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$
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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
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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)$
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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.
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