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Question: $y=e^{2 x}(a+b x)$ Solution: $y=e^{2 x}(a+b x)$ ...(1) Differentiating both sides with respect tox, we get: $y^{\prime}=2 e^{2 x}(a+b x)+e^{2 x} \cdot b$ $\Rightarrow y^{\prime}=e^{2 x}(2 a+2 b x+b)$ ...(2) Multiplying equation (1) with 2 and then subtracting it from equation (2), we get: $y^{\prime}-2 y=e^{2 x}(2 a+2 b x+b)-e^{2 x}(2 a+2 b x)$ $\Rightarrow y^{\prime}-2 y=b e^{2 x}$ ...(3) Differentiating both sides with respect tox, we get: $y^{\prime \prime}-2 y^{\prime}=2 b e^{2 x}$...
Read More →In a ∆A, B, C, D be the angles of a cyclic quadrilateral,
Question: In a ∆A, B, C, Dbe the angles of a cyclic quadrilateral, taken in order, prove that cos(180 A) + cos (180 +B) + cos (180 +C) sin (90 +D) = 0 Solution: $A, B, C$ and $D$ are the angles of a cyclic quadrilateral. $\therefore A+C=180^{\circ}$ and $B+D=180^{\circ}$ $\Rightarrow A=180-C$ and $B=180-D$ Now, LHS $=\cos \left(180^{\circ}-A\right)+\cos \left(180^{\circ}+B\right)+\cos \left(180^{\circ}+C\right)-\sin \left(90^{\circ}+D\right)$ $=-\cos A+[-\cos B]+[-\cos C]-\cos D$ $=-\cos A-\cos ...
Read More →In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm.
Question: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. (Take = 3.14) Solution: Given that Height of cylinder = length of cylindrical pipe = 28 m Radius(r) of circular end of pipe = 5/2 = 2.5 cm = 0.025 m Curved surface area of cylindrical pipe $=2 \pi r h=2 * 3.14 * 0.035 * 30=4.4 \mathrm{~m}^{2}$ Therefore the area of radiating surface of the system is $4.4 \mathrm{~m}^{2}$ or $44000 \mathrm{~cm}^{2}...
Read More →In a triangle ABC,
Question: In a triangle $A B C, N$ is a point on $A C$ such that $B N \perp A C$. If $B N^{2}=A N$. NC, prove that $\angle B=90^{\circ}$. Solution: In $\triangle A B C, B N \perp A C$ Also, $B N^{2}=A N \times N C$ We have to prove that $\angle B=90^{\circ}$. In trianglesABNandBNC, we have $A B^{2}=A N^{2}+B N^{2}$ $B C^{2}=B N^{2}+C N^{2}$ Adding above two equations, we get $A B^{2}+B C^{2}=A N^{2}+C N^{2}+2 B N^{2}$ Since $B N^{2}=A N \cdot N C$ So, $A B^{2}+B C^{2}=A N^{2}+C N^{2}+2 A N \time...
Read More →In a ∆ABC, prove that:
Question: In a ∆ABC,prove that: (i) cos (A+B) + cosC= 0 (ii) $\cos \left(\frac{A+B}{2}\right)=\sin \frac{C}{2}$ (iii) $\tan \frac{A+B}{2}=\cot \frac{C}{2}$ Solution: (i) In $\Delta A B C$ : $A+B+C=\pi$ $\therefore A+B=\pi-C$ Now, LHS $=\cos (A+B)+\cos C$ $=\cos (\pi-C)+\cos C$ $=-\cos (C)+\cos C \quad[\because \cos (\pi-C)=-\cos (C)]$ $=0$ = RHS Hence proved. (ii) In $\Delta A B C$ : $A+B+C=\pi$ $\Rightarrow A+B=\pi-C$ $\Rightarrow \frac{A+B}{2}=\frac{\pi-C}{2}$ $\Rightarrow \frac{A+B}{2}=\frac{...
Read More →Curved surface area of a right circular cylinder is 4.4 m2.
