Question 12 Marks
Show that the function $f(x) = x^3 – 3x^2 + 6x – 100$ is increasing on R.
Answer$f(x) = x^3 – 3x^2 + 6x – 100$
$f'(x) = 3x^2 – 6x + 6$
$= 3[x^2 – 2x + 2] = 3[(x – 1)^2 + 1]$
$\text{since f}'(\text{x)} > 0 \text{ }\forall \text{ x} \in$ R $\therefore$ f(x) is increasing on R.
View full question & answer→Question 22 Marks
The volume of a sphere is increasing at the rate of 8cm3/s. Find the rate at which its surface area is increasing when the radius of the sphere is 12cm.
Answer$\frac{\text{dv}}{\text{dt}}=8\text{cm}^3/\text{s},\text{Where V is the volume of sphere i.e.,}\text{ }\text{V}=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\frac{\text{dv}}{\text{dt}}=4\pi\text{r}^2\Rightarrow\frac{\text{dr}}{\text{dt}}=\frac{1}{4\pi\text{r}^2}.\frac{\text{dv}}{\text{dt}}$
$\text{S}=4\pi\text{r}^2\Rightarrow\frac{\text{dS}}{\text{dt}}=8\pi\text{r}\frac{\text{dr}}{\text{dt}}=8\pi\text{r}.\frac{1}{4\pi\text{r}^2}.8$
$=\frac{2\times8}{12}=\frac{\text{4}}{\text{3}}\text{cm}^2/\text{s}$
View full question & answer→Question 32 Marks
The length x, of a rectangle is decreasing at the rate of 5 cm/minute and the width y, is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of the area of the rectangle.
Answer$\text{given}\frac{\text{dx}}{\text{dt}}=-5\text{cm/}\text{m}.,\frac{\text{dy}}{\text{ dt}}=4\text{ cm/}\text{m}$
$\text{A}=\text{xy}\Rightarrow\frac{\text{dA}}{\text{dt}}=\text{x}\frac{\text{dy}}{\text{dt}}+\text{y}\frac{\text{dx}}{\text{dt}}$
= 8(4) + 6(–5) = 2
$\therefore\ $Area is increasing at the rate of $2 cm^2/minute$.
View full question & answer→Question 42 Marks
The volume of a cube is increasing at the rate of $9 cm^3/s$. How fast is its surface area increasing when the length of an edge is $10 cm$?
AnswerLet V be the volume of cube, then $\frac{\text{dV}}{\text{dt}} = 9 \text{ cm}^{3}/\text{s.}$
Surface area (S) of cube = $6x^2$, where x is the side.
$\text{then V} = \text{x}^{3} \Rightarrow \frac{\text{dV}}{\text{dt}} = \text{3x}^{2} \frac{\text{dx}}{\text{dt}} \Rightarrow \frac{\text{dx}}{\text{dt}} = \frac{1}{\text{3x}^{2}}. \frac{\text{dV}}{\text{dt}}$
$\text{S} = \text{6x}^{2} \Rightarrow \frac{\text{dS}}{\text{dt}} = \text{12x} \frac{\text{dx}}{\text{dt}} = 12\text{x}. \frac{1}{\text{3x}^{2 }} \frac{\text{dV}}{\text{dt}}$
$= 4.\frac{1}{10}.9 = 3.6 \text{ cm}^{2}/\text{s}$
View full question & answer→Question 52 Marks
Show that the function f given by $f(x) = \tan^{–1} (sin\ x + cos\ x)$ is decreasing for all $\text{x} \in \bigg(\frac{\pi}{4}, \frac{\pi}{2}\bigg).$
Answer$\text{f}'\text{(x)} = \frac{\cos \text{x} - \sin \text{x}}{1 + (\sin \text{x} + \cos \text{x})^{2}}$
$1 + ( \sin \text{x} + \cos \text{x})^{2} > 0, \forall \text{ x} \in \text{R}$
$\text{and }\frac{\pi}{4} < \text{x} < \frac{\pi}{2} \Rightarrow \cos \text{x} < \sin \text{x} \Rightarrow \cos \text{x} - \sin \text{x} < 0$
$\Rightarrow \text{f}' \text{(x)} < 0 \Rightarrow \text{f(x) is decreasing in} \bigg(\frac{\pi}{4}, \frac{\pi}{2}\bigg)$
View full question & answer→Question 62 Marks
The radius $r$ of a right circular cylinder is increasing uniformly at the rate of $0.3$ cm/s and its height $h$ is decreasing at the rate of rate of $0.3$ cm/s and its height $h$ is decreasing at the rate of $0.4$ cm/s. When $r = 3.5$ cm and $h = 7$ cm, find the rate of change of the curved surface area of the cylinder. $\text{[Use } \pi = \frac{22}{7}]$
AnswerCSA of cylinder,$\text{A}=2\pi\text{rh}$
$\Rightarrow \frac{\text{dA}}{\text{dt}}= 2\pi \bigg[\text{r}\frac{\text{dh}}{\text{dt}} + \text{h}\frac{\text{dr}}{\text{dt}} \bigg]$
$= 2\times\frac{22}{7} [3.5 \times ( -0.4) + 7(0.3) ] = 4.4 \text{ cm}^{2}/\text{min}$
$\therefore$ CSA is increasing at the rate of $4.4\ cm^2/min.$
View full question & answer→Question 72 Marks
The radius r of the base of a right circular cone is decreasing at the rate of $2 cm/min$. and its height h is increasing at the rate of $3 cm/min$. When $r = 3·5 cm$ and $h = 6 cm$, find the rate of change of the volume of the cone.$\text{[Use } \pi = \frac{22}{7}]$
Answer$\text{V}=\frac{1}{3}\pi\text{r}^2\text{h}$
$\Rightarrow\frac{\text{dV}}{\text{dt}}=\frac{1}{3}\pi\Big[\text{r}^2\frac{\text{dh}}{\text{dt}}+\text{h}.2\text{r}\frac{\text{dr}}{\text{dt}}\Big]$
$=\frac{1}{3}\times\frac{22}{7}\times[(3.5)^2\times3-2(3.5)(6)(2)]$
$= –49.5 cm^3/min$
$\therefore\ $Volume is decreasing at the rate of $49.5 cm^3/min$
View full question & answer→Question 82 Marks
The radius r of a right circular cylinder is decreasing at the rate of 3 cm/min. and its height h is increasing at the rate of 2 cm/min. When r = 7 cm and h = 2 cm, find the rate of change of the volume of cylinder. $\text{[Use } \pi = \frac{22}{7}]$
Answer$\text{V} = \pi \text{r}^{2} \text{h} \Rightarrow \frac{\text{dv}}{\text{dt}}= \pi \bigg(\text{r}^{2} \frac{\text{dh}}{\text{dt}} + \text{2r} \frac{\text{dr}}{\text{dt}} \text{h}\bigg)$
$= \frac{22}{7} [49 \times ( + 2) + 14 (2) (-3)] = 44 \text{ cm}^{3}/\text{min}$
$\therefore$ Volume is increasing at the rate of $44 cm^3/min$.
View full question & answer→Question 92 Marks
The volume of a sphere is increasing at the rate of 3 cubic centimeter per second. Find the rate of increase of its surface area, when the radius is 2 cm.
Answer$\text{V} = \frac{4}{3} \pi\text{r}^{3}$
$\Rightarrow \frac{\text{dv}}{\text{dt}} = 4\pi\text{r}^{2} \frac{\text{dr}}{\text{dt}} \Rightarrow \frac{\text{dr}}{\text{dt}} = \frac{3}{4\pi\text{r}^{2}}$
$\text{S} = 4\pi\text{r}^{2}$
$\Rightarrow \frac{\text{dS}}{\text{dt}} = 8\pi\text{r}.\frac{\text{dr}}{\text{dt}}$
$\Rightarrow \frac{\text{dS}}{\text{dt}}\bigg|_{\text{r = 2}} = 3\text{cm}^{2} \text{/s}$
View full question & answer→Question 102 Marks
For the curve $y = 5x – 2x^3$, if x increases at the rate of $2$ units/sec, then find the rate of change of the slope of the curve when $x = 3$.
