Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Question: Form the differential equation of the family of parabolas having vertex at origin and axis along positivey-axis. Solution: The equation of the parabola having the vertex at origin and the axis along the positivey-axis is: $x^{2}=4 a y$ ...(1) Differentiating equation (1) with respect tox, we get: $2 x=4 a y^{\prime}$ ...(2) Dividing equation (2) by equation (1), we get: $\frac{2 x}{x^{2}}=\frac{4 a y^{\prime}}{4 a y}$ $\Rightarrow \frac{2}{x}=\frac{y^{\prime}}{y}$ $\Rightarrow x y^{\pr...
Read More →If
Question: If $\frac{\pi}{2}x\frac{3 \pi}{2}$, then $\sqrt{\frac{1-\sin x}{1+\sin x}}$ is equal to (a) secx tanx (b) secx+ tanx (c) tanx secx (d) none of these Solution: (c) tanx secx $\sqrt{\frac{1-\sin x}{1+\sin x}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}$ $=\frac{(1-\sin x)}{-\cos x} \quad\left[\operatorname{as}, \frac{\pi}{2}x\frac{3 \pi}{2}\right.$, so $\cos \theta$ will be negative $]$ $=-(\sec x-\tan x)$ $=-\sec x+\tan x$...
Read More →If
Question: If $\frac{\pi}{2}x\frac{3 \pi}{2}$, then $\sqrt{\frac{1-\sin x}{1+\sin x}}$ is equal to (a) secx tanx (b) secx+ tanx (c) tanx secx (d) none of these Solution: (c) tanx secx $\sqrt{\frac{1-\sin x}{1+\sin x}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{1-\sin ^{2} x}}$ $=\sqrt{\frac{(1-\sin x)^{2}}{\cos ^{2} x}}$ $=\frac{(1-\sin x)}{-\cos x} \quad\left[\operatorname{as}, \frac{\pi}{2}x\frac{3 \pi}{2}\right.$, so $\cos \theta$ will be negative $]$ $=-(\sec x-\tan x)$ $=-\sec x+\tan x$...
Read More →The total surface of a hollow metal cylinder, open at both ends of an external radius of 8cm and height 10 cm
Question: The total surface of a hollow metal cylinder, open at both ends of an external radius of $8 \mathrm{~cm}$ and height $10 \mathrm{~cm}$ is $338 \mathrm{~cm}^{2}$. Take $r$ to be the inner radius, obtain an equation in $r$ and use it to obtain the thickness of the metal in the cylinder. Solution: Given that The external radius of the cylinder (R) = 8 cm Height of the cylinder (h) = 10 cm The total surface area of the hollow cylinder $(T . S . A)=338 \pi \mathrm{cm}^{2}$ As we know that, ...
Read More →In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm.
Question: In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ABC to the area of ∆ADE? Solution: GIVEN:DE is parallel to BC, AD = 1cm and BD = 2cm. TO FIND:Ratio of ΔABC to area of ΔADE According toBASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. $\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$ $\frac{1}{2+1}...
Read More →If sec x=x
Question: If $\sec x=x+\frac{1}{4 x}$, then $\sec x+\tan x=$ (a) $x, \frac{1}{x}$ (b) $2 x, \frac{1}{2 x}$ (c) $-2 x, \frac{1}{2 x}$ (d) $-\frac{1}{x}, x$ Solution: (b) $2 x, \frac{1}{2 x}$ We have, $\sec x=x+\frac{1}{4 x}$ $\Rightarrow \sec ^{2} x==x^{2}+\frac{1}{16 x^{2}}+\frac{1}{2}$ $\Rightarrow 1+\tan ^{2} x=1+x^{2}+\frac{1}{16 x^{2}}-\frac{1}{2}$ $\Rightarrow \tan ^{2} x=x^{2}+\frac{1}{16 x^{2}}-\frac{1}{2}$ $\Rightarrow \tan ^{2} x=\left(x-\frac{1}{4 x}\right)^{2}$ $\therefore \tan x=\pm\...