Question: Curved surface area of a right circular cylinder is $4.4 \mathrm{~m}^{2}$. If the radius of the base of the cylinder is $0.7 \mathrm{~m}$. Find its height. (Take $\pi=3.14$ ) Solution: Given that Radius of the base of the cylinder (r) = 0.7 m Curved surface area of cylinder (C.S.A) $=4.4 \mathrm{~m}^{2}$ Let the height of the cylinder be h The curved surface area of a cylinder is given by: 2rh $2 \pi \mathrm{rh}=4.4 \mathrm{~m}^{2}$ $2 * 3.14^{*} 0.7^{*} \mathrm{~h}=4.4 \mathrm{~m}^{2}...
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Question: $y=a e^{3 x}+b e^{-2 x}$ Solution: $y=a e^{3 x}+b e^{-2 x}$ ...(1) Differentiating both sides with respect tox, we get: $y^{\prime}=3 a e^{3 x}-2 b e^{-2 x}$ ...(2) Again, differentiating both sides with respect tox, we get: $y^{\prime \prime}=9 a e^{3 x}+4 b e^{-2 x}$ ...(3) Multiplying equation (1) with (2) and then adding it to equation (2), we get: $\left(2 a e^{3 x}+2 b e^{-2 x}\right)+\left(3 a e^{3 x}-2 b c^{-2 x}\right)=2 y+y^{\prime}$ $\Rightarrow 5 a e^{3 x}=2 y+y^{\prime}$ $...
Read More →In a quadrilateral ABCD,
Question: In a quadrilateral $\mathrm{ABCD}, \angle \mathrm{B}=90^{\circ}$. If $\mathrm{AD}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}$ then prove that $\angle \mathrm{ACD}=90^{\circ}$. Solution: In quadrilateral $A B C D$, we have $\angle B=90^{\circ}$ So, $A C^{2}=A B^{2}+B C^{2} \quad$ (Pythagoras theorem) and $A D^{2}=A B^{2}+B C^{2}+C D^{2} \quad$ (Given) So, $A D^{2}=A B^{2}+B C^{2}+C D^{2}$ $A D^{2}=A C^{2}+C D^{2}$ Hence, $\angle A C D=90^{\circ}$ (Converse of Pythagoras theorem...
Read More →In ∆ABC, ∠ABC = 135°.
Question: In $\triangle \mathrm{ABC}, \angle \mathrm{ABC}=135^{\circ}$. Prove that $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}+4$ ar $(\triangle \mathrm{ABC})$ Solution: We have the following figure. Here $\triangle A D B$ is a right triangle right angled at $D$. Therefore by Pythagoras theorem we have $A B^{2}=A D^{2}+D B^{2}$ Again $\triangle A D C$ is a right triangle right angled at $D$. Therefore, by Pythagoras theorem, we have $A C^{2}=A D^{2}+D C^{2}$ $A C^{2}=A D^{2}+(D B+B C)^{2}$ ...
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Question: $y^{2}=a\left(b^{2}-x^{2}\right)$ Solution: $y^{2}=a\left(b^{2}-x^{2}\right)$ Differentiating both sides with respect tox, we get: $2 y \frac{d y}{d x}=a(-2 x)$ $\Rightarrow 2 y y^{\prime}=-2 a x$ $\Rightarrow y y^{\prime}=-\alpha x$ ...(1) Again, differentiating both sides with respect tox, we get: $y^{\prime} \cdot y^{\prime}+y y^{\prime \prime}=-a$ $\Rightarrow\left(y^{\prime}\right)^{2}+y y^{\prime \prime}=-a$ ...(2) Dividing equation (2) by equation (1), we get: $\frac{\left(y^{\p...
Read More →A wall of length 10 m was to be built across an open ground. The height of the wall is 4m and thickness of the wall is 24 cm.
Question: A wall of length 10 m was to be built across an open ground. The height of the wall is 4m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm 12 cm 8 cm, how many bricks would be required? Solution: Given that, The wall with all its bricks makes up space occupies by it, we need to find the volume of the wall, which is nothing but cuboid. Here, length = 10 m = 1000 cm Thickness = 24 cm Height = 4 m = 400 cm Therefore, volume of the w...
Read More →Prove that in an equilateral triangle,
Question: Prove that in an equilateral triangle, three times the square of a side is equal to four times the square of its altitudes. Solution: Let $A B C$ be an equilateral triangle and let $A D \perp B C$. In $\triangle A D B$ and $\triangle A D C$ we have $A B=A C$ $\angle B=\angle C$ And $\angle A D B=\angle A D C$ $\Rightarrow \triangle A D B \cong \triangle A D C$ So, $B D=D C$ $\Rightarrow B D=D C=\frac{1}{2} B C$ Since $\triangle A D B$ is a right triangle right-angled at $D .$ So $A B^{...