Answer$\text{Given curve is y}=5\text{x}-2\text{x}^3$
$\Rightarrow\text{ }\frac{\text{dy}}{\text{dx}}=5-6\text{x}^2$
$\Rightarrow\text{ }\text{m}=5-6\text{x}^2$
$\frac{\text{dm}}{\text{dt}}=-12\text{x}\frac{\text{dx}}{\text{dt}}=-24\text{x}$
$\frac{\text{dm}}{\text{dt}}|_{\text{x}-3}=-72$
View full question & answer→Question 112 Marks
Show that the function $\text{f} (x) = 4x^{3} - 18x^{2} + 27x - 7$ is always increasing on IR.
Answer$f(x) = 4x^3 –18x^2 + 27x – 7$
$f’(x) = 12x^2 – 36x + 27$
$= 3(\text{2x - 3)}^{2} \geq 0 \text{ } \forall \text{ x} \text{}\in \text{R}$
$\Rightarrow$ f(x) is increasing on R.
View full question & answer→Question 122 Marks
The total cost $C(x)$ associated with the production of $x$ units of an item is given by $C(x) = 0.005x^3 – 0.02x^2 + 30x + 5000.$ Find the marginal cost when $3$ units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.
Answer$C(x) = 0.005x^3 - 0.02x^2 + 30x + 5000$
marginal cost (MC) $=\frac{\text{dC(x)}}{\text{dx}}$
$= (0.005)(3x^2) - 0.02(2x) + 30$
When $x = 3$
MC $= 0.005(3 \times 9) - 0.02(2 \times 3) + 30$
$= 30.015.$
View full question & answer→Question 132 Marks
Find the points on the curve $y = x^3 - 3x^2 - 4x$ at which the tangent lines are parallel to the line $4x + y - 3 = 0$.
AnswerLet $P(x_1, y_1)$ be requied point
Given curve is $y = x^3 - 3x^2 - 4x$
$\frac{\text{dy}}{\text{dx}}=3\text{x}^2-6\text{x}-4$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\text{x}_1,\text{y}_1}=3\text{x}^2_1-6\text{x}_1-4$
Since tangent of $(x_1, y_1)$ is parallel to line $\text{y}=4+\frac{\text{dy}}{\text{dx}}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(\text{x}_1,\text{y}_1)}=-4$
Now,
Slope of tangent = slope of line
$3x^2 - 6x - 4 = -4$
$3x^2 - 6x = 3x(x - 2)$
$\therefore x = D, x = 2$
Hence, point on the curves are $(0, 0), (2, -12)$.
View full question & answer→Question 142 Marks
Find the intervals in which the following function are strictly increasing or decreasing:
$6 - 9x - x^2$
AnswerGiven: $\text{f}\text{(x)} = 6-9\text{x} - \text{x}^2\ \Rightarrow\ \text{f}'\text{(x)} = -9 - 2\text{x}$
$\text{Now }-9-2\text{x} = 0\ \Rightarrow\ \text{x}=\frac{-9}{2}$
Therefore, we have three disjoint intervals $ \bigg(-\infty,\ \frac{-9}{2}\bigg)\text {and}\bigg(\frac{-9}{2},\ \infty\bigg).$
For interval $\bigg(-\infty,\ \frac{-9} {2}\bigg),\ \text{x}<\frac{-9}{2}$
Therefore, f is strictly increasing.
For interval $\bigg (\frac{-9}{2},\ \infty\bigg),\ \text{x}>\frac{-9}{2}$
Therefore, f is strictly decreasing.
View full question & answer→Question 152 Marks
Find the values of b for which the function $\text{f}(\text{x})=\sin\text{x}-\text{b}\text{x}+\text{c}$ is a decreasing function on R.
Answer$\text{f}(\text{x})=\sin\text{x}-\text{b}\text{x}+\text{c}$$\text{f}'(\text{x})=\cos\text{x}-\text{b}$
Given: f(x) is decreasing on R. $\Rightarrow\text{f}'(\text{x})<0\ \forall\ \text{x}\in\text{R}$ $\Rightarrow\cos\text{x}-\text{b}<0\ \forall\ \text{x}\in\text{R}$ $\Rightarrow\cos\text{x}<\text{b},\forall\ \text{x}\in\text{R}$ $\Rightarrow\text{b}\geq1$ $[\because\ -1\leq\cos\text{x}\leq1]$
View full question & answer→Question 162 Marks
If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating its surface area.
AnswerLet r be the radius of the sphere and $\Delta\text{r}$ be the error in measuring the radius.
$\therefore\text{ r = 9 m},\Delta\text{r}=0.03\text{ m}.$ Let S be the surface area of sphere.
$\therefore \text{S } = 4\pi \text{r}^2$
$\text{Now, ds} =\Big(\frac{\text{ds}}{\text{dr}}\Big) \Delta\text{r} = (8\pi\text{r}) \Delta\text{r}$
$= 8 \pi \times 9 \times 0.03 = 2.16 \pi\text{ m}^2 $
$\therefore$ approximate erros in calculating the surface area $=2.16\ \pi \text{ m}^2$
View full question & answer→Question 172 Marks
Find the rate of change of the area of a circle with respect to its radius r when:
- r = 3 cm.
- r = 4 cm.
AnswerLet x denote the area of the circle of variable radius r.
$\because$ Area of circle $ (\text{x})=\pi \text{r}^2$
$\therefore$ Rate of change of area w.r.t.r $=\frac{\text{dx}}{\text{dr}}=\pi(2\text{r})=2\pi \text{r}$
- When r = 3 cm, then $\frac{\text{dx}}{\text{dr}} = 2\pi(3)=6\pi \text{ sq. cm.}$
- When r = 4 cm, then $\frac{\text{dx}}{\text{dr}}=2\pi\ (4) =8\pi\ \text{sq. cm.}$
View full question & answer→Question 182 Marks
Write the set of values of a for which the function f(x) = ax + b is decreasing for all $\text{x}\in\text{R}.$
Answerf(x) = ax + b
f'(x) = a
For f(x) to be decreasing, we must have
f'(x) < 0
⇒ a < 0
$\Rightarrow\text{a}\in(-\infty,0)$
View full question & answer→Question 192 Marks
If a particle moves in a straight line such that the distance travelled in time t is given by $s = t^3 - 6t^2 + 9t + 8$. Find the initial velocity of the particle.
Answer$\text{S}=\text{t}^3-6\text{t}+9\text{t}+8$
$\frac{\text{dS}}{\text{dt}}=3\text{t}^2-12\text{t}+9$
Initial velocity t = 0
$\frac{\text{dS}}{\text{dt}}=3(0)^2-12(0)+9$
$\frac{\text{dS}}{\text{dt}}=9$ units/ units time.
View full question & answer→Question 202 Marks
Find the rate of change of the volume of a cone with respect to the radius of its base.
AnswerLet r be the radius, v be the volume of cone and h height.
$\text{V}=\frac{1}{3}\pi\text{r}^2\text{h}$
$\frac{\text{dv}}{\text{dr}}=\frac{1}{3}\pi\text{r}^2\text{h}.$
View full question & answer→Question 212 Marks
The sides of an equilateral triangle are increasing at the rate of 2cm/ sec. How far is the area increasing when the side is 10cms?
AnswerLet x be the side and A be the area of the equilateral triangle at any time t. Then,
$\text{A}=\frac{\sqrt{3}}{4}\text{x}^2$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\times\frac{\sqrt{3}}{4}\text{x}\frac{\text{dx}}{\text{dt}}$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=\frac{\sqrt{3}}{4}\times2\times10$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=10\sqrt{3}\text{cm}^2/\sec$
View full question & answer→Question 222 Marks
Find the equations of the tangent and normal to the given curves at the indicated points:
$\text{x} =\cos \text{t}, \text{y} = \sin \text{t at t}=\frac{\pi}{4} $
AnswerThe equation of the curve is $\text{x} = \cos \text{t, y} = \sin\text{t}.$
$\text{x}=\cos \text{t}\text{ and y} = \sin\text{t}$
$ \therefore \frac{\text{dx}}{\text{dt}}=-\sin\text{t},\frac{\text{dy}}{\text{dt}}=\cos \text{t}$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{\Big(\frac{\text{dy}}{\text{dt}}\Big)}{\Big(\frac{\text{dx}}{\text{dt}}\Big)}=\frac{\cos \text{t}}{-\sin \text{t}}=-\cot \text{t} $
$\frac{\text{dy}}{\text{dx}}\Big]_{\text{t}=\frac{\pi}{4}}=-\cot \text{t}=-1$
$\therefore$ The slope of the tangent at $\text{t}=\frac{\pi}{4}$
View full question & answer→Question 232 Marks
Write the value of $\Big(\frac{\text{dy}}{\text{dx}}\Big),$ if the normal to the curve y = f(x) at (x, y) is parallel to y-axis.
