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Question: $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}$ Solution: Let $I=\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x$ $\begin{aligned} I =\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \cdot \sin x d x \\ =\int_{0}^{\frac{\pi}{2}}\left(1-\cos ^{2} x\right) \sin x d x \\ =\int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \cdot \sin x d x \\ =[-\cos x]_{0}^{\frac{\pi}{2}}+\left[\frac{\cos ^{3} x}{3}\right]_{0}^{\frac{\pi}{2}} \\ =1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3} \end{...
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Question: $\int_{-1}^{1} x^{17} \cos ^{4} x d x=0$ Solution: Let $I=\int_{-1}^{1} x^{17} \cos ^{4} x d x$ Also, let $f(x)=x^{17} \cos ^{4} x$ $\Rightarrow f(-x)=(-x)^{17} \cos ^{4}(-x)=-x^{17} \cos ^{4} x=-f(x)$ Therefore, $f(x)$ is an odd function. It is known that if $f(x)$ is an odd function, then $\int_{-a}^{a} f(x) d x=0$ $\therefore I=\int_{-1}^{1} x^{17} \cos ^{4} x d x=0$ Hence, the given result is proved....
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Question: $\int_{0}^{1} x e^{x} d x=1$ Solution: Let $I=\int_{0}^{1} x e^{x} d x$ Integrating by parts, we obtain $\begin{aligned} I =x \int_{0}^{1} e^{x} d x-\int_{0}^{1}\left\{\left(\frac{d}{d x}(x)\right) \int e^{x} d x\right\} d x \\ =\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} e^{x} d x \\ =\left[x e^{x}\right]_{0}^{1}-\left[e^{x}\right]_{0}^{1} \\ =e-e+1 \\ =1 \end{aligned}$ Hence, the given result is proved....
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Question: $\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}$ Solution: Let $I=\int_{1}^{3} \frac{d x}{x^{2}(x+1)}$ Also, let $\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$ $\Rightarrow 1=A x(x+1)+B(x+1)+C\left(x^{2}\right)$ $\Rightarrow 1=A x^{2}+A x+B x+B+C x^{2}$ Equating the coefficients of $x^{2}, x$, and constant term, we obtain $A+C=0$ $A+B=0$ $B=1$ On solving these equations, we obtain $A=-1, C=1$, and $B=1$ $\therefore \frac{1}{x^{2}(x+1)}=\frac{-1}{x}+\...
Read More →A triangle has sides 5 cm, 12 cm and 13 cm.
Question: A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm. Solution: Since $B D \perp A C$ we obtained two right angled triangles, $\triangle A B D$ and $\triangle B D C$. In $\triangle A B C$ and $\triangle A B D$ $\angle A=\angle A \quad$ (Common angle) $\angle B=\angle D$ So, by AA-criterion $\triangle A B C \sim \triangle A D B$ $\therefore \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\...
Read More →The radius of the circle whose arc of length 15 π cm
Question: The radius of the circle whose arc of length 15 cm makes an angle of3434radian at the centre is (a) 10 cm (b) 20 cm (c) $11 \frac{1}{4} \mathrm{~cm}$ (d) $22 \frac{1}{2} \mathrm{~cm}$ Solution: (b) 20 cm $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Rightarrow \frac{3 \pi}{4}=\frac{15 \pi}{\text { Radius }}$ $\Rightarrow$ Radius $=\frac{60}{3}=20 \mathrm{~cm}$...
Read More →The radius of the circle whose arc of length 15 π cm
Question: The radius of the circle whose arc of length 15 cm makes an angle of3434radian at the centre is (a) 10 cm (b) 20 cm (c) $11 \frac{1}{4} \mathrm{~cm}$ (d) $22 \frac{1}{2} \mathrm{~cm}$ Solution: (b) 20 cm $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Rightarrow \frac{3 \pi}{4}=\frac{15 \pi}{\text { Radius }}$ $\Rightarrow$ Radius $=\frac{60}{3}=20 \mathrm{~cm}$...
Read More →A circular wire of radius 7 cm is cut and bent again into an arc of a circle of radius 12 cm.
Question: A circular wire of radius 7 cm is cut and bent again into an arc of a circle of radius 12 cm. The angle subtended by the arc at the centre is (a) 50 (b) 210 (c) 100 (d) 60 (e) 195 Solution: (b) 210 Length of the arc of radius = Circumference of the circle of radius $7 \mathrm{~cm}=2 \pi r=14 \pi$ Now, Angle subtended by the arc $=\frac{\text { Arc }}{\text { Radius }}=\frac{14 \pi}{12}=\left(\frac{14 \pi}{12} \times \frac{180}{\pi}\right)^{\circ}=210^{\circ}$...
Read More →A circular wire of radius 7 cm is cut and bent again into an arc of a circle of radius 12 cm.