Read More →A godown measures 40 m × 25 m × 10 m.
Question: A godown measures 40 m 25 m 10 m. Find the maximum number of wooden crates each measuring 1.5 m 1.25 m 0.5 m that can be stored in the godown. Solution: Given, Godown length (l1) = 40 m Godown breadth (b1) = 25 m Godown height (h1) = 10 m Volume of the godown = l1*b1*h1= 40 * 25 * 10 $=10000 \mathrm{~m}^{3}$ Wooden crate length (l2) = 1.5 m Wooden crate breadth (b2) = 1.25 m Wooden crate height (h2) = 0.5 m Volume of the wooden crate = l2* b2* h2= 1.5 * 1.25 * 0.5 $=0.9375 \mathrm{~m}^...
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Question: $\frac{x}{a}+\frac{y}{b}=1$ Solution: $\frac{x}{a}+\frac{y}{b}=1$ Differentiating both sides of the given equation with respect tox, we get: $\frac{1}{a}+\frac{1}{b} \frac{d y}{d x}=0$ $\Rightarrow \frac{1}{a}+\frac{1}{b} y^{\prime}=0$ Again, differentiating both sides with respect tox, we get: $0+\frac{1}{b} y^{\prime \prime}=0$ $\Rightarrow \frac{1}{b} y^{\prime \prime}=0$ $\Rightarrow y^{\prime \prime}=0$ Hence, the required differential equation of the given curve is $y^{\prime \pr...
Read More →A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in Fig.
Question: A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in Fig. 18.12. If the edge of each cube is 3cm, find the volume of the structure built by the child. Solution: Volume of each cube = edge * edge * edge = 3 * 3 * 3 $=27 \mathrm{~cm}^{3}$ Number of cubes in the structure = 15 Therefore, volume of the structure = 27*15 $=405 \mathrm{~cm}^{3}$...
Read More →The numbers of arbitrary constants in the particular solution of a differential equation of third order are
Question: The numbers of arbitrary constants in the particular solution of a differential equation of third order are: (A) 3 (B) 2 (C) 1 (D) 0 Solution: In a particular solution of a differential equation, there are no arbitrary constants. Hence, the correct answer is D...
Read More →In ∆ABC, AD is a median.
Question: In $\triangle \mathrm{ABC}, \mathrm{AD}$ is a median. Prove that $\mathrm{AB}^{2}+\mathrm{AC}^{2}=2 \mathrm{AD}^{2}+2 \mathrm{DC}^{2}$. Solution: We have the following figure. Since triangleABMandACMare right triangles right angled atM $A B^{2}=A M^{2}+B M^{2} \ldots \ldots(i)$ $A C^{2}=A M^{2}+C M^{2} \ldots \ldots(i i)$ Adding (i) and (ii), we get $A B^{2}+A C^{2}=2 A M^{2}+B M^{2}+C M^{2}$ Since in triangleADMwe have $A D^{2}=D M^{2}+A M^{2}$ So, $A B^{2}+A C^{2}=2\left(A D^{2}-D M^...
Read More →The numbers of arbitrary constants in the general solution
Question: The numbers of arbitrary constants in the general solution of a differential equation of fourth order are: (A)0(B)2(C)3(D)4 Solution: We know that the number of constants in the general solution of a differential equation of ordernis equal to its order. Therefore, the number of constants in the general equation of fourth order differential equation is four. Hence, the correct answer is D....
Read More →A village having a population of 4000 requires 150 liters of water per head per day.
Question: A village having a population of 4000 requires 150 liters of water per head per day. It has a tank measuring 20 m 15 m 6 m. For how many days will the water of this tank last? Solution: Given that, Length of the cuboidal tank (l) = 20 m Breadth of the cuboidal tank (b) = 15 m Height of the cuboidal tank (h) = 6 m Capacity of the tank = l * b * h = 20 * 15 * 6 $=1800 \mathrm{~m}^{3}$ = 1800000 litres Water consumed by the people of village in one day = 4000 *150 litres = 600000 litres L...