AnswerThe slope of the y-axis is $\infty.$Also, the tangent at a point (x, y) on the curve y = f(x) is parallel to the y-axis.
$\therefore$ Slope of the normal = Slope of the y-axis = $\infty$
$\Rightarrow\frac{\text{dy}}{\text{dx}}$ Slope of the tangent $=\frac{1}{\text{slope of the normal}}=\frac{-1}{\infty}=0$
View full question & answer→Question 242 Marks
Find the maximum and minimum values, if any, of the function given by:
$\text{h}\text{(x)} = \text{x} + 1, \text{x}\in (-1,\ 1)$
AnswerGiven: $\text{h}\text{(x)}=\text{x}+1, \text{x} \in (-1,\ 1)\ \dots\text{(i)}$
Since $-1\leq \text{x}\leq1$
Adding 1 to both sides, $-1 + 1 < \text{x} + 1 < 1 + 1 \Rightarrow\ \ 0<\text{h}\text{(x)}<2$
Therefore, neither minimum value not maximum value of h(x) exists.
View full question & answer→Question 252 Marks
Prove that the function $\text{f}(\text{x})=\log_{\text{e}}\text{x}$ is increasing on $(0,\infty).$
AnswerLet $\text{x}_1,\text{x}_2\in(0,\infty)$ such that $x_1 < x_2$. Then $x_1 < x_2$ Implies that $\log_{\text{e}}\text{x}_1<\log_{\text{e}}\text{x}_2$Implies that $f(x_1) < f(x_2)$
$\therefore x_1 < x_2$ Implies that $\text{f}(\text{x}_1)<\text{f}(\text{x}_2),\forall\ \text{x}_1,\text{x}_2\in(0,\infty)$Therefore, f(x) is increasing on $(0,\infty)$
View full question & answer→Question 262 Marks
Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Strictly increasing in $(\pi,2\pi)$
Answer$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
Here,
$\pi<\text{x}<2\pi$
$\Rightarrow\sin\text{x}<0$ $[\because$ sine function is negative in third and fourth quadrent$]$
$\Rightarrow-\sin\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in(\pi,2\pi)$
So, f(x) is strictly decreasing on $(\pi,2\pi).$
View full question & answer→Question 272 Marks
The radius of an air bubble is increasing at the rate of $\frac{1}{2}\text{cm/s.}$ At what rate is the volume of the bubble increasing when the radius is 1 cm?
AnswerLet x cm be the radius of the air bubble at time t.
According to question,$\frac{\text{dx}}{\text{dt}}$ is positive is $=\frac{1}{2}\text{ cm}/\sec\ \dots \text{(i)}$
Volume of air bubble $\text{(z)}=\frac {4\pi}{3}\text{x}^3 \ \Rightarrow \ \frac{\text{dz}}{\text{dt}}= \frac{4\pi}{3}\frac{\text{d}}{\text{dt}}\text{x}^3$$= \frac{4\pi}{3}.3\text{x}^{2} \frac{\text{dx}}{\text{dt}}=4{\pi}\text{x}^2\Big(\frac{1}{2}\Big)$
$\Rightarrow \ \frac{\text{dz}}{\text{dt}}=2{\pi}\text{x}^2= 2{\pi}(1)^2=2{\pi}$
Therefore, required rate of increase of volume of air bubble is $2 \pi \text{cm}^3/\sec.$
View full question & answer→Question 282 Marks
Find the surface area of a sphere when its volume is changing at the same rate as its radius.
AnswerLet r be the radius and V be the volume of the sphere at any time t then, $\text{V}=\frac{4}{3}\pi\text{r}^3$ Implies that $\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\Big(\frac{\text{dr}}{\text{dt}}\Big)$ Implies that $\frac{\text{dV}}{\text{dt}}=4\pi\text{r}^2\Big(\frac{\text{dV}}{\text{dt}}\Big)$ $\Big[\therefore\frac{\text{dV}}{\text{dt}}=\frac{\text{dr}}{\text{dt}}\Big]$ Implies that $4\pi\text{r}^2=1$Implies that that surface area of sphere = 1 square unit.
View full question & answer→Question 292 Marks
Find the intervals in which the following function are strictly increasing or decreasing:
$x^2 + 2x - 5$
AnswerGiven: $\text{f} \text{(x)} = \text{x}^2 + 2\text{x} - 5$$\Rightarrow\ \text{f'}\text{(x)} = 2\text{x} + 2 = 2\text{(x} + 1)\ \dots\text{(i)}$
$\text{Now } 2(\text{x} +1) = 0 \ \Rightarrow\ \text{x} = -1$
Therefore, we have two sub-intervals $(-\infty,\ -1)\text{ and }( -1,\ \infty).$
For interval $(- \infty,\ -1)$ taking x = -2 (say), from eq. (i), f'(x) = (-) < 0
Therefore, f is strictly decreasing.
For interval $(-1, \ \infty)$ taking x = 0 (say), from eq. (i). f'(x) = (+) > 0
Therefore, f is strictly increasing.
View full question & answer→Question 302 Marks
If the tangent to a curve at a point (x, y) is equally inclined to the co-ordinates axes then write the value of $\frac{\text{dy}}{\text{dx}}.$
AnswerSince. the tangent to a curve y = f(x) at (x, y) is equally inclined to the co-ordiante axes.
$\therefore\theta=45^\circ\text{or }\theta-45^\circ$
$\therefore\text{m}=\tan45^\circ\text{or }\text{m}=-\tan45^\circ=\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1\text{ or }\frac{\text{dy}}{\text{dx}}=-1$
View full question & answer→Question 312 Marks
Prove that the function given by $f(x) = x^3 - 3x^2 + 3x - 100$ is increasing in R.
AnswerGiven: $\text{f}\text{(x)} = \text{x}^{3} -3\text{x}^{2} + 3\text{x} -100 $
$ \Rightarrow \ \text{f}'\text{(x)} = 3\text{x}^{2} -6\text{x} + 3 = (\text{x}^2 -2\text{x}+1)$
$\Rightarrow\ \text{f}'\text{(x)} = (\text{x} -1)^{2}\geq 0 $ for all x in R.
Therefore, f(x) is increasing on R.
View full question & answer→Question 322 Marks
The total revenue in rupees received from the sale of x units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.
AnswerMarginal Revenue $(\text{MR})=\frac{\text{dR}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(13\text{x}^2+26\text{x}+15)=26\text{x}+26$
Now, when x = 7, MR = 26 × 7 + 26 = 208
Therefore, the required marginal revenue is Rs. 208.
View full question & answer→Question 332 Marks
Find the value(s) of a for which $f(x) = x^3 - ax$ is an increasing function on R.
Answer$f(x) = x^3 − ax$
$f'(x) = 3x^2 − a$
Given: f(x) is increasing on R.
$\Rightarrow\text{f}'(\text{x})\geq0\ \forall\ \text{x}\in\text{R}$
$\Rightarrow3\text{x}^2-\text{a}\geq0\ \forall\ \text{x}\in\text{R}$
$\Rightarrow\text{a}\leq3\text{x}^2\ \forall\ \text{x}\in\text{R}$
The least value of $3x^2$ is $0$.
$\therefore\ \text{a}\leq0$
View full question & answer→Question 342 Marks
Prove that the function $f(x) = x^3 - 6x^2 + 12x - 18$ is increasing on R.
