In ∆ABC, D and E are points on side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. If EA = 3.3 cm, then AC =
Question: In ∆ABC, D and E are points on side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. If EA = 3.3 cm, then AC = (a) 1.1 cm(b) 4 cm(c) 4.4 cm(d) 5.5 cm Solution: Given: In ΔABC, D and E are points on the side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. Also, EA = 3.3cm. To find: AC Using corollory of basic proportionality theorem, we have $\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{EA}}{\mathrm{AC}}$ $\frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{BD}}=\frac{3...
Read More →Show that
Question: $\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$ Solution: The given differential equation is: $\frac{d y}{d x}=\left(1+x^{2}\right)\left(1+y^{2}\right)$ $\Rightarrow \frac{d y}{1+y^{2}}=\left(1+x^{2}\right) d x$ Integrating both sides of this equation, we get: $\int \frac{d y}{1+y^{2}}=\int\left(1+x^{2}\right) d x$ $\Rightarrow \tan ^{-1} y=\int d x+\int x^{2} d x$ $\Rightarrow \tan ^{-1} y=x+\frac{x^{3}}{3}+\mathrm{C}$ This is the required general solution of the given diff...
Read More →The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3.
Question: The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. Calculate the ratio of their volumes and the ratio of their curved surfaces. Solution: $\frac{\text { Volume of cylinder } 1}{\text { Volume of cylinder } 2}=\frac{\pi(2 x)^{2} 5 y}{\pi(3 x)^{2} 3 y}=\frac{20}{27}$ $\frac{\text { Surface area of cylinder 1 }}{\text { Surface area of cylinder 2 }}=\frac{2 \pi \times 2 x \times 5 y}{2 \pi \times 3 x \times 3 y}=\frac{10}{9}$...
Read More →The value of tan1
Question: The value of $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$ is (a) 0 (b) 1 (c) $\frac{1}{2}$ (d) not defined Solution: We know that, $\tan \left(90^{\circ}-\theta\right)=\cot \theta$ So, $\tan 89^{\circ}=\tan \left(90^{\circ}-1^{\circ}\right)=\cot 1^{\circ}$ $\tan 88^{\circ}=\tan \left(90^{\circ}-2^{\circ}\right)=\cot 2^{\circ}$ $\tan 87^{\circ}=\tan \left(90^{\circ}-3^{\circ}\right)=\cot 3^{\circ}$ $\tan 46^{\circ}=\tan \left(90^{\circ}-44^{\circ}\right)=\cot 44...
Read More →The value of tan1
Question: The value of $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$ is (a) 0 (b) 1 (c) $\frac{1}{2}$ (d) not defined Solution: We know that, $\tan \left(90^{\circ}-\theta\right)=\cot \theta$ So, $\tan 89^{\circ}=\tan \left(90^{\circ}-1^{\circ}\right)=\cot 1^{\circ}$ $\tan 88^{\circ}=\tan \left(90^{\circ}-2^{\circ}\right)=\cot 2^{\circ}$ $\tan 87^{\circ}=\tan \left(90^{\circ}-3^{\circ}\right)=\cot 3^{\circ}$ $\tan 46^{\circ}=\tan \left(90^{\circ}-44^{\circ}\right)=\cot 44...
Read More →The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre
Question: The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is Rs 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places Solution: Let the radius of the tank be r dm Then, height = 6r dm Cost of painting for 50 paisa per $\mathrm{dm}^{2}=$ Rs 198 ⟹2r(r + h)= 198 ⟹2 22/7 r(r + 6r) 1/2= 198 ⟹ r = 3 dm Therefore, h = (6 3) dm = 18 dm Volume of the tank $=\pi r^{2} h=22 / ...
Read More →In ∆ABC, a line XY parallel to BC cuts AB at X and AC at Y. If BY bisects ∠XYC, then
Question: In ∆ABC, a line XY parallel to BC cuts AB at X and AC at Y. If BY bisects XYC, then (a) $\mathrm{BC}=\mathrm{CY}$ (b) $B C=B Y$ (c) $B C \neq C Y$ (d) $B C \neq B Y$ Solution: Given: XY||BC and BY is bisector ofXYC. Since XY||BC SoYBC =BYC (Alternate angles) Now, in triangle BYC two angles are equal. Therefore, the two corresponding sides will be equal. Hence, BC = CY Hence option (a) is correct....
