If sin x + cosec x = 2,
Question: If sinx+ cosecx= 2, then sin2x+ cosec2x= ___________ . Solution: Given sinx+ cosecx= 2 Squaring both sides,(sinx+ cosecx)2= 4 i. e $\sin ^{2} x+\operatorname{cosec}^{2} x+2 \sin x \operatorname{cosec} x=4$ i. e $\sin ^{2} x+\operatorname{cosec}^{2} x+2 \sin x \times \frac{1}{\sin x}=4$ i. e $\sin ^{2} x+\operatorname{cosec}^{2} x+2=4$ i. e $\sin ^{2} x+\operatorname{cosec}^{2} x=2$...
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Question: $x\left(x^{2}-1\right) \frac{d y}{d x}=1 ; y=0$ when $x=2$ Solution: $x\left(x^{2}-1\right) \frac{d y}{d x}=1$ $\Rightarrow d y=\frac{d x}{x\left(x^{2}-1\right)}$ $\Rightarrow d y=\frac{1}{x(x-1)(x+1)} d x$ Integrating both sides, we get: $\int d y=\int \frac{1}{x(x-1)(x+1)} d x$ ...(1) Let $\frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$. ...(2) $\Rightarrow \frac{1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)}$ $=\frac{(A+B+C) x^{2}+(B-C) x-A}{x(x-1)(...
Read More →If sin x=
Question: If $\sin x=\frac{-24}{25}$, then the value of $\tan x$ is __________________ . Solution: Given $\sin x=\frac{-24}{25}$ i.e $\mathrm{x}$ lies in III or IV quadrant $\sin x=\frac{A B}{A C}=\frac{-24}{25}$ Since $A C^{2}=A B^{2}+B C^{2}$ i. e $25^{2}=(24)^{2}+B C^{2}$ $\Rightarrow 625=576+B C^{2}$ $\Rightarrow B C^{2}=49$ $B C=\pm 7$ $\Rightarrow \tan x=\frac{24}{7}$ or $\tan x=\frac{-24}{7}$ sin...
Read More →If ABC is an isosceles triangle and D is a point of BC such that AD ⊥ BC, then
Question: If ABC is an isosceles triangle and D is a point of BC such that AD BC, then (a) $\mathrm{AB}^{2}-\mathrm{AD}^{2}=\mathrm{BD} . \mathrm{DC}$ (b) $A B^{2}-A D^{2}=B D^{2}-D C^{2}$ (c) $A B^{2}+A D^{2}=B D . D C$ (d) $A B^{2}+A D^{2}=B D^{2}-D C^{2}$ Solution: Given: $\triangle \mathrm{ABC}$ is an isosceles triangle, $\mathrm{D}$ is a point on $\mathrm{BC}$ such that $\mathrm{AD} \perp \mathrm{BC}$ We know that in an isosceles triangle the perpendicular from the vertex bisects the base. ...
Read More →How many cubic meters of earth must be dug out to sink a well 21m deep and 6 m diameter?
Question: How many cubic meters of earth must be dug out to sink a well $21 \mathrm{~m}$ deep and $6 \mathrm{~m}$ diameter? Find the cost of plastering the inner surface as well at Rs.9.50 per $\mathrm{m}^{2}$. Solution: Let, r be the radius h be the height here, h = 21m 2r = 6 ⟹r =6/2 = 3 m Volume of the cylinder $=r^{2} * h$ =22/7 * 3 * 3 * 21 = 66 * 9 $=594 \mathrm{~cm}^{3}$ Cost of plastering $=9.5$ per $\mathrm{m}^{3}$ Cost of plastering inner surface = Rs.(594 * 9.50) = Rs. 5643...
Read More →If cosec x+cot x=
Question: If $\operatorname{cosec} x+\cot x=\frac{11}{2}$, then the value of $\tan \mathrm{x}$ is ____________ . Solution: $\operatorname{cosec} x+\cot x=\frac{11}{2}$ ....(1) Since $\operatorname{cosec}^{2} x-\cot ^{2} x=1$ $\Rightarrow(\operatorname{cosec} x-\cot x)(\operatorname{cosec} x+\cot x)=1$ $\Rightarrow \operatorname{cosec} x-\cot x=\frac{2}{11}$ Subtrating (2) from (1) $\operatorname{cosec} x+\tan x-\operatorname{cosec} x+\cot x=\frac{11}{2}-\frac{2}{11}$ $2 \cot x=\frac{121-4}{22}=\...