Answer$f(x) = x^3 - 6x^2 + 12x - 18$
$f'(x) = 3x^2 - 12x + 12$
$= 3(x^2 - 4x + 4)$
$= 3(\text{x} - 2)^2\geq0,\forall\text{x}\in\text{R}$ $[3>0\ \&(\text{x}-2)^2\geq0]$
So, f(x) is increasing on R.
View full question & answer→Question 352 Marks
Show that $\text{f}(\text{x})=\tan\text{x}$ is an increasing function on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$
AnswerWe have,
$\text{f}(\text{x})=\tan\text{x}$
$\therefore\ \text{f}'(\text{x})=\sec^2\text{x}$
Now,
$\text{x}\in\Big(\frac{-\pi}{2}\frac{\pi}{2}\Big)$
$\Rightarrow\sec^2\text{x}>0$
$\Rightarrow\text{f}'(\text{x})>0$
Hence, f(x) is increasing function on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$
View full question & answer→Question 362 Marks
Prove that the function f given by $f(x) = x^2 - x + 1$ is neither strictly increasing nor strictly decreasing on $(-1, 1)$.
Answer$\text{f}\text{(x)} = \text{x}^{2} - \text{x} + 1\ \Rightarrow\ \text{f}'\text{(x)} = 2\text{x} - 1$
f(x) is strictly increasing if f'(x) > 0 $\Rightarrow\ 2\text{x} - 1 > 0 \ \Rightarrow\ \text{x}>\frac{1}{2}$
i.e., increasing on the interval $\Big(\frac{1}{2},\ 1\Big)$
f(x) is strictly decreasing if f'(x) < 0 $ \Rightarrow\ 2\text{x} - 1< 0 \ \Rightarrow \ \text{x} < \frac{1} {2}$
i.e., decreasing on the interval $\Big(-1,\ \frac{1}{2}\Big)$
hence, f(x) is neither strictly increasing nor decreasing on the interval (-1, 1 ).
View full question & answer→Question 372 Marks
The total revenue received from the sale of x units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.
AnswerSince the marginal revenue is the rate of change of total revenue with respect to its output,
Marginal Revenue (MR) $=\frac{\text{dR}(\text{x})}{\text{dx}}=\frac{\text{d}}{\text{dx}}(13\text{x}^2+26\text{x}+15)=26\text{x}+26$
When $x = 7$,
Marginal Revenue (MR) = 26(7) + 26 = 182 + 26 = Rs. 208
View full question & answer→Question 382 Marks
Write the equation of the tangent drawn to the curve $\text{y}=\sin\text{x}$ at the point (0, 0).
AnswerWe have,$\text{y}=\sin\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\cos\text{x}$
Slope at (0, 0) $=\text{m}=\Big[\frac{\text{dy}}{\text{dx}}\Big]_{\text{x}=0}=\cos0=1$
So, the equation of the tangent at (0, 0) is given by,
y = mx
putting m = 1, we get
The equation of the tangent is y = x.
View full question & answer→Question 392 Marks
The radius of a circle is increasing at the rate of 0.5cm/ sec. Find the rate of increase of its circumference.
AnswerLet r be radius and C be circumference of the circle.
Given, $\frac{\text{dr}}{\text{dt}}=0.5\text{cm}/\sec$
$\text{C}=2\pi\text{r}$
$\frac{\text{dC}}{\text{dt}}=2\pi\frac{\text{dr}}{\text{dt}}$
$=2\pi(0.5)$
$\frac{\text{dC}}{\text{dt}}=\pi\text{ cm}/\sec.$
View full question & answer→Question 402 Marks
Prove that the following function are increasing on R.
$f(x) = 4x^3 + 18x^2 + 27x - 27$
Answer$f(x) = 4x^3 + 18x^2 + 27x - 27$
$\Rightarrow f'(x) = 12x^2 + 36x + 27$
$\Rightarrow f'(x) = 3(4x^2 - 12x +9)$
$\Rightarrow\text{f}'(\text{x})=3(2\text{x}-3)^2>0,\forall\ \text{x}\in\text{R}$
So, f(x) increasing on R.
View full question & answer→Question 412 Marks
It is given that at $x = 1$, the function $x^4 - 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2]$. Find the value of a.
AnswerLet $\text{f}\text{(x)}=\text{x}^4-62\text{x}^2+a\text{x}+9$ $\Rightarrow\ \text{f}'\text{(x)}=4\text{x}^3-124\text{x}+\text{a}$
Since, f(x) attains its maximum value at x = 1 in the interval [0, 2], therefore f'(1) = 0
$\therefore\ \text{f}'(1)=4-124+\text{a}=0$ $\Rightarrow\ \text{a}-120=0\ \Rightarrow\ \text{a}=120$
View full question & answer→Question 422 Marks
Find the maximum and minimum values, if any, of the function given by:
f(x) = |sin 4x + 3|
AnswerGiven: $\text{f}\text{(x)} = |\sin 4\text{x} + 3|$
Since $-1\leq\sin 4\text{x}\leq 1\text{ for all }\text{x} \in \text{R}$
Adding 3 to all sides, $-1+3 \leq \sin 2\text{x}+5 \leq 1+3\ \Rightarrow \ 2\leq \text{f}\text{(x)}\leq4$
Therefore, minimum value of f(x) is 2 and maximum value is 4.
View full question & answer→Question 432 Marks
Find the slope of the tangent to curve $y = x^3 – x + 1$ at the point whose x-coordinate is $2$.
AnswerThe given curvw is $y = x^3 -x +1$.
$\therefore\frac{\text{dy}}{\text{dx}}=3\text{x}^{2} -1$
The slope of the tangent to a curve at $(x_0, y_0)$ is $\frac{\text{dy}}{\text{dx}}\Big]_{(\text{x}_0,\text{y}_0)}$
It is given that $x_0 = 2$.
Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by,
$\frac{\text{dy}}{\text{dx}}\bigg]_{\text{x}=2}=3\text{x}^2-1\big]_{\text{x}=2}=3(2)^2-1=12-1=11$
View full question & answer→Question 442 Marks
Find the approximate value of $(1.999)^5$.
AnswerLet $x = 2$
And $\triangle\text{x}=-0.001$ $[\because2-0.001=1.999]$
Let $y = x^5$
On differentiating both sides w.r.t. x, we get
$\frac{\text{dy}}{\text{dx}}=5\text{x}^4$
Now, $\triangle\text{y}=\frac{\text{dy}}{\text{dx}}\triangle\text{x}=5\text{x}^4\times\triangle\text{x}=5\times2^4\times[-0.001]=-80\times0.001=-0.080$
$\therefore\ (1.999)^5-\text{y}+\triangle\text{y}$
$=2^5+(-0.080)$
$=32-0.080=31.920$
View full question & answer→Question 452 Marks
If f(x) attains a local minimum at x = c, then write the values of f' (c) and f'' (c).
AnswerIf f(x) attains a local minimum at x = c, then the first order derivative of the function at the given point must be equal to zero, i.e.
f'(x) = 0 at x = c
⇒ f'(c) = 0
The second order derivative of the function at the given point must be greater than zero, i.e.
f''(c) > 0
View full question & answer→Question 462 Marks
Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Strictly decreasing in $(0,\pi).$
Answer$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
Here,
$0<\text{x}<\pi$
$\Rightarrow\sin\text{x}>0$ $[\because$ sine function is positive in first and second quadrent$]$
$\Rightarrow-\sin\text{x}<0$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in(0,\pi)$
So, f(x) is strictly decreasing on $(0,\pi).$
View full question & answer→Question 472 Marks
Find the maximum and minimum values, if any, of the function given by:
h(x) = Sin (2x) + 5
AnswerGiven: $\text{h}\text{(x)} = \sin (2\text{x)} + 5\ \dots\text{(i)}$
Since $-1\leq\sin2\text{x}\leq 1\text{ for all } \text{x} \in \text{R}$
Adding 5 to all sides, $-1 + 5 \leq \sin 2\text{x} + 5\leq 1+ 5\ \Rightarrow 4\leq \text{h}\text{(x)} \leq6$
Therefore, minimum value of h(x) is 4 and maximum value is 6.