Read More →Show that
Question: $\left(e^{x}+e^{-x}\right) d y-\left(e^{x}-e^{-x}\right) d x=0$ Solution: The given differential equation is: $\left(e^{x}+e^{-x}\right) d y-\left(e^{x}-e^{-x}\right) d x=0$ $\Rightarrow\left(e^{x}+e^{-x}\right) d y=\left(e^{x}-e^{-x}\right) d x$ $\Rightarrow d y=\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x$ Integrating both sides of this equation, we get: $\int d y=\int\left[\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\right] d x+\mathrm{C}$ $\Rightarrow y=\int\left[\frac{e^{x}-e^{-x}}{e^{...
Read More →A solid cylinder has a total surface area of 231cm2.
Question: A solid cylinder has a total surface area of $231 \mathrm{~cm}^{2}$. Its curved surface area is 3 of the total surface area. Find the volume of the cylinder. Solution: Given, Total surface area $=231 \mathrm{~cm}^{2}$ Curved surface area =2/3 * (total surface area) =2/3* 231 = 154 We know that, $2 \pi r h+2 \pi r^{2}=231 \ldots .1$ Here $2 \pi r h$ is the curved surface area, so substitute the value of CSA in eq 1 $\Rightarrow 154+2 \pi r^{2}=231$ $\Rightarrow 2 \pi r^{2}=231-154$ $\Ri...
Read More →Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is
Question: Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is (a) 12 m(b) 14 m(c) 13 m(d) 11 m Solution: Given: Two poles of heights 6m and 11m stand vertically upright on a plane ground. Distance between their foot is 12 m. To find: Distance between their tops. Let CD be the pole with height 6m. AB is the pole with height 11m, distance between their foot i.e. DB is 12 m. Let us assume a point...
Read More →The value of cos1°
Question: The value of $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ}\|.\| \cos 179^{\circ}$ is (a) $\frac{1}{\sqrt{2}}$ (b) 0 (c) 1 (d) $-1$ Solution: $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 179^{\circ}$ $=\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots \cos 90^{\circ} \ldots \cos 179^{\circ}$ $=0 \quad\left(\cos 90^{\circ}=0\right)$ Hence, the correct answer is option B....
Read More →Show that
Question: $\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$ Solution: The given differential equation is: $\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y=0$ $\Rightarrow \frac{\sec ^{2} x \tan y d x+\sec ^{2} y \tan x d y}{\tan x \tan y}=0$ $\Rightarrow \frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0$ $\Rightarrow \frac{\sec ^{2} x}{\tan x} d x=-\frac{\sec ^{2} y}{\tan y} d y$ Integrating both sides of this equation, we get: $\int \frac{\sec ^{2} x}{\tan x} d x=-\int \frac{\sec ^{...
Read More →Which of the following is incorrect?
Question: Which of the following is incorrect? (a) $\sin x=-\frac{1}{5}$ (b) $\cos x=1$ (c) $\sec x=\frac{1}{2}$ (d) $\tan x=20$ Solution: (a) $\sin x=-\frac{1}{5}$ is correct as $-1 \leq \sin x \leq 1$ (b) $\cos x=1$ is correct as $-1 \leq \cos x \leq 1$ (c) $\sec x=\frac{1}{2}$ is not correct as sec $x \in(-\infty,-1] \cup[1, \infty)$ (d) $\tan x=20$ is correct as $\tan x$ can take any real value. Hence, the correct answer is option C....
Read More →Which of the following is incorrect?
Question: Which of the following is incorrect? (a) $\sin x=-\frac{1}{5}$ (b) $\cos x=1$ (c) $\sec x=\frac{1}{2}$ (d) $\tan x=20$ Solution: (a) $\sin x=-\frac{1}{5}$ is correct as $-1 \leq \sin x \leq 1$ (b) $\cos x=1$ is correct as $-1 \leq \cos x \leq 1$ (c) $\sec x=\frac{1}{2}$ is not correct as sec $x \in(-\infty,-1] \cup[1, \infty)$ (d) $\tan x=20$ is correct as $\tan x$ can take any real value. Hence, the correct answer is option C....