Read More →Show that
Question: Form the differential equation of the family of circles touching they-axis at the origin. Solution: The centre of the circle touching they-axis at origin lies on thex-axis. Let (a, 0) be the centre of the circle. Since it touches they-axis at origin, its radius isa. Now, the equation of the circle with centre (a, 0) and radius (a)is $(x-a)^{2}+y^{2}=a^{2}$ $\Rightarrow x^{2}+y^{2}=2 a x$ ...(1) Differentiating equation (1) with respect tox, we get: $2 x+2 y y^{\prime}=2 a$ $\Rightarrow...
Read More →Find the ratio between the total surface area of a cylinder
Question: Find the ratio between the total surface area of a cylinder to its curved surface area, given that height and radius of the tank are 7.5 m and 3.5 m respectively. Solution: Given that, Radius of the cylinder (r) = 3.5 m Height of the cylinder (h) = 7.5 m Total Surface Area of cylinder (T. S. A) =2r(r + h) Curved surface area of a cylinder (C.S.A) =2rh Now, $\frac{\text { T.S. A }}{\text { C.S. A }}=\frac{2 \pi r(r+h)}{2 \pi r h}$ $=\frac{h+r}{h}$ Putting the values in eq (1) $=\frac{7....
Read More →State SAS similarity criterion.
Question: State SAS similarity criterion. Solution: SAS Similarity Criterion: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then two triangles are similar. In ΔABC and ΔDEF, if $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}$ and $\angle \mathrm{A}=\angle \mathrm{D}$ Then, $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$...
Read More →If tan x = x
Question: If $\tan x=x-\frac{1}{4 x}$, then $\sec x-\tan x$ is equal to (a) $-2 x, \frac{1}{2 x}$ (b) $-\frac{1}{2 x}, 2 x$ (c) $2 x$ (d) $2 x, \frac{1}{2 x}$ Solution: (a) $-2 x, \frac{1}{2 x}$ We have, $\tan x=x-\frac{1}{4 x}$ $\Rightarrow \sec ^{2} x=1+\tan ^{2} x$ $\Rightarrow \sec ^{2} x=1+\left(x-\frac{1}{4 x}\right)^{2}$ $\Rightarrow \sec ^{2} x=x^{2}+\frac{1}{16 x^{2}}+\frac{1}{2}$ $\Rightarrow \sec ^{2} x=\left(x+\frac{1}{4 x}\right)^{2}$ $\therefore \sec x=\pm\left(x+\frac{1}{4 x}\righ...
Read More →State SSS similarity criterion.
Question: State SSS similarity criterion. Solution: SSS Similarity Criterion: If the corresponding sides of two triangles are proportional, then they are similar. In ΔABC and ΔDEF, if $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$ Then, $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$...
Read More →State AAA similarity criterion.
Question: StateAAAsimilarity criterion. Solution: AAA Similarity Criterion: If two triangles are equiangular, then they are similar. In ΔABC and ΔDEF, if $\angle \mathrm{A}=\angle \mathrm{D}$ $\angle \mathrm{B}=\angle \mathrm{E}$ $\angle \mathrm{C}=\angle \mathrm{F}$ Then, $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$...
Read More →The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.
Question: The total surface area of a hollow cylinder which is open on both the sides is 4620 sq.cm and the area of the base ring is 115.5 sq.cm and height is 7cm. Find the thickness of the cylinder. Solution: Let the inner radius of the hollow cylinder ber1cm The outer radius of the hollow cylinder ber2cm Then, $2 \pi r_{1} * h+2 \pi r_{2} * h+2 \pi r_{2}^{2}-2 \pi r_{1}^{2}=4620 \ldots$ (a) $\pi r_{1}^{2}-\pi r_{2}^{2}=115.5 \ldots$ Now solving eq (a) $2 \pi r_{1} h+2 \pi r_{2} h+2 \pi r_{2}^{...