Question: A circular wire of radius 7 cm is cut and bent again into an arc of a circle of radius 12 cm. The angle subtended by the arc at the centre is (a) 50 (b) 210 (c) 100 (d) 60 (e) 195 Solution: (b) 210 Length of the arc of radius = Circumference of the circle of radius $7 \mathrm{~cm}=2 \pi r=14 \pi$ Now, Angle subtended by the arc $=\frac{\text { Arc }}{\text { Radius }}=\frac{14 \pi}{12}=\left(\frac{14 \pi}{12} \times \frac{180}{\pi}\right)^{\circ}=210^{\circ}$...
Read More →Using Pythagoras theorem determine the length of AD in terms of b and c shown in the given figure.
Question: Using Pythagoras theorem determine the length of AD in terms of b and c shown in the given figure. Solution: In $\triangle A B C$ and $\triangle D B A$, $\angle A=\angle D \quad\left(90^{\circ}\right.$ each $)$ $\angle B=\angle B \quad$ (Common) Therefore, by AA-criterion for similarity, we have $\triangle A B C \sim \triangle D B A$. $\therefore \frac{A B}{B D}=\frac{B C}{B A}=\frac{A C}{A D}$ Now we will substitute the values of AC and AB $\therefore \frac{c}{B D}=\frac{B C}{c}=\frac...
Read More →If OP makes 4 revolutions in one second,
Question: IfOPmakes 4 revolutions in one second, the angular velocity in radians per second is (a) (b) 2 (c) 4 (d) 8 Solution: (d) 8 Angular velocity $=\frac{\text { Distance }}{\text { Time }}$ $=\frac{4 \text { revolutions }}{1 \text { second }}$ $=\frac{4 \times 2 \pi}{1} \quad(\because 1$ revolution $=2 \pi$ radians $)$ $=8 \pi$ radians per second...
Read More →If the arcs of the same length in two circles subtend angles 65° and 110° at the centre,
Question: If the arcs of the same length in two circles subtend angles 65 and 110 at the centre, than the ratio of the radii of the circles is (a) 22 : 13 (b) 11 : 13 (c) 22 : 15 (d) 21 : 13 Solution: (a) 22 : 13 Let the angles subtended at the centres by the arcs and radii of the first and second circles be $\theta_{1}$ and $r_{1}$ and $\theta_{2}$ and $r_{2}$, respectively. We have: $\theta_{1}=65^{\circ}=\left(65 \times \frac{\pi}{180}\right)$ radian $\theta_{2}=65^{\circ}=\left(110 \times \f...
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Question: Solution: $I_{1}=\int_{1}^{4}|x-1| d x$ $(x-1) \geq 0$ for $1 \leq x \leq 4$ $\therefore I_{1}=\int_{1}^{4}(x-1) d x$ $\Rightarrow I_{1}=\left[\frac{x^{2}}{x}-x\right]_{1}^{4}$ $\Rightarrow I_{1}=\left[8-4-\frac{1}{2}+1\right]=\frac{9}{2}$ ...(2) $I_{2}=\int_{1}^{4}|x-2| d x$ $x-2 \geq 0$ for $2 \leq x \leq 4$ and $x-2 \leq 0$ for $1 \leq x \leq 2$ $\therefore I_{2}=\int_{1}^{2}(2-x) d x+\int_{2}^{4}(x-2) d x$ $\Rightarrow I_{2}=\left[2 x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2...
Read More →Two poles of height 9 m and 14 m stand on a plane ground.
Question: Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. Solution: Let us draw the diagram from the given information. As we are given that distance between their feet is 12 m $\therefore D C=12$. Now we get a right angled triangle DCE. Let us applying the Pythagoras theorem we get, $D E^{2}=D C^{2}+E C^{2}$ Substituting the values we get, $D E^{2}=12^{2}+5^{2}$ $D E^{2}=144+25$ $\therefore D E^{2}=169$ ...
Read More →At 3:40, the hour and minute hands of a clock are inclined at
Question: At 3:40, the hour and minute hands of a clock are inclined at (a) $\frac{2 \pi^{c}}{3}$ (b) $\frac{7 \pi^{\mathrm{c}}}{12}$ (C) $\frac{13 \pi_{\mathrm{c}}}{18}$ (d) $\frac{13 \pi_{c}}{4}$ Solution: (C) $\frac{13 \pi_{\mathrm{c}}}{18}$ We know that the hour hand of a clock completes one rotation in 12 hours. Angle traced by the hour hand in 12 hours = 360 Now, Angle traced by the hour hand in 3 hours 40 minutes, i.e., $\frac{11}{3}=\left(\frac{360}{12} \times \frac{11}{3}\right)^{\circ}...
Read More →The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground.
Question: The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach? Solution: The given information can be represented as follows. Here, A is the position of the window and AC is the ladder. Also, DE is the same ladder when it is shifted. C and E are the original and final position of the foot of the ladder. Now, applying Pythagoras theorem inΔA...
Read More →The angle between the minute and hour hands of a clock at 8:30 is
Question: The angle between the minute and hour hands of a clock at 8:30 is (a) 80 (b) 75 (c) 60 (d) 105 Solution: (b) 75 We know that the hour hand of a clock completes one rotation in 12 hours. Angle traced by the hour hand in 12 hours = 360 Now, Angle traced by the hour hand in 8 hours 30 minutes, i.e., $\frac{17}{2}=\left(\frac{360}{12} \times \frac{17}{2}\right)^{\circ}=255^{\circ}$ We also know that the minute hand of a clock completes one rotation in 60 minutes. Angle traced by the minute...
