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Question: $x+y=\tan ^{-1} y \quad: y^{2} y^{\prime}+y^{2}+1=0$ Solution: $x+y=\tan ^{-1} y$ Differentiating both sides of this equation with respect tox, we get: $\frac{d}{d x}(x+y)=\frac{d}{d x}\left(\tan ^{-1} y\right)$ $\Rightarrow 1+y^{\prime}=\left[\frac{1}{1+y^{2}}\right] y^{\prime}$ $\Rightarrow y^{\prime}\left[\frac{1}{1+y^{2}}-1\right]=1$ $\Rightarrow y^{\prime}\left[\frac{1-\left(1+y^{2}\right)}{1+y^{2}}\right]=1$ $\Rightarrow y^{\prime}\left[\frac{-y^{2}}{1+y^{2}}\right]=1$ $\Rightarr...
Read More →In a quadrilateral ABCD, given that ∠A + ∠D = 90°.
Question: In a quadrilateral $\mathrm{ABCD}$, given that $\angle \mathrm{A}+\angle \mathrm{D}=90^{\circ}$. Prove that $\mathrm{AC}^{2}+\mathrm{BD}^{2}=\mathrm{AD}^{2}+\mathrm{BC}^{2}$. Solution: Given: A quadrilateral $\mathrm{ABCD}$ where $\angle \mathrm{A}+\angle \mathrm{D}=90^{\circ}$. To prove: $\mathrm{AC}^{2}+\mathrm{BD}^{2}=\mathrm{AD}^{2}+\mathrm{BC}^{2}$ Construction: Extend AB and CD to intersect at O. Proof: In $\triangle \mathrm{AOD}, \angle \mathrm{A}+\angle \mathrm{O}+\angle \mathr...
Read More →A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field.
Question: A field is 200 m long and 150 m broad. There is a plot, 50 m long and 40 m broad, near the field. The plot is dug 7m deep and the earth taken out is spread evenly on the field. By how many meters is the level of the field raised? Give the answer to the second place of decimal. Solution: Volume of the earth dug out $=50 * 40 * 7=14000 \mathrm{~m}^{3}$ Let 'h' be the rise in the height of the field Therefore, volume of the field (cuboidal) = Volume of the earth dug out ⇒ 200 150 h = 1400...
Read More →A rectangular container, whose base is a square of side 5cm, stands on a horizontal table,
Question: A rectangular container, whose base is a square of side 5cm, stands on a horizontal table, and holds water up to 1cm from the top. When a solid cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge. Solution: Let the length of each edge of the cube be xcm Then, volume of the cube = Volume of water inside the tank + Volume of water that overflowed $x^{3}=(5 * ...
Read More →In ∆ABC, if BD ⊥ AC and BC2 = 2 AC .
Question: In $\triangle \mathrm{ABC}$, if $\mathrm{BD} \perp \mathrm{AC}$ and $\mathrm{BC}^{2}=2 \mathrm{AC}$. $\mathrm{CD}$, then prove that $\mathrm{AB}=\mathrm{AC}$. Solution: Since $\triangle A D B$ is right triangle right angled at $D$ $A B^{2}=A D^{2}+B D^{2}$ In right $\triangle B D C$, we have $C D^{2}+B D^{2}=B C^{2}$ Since $2 A C \cdot D C=B C^{2}$ $\Rightarrow D C^{2}+B D^{2}=2 A C . D C$ $2 A C \cdot D C=A C^{2}-A C^{2}+D C^{2}+B D^{2}$ $A C^{2}=A C^{2}+D C^{2}-2 A C . D C+B D^{2}$ $...
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Question: $y-\cos y=x \quad:(y \sin y+\cos y+x) y^{\prime}=y$ Solution: $y-\cos y=x$ ...(1) Differentiating both sides of the equation with respect tox, we get: $\frac{d y}{d x}-\frac{d}{d x}(\cos y)=\frac{d}{d x}(x)$ $\Rightarrow y^{\prime}+\sin y \cdot y^{\prime}=1$ $\Rightarrow y^{\prime}(1+\sin y)=1$ $\Rightarrow y^{\prime}=\frac{1}{1+\sin y}$ Substituting the value of $y^{\prime}$ in equation (1), we get: L.H.S. $=(y \sin y+\cos y+x) y^{\prime}$ $=(y \sin y+\cos y+y-\cos y) \times \frac{1}{...