Read More →The height of a right circular cylinder is 10.5 m.
Question: The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder. Solution: Let, r be the radius of the right circular cylinder h be the height of the right circular cylinder h = 10.5 cm $\Rightarrow 3\left(2 \pi r^{2}\right)=2(2 \pi r h)$ ⟹ 3r = 2h ⟹r =2/3 h ⟹r =2/3 10.5 ⟹ r = 7cm Volume of the cylinder $=r^{2}{ }^{*} h$ =22/7 * 7 * 7 * 10.5 = 154 * 10.5 $=1617 \mathrm...
Read More →In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC
Question: In a ∆ABC, AD is the bisector of BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC (a) 4 cm(b) 6 cm(c) 3 cm(d) 8 cm Solution: Given: In a ΔABC, AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm. To find: AC We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Hence, $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$ $\frac{8}{\mathrm{AC}}=\frac{6}{3}$ $\mathrm{A...
Read More →In a ∆ABC, AD is the bisector of ∠BAC. If AB = 6 cm, AC = 5 cm and BD = 3 cm, then DC =
Question: In a ∆ABC, AD is the bisector of BAC. If AB = 6 cm, AC = 5 cm and BD = 3 cm, then DC = (a) 11.3 cm(b) 2.5 cm(c) 3 : 5 cm(d) None of these Solution: Given: In a $\triangle \mathrm{ABC}, \mathrm{AD}$ is the bisector of $\angle \mathrm{BAC} . \mathrm{AB}=6 \mathrm{~cm}$ and $\mathrm{AC}=5 \mathrm{~cm}$ and $\mathrm{BD}=3 \mathrm{~cm}$. To find: DC We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. ...
Read More →If cosec x + cot x = α,
Question: If cosecx+ cotx= , then sinx= _________. Solution: Given cosecx+ cotx=a ...(1) Since cosec2x cot2x= 1 (cosecx cotx) (cosecx+ cotx) = 1 i.e (cosecx cotx)a= 1 i. $\mathrm{e} \operatorname{cosec} x-\cot x=\frac{1}{a}$ ..(2) adding (1) and (2) $2 \operatorname{cosec} x=a+\frac{1}{a}$ $2 \operatorname{cosec} x=\frac{a^{2}+1}{a}$ $\frac{2 a}{a^{2}+1}=\frac{1}{\operatorname{cosec} x}$ i. e $\sin x=\frac{2 a}{a^{2}+1}$...
Read More →If cosec x + cot x = α,
Question: If cosecx+ cotx= , then sinx= _________. Solution: Given cosecx+ cotx=a ...(1) Since cosec2x cot2x= 1 (cosecx cotx) (cosecx+ cotx) = 1 i.e (cosecx cotx)a= 1 i. $\mathrm{e} \operatorname{cosec} x-\cot x=\frac{1}{a}$ ...92) adding (1) and (2) $2 \operatorname{cosec} x=a+\frac{1}{a}$ $2 \operatorname{cosec} x=\frac{a^{2}+1}{a}$ $\frac{2 a}{a^{2}+1}=\frac{1}{\operatorname{cosec} x}$ i. e $\sin x=\frac{2 a}{a^{2}+1}$...
Read More →Two circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of two radii.
Question: Two circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of two radii. Solution: Let,r1, r2be the radii of the cylinder h1, h2be the height of the cylinder v1, v2be the volume of the cylinder h1/h2=1/2and v1= v2 $\Rightarrow \mathrm{v}_{1} / \mathrm{v}_{2}=\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right)^{2} *\left(\mathrm{~h}_{1} / \mathrm{h}_{2}\right)$ Since,v1= v2 $\Rightarrow \mathrm{v}_{1} / \mathrm{v}_{1}=\left(\mathrm{r}_{1} / \mathrm{r}_{2}\right...
Read More →Show that
Question: $\left(x^{3}+x^{2}+x+1\right) \frac{d y}{d x}=2 x^{2}+x ; y=1$ when $x=0$ Solution: The given differential equation is: $\left(x^{3}+x^{2}+x+1\right) \frac{d y}{d x}=2 x^{2}+x$ $\Rightarrow \frac{d y}{d x}=\frac{2 x^{2}+x}{\left(x^{3}+x^{2}+x+1\right)}$ $\Rightarrow d y=\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x$ Integrating both sides, we get: $\int d y=\int \frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)} d x$ ...(1) Let $\frac{2 x^{2}+x}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\fra...