Read More →XY is drawn parallel to the base BC of a ∆ABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY =
Question: XY is drawn parallel to the base BC of a ∆ABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY = (a) 2 cm(b) 4 cm(c) 6 cm(d) 8 cm Solution: Given: XY is drawn parallel to the base BC of a ΔABC cutting AB at X and AC at Y. AB = 4BX and YC = 2 cm. To find: AY In ΔAXY and ΔABC,AXY=BCorrespondinganglesA=ACommon∆AXY~∆ABCAAsimilarity We know that if two triangles are similar, then their sides are proportional.It is given that AB = 4BX.Let AB = 4xand BX =x.Then, AX = 3x $\frac...
Read More →If f(x) =
Question: If $f(x)=\cos ^{2} x+\sec ^{2} x$, then (a)f(x) 1 (b)f(x) = 1 (c) 1 f(x) 2 (d)f(x) 2 Solution: $f(x)=\cos ^{2} x+\sec ^{2} x$ $=\cos ^{2} x+\sec ^{2} x-2 \cos x \sec x+2 \cos x \sec x$ $=(\sec x-\cos x)^{2}+2$ $\therefore f(x) \geq 2 \forall x \quad\left[(\sec x-\cos x)^{2} \geq 0 \forall x\right]$ Hence, the correct option is answer D....
Read More →∆ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ∆DEF ∼ ∆ABC and EF = 4 cm, then perimeter of ∆DEF is
Question: ∆ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ∆DEF ∆ABC and EF = 4 cm, then perimeter of ∆DEF is (a) 7.5 cm(b) 15 cm(c) 22.5 cm(d) 30 cm Solution: Given: In $\triangle A B C, A B=3 \mathrm{~cm}, B C=2 \mathrm{~cm}, C A=2.5 \mathrm{~cm} . \triangle D E F \sim \triangle A B C$ and $E F=4 \mathrm{~cm} .$ To find: Perimeter of ΔDEF. We know that if two triangles are similar, then their sides are proportional Since ΔABC and ΔDEF are similar, $\frac{\mathrm{AB}}{\mathrm{DE}}=\fr...
Read More →If sec x + tan x = k, cos x =
Question: If secx+ tanx=k, cosx= (a) $\frac{k^{2}+1}{2 k}$ (b) $\frac{2 k}{k^{2}+1}$ (c) $\frac{k}{k^{2}+1}$ (d) $\frac{k}{k^{2}-1}$ Solution: (b) $\frac{2 k}{k^{2}+1}$ We have: $\sec x+\tan x=k$ ...(1) $\Rightarrow \frac{1}{\sec x+\tan x}=\frac{1}{k}$ $\Rightarrow \frac{\sec ^{2} x-\tan ^{2} x}{\sec x+\tan x}=\frac{1}{k}$ $\Rightarrow \frac{(\sec x+\tan x)(\sec x-\tan x)}{(\sec x+\tan x)}=\frac{1}{k}$ $\therefore \sec x-\tan x=\frac{1}{k}$ ...(2) Adding (1) and (2): $2 \sec x=k+\frac{1}{k}$ $\R...
Read More →If sec x + tan x = k, cos x =
Question: If secx+ tanx=k, cosx= (a) $\frac{k^{2}+1}{2 k}$ (b) $\frac{2 k}{k^{2}+1}$ (c) $\frac{k}{k^{2}+1}$ (d) $\frac{k}{k^{2}-1}$ Solution: (b) $\frac{2 k}{k^{2}+1}$ We have: $\sec x+\tan x=k$ ...(1) $\Rightarrow \frac{1}{\sec x+\tan x}=\frac{1}{k}$ $\Rightarrow \frac{\sec ^{2} x-\tan ^{2} x}{\sec x+\tan x}=\frac{1}{k}$ $\Rightarrow \frac{(\sec x+\tan x)(\sec x-\tan x)}{(\sec x+\tan x)}=\frac{1}{k}$ $\therefore \sec x-\tan x=\frac{1}{k}$ ...(2) Adding (1) and (2): $2 \sec x=k+\frac{1}{k}$ $\R...
Read More →If ∆ABC and ∆DEF are two triangles such that ABDE=BCEF=CAFD=25, then Area (∆ABC) : Area (∆DEF) =
Question: If ∆ABC and ∆DEF are two triangles such thatABDE=BCEF=CAFD=25, then Area (∆ABC) : Area (∆DEF) = (a) 2 : 5(b) 4 : 25(c) 4 : 15(d) 8 : 125 Solution: Given: $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles such that $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}=\frac{2}{5}$. To find: $\operatorname{Ar}(\triangle \mathrm{ABC}): \operatorname{Ar}(\triangle \mathrm{DEF})$ We know that if the sides of two triangles are ...
