sec
Question: $\sec ^{2} x=\frac{4 x y}{(x+y)^{2}}$ is true if and only if (a)x+y 0 (b)x=y,x 0 (c)x=y (d)x0,y 0 Solution: (b)x=y,x 0 We hsve: $\sec ^{2} \mathrm{x}=\frac{4 x y}{(x+y)^{2}}$ $\Rightarrow \frac{4 x y}{(x+y)^{2}} \geq 1 \quad\left[\because \sec ^{2} \mathrm{x} \geq 1\right]$ $\Rightarrow 4 x y \geq(x+y)^{2}$ $\Rightarrow 4 x y \geq x^{2}+y^{2}+2 x y$ $\Rightarrow 2 x y \geq x^{2}+y^{2}$ $\Rightarrow(x-y)^{2} \leq 0$ $\Rightarrow(x-y) \leq 0$ $\Rightarrow x=y$ For $x=0, \sec ^{2} x$ will...
Read More →If the lateral surface of a cylinder is 94.2 cm2
Question: If the lateral surface of a cylinder is $94.2 \mathrm{~cm}^{2}$ and its height is $5 \mathrm{~cm}$, find: (i) radius of its base (ii) volume of the cylinder [Use pi = 3.141] Solution: (i) Given, Height of the cylinder = 5 cm Let radius of cylinder be 'r' Curved surface of the cylinder $=94.2 \mathrm{~cm}^{2}$ $2 \pi \mathrm{rh}=94.2 \mathrm{~cm}^{2}$ $r=3 \mathrm{~cm} \quad[\pi=31.4, h=5 \mathrm{~cm}]$ (ii) Volume of the cylinder $=\pi r^{2} h=\left(3.14 \times 3^{2} \times 5\right) \m...
Read More →Which of the following differential equation has
Question: Which of the following differential equation has $y=x$ as one of its particular solution? A. $\frac{d^{2} y}{d x^{2}}-x^{2} \frac{d y}{d x}+x y=x$ B. $\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+x y=x$ C. $\frac{d^{2} y}{d x^{2}}-x^{2} \frac{d y}{d x}+x y=0$ D. $\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+x y=0$ Solution: The given equation of curve isy=x. Differentiating with respect tox, we get: $\frac{d y}{d x}=1$ ...(1) Again, differentiating with respect tox, we get: $\frac{d^{2} y}{d...
Read More →If cosec x
Question: If $\operatorname{cosec} x+\cot x=\frac{11}{2}$, then $\tan x=$ (a) $\frac{21}{22}$ (b) $\frac{15}{16}$ (c) $\frac{44}{117}$ (d) $\frac{117}{44}$ 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)...
Read More →In the given figure, DE || BC in ∆ABC such that BC = 8 cm, AB = 6 cm and DA = 1.5 cm. Find DE.
Question: In the given figure, DE || BC in ∆ABC such that BC = 8 cm, AB = 6 cm and DA = 1.5 cm. Find DE. Solution: Given: In ∆ABC, DE || BC. BC = 8 cm, AB = 6 cm and DA = 1.5 cm. To find: DE In ∆ABC and ∆ADEB=ADECorrespondinganglesA=ACommon∆ABC~∆ADEAASimilaritySo, $\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{DA}}$ $\frac{8}{D E}=\frac{6}{1.5}$ $\mathrm{DE}=\frac{8 \times 1.5}{6}$ $\mathrm{DE}=2 \mathrm{~cm}$...
Read More →If cosec x
Question: If $\operatorname{cosec} x+\cot x=\frac{11}{2}$, then $\tan x=$ (a) $\frac{21}{22}$ (b) $\frac{15}{16}$ (c) $\frac{44}{117}$ (d) $\frac{117}{44}$ 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)...
Read More →The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm.