Read More →Water in a rectangular reservoir having base 80 m by 60 m is 6.5m deep.
Question: Water in a rectangular reservoir having base 80 m by 60 m is 6.5m deep. In what time can the water be pumped by a pipe of which the cross-section is a square of side 20 cm if the water runs through the pipe at the rate of 15km/hr. Solution: Flow of water = 15 km/hr = 15000 m/hr Volume of water coming out of the pipe in one hour, $\Rightarrow 20 / 100 * 20 / 100 * 15000=600 \mathrm{~m}^{3}$ Volume of the tank = 80 * 60 * 6.5 $=31200 \mathrm{~m}^{3}$ Time taken to empty the tank $=\frac{...
Read More →A rectangular tank is 80 m long and 25 m broad.
Question: A rectangular tank is $80 \mathrm{~m}$ long and $25 \mathrm{~m}$ broad. Water flows into it through a pipe whose cross-section is $25 \mathrm{~cm}^{2}$, at the rate of $16 \mathrm{~km}$ per hour. How much the level of the water rises in the tank in 45 minutes? Solution: Consider 'h' be the rise in water level. Volume of water in rectangular tank $=8000 * 2500 * \mathrm{~h} \mathrm{~cm}^{2}$ Cross-sectional area of the pipe $=25 \mathrm{~cm}^{2}$ Water coming out of the pipe forms a cub...
Read More →ABCD is a rectangle. Points M and N are on BD such that AM ⊥ BD and CN ⊥ BD.
Question: $A B C D$ is a rectangle. Points $M$ and $N$ are on $B D$ such that $A M \perp B D$ and $C N \perp B D$. Prove that $B M^{2}+B N^{2}=D M^{2}+D N^{2}$. Solution: Given: A rectangle ABCD where AMBD and CNBD. To prove: $\mathrm{BM}^{2}+\mathrm{BN}^{2}=\mathrm{DM}^{2}+\mathrm{DN}^{2}$ Proof: Apply Pythagoras Theorem in $\triangle \mathrm{AMB}$ and $\triangle \mathrm{CND}$, $\mathrm{AB}^{2}=\mathrm{AM}^{2}+\mathrm{MB}^{2}$ $\mathrm{CD}^{2}=\mathrm{CN}^{2}+\mathrm{ND}^{2}$ Since $A B=C D, A ...
Read More →A field is in the form of a rectangular length 18m and width 15m.
Question: A field is in the form of a rectangular length 18m and width 15m. A pit 7.5m long, 6m broad and 0.8m deep, is dug in a corner of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised. Solution: Let 'h' metres be the rise in the level of field Volume of earth taken out from the pit $=7.5^{*} 6^{*} 0.8=36 \mathrm{~m}^{3}$ Area of the field on which the earth taken out is to be spread $=18$ * $...
Read More →In ∆ABC, given that AB = AC and BD ⊥ AC
Question: In $\triangle A B C$, given that $A B=A C$ and $B D \perp A C$. Prove that $B C^{2}=2 A C . C D$ Solution: Since $\triangle A D B$ is right triangle right angled at $D$ $A B^{2}=A D^{2}+B D^{2}$ Substitute $A B=A C$ $A C^{2}=A D^{2}+B D^{2}$ $A C^{2}=(A C-D C)^{2}+B D^{2}$ $A C^{2}=A C^{2}+D C^{2}-2 A C \cdot D C+B D^{2}$ $2 A C \cdot D C=A C^{2}-A C^{2}+D C^{2}+B D^{2}$ $2 A C \cdot D C=D C^{2}+B D^{2}$ Now, in $\triangle B D C$, we have $C D^{2}+B D^{2}=B C^{2}$ Therefore, $2 A C \cd...
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Question: $y=\sqrt{a^{2}-x^{2}} x \in(-a, a) \quad: x+y \frac{d y}{d x}=0(y \neq 0)$ Solution: $y=\sqrt{a^{2}-x^{2}}$ Differentiating both sides of this equation with respect tox, we get: $\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^{2}-x^{2}}\right)$ $\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}\left(a^{2}-x^{2}\right)$ $=\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x)$ $=\frac{-x}{\sqrt{a^{2}-x^{2}}}$ Substituting the value of $\frac{d y}{d x}$ in the given differential e...
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