Read More →If tan θ + sec θ =e
Question: If tan + sec =ex, then cos equals (a) $\frac{e^{x}+e^{-x}}{2}$ (b) $\frac{2}{e^{x}+e^{-x}}$ (C) $\frac{e^{x}-e^{-x}}{2}$ (d) $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ Solution: (b) $\frac{2}{e^{x}+e^{-x}}$ We have: $\tan \theta+\sec \theta=\mathrm{e}^{x}$ $\sec \theta+\tan \theta=\mathrm{e}^{x}$ ...(1) $\Rightarrow \frac{1}{\sec \theta+\tan \theta}=\frac{1}{\mathrm{e}^{x}}$ $\Rightarrow \frac{\sec ^{2} \theta-\tan ^{2} \theta}{\sec \theta+\tan \theta}=\frac{1}{\mathrm{e}^{x}}$ $\Rightarrow \...
Read More →If tan θ + sec θ =e
Question: If tan + sec =ex, then cos equals (a) $\frac{e^{x}+e^{-x}}{2}$ (b) $\frac{2}{e^{x}+e^{-x}}$ (C) $\frac{e^{x}-e^{-x}}{2}$ (d) $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ Solution: (b) $\frac{2}{e^{x}+e^{-x}}$ We have: $\tan \theta+\sec \theta=\mathrm{e}^{x}$ $\sec \theta+\tan \theta=\mathrm{e}^{x}$ ...(1) $\Rightarrow \frac{1}{\sec \theta+\tan \theta}=\frac{1}{\mathrm{e}^{x}}$ $\Rightarrow \frac{\sec ^{2} \theta-\tan ^{2} \theta}{\sec \theta+\tan \theta}=\frac{1}{\mathrm{e}^{x}}$ $\Rightarrow \...
Read More →If cosec x
Question: If $\cos e c x+\cot x=\frac{11}{2}$, then $\tan x=$ (a) $\frac{21}{22}$ (b) $\frac{15}{16}$ (c) $\frac{44}{117}$ (d) $\frac{117}{43}$ Solution: (c) $\frac{44}{117}$ We have: $\operatorname{cosec} x+\cot x=\frac{11}{2}$ ....(i) $\Rightarrow \frac{1}{\operatorname{cosec} x+\cot x}=\frac{2}{11}$ $\Rightarrow \frac{\operatorname{cosec}^{2} x-\cot ^{2} x}{\operatorname{cosec} x+\cot x}=\frac{2}{11}$ $\Rightarrow \frac{(\operatorname{cosec} x+\cot x)(\operatorname{cosec} x-\cot x)}{(\operato...
Read More →If cosec x
Question: If $\cos e c x+\cot x=\frac{11}{2}$, then $\tan x=$ (a) $\frac{21}{22}$ (b) $\frac{15}{16}$ (c) $\frac{44}{117}$ (d) $\frac{117}{43}$ Solution: (c) $\frac{44}{117}$ We have: $\operatorname{cosec} x+\cot x=\frac{11}{2}$ ....(i) $\Rightarrow \frac{1}{\operatorname{cosec} x+\cot x}=\frac{2}{11}$ $\Rightarrow \frac{\operatorname{cosec}^{2} x-\cot ^{2} x}{\operatorname{cosec} x+\cot x}=\frac{2}{11}$ $\Rightarrow \frac{(\operatorname{cosec} x+\cot x)(\operatorname{cosec} x-\cot x)}{(\operato...
Read More →If A lies in second quadrant 3tanA + 4 = 0,
Question: IfAlies in second quadrant 3tanA+ 4 = 0, then the value of 2cotA 5cosA+ sinAis equal to (a) $-\frac{53}{10}$ (b) $\frac{23}{10}$ (c) $\frac{37}{10}$ (d) $\frac{7}{10}$ Solution: It is given that $\frac{\pi}{2}A\pi$. $3 \tan A+4=0$ $\Rightarrow \tan A=-\frac{4}{3}$ $\Rightarrow \cot A=-\frac{3}{4}$ Now, $\sec A=\pm \sqrt{1+\tan ^{2} A}=\pm \sqrt{1+\frac{16}{9}}=\pm \sqrt{\frac{25}{9}}=\pm \frac{5}{3}$ $\therefore \sec A=-\frac{5}{3} \quad$ (A lies in 2 nd quadrant) $\Rightarrow \cos A=-...
Read More →