Question: The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 gm. Solution: Given, Inner radius (r1) of a cylindrical pipe =24/2= 12 cm Outer radius (r2) of a cylindrical pipe =24/2= 14 cm Height of pipe (h) = length of pipe = 35 cm Mass of pipe $=$ volume $\times$ density $=\pi\left(\mathrm{r}_{2}^{2}-\mathrm{r}_{1}^{2}\right) \mathrm{h}$ $=22 / 7\left(14^{2}-12^{...
Read More →In the given figure, ∆AHK is similar to ∆ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC.
Question: In the given figure, ∆AHK is similar to ∆ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC. Solution: Given: $\triangle \mathrm{AHK} \sim \triangle \mathrm{ABC}$ AK = 10 cm BC = 3.5 cm HK = 7 cm To find: AC Since $_{\triangle \mathrm{AHK}} \sim \Delta \mathrm{ABC}$, so their corresponding sides are proportional. $\frac{\mathrm{AC}}{\mathrm{AK}}=\frac{\mathrm{BC}}{\mathrm{HK}}$ $\frac{\mathrm{AC}}{10}=\frac{3.5}{7}$ $\mathrm{AC}=5 \mathrm{~cm}$...
Read More →If cosec x−cot x
Question: If $\operatorname{cosec} x-\cot x=\frac{1}{2}, 0x\frac{\pi}{2}$, then $\cos x$ is equal to (a) $\frac{5}{3}$ (b) $\frac{3}{5}$ (c) $-\frac{3}{5}$ (d) $-\frac{5}{3}$ Solution: (b) $\frac{3}{5}$ We have: $\operatorname{cosec} x-\cot x=\frac{1}{2}$ ...(i) $\Rightarrow \frac{1}{\operatorname{cosec} x-\cot x}=2$ $\Rightarrow \frac{\operatorname{cosec}^{2} x-\cot ^{2} x}{\operatorname{cosec} x-\cot x}=2$ $\Rightarrow \frac{(\operatorname{cosec} x+\cot x)(\operatorname{cosec} x-\cot x)}{(\ope...
Read More →The pillars of a temple are cylindrically shaped.
Question: The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m, how much concrete mixture would be required to build 14 such pillars? Solution: Given, The concrete mixture is used to build up the pillars is required for the entire space of the pillar i.e, we need to find the volume of the cylinders. Radius of the base of a cylinder = 20 cm Volume of the cylindrical pillar $=\pi R^{2} H$ $=\left(22 / 7 \times 20^{2} \times 1000\right...
Read More →In the given figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm,
Question: In the given figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of ∆PST and ∆PQR. Solution: Given: In ΔPQR, S and T are the points on the sides PQ and PR respectively such that PT = 2cm, TR = 4cm and ST is parallel to QR. To find: Ratio of areas of ΔPST and ΔPQR In∆PSTand∆PQR,PST=QCorrespondinganglesP=PCommon∆PST~∆PQRAASimilarity Now, we know that the areas of two similar triangles ...
Read More →Which of the following differential equations has
Question: Which of the following differential equations has $y=c_{1} e^{x}+c_{2} e^{-x}$ as the general solution? A. $\frac{d^{2} y}{d x^{2}}+y=0$ B. $\frac{d^{2} y}{d x^{2}}-y=0$ C. $\frac{d^{2} y}{d x^{2}}+1=0$ D. $\frac{d^{2} y}{d x^{2}}-1=0$ Solution: The given equation is: $y=c_{1} e^{x}+c_{2} e^{-x}$ ...(1) Differentiating with respect tox, we get: $\frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}$ Again, differentiating with respect tox, we get: $\frac{d^{2} y}{d x^{2}}=c_{1} e^{x}+c_{2} e^{-x}$ ...
Read More →sin
Question: sin6A+ cos6A+ 3 sin2Acos2A= (a) 0 (b) 1 (c) 2 (d) 3 Solution: (b) 1 We have: $\sin ^{6} A+\cos ^{6} A+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)$ $=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right) \times 1$ $=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)\left(\sin ^{2} A+\cos ^{2} A\right)$ $=\left(\sin ^{2} A+\cos ^{2} A\right)^{3}$ $=1^{3}=1$...