Read More →A cube of 9 cm edge is immersed completely in a rectangular vessel containing water.
Question: A cube of 9 cm edge is immersed completely in a rectangular vessel containing water. If the dimensions of the base are 15 cm and 12 cm, find the rise in water level in the vessel. Solution: Volume of the cube $=\mathrm{s}^{3}=9^{3}=729 \mathrm{~cm}^{3}$ Area of the base $=\mathrm{l}^{*} \mathrm{~b}=15^{*} 12=180 \mathrm{~cm}^{2}$ Rise in water level $=\frac{\text { Volume of the cube }}{\text { Area of base of rectangular vessel }}$ = 729/180 = 4.05 cm...
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Question: $x y=\log y+\mathrm{C} \quad: y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)$ Solution: $x y=\log y+\mathrm{C}$ Differentiating both sides of this equation with respect tox, we get: $\frac{d}{d x}(x y)=\frac{d}{d x}(\log y)$ $\Rightarrow y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}$ $\Rightarrow y+x y^{\prime}=\frac{1}{y} y^{\prime}$ $\Rightarrow y^{2}+x y y^{\prime}=y^{\prime}$ $\Rightarrow(x y-1) y^{\prime}=-y^{2}$ $\Rightarrow y^{\prime}=\frac{y^{2}}{1-x ...
Read More →A point D is on the side BC of an equilateral triangle
Question: A point $D$ is on the side $B C$ of an equilateral triangle $A B C$ such that $D C=14 B C$. Prove that $A D^{2}=13 C D^{2}$. Solution: We are given $A B C$ is an equilateral triangle with $C D=\frac{1}{4} B C$ We have to prove $A D^{2}=13 D C^{2}$ Draw $A E \perp B C$ In $\triangle A E B$ and $\triangle A E D$ we have $A B=A C$ $\angle A E B=\angle A E C=90^{\circ}$ $A E=A E$ So by right side criterion of similarity we have Thus we have $D C=\frac{1}{4} B C$ and $B E=E C=\frac{1}{2} B ...
Read More →How many cubic centimeters of iron are there in an open box whose external dimensions are 36 cm,
Question: How many cubic centimeters of iron are there in an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cubic cm of iron weighs 15 gms. Find the weight of the empty box in kg. Solution: Given, Outer dimensions of iron: Length (l) = 36 cm Breadth (b) = 25 cm Height (h) = 16.5 cm Inner dimensions of iron: Length (l) = 36 - (2*1.5) = 33 cm Breadth (b) = 25 - (2*1.5) = 22 cm Height (h) = 16.5 - 1.5 = 15 cm Volume of Iron = Outer volu...
Read More →The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm.
Question: The external dimensions of a closed wooden box are 48 cm, 36 cm, 30 cm. The box is made of 1.5 cm thick wood. How many bricks of size 6 cm 3 cm 0.75 cm can be put in this box? Solution: Given that, The external dimensions of the wooden box are as follows: Length (l) = 48 cm, Breadth (b) = 36 cm, Height (h) = 30 cm Now, the internal dimensions of the wooden box are as follows: Length (l) = 48 - (2 * 1.5) = 45 cm Breadth (b) = 36 - (2 * 1.5) = 33 cm Height (h) = 30 - (2 * 1.5) = 27 cm In...
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Question: $y=x \sin x \quad: x y^{\prime}=y+x \sqrt{x^{2}-y^{2}}(x \neq 0$ and $xy$ or $x-y)$ Solution: $y=x \sin x$ Differentiating both sides of this equation with respect tox, we get: $y^{\prime}=\frac{d}{d x}(x \sin x)$ $\Rightarrow y^{\prime}=\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)$ $\Rightarrow y^{\prime}=\sin x+x \cos x$ Substituting the value ofin the given differential equation, we get: L.H.S. $=x y^{\prime}=x(\sin x+x \cos x)$ $=x \sin x+x^{2} \cos x$ $=y+x^{2} \cdo...
Read More →A box with lid is made of 2 cm thick wood. Its external length,
Question: A box with lid is made of 2 cm thick wood. Its external length, breadth and height are 25 cm, 18 cm and 15 cm respectively. How much cubic cm of a fluid can be placed in it? Also, find the volume of the wood used in it. Solution: Given, The external dimensions of cuboid are as follows Length (l) = 25 cm Breadth (b) = 18 cm Height (h) = 15 cm External volume of the case with cover (cuboid) $=1^{*} \mathrm{~b}^{*} \mathrm{~h} \mathrm{~cm}^{3}$ $=25^{*} 18^{*} 15 \mathrm{~cm}^{3}$ $=6750 ...
