text{y + 1}}{-4} = \frac{\text{z - 1}}{-7} \text{[cartesian form]}$
Vector form, $\overrightarrow{\text{r}} = \bigg(\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}}\bigg) + \lambda \bigg(10\hat{\text{i}} - 4\hat{\text{j}} -7 \hat{\text{k}}\bigg)$
View full question & answer→Question 151 Mark
If $\hat{\text{a}}, \hat{\text{b}}$ and $\hat{\text{c}}$are mutually perpendicular unit vectors, then find the value of $|\hat{2\text{a}} + \hat{\text{b}} + \hat{\text{c}}|.$
Answer$|2\hat{\text{a}} + \hat{\text{b}} + \hat{\text{c}}|^{2} = (2\hat{\text{a}})^{2} + (\hat{\text{b}})^{2} + (\hat{\text{c}})^{2} +2(2\hat{\text{a}}. \hat{\text{b}} + \hat{\text{b}}. \hat{\text{c}} + \hat{\text{c}}. 2\hat{\text{a}})$
$\therefore|2\hat{\text{a}} + \hat{\text{b}} + \hat{\text{c}}| = \sqrt{6}$
View full question & answer→Question 161 Mark
Write the value of $\hat{\text{i}}.(\hat{\text{j}}\times\hat{\text{k}}).+\hat{\text{j}}. (\hat{\text{k}}\times\hat{\text{i}}).+\hat{\text{k}}.(\hat{\text{i}} \times\hat{\text{j}}).$
View full question & answer→Question 171 Mark
The money to be spent for the welfare of the employees of a firm is proportional to the rate of change of its total revenue (marginal revenue). If the total revenue (in rupees) received from the sale of x units of a product is given by $R(x) =3x^2 +36x + 5$, find the marginal revenue when $x = 5$, and write which value does the question indicate.
AnswerGiven: $R(x) = 3x^2 + 36x + 5$
$\Rightarrow\text{ R}'\text{(x)} = 6\text{x} + 36$
$\therefore\text{ Marginal revenue (when x} = 5 ) = \text{R}'\text{(x)}]_{\text{x} = 5 } = 6\times5 + 36 =\text{ Rs.} 66.$
View full question & answer→Question 181 Mark
The amount of pollution content added in air in a city due to $x-$diesel vehicles is given by $P(x) = 0.005x^3 + 0.02x^2 + 30x.$ Find the marginal increase in pollution content when $3$ diesel vehicles are added and write which value is indicated in the above question.
AnswerWe have to find $[P'(x)]_{x = 3}$_
Now, $P(x) = 0.005x^3 + 0.02x^2 + 30x$
$\therefore P'(x) = 0.015x^2 + 0.04x + 30$
$\Rightarrow [P'(x)]_{x = 3} = 0.015 \times 9 + 0.04\times 3 + 30$
$= 0.135 + 0.12 + 30 = 30.255$
This question indicates “how increase in number of diesel vehicles increase the air pollution, which is harmful for living body.”
View full question & answer→MCQ 191 Mark
Choose the correct answer in the Exercise : The slope of the tangent to the curve $x = t^2 + 3t - 8, y = 2t^2 - 2t - 5$ at the point $(2, –1)$ is
- A
$\frac{22}{7}$
- ✓
$\frac{6}{7}$
- C
$\frac{7}{6}$
- D
$\frac{-6}{7}$
AnswerCorrect option: B. $\frac{6}{7}$
The given curve is $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$,
$\therefore\ \frac{\text{dx}}{\text{dt}}=2\text{t}+3$ and $\frac{\text{dy}}{\text{dt}}=4\text{y}-2$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}.\frac{\text{dt}}{\text{dx}}=\frac{4\text{t}-2}{2\text{t}+3}$
The given point is $(2, -1)$.
At $x = 2,$ we have:
$t^2 + 3t - 8 = 2$
$\Rightarrow\ t^2 + 3t - 10 = 0$
$\Rightarrow\ (t - 2)(t + 5) = 0$
$\Rightarrow\ t = 2$ or $t = -5$
At $y = - 1,$ we have:
$2t^2 - 2t - 5 = -1$
$\Rightarrow\ 2t^2 -2t - 4 =0$
$\Rightarrow\ 2(t^2 - t - 2) = 0$
$\Rightarrow\ (t - 2)(t + 1) = 0$
$\Rightarrow\ t = 2$ or $t = -1$
The common value of $t$ is $2$.
Hence, the slope of the tangent to the given curve at point $(2, - 1)$ is
$\frac{\text{dy}}{\text{dx}}\Big]_{t=2}=\frac{4(2)-2}{2(2)+3}=\frac{8-2}{4+3}=\frac{6}{7}.$
View full question & answer→Question 201 Mark
Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
AnswerGiven: f(x) = 3x + 17
$\therefore$ f'(x) = 3(1) + 0 = 3 >0 i.e., positive for all $\text{x}\in \text{R}$
Therefore, f(x) is strictly increasing on R.