Read More →sin
Question: sin6A+ cos6A+ 3 sin2Acos2A= (a) 0 (b) 1 (c) 2 (d) 3 Solution: (b) 1 We have: $\sin ^{6} A+\cos ^{6} A+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)$ $=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right) \times 1$ $=\left(\sin ^{2} A\right)^{3}+\left(\cos ^{2} A\right)^{3}+3\left(\sin ^{2} A\right)\left(\cos ^{2} A\right)\left(\sin ^{2} A+\cos ^{2} A\right)$ $=\left(\sin ^{2} A+\cos ^{2} A\right)^{3}$ $=1^{3}=1$...
Read More →In the given figure, LM = LN = 46°.
Question: In the given figure, LM = LN = 46. Expressxin terms ofa,bandcwherea,b,care lengths of LM, MN and NK respectively. Solution: Given: In the given figure $\angle \mathrm{LMN}=\angle \mathrm{PNK}=46^{\circ}$ TO EXPRESS:xin terms ofa,b,cwherea,b, andcare the lengths of LM, MN and NK respectively. Here we can see that $\angle \mathrm{LMN}=\angle \mathrm{PNK}=46^{\circ}$. It forms a pair of corresponding angles. Hence, LM || PN In∆LMKand∆PNK,LMK=PNKCorrespondinganglesLKM=PKNCommon∆LMK~∆PNKAAS...
Read More →A soft drink is available in two packs-
Question: A soft drink is available in two packs- (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm, Which container has greater capacity and by how much? Solution: Given, The tin can will be cubical in shape. Length (L) of tin can = 5 cm Breadth (B) of tin can = 4 cm Height (H) of tin can = 15 cm Capacity of the tin can $=1 \times \mathrm{b} \times \mathrm{h}=(5 \times 4 ...
Read More →If
Question: If $\frac{3 \pi}{4}\alpha\pi$, then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ is equal to (a) 1 cot (b) 1 + cot (c) 1 + cot (d) 1 cot Solution: (d) 1 cot We have: $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{2 \cos \alpha}{\sin \alpha}+\frac{1}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{2 \sin \alpha \cos \alpha+1}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{2 \sin \alpha \cos \alpha+\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{(\sin \alpha+\cos \alpha)^...
Read More →If
Question: If $\frac{3 \pi}{4}\alpha\pi$, then $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ is equal to (a) 1 cot (b) 1 + cot (c) 1 + cot (d) 1 cot Solution: (d) 1 cot We have: $\sqrt{2 \cot \alpha+\frac{1}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{2 \cos \alpha}{\sin \alpha}+\frac{1}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{2 \sin \alpha \cos \alpha+1}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{2 \sin \alpha \cos \alpha+\sin ^{2} \alpha+\cos ^{2} \alpha}{\sin ^{2} \alpha}}$ $=\sqrt{\frac{(\sin \alpha+\cos \alpha)^...
Read More →The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus.
Question: The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus. Solution: GIVEN: the lengths of the diagonals of a rhombus are 30 cm and 40 cm. TO FIND: side of the rhombus. Let the diagonals AC and CD of the rhombus ABCD meet at point O. We know that the diagonals of the rhombus bisect each other perpendicularly. Hence in right triangle AOD, by Pythagoras theorem hypotenuse $^{2}=$ perpendicular $^{2}+$ base $^{2}$ $=15^{2}+20^{2}$ $=225+400$ $=625$ hypote...
Read More →Twenty cylindrical pillars of the Parliament House are to be cleaned.
