If
Question: If $\frac{\pi}{2}x\pi$ and $\sqrt{\frac{1+\sin x}{1-\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x}}=k$ sec $x$, then $k=$ Solution: If $\frac{\pi}{2}x\pi$ $\sqrt{\frac{1+\sin x}{1-\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x}}=k \sec x \quad$ (given) L.H. $\mathrm{S}=\sqrt{\frac{1+\sin x}{1-\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x}}$ $=\sqrt{\frac{1+\sin x}{1-\sin x} \times \frac{1+\sin x}{1+\sin x}}+\sqrt{\frac{1-\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}}$ $=\sqrt{\left[\frac{(1+\sin x)^...
Read More →∆ABC ∼ ∆DEF, ar(∆ABC) = 9 cm2, ar(∆DEF) = 16 cm2.
Question: $\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}, \operatorname{ar}(\triangle \mathrm{ABC})=9 \mathrm{~cm}^{2}, \operatorname{ar}(\triangle \mathrm{DEF})=16 \mathrm{~cm}^{2} .$ If $\mathrm{BC}=2.1 \mathrm{~cm}$, then the measure of $\mathrm{EF}$ is (a) 2.8 cm(b) 4.2 cm(c) 2.5 cm(d) 4.1 cm Solution: Given: $\operatorname{Ar}(\triangle \mathrm{ABC})=9 \mathrm{~cm}^{2}, \operatorname{Ar}(\triangle \mathrm{DEF})=16 \mathrm{~cm}^{2}$ and $\mathrm{BC}=2.1 \mathrm{~cm}$ To find: measure of...
Read More →A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid
Question: A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm 22 cm 14 cm. Find the rise in the level of the water when the solid is completely submerged. Solution: Given data is as follows: Diameter of cylinder = 56 cm Dimensions of rectangular block = 32 cm22 cm14 cm We have to find the raise in the level of water in the cylinder. First let us find the raise in the level of water in the cylinder. Diameter i...
Read More →If
Question: If $\frac{\pi}{2}x\pi$ and $\sqrt{\frac{1+\sin x}{1-\sin x}}=k$ sec $x$, then $k=$ Solution: If $\frac{\pi}{2}x\pi$ Given $\sqrt{\frac{1+\sin x}{1-\sin x}}=k$ sec $x$ L.H.S $\sqrt{\frac{1+\sin x}{1-\sin x}}=\sqrt{\frac{1+\sin x}{1-\sin x} \times \frac{1+\sin x}{1+\sin x}}$ (By multiplying dividing by $1+\sin x$ ) $=\sqrt{\frac{(1+\sin x)}{1-\sin ^{2} x}}$(correction)...
Read More →If in two triangles ABC and DEF, ABDE=BCFE=CAFD, then
Question: If in two triangles ABC and DEF,ABDE=BCFE=CAFD, then (a) ∆FDE ∆CAB(b) ∆FDE ∆ABC(c) ∆CBA ∆FDE(d) ∆BCA ∆FDE Solution: We know that if two triangles are similar if their corresponding sides are proportional. It is given that $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles such that $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{CA}}{\mathrm{FD}}$. A=DB=EC=F $\therefore \triangle \mathrm{CAB} \sim \triangle \mathrm{FDE}$ Hence the cor...
Read More →Water flows out through a circular pipe whose internal diameter is 2 cm,
Question: Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into a cylindrical tank. The water is collected in a cylindrical vessel radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes? Solution: Given data is as follows: Internal diameter of the pipe = 2 cm Water flow rate through the pipe =6 m/sec Radius of the tank = 60 cm Time = 30 minutes The volume of water that flows for 1 sec through the pipe at the rat...
Read More →The value of tan 1° tan 2° tan 3°
Question: The value of tan 1 tan 2 tan 3 _________ tan 89 is _________. Solution: tan 1 tan 2 tan 3 ........ tan 89 Since 89 = 90 1 88 = 90 2 ........... 46 = 90 44 and tan 89 = tan (90 1) = cot 1 tan 88 = tan (90 2) = cot 2 tan 46 =tan (90 44) = cot 44 tan 45 = 1 tan 1 tan 2 tan 3 .......... tan 45 cot 44 .....cot 3 cot 2 cot 1 $=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \frac{1}{\tan 4.1^{\circ}} \cdots \frac{1}{\tan 1^{\circ}}$ = 1 Hence, tan 1 tan 2 ....... tan 89 = 1....
Read More →If in ∆ABC and ∆DEF, ABDE=BCFD, then ∆ABC ∼ ∆DEF when
Question: If in ∆ABC and ∆DEF,ABDE=BCFD, then ∆ABC ∆DEF when (a) A = F(b) A = D(c) B = D(d) B = E Solution: Given: In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}, \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{FD}}$. We know that if in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar. Then,$\angle \mathrm{B}=\angle \mathrm{D}$ Hence, $\triangle \mathrm{ABC}$ is similar to $\triangle \mathrm{...