View full question & answer→Question 482 Marks
If g (x) is a decreasing function on R and $\text{f}(\text{x})=\tan^{-1}[\text{g}(\text{x})].$ State whether f(x) is increasing or decreasing on R.
AnswerGiven: g(x) is decreasing on R.
$\Rightarrow\text{x}_1<\text{x}_2$
$\Rightarrow\text{g}(\text{x}_1)>\text{g}(\text{x}_2)$
Applying $\tan^{-1}$ on both sides we get,
$\Rightarrow\tan^{-1}\{\text{g}(\text{x}_1)\}>\tan^{-1}\{\text{g}(\text{x}_2)\}$
$\Rightarrow\text{f}(\text{x}_1)>\text{f}(\text{x}_2)$
View full question & answer→Question 492 Marks
State whether $\text{f}(\text{x})=\tan\text{x}-\text{x}$ is increasing or decreasing its domain.
Answer$\text{f}(\text{x})=\tan\text{x}-\text{x}$$\text{f}'(\text{x})=\sec^2\text{x}-1$
$\tan^{2}\text{x}\geq0,\forall\ \text{x}\in[0,2\pi]$
So, f(x) is increasing in its domain.
View full question & answer→Question 502 Marks
Find the equation of all lines having slope 2 which are tangents to the curve $\text{y} = \frac{1}{\text{x}-3}, \text{x} \neq 3.$
AnswerThe equation of the given curve is $\text{y} = \frac{1}{\text{x}-3},\text{x}\neq3.$
The slope of the tangent to the given curve at any point (x, y) is given by,
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{(\text{x}-3)^2}$
If the slope of the tangent is 2, then we have:
$\frac{-1}{(\text{x}-3)^2}=2$
$\Rightarrow\ 2(\text{x}-3)2=-1$
$\Rightarrow\ (\text{x}-3)^2 =\frac{-1}{2}$
This is not possible since the L.H.S. is positive while the R.H.S. is negative.
Hence, thers is no tangent to the given curve having slope 2.
View full question & answer→Question 512 Marks
Show that $f(x) = x^3 - 15x^2 + 75x - 50$ is an increasing function for all $\text{x}\in\text{R}.$
Answer$f(x) = x^3 - 15x^2 + 75x - 50$
$f'(x) = 3x^2 - 30x + 75$
$= 3(x^2- 10x + 25)$
$=3(\text{x}-5)^2>0,\forall\ \text{x}\in\text{R}$ $[\because$ Square of any function is always greater than zero$]$
So, f(x) is an increasing function for all $\text{x}\in\text{R}.$
View full question & answer→Question 522 Marks
Write the slop of the normal to the curve $\text{y}=\frac{1}{\text{x}}$ at the point $\Big(3,\frac{1}{3}\Big).$
Answer$\text{y}=\frac{1}{\text{x}}$
On differentiating both sides w.r.t. x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{-1}{\text{x}^2}$
Now,
Slope of the tangent $=\Big(\frac{\text{dy}}{\text{dx}}\Big)_{\big(3,\frac{1}{3}\big)}=\frac{-1}{9}$
Slope of the normal $=\frac{-1}{\text{slope of tangent}}=\frac{-1}{\big(\frac{-1}{9}\big)}=9$
View full question & answer→Question 532 Marks
Find the slope of the tangent to the curve $\text{y} = \frac{\text{x}-1}{\text{x}-2},\ \text{x} \neq \text{at}\ \text{x} = 10.$
AnswerThe given curve is y = $\frac{\text{x}-1}{\text{x}-2}.$
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}-2)(1)-(\text{x}-1)(1)}{(\text{x}-2)^2}$
$= \frac{\text{x}-2-\text{x}+1}{(\text{x}-2)^2}=\frac{-1}{(\text{x}-2)^2}$
Thus, the slope of the tangent at x = 10 is given by,
$\frac{\text{dy}}{\text{dx}}\Big]_{\text{x}=10}=\frac{-1}{(\text{x}-2)^2 }\Bigg]_{\text{x}=10}=\frac{-1}{(10-2)^2}=\frac{-1}{64}$
Hence, the slope of the tangent at x = 10 is $\frac{-1}{64}.$
View full question & answer→Question 542 Marks
If the relative error in measuring the radius of a circular plane is $\alpha,$ find the relative error in measuring its area.
AnswerLet x be the radius and y be the area of the circular plane.
We have $\frac{\triangle\text{x}}{\text{x}}=\alpha\ \text{and}\ \text{y}=\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\text{x}$
$\Rightarrow\frac{\triangle\text{y}}{\text{y}}=\frac{2\text{x}}{\text{y}}\text{dx}=\frac{2\text{x}}{\text{x}^2}\times\text{ax}=2\alpha$
Hence, the relative error in the area of the circular plane is $2\alpha$
View full question & answer→Question 552 Marks
Find the intervals in which the function f given by $f(x) = 2x^2 - 3x$ is (a) strictly increasing, (b) strictly decreasin.
AnswerGiven: $\text{f}\text{(x)} = 2x^2 - 3\text{x }\Rightarrow \text{f}'\text{(x)} = 4\text{x}-3\ \dots\text{(i)}$
$\text{Now}\ 4\text{x}-3=0\ \Rightarrow \ \text{x}=\frac{3}{4}$
Therefore, we have two intervals $\bigg(-\infty,\frac{3}{4}\bigg)\text{ and } \bigg(\frac{3}{4},\infty \bigg).$
- For interval $\bigg(\frac{3}{4},\infty\bigg),$ taking x = 1, (say), then from eq. (i), f'(x) > 0.
Therefore, f is strictly increasing in $\bigg(\frac{3}{4},\infty\bigg).$
- For interval $\bigg(-\infty,\frac{3}{4}\bigg),$ taking x = 0.5, (say), then from eq. (i), f'(x) < 0.
Therefore, f is strictly decreasing in $\bigg(-\infty,\frac{3}{4}\bigg).$ View full question & answer→Question 562 Marks
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of its circumference?
AnswerLet x cm be the radius of the circle at time t.Rate of increase of radius of circle = 0.7 cm/sec
$\Rightarrow\ \frac{\text{dy}}{\text{dt}}$ is positive and = 0.7 cm/sec
Let y be the circumference of the circle $\Rightarrow\ \text{y}=2\pi \text{x}$
$\therefore$ Rate of change of circumference of circle
$= \frac{\text{dy}}{\text{dt}} = 2{\pi} \frac{\text{d}}{\text{dt}}\text{x} = 2{\pi}(0.7) =1.4{\pi} \text{ cm}/\sec$
View full question & answer→Question 572 Marks
Find the intervals in which the following function are strictly increasing or decreasing:
$(x + 1)^3 (x - 3)^3$
AnswerGiven: $\text{f}\text{(x)} = (\text{x} + 1)^3(\text{x}-3)^3 $
$\Rightarrow\ \text{f}'\text{(x)} = (\text{x} + 1)^3.3(\text{x} - 3)^2 + (\text{x} -3)^3.3 (\text{x} + 1)^2 $
$\Rightarrow\ \text{f}'\text{(x)} = 3(\text{x}+1)^2(\text{x}- 3)^2(\text{x}+1+\text{x}-3)$
$\Rightarrow\ \text{f}'\text{(x)} =3(\text{x} + 1)^2 (\text{x}-3)^2(2\text{x}-2)$
$\Rightarrow\ \text{f}'\text{(x)} = 6(\text{x}+1)^2(\text{x}-3)^2(\text{x}-1)$
Here, factors $(\text{x} +1)^2 \text{ and } (\text{x}-3)^2$ are non-negative for all x.