Question: Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50m and height is 4m.What will be the cost of cleaning them at the rate of Rs 2.50 per square meter? Solution: Diameter of each pillar = 0.5, Radius of each pillar(r) =d/2 = 0.5/2= 0.25 m Height of each pillar = 4 m Lateral surface area of one pillar $=2 \pi r h=2 * 3.14 * 0.25 * 4=44 / 7 \mathrm{~m}^{2}$ Lateral surface area of 20 pillars $=20 * 44 / 7 \mathrm{~m}^{2}$ Cost of cle...
Read More →State Pythagoras theorem and its converse.
Question: State Pythagoras theorem and its converse. Solution: TO STATE: Pythagoras theorem and its converse. PYTHAGORAS THEOREM: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. hypotenuse $^{2}=$ perpendicular $^{2}+$ base $^{2}$ CONVERSE OF PYTHAGORAS THEOREM: In a triangle, if the square of one side is equal to the sum of the square of the other two sides then the angle opposite to the greatest side is a right angle....
Read More →If tan x
Question: If $\tan x=-\frac{1}{\sqrt{5}}$ and $\theta$ lies in the IV quadrant, then the value of $\cos x$ is (a) $\frac{\sqrt{5}}{\sqrt{6}}$ (b) $\frac{2}{\sqrt{6}}$ (c) $\frac{1}{2}$ (d) $\frac{1}{\sqrt{6}}$ Solution: (a) $\frac{\sqrt{5}}{\sqrt{6}}$ In the fourth quadrant, $\cos x$ and $\sec x$ are positive. $\cos x=\frac{1}{\sec x}$ $=\frac{1}{\sqrt{\sec ^{2} x}}$ $=\frac{1}{\sqrt{1+\tan ^{2} x}}$ $=\frac{1}{\sqrt{1+\left(-\frac{1}{\sqrt{5}}\right)^{2}}}$ $=\frac{1}{\sqrt{\frac{6}{5}}}$ $=\fr...
Read More →If tan x
Question: If $\tan x=-\frac{1}{\sqrt{5}}$ and $\theta$ lies in the IV quadrant, then the value of $\cos x$ is (a) $\frac{\sqrt{5}}{\sqrt{6}}$ (b) $\frac{2}{\sqrt{6}}$ (c) $\frac{1}{2}$ (d) $\frac{1}{\sqrt{6}}$ Solution: (a) $\frac{\sqrt{5}}{\sqrt{6}}$ In the fourth quadrant, $\cos x$ and $\sec x$ are positive. $\cos x=\frac{1}{\sec x}$ $=\frac{1}{\sqrt{\sec ^{2} x}}$ $=\frac{1}{\sqrt{1+\tan ^{2} x}}$ $=\frac{1}{\sqrt{1+\left(-\frac{1}{\sqrt{5}}\right)^{2}}}$ $=\frac{1}{\sqrt{\frac{6}{5}}}$ $=\fr...
Read More →Form the differential equation of the family of circles having centre on y-axis and radius 3 units.
Question: Form the differential equation of the family of circles having centre ony-axis and radius 3 units. Solution: Let the centre of the circle ony-axis be (0,b). The differential equation of the family of circles with centre at (0,b) and radius 3 is as follows: $x^{2}+(y-b)^{2}=3^{2}$ $\Rightarrow x^{2}+(y-b)^{2}=9$ ...(1) Differentiating equation (1) with respect tox, we get: $2 x+2(y-b) \cdot y^{\prime}=0$ $\Rightarrow(y-b) \cdot y^{\prime}=-x$ $\Rightarrow y-b=\frac{-x}{y^{\prime}}$ Subs...
Read More →If ∆ABC and ∆DEF are similar triangles such that AB = 3 cm,
Question: If ∆ABC and ∆DEF are similar triangles such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm and EF = 4 cm, write the perimeter of ∆DEF. Solution: GIVEN: ΔABC and ΔDEF are similar triangles such that AB = 3cm, BC = 2cm, CA = 2.5cm and EF = 4cm. TO FIND: Perimeter of ΔDEF. We know that if two triangles are similar then their corresponding sides are proportional. Hence, $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}$ Substituting the values, we get...
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