Read More →Prove that:
Question: Prove that: (i) $\tan 4 \pi-\cos \frac{3 \pi}{2}-\sin \frac{5 \pi}{6} \cos \frac{2 \pi}{3}=\frac{1}{4}$ (ii) $\sin \frac{13 \pi}{3} \sin \frac{8 \pi}{3}+\cos \frac{2 \pi}{3} \sin \frac{5 \pi}{6}=\frac{1}{2}$ (iii) $\sin \frac{13 \pi}{3} \sin \frac{2 \pi}{3}+\cos \frac{4 \pi}{3} \sin \frac{13 \pi}{6}=\frac{1}{2}$ (iv) $\sin \frac{10 \pi}{3} \cos \frac{13 \pi}{6}+\cos \frac{8 \pi}{3} \sin \frac{5 \pi}{6}=-1$ (v) $\tan \frac{5 \pi}{4} \cot \frac{9 \pi}{4}+\tan \frac{17 \pi}{4} \cot \frac{...
Read More →Prove that:
Question: Prove that: (i) $\tan 4 \pi-\cos \frac{3 \pi}{2}-\sin \frac{5 \pi}{6} \cos \frac{2 \pi}{3}=\frac{1}{4}$ (ii) $\sin \frac{13 \pi}{3} \sin \frac{8 \pi}{3}+\cos \frac{2 \pi}{3} \sin \frac{5 \pi}{6}=\frac{1}{2}$ (iii) $\sin \frac{13 \pi}{3} \sin \frac{2 \pi}{3}+\cos \frac{4 \pi}{3} \sin \frac{13 \pi}{6}=\frac{1}{2}$ (iv) $\sin \frac{10 \pi}{3} \cos \frac{13 \pi}{6}+\cos \frac{8 \pi}{3} \sin \frac{5 \pi}{6}=-1$ (v) $\tan \frac{5 \pi}{4} \cot \frac{9 \pi}{4}+\tan \frac{17 \pi}{4} \cot \frac{...
Read More →Show that
Question: $y=e^{x}(a \cos x+b \sin x)$ Solution: $y=e^{x}(a \cos x+b \sin x)$ ...(1) Differentiating both sides with respect tox, we get: $y^{\prime}=e^{x}(a \cos x+b \sin x)+e^{x}(-a \sin x+b \cos x)$ $\Rightarrow y^{\prime}=e^{x}[(a+b) \cos x-(a-b) \sin x]$ ...(2) Again, differentiating with respect tox, we get: $y^{\prime \prime}=e^{x}[(a+b) \cos x-(a-b) \sin x]+e^{x}[-(a+b) \sin x-(a-b) \cos x]$ $y^{\prime \prime}=e^{x}[2 b \cos x-2 a \sin x]$ $y^{\prime \prime}=2 e^{x}(b \cos x-a \sin x)$ $...
Read More →In the given figure, PQ || BC and AP : PB = 1 : 2. Find area ∆APQarea ∆ABC.
Question: In the given figure, PQ || BC and AP : PB = 1 : 2. Findarea∆APQarea∆ABC. Solution: GIVEN: In the given figure PQ || BC, and AP: PB = 1:2 TO FIND: $\frac{\text { Area }(\mathrm{APQ})}{\text { Area }(\mathrm{ABC})}$ We know that according to basic proportionality theorem if a line is drawn parallel to one side of a triangle intersecting the other side, then it divides the two sides in the same ratio. Since triangle APQ and ABC are similar Hence, $\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\ma...
Read More →In the adjoining figure, if AD is the bisector of ∠A, what is AC?
Question: In the adjoining figure, if AD is the bisector of A, what is AC? Solution: GIVEN: $A B=6 \mathrm{~cm}, B D=3 \mathrm{~cm}$ and $D C=2 \mathrm{~cm} .$ Also, $A D$ is the bisector of $\angle A$. TO FIND:AC SOLUTION:We know that the internal bisector of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Therefore, $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$ $\frac{6}{\mathrm{AC}}=\frac{3}{2}$ $\mathrm{AC}=\frac{6 \times 2}...