Therefore, f(x) is strictly increasing if $\text{f}'\text{(x)} >0 \ \Rightarrow\ \text{x} - 1>0 \Rightarrow \text{x}> 1$
And f(x) is strictly decreasing if $\text{f}'\text{(x)} < 0\Rightarrow \ \text{x}-1< 0\ \Rightarrow \ \text{x} < 1$
Hence, f is strictly increasing in $(1,\ \infty)$ and f is strictly decreasing in $(-\infty,\ 1).$
View full question & answer→Question 582 Marks
Let f defined on [0, 1] be twice differentiable such that $|\text{f}'(\text{x})|\leq1$ for all $\text{x}\in[0,1].$ If f(0) = f(1), then show that $|\text{f}'(\text{x})|<1$ for all $\text{x}\in[0,1].$
AnswerIf a function is continuous and differentiable and f(0) = f(1) in given domain $\text{x}\in[0,1],$
Then by Rolle's theorem:
f'(x) = 0 for some $\text{x}\in[0,1]$
Given: $|\text{f}'(\text{x})|\leq1$
On integrating boty sides we get,
$|\text{f}'(\text{x})|\leq\text{x}$
Now, within interval $\text{x}\in[0,1]$
We get, $|\text{f}'(\text{x})|<1.$
View full question & answer→Question 592 Marks
Find an angle $\theta,0<\theta<\frac{\pi}{2},$ which increases twice as fast as its sine.
AnswerLet $\theta$ increases twice as fast as its sine
$\Rightarrow\ \theta=2\sin\theta$
$\Rightarrow\ \frac{\text{d}\theta}{\text{dt}}=2\cdot\cos\theta\cdot\frac{\text{d}\theta}{\text{dt}}$
$\Rightarrow\ 1=2\cos\theta$
$\Rightarrow\ \frac{1}{2}=\cos\theta$
$\Rightarrow\ \cos\theta=\cos\frac{\pi}{3}$
$\therefore\ \theta=\frac{\pi}{3}$
So, the required angle is $\frac{\pi}{3}$
View full question & answer→Question 602 Marks
If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.
AnswerLet r be the radius of the sphere and $\Delta\text{r}$ be the error in measuring the radius.
$\therefore\ \text{r = 7m. } \Delta\text{r }= 0.02\text{ m}$
Let V be volume of sphere
$\therefore\ \text{V }= \frac{4}{3}\pi\text{r}^3 $
$\text{Now, dV}\ =\Big(\frac{\text{dV}}{\text{dr}}\Big)\Delta\text{r}$
$= 4\pi\text{r}^2 \Delta\text{r } = 4\pi(7)^2 (0.02) $
$= 4\pi \times 49 \times 0.02$
$= 3.92 \ \pi\text{m}^3$
$\therefore$ approximate error in caluclating the volume $=3.92\ \pi \text{m}^3.$
View full question & answer→Question 612 Marks
If the tangent line at a point (x, y) on the curve y = f(x) is parallel to y-axis, find the value of $\frac{\text{dy}}{\text{dx}}.$
AnswerThe slope of the x-axis is $\infty.$
Also, the tangent at a point (x, y) on the curve y = f(x) is parallel to the x-axis.
$\therefore$ Slope of the tangent $\Big(\frac{\text{dy}}{\text{dx}}\Big)$ = Slope of the x-axis = $\infty.$
$\frac{\text{dx}}{\text{dy}}=\frac{1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)}=\frac{1}{\infty}=0$
View full question & answer→Question 622 Marks
The side of an equilataral triangle is inreasing at the rate of $\frac{1}{3}\text{cm}/\sec.$ Find the rate of increase of his primeter.
AnswerLet x be the side and P be the perimeter of the equilateral triangle at any time t. Then,
$\text{P}=3\text{x}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=3\frac{\text{dx}}{\text{dt}}$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=3\times\frac{1}{3}$$\Big[\therefore\frac{\text{dx}}{\text{dt}}=\frac{1}{3}\text{cm}/\sec\Big]$
$\Rightarrow\frac{\text{dP}}{\text{dt}}=1\text{cm}/\sec$
View full question & answer→Question 632 Marks
Find an angle $\theta$
Whose rate of increase twice is twice the rate of decrease of its cosine.
Answer$\frac{\text{d}\theta}{\text{dt}}=-2\frac{\text{d}}{\text{dt}}(\cos\theta)$
$\frac{\text{d}\theta}{\text{dt}}=-2(-\sin\theta)\frac{\text{d}\theta}{\text{dt}}$
$1=2\sin\theta$
$\sin\theta=\frac{1}{2}$
$\theta=\frac{\pi}{6}$
View full question & answer→Question 642 Marks
Prove that the following function are increasing on R.
$f(x) = 3x^5 + 40x^3 + 240x$
Answer$f(x) = 3x^5 + 40x^3 + 240x$
$f'(x) = 15x^4 + 120x^2 + 240$
$= 15(x^4 + 8x^2 +16)$
$=15(\text{x}^2+4)^2>0,\forall\ \text{x}\in\text{R}$ $[\because\ 15>0\text{ and }(\text{x}^2+4)^2>0]$
So, f(x) increasing on R.
View full question & answer→Question 652 Marks
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:
$g(x) = x^3 - 3x$
AnswerGiven: $\text{g}\text{(x)} = \text{x}^3 - 3\text{x}$
$\therefore \ \text{g}'\text{(x)} = 3\text{x}^2-3\ \text{ and g}"\text{(x)} = 6\text{x}$
$ \text{Now g}'\text{(x)}=0\ \Rightarrow 3\text{x}^2-3=0$
$\Rightarrow\ 3(\text{x}^2-1)=0\ \Rightarrow\ \ 3(\text{x}+1)(\text{x}-1)=0$
$\Rightarrow\ \text{x}=-1 \text{ or }\text{x}=1 \ \ [\text{Turning points}]$
Again, when $\text{x}=-1, \text{g}"\text{(x)}=6\text{x}=6(-1)=-6\ \ [\text{Negative}]$
$\therefore\ \text{x}=-1 $ is a point of local maxima and local maximum value
$\text{g}(-1) =(-1)^3-3(-1)=2$
And when $\text{x}=1 \ \text{ g}"\text{(x)}= 6\text{x}=6(1)=6\ \ [\text{Positive}]$
$\therefore\ \text{x}= 1$ is a point of local minima and local minimum value $\text{g}(1)=(1)^3-3(1)=-2$
View full question & answer→Question 662 Marks
Write sufficient condition for a point x = c to be an extreme point of the function f(x).
AnswerLet f(x) be a function.
f(x) will attain its extreme value at a point x = c if f'(c) = 0
The point x = c is known as point of local maxima if f''(x) < 0
View full question & answer→Question 672 Marks
The side of a square is increasing at the rate of 0.1cm/ sec. Find the rate of increase of its perimeter.
AnswerLet x be the side and P be the perimeter of the square at any time t.
Then, $\text{P}=4\text{x}$
Implies that $\frac{\text{dP}}{\text{dt}}=4\frac{\text{dx}}{\text{dt}}$
Implies that $\frac{\text{dP}}{\text{dt}}=4\times0.1$
$\Big[\therefore\frac{\text{dx}}{\text{dt}}\text{dt}=0.1\text{cm}/\sec\Big]$
Implies that $\frac{\text{dP}}{\text{dt}}=0.4\text{cm}/\sec$
View full question & answer→Question 682 Marks
If the tangent line at a point (x, y) on the curve y = f(x) is parallel to x-axis, then write the value of $\Big(\frac{\text{dy}}{\text{dx}}\Big).$
AnswerThe slope of the x-axis is 0.
Also, the tangent at a point (x, y) on the curve y = f(x) is parallel to the x-axis.