Read More →Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.
Question: Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm. Solution: Given that, Radius of the circle (OA) = 6 cm Distance (OC) = 4 cm In ΔOCA, by Pythagoras theorem $\mathrm{AC}^{2}+\mathrm{OC}^{2}=\mathrm{OA}^{2}$ $\Rightarrow \mathrm{AC}^{2}+4^{2}=6^{2}$ $\Rightarrow \mathrm{AC}^{2}=36-16$ $\Rightarrow \mathrm{AC}^{2}=20$ $\Rightarrow \mathrm{AC}=\sqrt{20}$ ⟹ AC = 4.47 cm We know that the perpendicular distance from centre to chord bisects ...
Read More →In an isosceles triangle ABC, AB = AC = 25 cm,
Question: In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm, Calculate the altitude from A on BC. Solution: We know that altitude that is a perpendicular drawn on the unequal side of the isosceles triangle bisects that side. Therefore, BD = DC = 7 cm. Let us use the Pythagoras theorem in right angled triangle ADB we get, $A B^{2}=A D^{2}+B D^{2}$ Substituting the values we get, $25^{2}=A D^{2}+7^{2}$ $\therefore 625=A D^{2}+49$ Subtracting 49 from both the sides we get, $625-49=A D^{2}$ ...
Read More →Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.
Question: Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm. Solution: Given that, Distance (OC) = 5 cm Radius of the circle (OA) = 10 cm In ΔOCA, by Pythagoras theorem $O C^{2}+A C^{2}=O A^{2}$ $\Rightarrow 5^{2}+A C^{2}=10^{2}$ $\Rightarrow 25+A C^{2}=100$ $\Rightarrow A C^{2}=100-25$ $\Rightarrow A C^{2}=75$ $\Rightarrow \mathrm{AC}=\sqrt{75}$ We know that, the perpendicular from the centre to chord bisects the chord Therefore, AC = BC = 8.6...
Read More →If the angles of a triangle are in A.P.,
Question: If the angles of a triangle are in A.P., then the measures of one of the angles in radians is (a) $\frac{\pi}{6}$ (b) $\frac{\pi}{3}$ (c) $\frac{\pi}{2}$ (d) $\frac{2 \pi}{3}$ Solution: (b) $\frac{\pi}{3}$ Let the angles of the triangle be $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$. Thus, we have: $a-d+a+a+d=180$ $\Rightarrow 3 a=180$ $\Rightarrow a=60$ Hence, the angles are $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$, i.e., $(60-d)^{\circ}, 60^{\circ}$ and $(60+d)^{\circ}$. $...
Read More →Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m,
Question: Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. Solution: Let us draw the diagram from the given information. Let us draw a perpendicular from B on CD which meets CD at P. It is clear that BP = 12 m because it is given that distance between feet of the two poles is 12 m. After drawing the perpendicular we get a rectangle BACP such that AB = PC and BP = AC. Because of this construction we also o...
Read More →If the angles of a triangle are in A.P.,
Question: If the angles of a triangle are in A.P., then the measures of one of the angles in radians is (a) $\frac{\pi}{6}$ (b) $\frac{\pi}{3}$ (c) $\frac{\pi}{2}$ (d) $\frac{2 \pi}{3}$ Solution: (b) $\frac{\pi}{3}$ Let the angles of the triangle be $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$. Thus, we have: $a-d+a+a+d=180$ $\Rightarrow 3 a=180$ $\Rightarrow a=60$ Hence, the angles are $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$, i.e., $(60-d)^{\circ}, 60^{\circ}$ and $(60+d)^{\circ}$. $...
Read More →The radius of a circle is 8 cm and the length of one of its chords is 12 cm.
Question: The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre. Solution: Given that, Radius of circle (OA) = 8 cm Chord (AB) = 12 cm DrawOCAB We know that The perpendicular from centre to chord bisects the chord $\mathrm{AC}^{2}+\mathrm{OC}^{2}=\mathrm{OA}^{2}$ $\Rightarrow 6^{2}+O C^{2}=8^{2}$ $\Rightarrow 36+O C^{2}=64$ $\Rightarrow O C^{2}=64-36$ $\Rightarrow O C^{2}=28$ $\Rightarrow O C=\sqrt{28}$ $\Rightarrow O C=5.291 ...
Read More →A ladder 17 m long reaches a window of a building 15 m above the ground.
Question: A ladder $17 \mathrm{~m}$ long reaches a window of a building $15 \mathrm{~m}$ above the ground. Find the distance of the foot of the ladder from the building. Solution: Let us draw the diagram from the given information we get a right angled triangle ABC as shown below, Let the window be at the point A. We know that angle formed between the building and ground is always 90. Given: AB = 15 m and CA = 17 m Now we will use Pythagoras theorem to find $l(B C)$. $\therefore A C^{2}=A B^{2}+...
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