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Question: $y=\mathrm{A} x \quad: \quad x y^{\prime}=y(x \neq 0)$ Solution: $y=\mathrm{A} x$ Differentiating both sides with respect tox, we get: $y^{\prime}=\frac{d}{d x}(\mathrm{Ax})$ $\Rightarrow y^{\prime}=\mathrm{A}$ Substituting the value ofin the given differential equation, we get: L.H.S. $=x y^{\prime}=x \cdot \mathrm{A}=\mathrm{A} x=y=$ R.H.S. Hence, the given function is the solution of the corresponding differential equation....
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Question: $y=\sqrt{1+x^{2}} \quad: y^{\prime}=\frac{x y}{1+x^{2}}$ Solution: $y=\sqrt{1+x^{2}}$ Differentiating both sides of the equation with respect tox, we get: $y^{\prime}=\frac{d}{d x}\left(\sqrt{1+x^{2}}\right)$ $y^{\prime}=\frac{1}{2 \sqrt{1+x^{2}}} \cdot \frac{d}{d x}\left(1+x^{2}\right)$ $y^{\prime}=\frac{2 x}{2 \sqrt{1+x^{2}}}$ $y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}$ $\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \times \sqrt{1+x^{2}}$ $\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \cdot y$ $\Rig...
Read More →Given that 1 cubic cm of marble weighs 0.25 kg,
Question: Given that 1 cubic cm of marble weighs 0.25 kg, the weight of marble block 28 cm in width and 5 cm thick is 112 kg. Find the length of the block. Solution: Let the length of the marble block be 'l' cm Volume of the marble block $=1$ * $b$ * $h \mathrm{~cm}^{3}$ $=1^{*} 28^{*} 5 \mathrm{~cm}^{3}$ Therefore, weight of the marble square = 140l * 0.25 kg As mentioned in the question, weight of the marble = 112 kgs Therefore, $=112=1401 * 0.25$ $\Rightarrow 1=\frac{112}{140 * 0.25}=3.2 \mat...
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Question: $y=\cos x+\mathrm{C} \quad: y^{\prime}+\sin x=0$ Solution: $y=\cos x+C$ Differentiating both sides of this equation with respect tox, we get: $y^{\prime}=\frac{d}{d x}(\cos x+\mathrm{C})$ $\Rightarrow y^{\prime}=-\sin x$ Substituting the value of $y^{\prime}$ in the given differential equation, we get: L.H.S. $=y^{\prime}+\sin x=-\sin x+\sin x=0=$ R.H.S. Hence, the given function is the solution of the corresponding differential equation....
Read More →The dimensions of a cinema hall are 100 m, 50 m, 18 m.
Question: The dimensions of a cinema hall are $100 \mathrm{~m}, 50 \mathrm{~m}, 18 \mathrm{~m}$. How many persons can sit in the hall, if each person requires $150 \mathrm{~m}$ of air? Solution: Given that Volume of cinema hall $=100 * 50 * 18 \mathrm{~m}^{3}$ Volume of air required by each person $=150 \mathrm{~m}^{3}$ Number of persons who sit in the hall $=\frac{\text { Volume of cinema hall }}{\text { Volume of air required by each person }}$ $=\frac{100 * 50 * 18}{150}=600[$ Since, $\mathrm...
Read More →In ∆ABC, ∠C is an obtuse angle. AD ⊥ BC and AB2 = AC2 + 3 BC2
Question: In $\triangle \mathrm{ABC}, \angle \mathrm{C}$ is an obtuse angle. $\mathrm{AD} \perp \mathrm{BC}$ and $\mathrm{AB}^{2}=\mathrm{AC}^{2}+3 \mathrm{BC}^{2}$. Prove that $\mathrm{BC}=\mathrm{CD}$. Solution: Given: $\triangle A B C$ where $\angle C$ is an obtuse angle, $A D \perp B C$ and $A B^{2}=A C^{2}+3 B C^{2}$ To prove: BC = CDProof: In $\triangle \mathrm{ABC}, \angle \mathrm{C}$ is obtuse. Therefore, $\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}+2 \mathrm{BC} \times \mathrm{DC}$ ...