Read More →In a ∆ABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE : BC = 3 : 5, then Area (∆ ADE) : Area (◻BCED) =
Question: In a ∆ABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE : BC = 3 : 5, then Area (∆ ADE) : Area (◻BCED) = (a) 3 : 4(b) 9 : 16(c) 3 : 5(d) 9 : 25 Solution: Given: In ΔABC, D is on side AB and point E is on side AC, such that BCED is a trapezium. DE: BC = 3:5. To find: Calculate the ratio of the areas of ΔADE and the trapezium BCED. In ΔADE and ΔABC,ADE=BCorrespondinganglesA=ACommon∆ADE~∆ABCAASimilarity $\frac{\operatorname{Ar}(\Delta \mathrm{ADE})...
Read More →If sec x
Question: If $\sec x-\tan x=\frac{2}{3}$, then $\tan x=$ _____________ , Solution: Given $\sec x-\tan x=\frac{2}{3}$ .....(1) Since $\sec ^{2} x-\tan ^{2} x=1$ $\Rightarrow(\sec x-\tan x)(\sec x+\tan x)=1$ $\Rightarrow \frac{2}{3}(\sec x+\tan x)=1$ $\Rightarrow \sec x+\tan x=\frac{3}{2}$ ....(2) Subtracting (1) from (2) $\sec x+\tan x-\sec x+\tan x=\frac{3}{2}-\frac{2}{3}$ $\Rightarrow 2 \tan x=\frac{9-4}{6}$ $\Rightarrow 2 \tan x=\frac{5}{6}$ $\Rightarrow \tan x=\frac{5}{12}$...
Read More →If sec x
Question: If $\sec x-\tan x=\frac{2}{3}$, then $\tan x=$ _____________ , Solution: Given $\sec x-\tan x=\frac{2}{3}$ .....(1) Since $\sec ^{2} x-\tan ^{2} x=1$ $\Rightarrow(\sec x-\tan x)(\sec x+\tan x)=1$ $\Rightarrow \frac{2}{3}(\sec x+\tan x)=1$ $\Rightarrow \sec x+\tan x=\frac{3}{2}$ ....(2) Subtracting (1) from (2) $\sec x+\tan x-\sec x+\tan x=\frac{3}{2}-\frac{2}{3}$ $\Rightarrow 2 \tan x=\frac{9-4}{6}$ $\Rightarrow 2 \tan x=\frac{5}{6}$ $\Rightarrow \tan x=\frac{5}{12}$...
Read More →If sec x+ tan x
Question: If $\sec x+\tan x=\sqrt{3}$, then $\sec x-\tan x=$ ______________ . Solution: Given $\sec x+\tan x=\sqrt{3}$ Since $\sec ^{2} x-\tan ^{2} x=1$ (using identity) $\Rightarrow(\sec x+\tan x)(\sec x-\tan x)=1$ $\Rightarrow \sqrt{3}(\sec x-\tan x)=1$ $\Rightarrow \sec x-\tan x=\frac{1}{\sqrt{3}}$...
Read More →The curved surface area of cylindrical pillar is 264 m2
Question: The curved surface area of cylindrical pillar is $264 \mathrm{~m}^{2}$ and its volume is $924 \mathrm{~m}^{3}$. Find the diameter and the height of the pillar. Solution: Let, r be the radius of the cylindrical pillar h be the height of the cylindrical pillar $C S A=264 \mathrm{~m}^{2}$ $2 \pi r h=264 m^{2} \ldots .1$ $\Rightarrow$ volume of the cylinder $=924 \mathrm{~m}^{2}$ $\Pi^{*} r^{2} * h=924$ rh(r) = 924 rh =924/r Substitute rh in eq 1 2 * 924/r= 264 r =1848/264 r = 7 m Substitu...
Read More →In a ∆ABC, perpendicular AD from A and BC meets BC at D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then
Question: In a ∆ABC, perpendicular AD from A and BC meets BC at D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then (a) ∆ABC is isosceles(b) ∆ABC is equilateral(c) AC = 2AB(d) ∆ABC is right-angled at A Solution: Given: $\ln \triangle \mathrm{ABC}, \mathrm{AD} \perp \mathrm{BC}, \mathrm{BD}=8 \mathrm{~cm}, \mathrm{DC}=2 \mathrm{~cm}$ and $\mathrm{AD}=4 \mathrm{~cm}$. In ΔADC, $\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}$ $\mathrm{AC}^{2}=4^{2}+2^{2}$ $\mathrm{AC}^{2}=20$......(1) Similarly, in ΔAD...