$\therefore$ Slope of the tangent $\Big(\frac{\text{dy}}{\text{dx}}\Big)$ = Slope of the x-axis = 0
View full question & answer→Question 692 Marks
Find the absolute maximum value and the absolute minimum value of the following function in the given intervals:
$\text{f}\text{(x)}=\text{x}^3, \text{x}\in[-2, 2]$
AnswerGiven: $\text{f}\text{(x)} =\text{x}^3, \text{x}\in [-2,\ 2]\ \Rightarrow\ \ \text{f}'\text{(x)}=3\text{x}^2$
$\text{Now }\text{f}\text{'(x)} =0\ \Rightarrow\ \ 3\text{x}^2=0\ \Rightarrow\ \ \text{x}=0\in[-2,\ 2]$
| At $x = 0,$ |
$f(0) = 0$ |
| At $x = -2,$ |
$f(-2) = (-2)^3 = -8$ |
| At $x = 2,$ |
$f(2) = (2)^3 = 8$ |
Therefore, absolute minimum value of $f(x)$ is $-8$ and absolute maximum value is $8.$ View full question & answer→Question 702 Marks
Find the maximum and minimum values, if any, of the function given by:
$f(x) = (2x - 1)^2 + 3$
AnswerGiven: $f(x) = (2x - 1)^2 + 3$
$\text{since}\ \ (2\text{x}-1)^2\geq\ 0\text{ for all }\text{x}\in \text{R}$
Adding 3 both sides, $(2\text{x} - 1)^2 + 3 \geq 0+3\ \Rightarrow \ \text{f}(\text{x}) \geq3$
Therefore, the minimum value of $\text{f}(\text{x}) \text{ is 3 when }2\text{x} -1 = 0, \text{i.e.,} \text{ x} =\frac{1}{2}$
This function does not have a maximum value.
View full question & answer→Question 712 Marks
Find the maximum and minimum values, if any, of the function given by:
f(x) = |x + 2| -1
AnswerGiven: $\text{f}\text{(x)} = |\text{x} + 2| -1\ \dots\text{(i)}$
Sicen $|\text{x} + 2|\geq0\text{ for all }\text{x}\in \text{R}$
Subtracting 1 from both sides, $|\text{x} + 2| -2\geq -1\ \Rightarrow\ \ \text{f}\text{(x)}\geq-1$
Therefore, minimum value of f(x) is -1 which is obtained when x + 2 = 0 i. e., x = -2
From eq. (i), maximum value of f(x) $\rightarrow \infty $ hence it does not exist.
View full question & answer→Question 722 Marks
Prove that the function $\text{f}(\text{x})=\cos\text{x}$ is:
Neither increasing nor decreasing in $(0,2\pi)$
Answer$\text{f}(\text{x})=\cos\text{x}$
$\text{f}'(\text{x})=-\sin\text{x}$
$\Rightarrow\text{f}'(\text{x})<0,\forall\ \text{x}\in(0,\pi)\ ....(1)$
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in(\pi,2\pi)\ ....(2)$
From eqs. (1) and (2), we get
f(x) is strictly decreasing on $(0,\pi)$ and is strictly increasing on $(\pi,2\pi).$
So, f(x) Neither increasing nor decreasing on $(0,2\pi).$
View full question & answer→Question 732 Marks
A balloon, which always remains spherical has a variables radius. Find the rate at which its volume is increasing with the radius when the later is $10 cm$.
AnswerSince, v $=\frac{4}{3}{\pi \text{x}^3}$
$ \therefore \ \frac{\text{dV}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{4}{3}{\pi}\text{x}^3 \Big) =\frac{4}{3}{\pi.}3\text{x}^2 = 4{\pi}\text{x}^2$
$ \Rightarrow \ \frac{\text{dV}}{\text{dx}}=4{\pi}(10)^2 =400{\pi}$
Therefore, the volume is increasing at the rate of $400p\ cm^3/\sec$.
View full question & answer→Question 742 Marks
Write the point where $\text{f}(\text{x})=\text{x}\log_{\text{e}}\text{x}$ attains minimum value.
AnswerGiven, $\text{f}(\text{x})=\text{x}\log_{\text{e}}\text{x}$
$\Rightarrow\text{f}'(\text{x})=\log_{\text{e}}\text{x}+1$
For a local maxima or a minima, we must have $\text{f}(\text{x})=0$
$\Rightarrow\log_{\text{e}}\text{x}+1=0$
$\Rightarrow\log_{\text{e}}\text{x}=-1$
$\Rightarrow\text{x}=\frac{1}{\text{e}}$
$\Rightarrow\text{f}\Big(\frac{1}{\text{e}}\Big)=\frac{1}{\text{e}}\ \log_{\text{e}}\Big(\frac{1}{\text{e}}\Big)=-\frac{1}{\text{e}}$
Now, $\text{f}''(\text{x})=\frac{1}{\text{x}}$
At, $\text{x}=\frac{1}{\text{e}}$
$\text{f}''\Big(\frac{1}{\text{e}}\Big)=\frac{1}{\frac{1}{\text{e}}}=\text{e}>0$
So, $\Big(\frac{1}{\text{e}},-\frac{1}{\text{e}}\Big)$ is a point of local minimum.
View full question & answer→Question 752 Marks
Show that $\text{f}(\text{x})=\log_{\text{a}}\text{x},0<\text{a}<1$ is a decreasing function for all x > 0.
Answer$\text{f}(\text{x})=\log_{\text{a}}\text{x}$
$=\frac{\log\text{x}}{\log\text{a}}$
$\text{f}'(\text{x})=\frac{1}{\text{x}\log\text{a}}$
Since, $0<\text{a}<1$ and $\text{f}'(\text{x})=\frac{1}{\text{x}\log\text{a}}<0.$
Hence, f(x) is decreasing function for all x > 0.
View full question & answer→Question 762 Marks
Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by $1\%$.
AnswerLet S be surface area of cude of side x metre.
$\therefore S = 6x^2$
$\text{Now, ds}=\Big(\frac{\text{dS}}{\text{dx}}\Big)\Delta\text{ x}=12 \times \Delta\text{x}$
$= 12 \times (-0.01\times)\ [\because 1\% \text{ of }\times \text{ is }0.01 \times]$
$= -0.12x^2$
$\therefore$ approximate change in volume = $0.12x^2m^2$ (numerically).
View full question & answer→Question 772 Marks
Show that $\text{f}(\text{x})=\text{x}+\cos\text{x}-\text{a}$ is an increasing function on R for all values of a.
AnswerWe have, $\text{f}(\text{x})=\text{x}+\cos\text{x}-\text{a}$$\therefore\ \text{f}'(\text{x})=1-\sin\text{x}=\frac{2\cos^2\text{x}}{2}$
Now, $\text{x}\in\text{R}$ $\Rightarrow\frac{\cos^2\text{x}}{2}>0$ $\Rightarrow\frac{2\cos^2\text{x}}{2}>0$ $\Rightarrow\text{f}'(\text{x})>0$ Hence, f(x) is an increasing function for $\text{x}\in\text{R}.$
View full question & answer→Question 782 Marks
Find the maximum and minimum values, if any, of the function given by:
$f(x) = 9x^2 + 12x + 2$
AnswerGiven: $\text{f}\text{(x)}=9\text{x}^2+12+2$
$\Rightarrow\ \text{f}\text{(x)}=9\Big(\text{x}^2+\frac{4\text{x}}{3}+\frac{2}{9}\Big)$
$\Rightarrow \ \text{f}\text{(x})=9\Bigg(\text{x}^2+\frac{4\text{x}}{3}+\Big(\frac{2}{3}\Big)^2-\Big(\frac{2}{3}\Big)^2+\frac{2}{9}\Bigg)$
$=9\Bigg[\Big(\text{x}+\frac{2}{3}\Big)^2-\frac{4}{9}+\frac{2}{9}\Bigg]$
$\Rightarrow \ \text{f}\text{(x)}=9\Big(\text{x}+\frac{2}{3}\Big)^2-2\ \dots\text{(i)}$
$\text{Since }\ 9\Big(\text{x}+\frac{2}{3}\Big)^2 \geq0\text{ for all } \text{x}\in \text{R}$
Subtracting 2 from both sides, $9\Big(\text{x}+\frac{2}{3}\Big) -2\geq0-2\ \Rightarrow\ \text{f}\text{(x)}\geq-2$
Therefore, minimum value of f(x) is -2 and is obtained when $\text{x}+\frac{2}{3}=0, \text{i. r.,}\text{ x}= \frac{-2}{3}$
And this function does not have a maximum value.
View full question & answer→Question 792 Marks
Find the slope of the tangent to the curve $y = x^3 – 3x + 2$ at the point whose x-coordinate is $3$.
AnswerThe given curve is $y = x^3 - 3x + 2$.