Read More →A metal cube of edge 12 cm is melted and formed into three smaller cubes.
Question: A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of the two smaller cubes are 6cm and 8cm, find the edge of the third smaller cube. Solution: Volume of the large cube =v1+ v2+ v3 Let the edge of the third cube be 'x' cm $12^{3}=6^{3}+8^{3}+a^{3}$ [Volume of cube $=$ side $^{3}$ ] $1728=216+512+x^{3}$ $\Rightarrow x^{3}=1728-728=1000$ ⇒ x = 10 cm Therefore, Side of third side = 10 cm...
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Question: $y=x^{2}+2 x+\mathrm{C} \quad: y^{\prime}-2 x-2=0$ Solution: $y=x^{2}+2 x+\mathrm{C}$ Differentiating both sides of this equation with respect tox, we get: $y^{\prime}=\frac{d}{d x}\left(x^{2}+2 x+\mathrm{C}\right)$ $\Rightarrow y^{\prime}=2 x+2$ Substituting the value of $y^{\prime}$ in the given differential equation, we get: L.H.S. $=y^{\prime}-2 x-2=2 x+2-2 x-2=0=$ R.H.S. Hence, the given function is the solution of the corresponding differential equation....
Read More →In ∆ABC, ∠A = 60°. Prove that BC2 = AB2 + AC2 − AB . AC. ANSWER:
Question: In $\triangle A B C, \angle A=60^{\circ}$. Prove that $B C^{2}=A B^{2}+A C^{2}-A B \cdot A C$. Solution: In $\triangle A B C$ in which $\angle A$ is an acute angle with $60^{\circ}$. $\sin 60^{\circ}=\frac{C D}{A C}=\frac{\sqrt{3}}{2}$ $\Rightarrow C D=\frac{\sqrt{3}}{2} A C$....(1) $\cos 60^{\circ}=\frac{A D}{A C}=\frac{1}{2}$ $\Rightarrow A D=\frac{1}{2} A C$.....$(2)$ Now apply Pythagoras theorem in triangleBCD $B C^{2}=C D^{2}+B D^{2}$ $=C D^{2}+(A B-A D)^{2}$ $=\left(\frac{\sqrt{3...
Read More →Half cubic meter of gold-sheet is extended
Question: Half cubic meter of gold-sheet is extended by hammering so as to cover an area of 1 hectare. Find the thickness of the gold-sheet. Solution: Given that, Volume of gold-sheet $=0.5 \mathrm{~m}^{3}$ Area of the gold-sheet $=1$ hectare $=1^{\star} 10000=10000 \mathrm{~m}^{2}$ Therefore, Thickness of gold sheet $=\frac{\text { Volume of solid }}{\text { Area of gold sheet }}$ $\Rightarrow 0.5 \mathrm{~m}^{3} / 1$ Hectare $\Rightarrow 0.5 \mathrm{~m}^{3} / 10000 \mathrm{~m}^{2}$ ⇒ 100 m/200...
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Question: $y=e^{x}+1 \quad: \quad y^{\prime \prime}-y^{\prime}=0$ Solution: $y=e^{x}+1$ Differentiating both sides of this equation with respect tox, we get: $\frac{d y}{d x}=\frac{d}{d x}\left(e^{x}+1\right)$ $\Rightarrow y^{\prime}=e^{x}$ ...(1) Now, differentiating equation (1) with respect tox, we get: $\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^{x}\right)$ $\Rightarrow y^{\prime \prime}=e^{x}$ Substituting the values of $y^{\prime}$ and $y^{\prime \prime}$ in the given differ...
Read More →Two cubes, each of volume 512 cm3
Question: Two cubes, each of volume $512 \mathrm{~cm}^{3}$ are joined end to end. Find the surface area of the resulting cuboid. Solution: Given that, Volume of the cube $=512 \mathrm{~cm}^{3}$ $\Rightarrow$ side $^{3}=512$ $\Rightarrow$ side $^{3}=8^{3}$ ⇒ side = 8 cm Dimensions of the new cuboid formed Length (l) = 8 + 8 = 16 cm, Breadth (b) = 8 cm, Height (h) = 8 cm Surface area = 2(lb + bh + hl) = 2(16 * 8 + 8 * 8 + 16 * 8) $=640 \mathrm{~cm}^{2}$ Therefore, Surface area is $640 \mathrm{~cm}...
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