Read More →If sec x+ tan x
Question: Ifsecx+tanx=3,secx+tanx=3,then secx tanx= ___________. Solution: Given $\sec x+\tan x=\sqrt{3}$ Since $\sec ^{2} x-\tan ^{2} x=1$ (using identity) $\Rightarrow(\sec x+\tan x)(\sec x-\tan x)=1$ $\Rightarrow \sqrt{3}(\sec x-\tan x)=1$ $\Rightarrow \sec x-\tan x=\frac{1}{\sqrt{3}}$...
Read More →If
Question: If $-\frac{\pi}{2}x\frac{\pi}{2}$, then $\sqrt{\frac{1-\sin x}{1+\sin x}}$ is equal to _____________________ . Solution: If $-\frac{\pi}{2}x\frac{\pi}{2}$ is given, $\sqrt{\frac{1-\sin x}{1+\sin x}}=\sqrt{\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}}$ (using identics : $-\sin ^{2} \theta+\cos ^{2} \theta-1 \sin 2 \theta=2 \sin \theta \cos \theta$ ) $=\sqrt{\frac{\l...
Read More →If ∆ABC is an equilateral triangle such that AD ⊥ BC, then AD2 =
Question: If $\triangle \mathrm{ABC}$ is an equilateral triangle such that $\mathrm{AD} \perp \mathrm{BC}$, then $\mathrm{AD}^{2}=$ (a) $32 \mathrm{DC} 2$ (b) $2 \mathrm{DC}^{2}$ (c) $3 \mathrm{CD}^{2}$ (d) $4 \mathrm{DC}^{2}$ Solution: Given: In an equilateral $\triangle \mathrm{ABC}, \mathrm{AD} \perp \mathrm{BC}$. Since $\mathrm{AD} \perp \mathrm{BC}, \mathrm{BD}=\mathrm{CD}=\mathrm{BC} 2$ Applying Pythagoras theorem, $\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}$ $\mathrm{BC}^{2}=\mathrm{...
Read More →If sin x + cos x = a
Question: If sinx+ cosx=a, then sinx cosx= __________. Solution: Given sinx+ cosx=a i.e (sinx cosx)2=a2 (squaring both side) $\Rightarrow \sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2}$ $\Rightarrow 1+2 \sin x \cos x=a^{2}$ i. e $2 \sin x \cos x=a^{2}-1$ Now, consider $(\sin x-\cos x)^{2}=\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x$ $=1-\left(a^{2}-1\right)$ $=1-a^{2}+1$ $(\sin x-\cos x)^{2}=2-a^{2}$ i. e $\sin x-\cos x=\pm \sqrt{2-a^{2}}$...
Read More →A rectangular sheet of paper, 44 cm * 20 cm, is rolled along its length of form cylinder.
Question: A rectangular sheet of paper, 44 cm * 20 cm, is rolled along its length of form cylinder. Find the volume of the cylinder so formed. Solution: Given, the dimensions of the sheet are 44 cm * 20 cm Here, length = 44 cm Height = 20 cm 2r = 44 r =44/2 r = 44/2 $r=\frac{44 * 7}{2 * 22}$ r = 7 cm Volume of cylinder $=r^{2} * h$ = 22/7 * 7 * 7 * 20 $=154^{*} 20=3080 \mathrm{~cm}^{3}$...
Read More →If sin x + cos x = a
Question: If sinx+ cosx=a, then sinx cosx= __________. Solution: Given sinx+ cosx=a i.e (sinx cosx)2=a2 (squaring both side) $\Rightarrow \sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2}$ $\Rightarrow 1+2 \sin x \cos x=a^{2}$ i. e $2 \sin x \cos x=a^{2}-1$ Now, consider $(\sin x-\cos x)^{2}=\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x$ $=1-\left(a^{2}-1\right)$ $=1-a^{2}+1$ $(\sin x-\cos x)^{2}=2-a^{2}$ i. e $\sin x-\cos x=\pm \sqrt{2-a^{2}}$...
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