$\therefore\frac{\text{dy}}{\text{dx}} = 3\text{x}^2 -3$
The slope of the tangent to a curve at $(x_0, y_0)$ is $\frac{\text{dy}}{\text{dx}}\Big]_{(\text{x}_0,\ \text{y}_0)}$
Hence, the slope of the tangent at the point where the x-coordinate is 3 is given
$\frac{\text{dy}}{\text{dx}}\bigg]_{\text{x}=3}=3\text{x}^2-3\bigg]_{\text{x}=3}=3(3)^2-3=27-3=24$
View full question & answer→Question 802 Marks
Show that $\text{f}(\text{x})=\text{x}^2-\text{x}\sin\text{x}$ is an increasing function on $\Big(0,\frac{\pi}{2}\Big).$
AnswerWe have,
$\text{f}(\text{x})=\text{x}^2-\text{x}\sin\text{x}$
$\therefore\ \text{f}'(\text{x})=2\text{x}-\sin\text{x}-\text{x}\cos\text{x}$
Now,
$\text{x}\in\Big(0,\frac{\pi}{2}\Big)$
$\Rightarrow0\leq\sin\text{x}\leq1,0\leq\cos\text{x}\leq1$
$\Rightarrow2\text{x}-\sin\text{x}-\text{x}\cos\text{x}>0$
$\Rightarrow\text{f}'(\text{x})\geq0$
So, f(x) is strictly increasing function on $\Big(0,\frac{\pi}{2}\Big).$
View full question & answer→Question 812 Marks
Write necessary condition for a point x = c to be an extreme point of the function f(x).
AnswerWe know that at the extreme points of a function f(x), the first order derivative of the function is equal to zero, i.e.
f'(x) = 0 at x = c
⇒ f'(c) = 0
View full question & answer→Question 822 Marks
The total cost $C(x)$ in rupees associated with the production of $x$ units of an item given by $C(x) = 0.007x^3- 0.003x^2 + 15x + 4000$. Find the marginal cost when $17$ units are produced.
AnswerMarginal cost
$= \frac{\text{dC}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(0.007\text{x}^3 -0.003\text{x}^2+15\text{x}+4000)$
$=0.021\text{x}^2-0.006\text{x}+15$
Now, when $\text{x}=17,\ $$\text{MC} = 0.021(17)^2-0.006\times17+15-6.069-0.102+15=20.967$
Therefore, required Marginal cost is Rs. 20.97.
View full question & answer→Question 832 Marks
Find the approximate change in the volume $V$ of a cube of side $x$ metres caused by increasing the side by $1\%$.
Answer$V = x^3$
$\text{Now dV}\ =\Big(\frac{\text{dV}}{\text{dx}}\Big)\Delta \text{x = 3 x} ^2\Delta\text{x}$
$= 3x^2 (0.01 x) [\because 1\% \text{ of x is }0.01 \text{ x }]$
$= 0.03 x^3 m^3$
$\therefore$ approximate change in volume $= 0.03 x^3m^3$.
View full question & answer→Question 842 Marks
Find the intervals in which the following function are strictly increasing or decreasing:
$10 - 6x - 2x^2$
AnswerGiven: $\text{f}\text{(x)} = 10-6\text{x} - 2\text{x}^2$ $\Rightarrow\ \text{f}'\text{(x)}=-6 -4\text{x} = -2(3 + 2\text{x)}\ \dots\text{(i)}$ $\text{Now} -2(3+2\text{x}) = 0\ \Rightarrow\ \text{x}= \frac{-3}{2}$
Therefore, we have two sub-intervals $\bigg(-\infty,\ \frac{-3} {2}\bigg)\text{and}\bigg(\frac{-3} {2},\ \infty\bigg)$. For interval $\bigg(- \infty,\ \frac{-3}{2}\bigg)$ taking x = -2 (say), from eq. (i), f'(x) = (-) (-) (+) > 0
Therefore, f is strictly increasing. For interval $ \bigg(\frac{-3}{2},\ \infty \bigg)$ taking x = -1 (say), from eq. (i), f'(x) = (-) (+) = (-) < 0Therefore, f is strictly decreasing.
View full question & answer→Question 852 Marks
Show that $\text{f}(\text{x})=\sin\text{x}$ is an increasing function on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$
Answer$\text{f}(\text{x})=\sin\text{x}$
$\text{f}'(\text{x})=\cos\text{x}>0\ \forall\ \text{x}\in\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)$
$[\because$ Cos function is positive in first and fourth quadrant$]$
So, f(x) is increasing on $\Big(\frac{-\pi}{2},\frac{\pi}{2}\Big).$
View full question & answer→Question 862 Marks
Find the intervals in which the following function are strictly increasing or decreasing:
$-2x^3 - 9x^2 - 12x + 1$
AnswerGiven:$\text{f}\text{(x)} = -2\text{x}^3 -9\text{x}^2 - 12\text{x} + 1$ $\Rightarrow\ \text{f}'\text{(x)} = -6\text{x}^2 - 18\text{x} - 12$
$\Rightarrow\ \text{f}'\text{(x)} = -6(\text{x}^2 + 3\text{x} + 2) = -6(\text{x} + 1)(\text{x} + 2)\ \dots\text{(i)}$
$\text{Now } -6(\text{x} + 1)(\text{x} + 2) = 0\ \Rightarrow\ \text{x} = -1\text{ or } \text{x} = -2$
Therefore, we have three disjoint intervals $(-\infty,\ -2), (-2,\ -1)\text{ and }(-1, \ \infty).$
For interval $(-\infty,\ -2),$ from eq. $(i), f'(x), = (-) (-) (-) = (-) < 0$
Therefore, $f$ is strictly decreasing.
For interval $(-2, -1)$ from eq. $(i), f'(x) = (-) (-) (+) = (+) > 0$
Therefore, $f$ is strictly increasing
For interval $(-1, \ \infty) $ from eq. $(i), f'(x) = (-) (+) (+) = (-) < 0$
Therefore, $f$ is strictly decreasing.
View full question & answer→Question 872 Marks
A stone is dropped into a quiet lake and waves move in circles at a speed of 4cm/ sec. At the instant when the radius of the circular wave is 10cm, how fast is the enclosed area increasing?
AnswerLet r be the radius and A be the area of the circle at any times t. Then,
$\text{A}=\pi\text{r}^2$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\text{r}\frac{\text{dr}}{\text{dt}}$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=2\pi\times4\times10$ $\big[\therefore\text{r}=4\text{cm}\text{ and }\frac{\text{dr}}{\text{dt}}=10\text{cm}/\sec\big]$
$\Rightarrow\frac{\text{dA}}{\text{dt}}=80\pi\text{ cm}^2/\sec$
View full question & answer→Question 882 Marks
Find the approximate value of $(1.999)^5.$
AnswerLet $x = 2$
And $\triangle\text{x}=-0.001$ $[\because2-0.001=1.999]$
Let $y = x^5$
On differentiating both sides w.r.t. x$,$ we get
$\frac{\text{dy}}{\text{dx}}=5\text{x}^4$
Now$, \triangle\text{y}=\frac{\text{dy}}{\text{dx}}\triangle\text{x}=5\text{x}^4\times\triangle\text{x}=5\times2^4\times[-0.001]=-80\times0.001=-0.080$
$\therefore\ (1.999)^5-\text{y}+\triangle\text{y}$
$=2^5+(-0.080)$
$=32-0.080=31.920$
View full question & answer→Question 892 Marks
Find an angle $\theta,0<\theta<\frac{\pi}{2},$ which increases twice as fast as its sine.
AnswerLet $\theta$ increases twice as fast as its sine
$\Rightarrow\ \theta=2\sin\theta$
$\Rightarrow\ \frac{\text{d}\theta}{\text{dt}}=2\cdot\cos\theta\cdot\frac{\text{d}\theta}{\text{dt}}$
$\Rightarrow\ 1=2\cos\theta$
$\Rightarrow\ \frac{1}{2}=\cos\theta$
$\Rightarrow\ \cos\theta=\cos\frac{\pi}{3}$
$\therefore\ \theta=\frac{\pi}{3}$
So, the required angle is $\frac{\pi}{3}$
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