View full question & answer→MCQ 211 Mark
Choose the correct answer in the Exercise : The line $y = mx + 1$ is a tangent to the curve $y^2 = 4x$ if the value of $m$ is:
- ✓
$1$
- B
$2$
- C
$3$
- D
$\frac{1}{2}$
AnswerThe equation of the tangent to the given curve is $y = mx + 1.$
Now, substituting $y = mx + 1$ in $y^2 = 4x$. we get:
$\Rightarrow\ (mx + 1)^2 = 4x$
$\Rightarrow\ m^2 x^2 + 1 + 2mx - 4x = 0$
$\Rightarrow\ m^2 x^2 + x(2m - 4) + 1 = 0 ....(i)$
Since a tangent touches the curve at one point, the roots of equation $(i)$ must be equal.
Therefore, we have:
Discriminant $= 0$
$(2m - 4)^2 -4(m^2)(1) = 0$
$\Rightarrow\ 4m^2 + 16 - 16m - 4m^2 = 0$
$\Rightarrow\ 16 - 16m = 0 $
$\Rightarrow\ m = 1$
Hence, the required value of $m$ is $1.$
View full question & answer→Question 221 Mark
Find the rate of change of the area of a circular disc with respect to its circumference when the radius is 3cm.
AnswerLet A be the area of the circular disc.
Then,
$\text{A}=\pi\text{r}^2$
Implies that $\frac{\text{dA}}{\text{dr}}=2\pi\text{r}$
Let C be the circumference of the circular disc.
Then,
$\text{C}=2\pi\text{r}$
Implies that $\frac{\text{dC}}{\text{dr}}=2\pi$
$\therefore\frac{\text{dA}}{\text{dC}}=\frac{\text{dA}/\text{dr}}{\text{dC}/\text{dr}}$
$\frac{\text{dA}}{\text{dC}}=\frac{2\pi\text{r}}{2\pi\text{r}}=\text{r}$
$\Big(\frac{\text{dA}}{\text{dC}}\Big)_\text{r=3}=3\text{cm}$
View full question & answer→Question 231 Mark
Show that the function given by $f(x) = e^{2x}$ is strictly increasing on R.
AnswerGiven: $f(x) = e^{2x}$
$\therefore\ \text{f} '(\text{x})=e^{2\text{x}}\frac{\text{d}}{\text{dx}}\ 2\text{x}=\text{e}^{2\text{x}}(2)=2\text{e}^{2\text{x}}>0$ i.e., positive for all $\text{x}\in \text{R}$
Therefore, f(x) is strictly increasing on R.
View full question & answer→Question 241 Mark
Find the maximum and minimum values, if any, of the function given by: $g(x) = x^3 + 1$
AnswerGiven: $g(x) = x^3 + 1$
$\text{As} \ \ \text{x}\rightarrow\infty\ \ \text{g}(\text{x)}\rightarrow\infty$
$\text{As}\ \ \text{x}\rightarrow-\infty\ \ \text{g}\text{(x)}\rightarrow-\infty$
Therefore, maximum value and minimum value of g(x) do not exist.
View full question & answer→MCQ 251 Mark
Choose the correct answer in the Exercise : A cylindrical tank of radius $10 m$ is being filled with wheat at the rate of $314$ cubic metre per hour. Then the depth of the wheat is increasing at the rate of :
- ✓
$1 m^3/ h$
- B
$0.1 m^3/ h$
- C
$1.1 m^3/ h$
- D
$0.5 m^3/ h$
AnswerCorrect option: A. $1 m^3/ h$
Let $r$ be the radius of the cylinder.
Then, volume $(V)$ of the cylinder is given by,
$\text{V}=\pi(\text{radius)}^2\times\text{height}$
$=\pi(10)^\text{h}\ \ (\text{radius}=10\text{m})$
$=100\pi\text{h}$
Differentiating with respect to time $t,$ we have :
$\frac{\text{dV}}{\text{dt}}=100\pi\frac{\text{dh}}{\text{dt}}$
The tank is being filled with wheat at the rate of $314$ cubic metres per hour.