Read More →A solid cylinder has a total surface area of 462 cm2.
Question: A solid cylinder has a total surface area of $462 \mathrm{~cm}^{2}$. |its curved surface area is one-third of its total surface area. Find the radius and height of the cylinder. (Take $\pi=3.14$ ). Solution: Given that Curved or lateral surface area = 13 * total surface area $2 \pi r h=1 / 3\left(2 \pi r h+2 \pi r^{2}\right)$ $4 \pi r h=2 \pi r^{2}$ 2h = r Total surface area $=462 \mathrm{~cm}^{2}$ Curved surface area = 1/3 462 2rh = 154 $2 * 3.14 * 2 * h^{2}=154$ $h^{2}=49 / 4$ h = 49...
Read More →In the adjoining figure, find AC.
Question: In the adjoining figure, find AC. Solution: GIVEN:In the figure we are given AD = 6cm, BD = 9cm, AE = 8cm TO FIND:AC According toBASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. In ∆ABC, DE || BC. So, $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ $\frac{6}{9}=\frac{8}{\mathrm{EC}}$ $\mathrm{EC}=\frac{8 \times 9}{6}$ $\mathrm{EC}=12 \mathrm{~cm}$ Now...
Read More →Find x from the following equations:
Question: Findxfrom the following equations: (i) $\operatorname{cosec}\left(\frac{\pi}{2}+\theta\right)+x \cos \theta \cot \left(\frac{\pi}{2}+\theta\right)=\sin \left(\frac{\pi}{2}+\theta\right)$ (ii) $x \cot \left(\frac{\pi}{2}+\theta\right)+\tan \left(\frac{\pi}{2}+\theta\right) \sin \theta+\operatorname{cosec}\left(\frac{\pi}{2}+\theta\right)=0$ Solution: $90^{\circ}=\frac{\pi}{2}$ (i) We have, $\operatorname{cosec}\left(90^{\circ}+\theta\right)+x \cos \theta \cot \left(90^{\circ}+\theta\rig...
Read More →State basic proportionality theorem and its converse.
Question: State basic proportionality theorem and its converse. Solution: TO STATE: The basic proportionality theorem and its converse. BASIC PROPORTIONALITY THEOREM: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. CONVERSE OF BASIC PROPORTIONALITY THEOREM: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side....
Read More →Nazima is fly fishing in a stream.
Question: Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the road. Assuming that her string (from the tip of her road to the fly) is taut, how much string does she have out (in the given figure)? If she pulls the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds. Solut...
Read More →It is required to make a closed cylindrical tank of height 1m and the base diameter of 140 cm from a metal sheet.
Question: It is required to make a closed cylindrical tank of height 1m and the base diameter of 140 cm from a metal sheet. How many square meters of the sheet are required for the same? (Take = 3.14). Solution: Height of the cylindrical tank (h) = 1 m Base radius of cylindrical tank (r) =140/2= 70 cm = 0.7 m Area of sheet required =total surface area of tank =2rh = 2 * 3.14 * 0.7(0.7 + 1) $=4.4^{*} 1.7=7.48 \mathrm{~m}^{2}$ Therefore it will require $7.48 \mathrm{~m}^{2}$ of metal sheet....
Read More →A cylindrical pillar is 50 cm in diameter and 3.5 m in height.
Question: A cylindrical pillar is $50 \mathrm{~cm}$ in diameter and $3.5 \mathrm{~m}$ in height. Find the cost of painting the curved surface of the pillar at the rate of Rs $12.50$ per $\mathrm{m}^{2}$. (Take $\pi=3.14$ ) Solution: Given that Height of cylindrical pillar = 3.5 m Radius(r) of circular end of pillar = 50/2 = 25 cm = 0.25 m Curved surface area of cylindrical pillar = 2 = 2 * 3.14 * 0.25 * 3.5 $=5.5 \mathrm{~m}^{2}$ The cost of whitewashing $1 \mathrm{~m}^{2}$ is Rs $12.50$ Cost of...
Read More →