$\therefore\ \frac{\text{dV}}{\text{dt}}=314\text{m}^3/\text{ h}$
Thus,we have :
$314=100\pi\frac{\text{dh}}{\text{dt}}$
$\Rightarrow\ \frac{\text{dh}}{\text{dt}}=\frac{314}{100(3.14)}=\frac{314}{314}=1$
Hence, the depth of wheat is increasing at the rate of $1 m/ h,$
View full question & answer→Question 261 Mark
For the function $y = x^2$, if $x = 10$ and $\triangle\text{x}=0.1$. Find $\triangle\text{y}.$
Answer$\text{y}=\text{x}^2$
$\triangle\text{x}=0.1$
$\text{x}=10$
$\frac{\text{dy}}{\text{dx}}=2\text{x}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}=10}=20$
$\Rightarrow\triangle\text{y}=\text{dy}=\frac{\text{dy}}{\text{dx}}\text{dx}=20\times0.1=2$
View full question & answer→Question 271 Mark
Prove that the logarithmic function is strictly increasing on $(0,\infty).$
AnswerGiven: $\text{f}\text{(x)} = \log \text{x}\ \Rightarrow\ \text{f}'\text{(x)} = \frac{1}{\text{x}} \text{for all x}\text{ in} (0,\ \infty).$
Therefore, f(x) is strictly increasing on $(0,\ \infty).$
View full question & answer→MCQ 281 Mark
Choose the correct answer in the Exercise : The normal at the point $(1, 1)$ on the curve $2y + x^2 = 3$ is :
- A
$x + y = 0$
- ✓
$x - y = 0$
- C
$x + y + 1 = 0$
- D
$x - y = 0$
AnswerCorrect option: B. $x - y = 0$
The equation of the given curve is $2y + x^2 = 3$.
Differentiating with respect to $x,$ we have:
$\frac{2\text{dy}}{\text{dx}}+2\text{x}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\text{x}$
$\therefore\ \frac{\text{dy}}{\text{dx}}\Big]_{(1.1)}=-1$
The slope of the normal to the given curve at point $(1,1)$ is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}\Big]_{(1.1)}}=1.$
Hence, the equation of the nomal to the given curve at $(1, 1)$ is given as :
$\Rightarrow\ y - 1 = 1(x - 1)$
$\Rightarrow\ y - 1 = x - 1$
$\Rightarrow\ x - y = 0$
View full question & answer→Question 291 Mark
Find the slope of the tangent to the curve $y = 3x^4 - 4x$ at $x = 4$.
AnswerThe given curve is $y = 3x^4- 4x$.
Then, the slope of the tangent to the given curve at x = 4 is given by,
$\frac{\text{dy}}{\text{dx}}\Big]_{\text{x}=4}=12\text{x}^3-4\big]_{\text{x}=4 }= 12(4)^3-4=12(64)-4=764$
View full question & answer→MCQ 301 Mark
Choose the correct answer in the Exercise : The points on the curve $9y^2 = x^3,$ where the normal to the curve makes equal intercepts with the axes are :
- ✓
$\Big(4,\pm\frac{8}{3}\Big)$
- B
$\Big(4,\frac{-8}{3}\Big)$
- C
$\Big(4,\pm\frac{3}{8}\Big)$
- D
$\Big(\pm4,\frac{3}{8}\Big)$
AnswerCorrect option: A. $\Big(4,\pm\frac{8}{3}\Big)$
The equation of the given curve is $9y^2 = x^3$.
Differentiating with respect to $x,$ we have :
$9(2\text{y})\frac{\text{dy}}{\text{dx}}=3\text{x}^2$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2}{6\text{y}}$
The slope of the normal to given curve at point $(x_1, k_1)$ is
$\frac{-1}{\frac{\text{dy}}{\text{dx}}\Big]_{(\text{x}_{1}, \text{y}_{1)}}}=-\frac{6\text{y}_1}{\text{x}_1^2}.$
$\therefore\ $The equation of the normal to the curve at $(x_1, y_1)$ is
$\text{y}-\text{y}_1=\frac{-6\text{y}_1}{\text{x}_1^2}(\text{x}-\text{x}_1).$
$\Rightarrow\ \text{x}_1^2\text{y}-\text{x}_1^2\text{y}_1=-6\text{xy}_1+6\text{x}_1\text{y}_1$
$\Rightarrow\ 6\text{xy}_1+\text{x}_1^2\text{y}=6\text{x}_1\text{y}_1+\text{x}_1^2\text{y}_1$
$\Rightarrow\ \frac{6\text{xy}_1}{6\text{x}_1\text{y}_1+\text{x}_1^2\text{y}_1}+\frac{\text{x}_1^2\text{y}}{6\text{x}_1\text{y}_1+\text{x}_1^2\text{y}_1}=1$
$\Rightarrow\ \frac{\text{x}}{\frac{\text{x}_1(6+\text{x}_1)}{6}}+\frac{\text{y}}{\frac{\text{y}_1(6+\text{x}_1)}{\text{x}_1}}=1$
It is given that the normal makes equal intercepts with the axes.
Therefore, we have :
$\therefore\ \frac{\text{x}_1(6+\text{x}_1)}{6}=\frac{\text{y}_1(6+\text{x}_1)}{\text{x}_1}$
$\Rightarrow\ \frac{\text{x}_1}{6}=\frac{\text{y}_1}{\text{x}_1}$
$\Rightarrow\ \text{x}_1^2=6\text{y}_1 \dots\text{(i)} $
Also, the point $(x_1, y_1)$ lies on the curve, so we have
$9\text{y}_1^2=\pi_1^3 \dots\text{(ii)}$
From $(i)$ and $(ii),$ we have:
$9\Big(\frac{\text{x}_1^2}{6}\Big)^2=\text{x}_1^3 $
$\Rightarrow\ \frac{\text{x}_1^4}{4}=\text{x}_1^3$
$ \Rightarrow\ \text{x}_1=4$
From $(ii),$ we have :
$9\text{y}_1^2=(4)^3=64$
$\Rightarrow\ \text{y}_1^2=\frac{64}{9}$
$\Rightarrow\ \text{y}_1=\pm\frac{8}{3}$
Hence, the required points are $\Big(4,\pm\frac{8}{3}\Big).$
View full question & answer→MCQ 311 Mark
Choose the correct answer in the Exercise : The normal to the curve $x^2 = 4y$ passing $(1,2)$ is :
- ✓
$x + y = 3$
- B
$x - y = 3$
- C
$x + y = 1$
- D
$x - y = 1$
AnswerCorrect option: A. $x + y = 3$
The equation of the given curve is $x^2 = 4y$.
Differentiating with respect to $x,$ we have :
$2\text{x}=4.\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2}$
The slope of the normal to given curve at point $(h, k)$ is given by,
$\frac{-1}{\frac{\text{dy}}{\text{dx}}\Big]_{(\text{h,k)}}}=-\frac{2}{\text{h}}$
$\therefore$ Equation of the normal at point $(h, k)$ is given as :
$\text{y}-\text{k}=\frac{-2}{\text{h}}(\text{x}-\text{h})$
Now, it is given that the normal passes through the point $(1,2).$
Therefore, we have :
$2-\text{k}=\frac{-2}{\text{h}}(1-\text{h})$ or ${k}=2+\frac{2}{\text{h}}(1-\text{h})\dots(\text{i})$
Since $(h, k)$ lies on the curve $x^2 = 4y$, we have $h^2 = 4k$.
$\Rightarrow\ \text{k}=\frac{\text{h}^2}{4}$
From equation $(i),$ we have :
$\frac{\text{h}^2}{4}=2+\frac{2}{\text{h}}(1-\text{h})$
$\Rightarrow\ \frac{\text{h}^2}{4}=2\text{h}+2-2\text{h}=2$
$\Rightarrow\ \text{h}^3=8$
$\Rightarrow\ \text{h}=2$
$\therefore\ \text{k}=\frac{\text{h}^2}{4}$
$\Rightarrow\ \text{k}=1$
Hence, the equation of the normal is given as :
$\Rightarrow\ \text{y}-1=\frac{-2}{2}(\text{x}-2)$
$\Rightarrow\ \text{y}-1=-(\text{x}-2)$
$\Rightarrow\ \text{x}+\text{y}=3$
View full question & answer→Question 321 Mark
Find the rate of change of the volume of a sphere with respect to its surface area when the radius is 2cm.
AnswerLet V be the volume of the sphere. Then,
$\text{V}=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\frac{\text{dv}}{\text{dr}}=4\pi\text{r}^2$
Let S be the total surface area of sphere. Then,
$\text{S}=4\pi\text{r}^2$
$\Rightarrow\frac{\text{dS}}{\text{dr}}=8\pi\text{r}$
$\therefore\frac{\text{dv}}{\text{dS}}=\frac{\text{dv}}{\text{dS}}/\frac{\text{dS}}{\text{dr}}$
$\Rightarrow\frac{\text{dV}}{\text{dS}}=\frac{4\pi\text{r}^2}{8\pi\text{r}}=\frac{\text{r}}{2}$
$\Rightarrow\Big(\frac{\text{dV}}{\text{dS}}\Big)_{\text{r}=2}=\frac{2}{2}$
$=1\text{cm}$
View full question & answer→Question 331 Mark
Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:
$f(x) = x^2$
AnswerGiven: $\text{f}\text{(x)} = \text{x}^2$
$\therefore\ \text{f}' \text{(x)} = 2\text{x} \text{ and } \ \text{f}"\text{(x)} =2$
Now $\text{f}'\text{(x)} = 0\ \Rightarrow\ \ \text{x} =0\ \ [\text{Turing point}]$
Again, when $\text{x} = 0, \text{f}"\text{(x)}= 2\ \ [\text{Positive}]$
Therefore, x = 0 is a point of local minima and local minimum value $= f(0) = (0)^2 = 0$
View full question & answer→