Read More →The value of cos 1° cos 2° cos 3°
Question: The value of cos 1 cos 2 cos 3 _______cos 179 is ____________. Solution: cos 1 cos 2 cos 3 ....... cos 179 = cos 1 cos 2 cos 3 ......... cos 90 cos 91 ......... cos 179 = cos 1 cos 2 cos 3 ..... 0 cos 91 .......... cos 179 (∵ cos 90 = 0) = 0 finite = 0 Hence, cos 1 cos 2 cos 3 ......... cos 179 = 0...
Read More →If sin x + cos x = a,
Question: If sinx+ cosx=a, then sin6x+ cos6x= __________. Solution: Given sinx+ cosx=a Squaring both sides, (sinx+ cosx)2=a2 $\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2}$ i. e $1+2 \sin x \cos x=a^{2}$ i. e $2 \sin x \cos x=a^{2}-1$ ...(1) Using identity, we have $a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$ $\sin ^{6} x+\cos ^{6} x=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}$ $=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \cos ^{2} x \sin ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)$ $=(1)^{3}...
Read More →In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then
Question: In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then (a) $\mathrm{AQ}^{2}+\mathrm{CP}^{2}=2\left(\mathrm{AC}^{2}+\mathrm{PQ}^{2}\right)$ (b) $2\left(\mathrm{AQ}^{2}+\mathrm{CP}^{2}\right)=\mathrm{AC}^{2}+\mathrm{PQ}^{2}$ (c) $\mathrm{AQ}^{2}+\mathrm{CP}^{2}=\mathrm{AC}^{2}+\mathrm{PQ}^{2}$ (d) $\mathrm{AQ}+\mathrm{CP}=12 \mathrm{AC}+\mathrm{PQ}$ Solution: Disclaimer: There is mistake in the problem. The question should be "In a righ...
Read More →If sin x + cos x = a,
Question: If sinx+ cosx=a, then sin6x+ cos6x= __________. Solution: Given sinx+ cosx=a Squaring both sides, (sinx+ cosx)2=a2 $\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x=a^{2}$ i. e $1+2 \sin x \cos x=a^{2}$ i. e $2 \sin x \cos x=a^{2}-1$ ...(1) Using identity, we have $a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b)$ $\sin ^{6} x+\cos ^{6} x=\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}$ $=\left(\sin ^{2} x+\cos ^{2} x\right)^{3}-3 \cos ^{2} x \sin ^{2} x\left(\sin ^{2} x+\cos ^{2} x\right)$ $=(1)^{3}...
Read More →The difference between inside and outside surfaces of a cylindrical tube is 14 cm long is 88 sq.cm.
Question: The difference between inside and outside surfaces of a cylindrical tube is 14 cm long is 88 sq.cm. If the volume of the tube is 176 cubic cm, Find the inner and outer radii of the tube. Solution: Let, R be the outer radius R be the inner radius Here, h = 14 cm 2Rh - 2rh = 88 ⟹ 2h(R r) = 88 ⟹ 2 * 22/7 * 14(R r) = 88 ⟹ (R r) = 1cm .....1 Volume of tube $=\pi * R^{2} * h-\pi * r^{2} \star h$ $176=\pi h\left(R^{2}-r^{2}\right)$ $176=22 / 7^{*} 14\left(R^{2}-r^{2}\right)$ $\Rightarrow\left...
Read More →If E is a point on side CA of an equilateral triangle ABC such that BE ⊥ CA,
Question: If $E$ is a point on side $C A$ of an equilateral triangle $A B C$ such that $B E \perp C A$, then $A B^{2}+B C^{2}+C A^{2}=$ (a) $2 \mathrm{BE}^{2}$ (b) $3 \mathrm{BE}^{2}$ (c) $4 \mathrm{BE}^{2}$ (d) $6 \mathrm{BE}^{2}$ Solution: In triangle $\mathrm{ABC}, \mathrm{E}$ is a point on $\mathrm{AC}$ such that $B E \perp A C$. We need to find $A B^{2}+B C^{2}+A C^{2}$. Since $B E \perp A C, C E=A E=\mathrm{AC} 2$ (In a equilateral triangle, the perpendicular from the vertex bisects the ba...
Read More →Given x > 0,
Question: Given $x0$, the value of $f(x)=-3 \cos \sqrt{3+x+x^{2}}$ lie in the interval ______________ . Solution: Given $x0$, $f(x)=3 \cos \sqrt{3+x+x^{2}}$ Since $-1 \leq \cos \theta \leq 1$ for any $\theta$ i. e $-1 \leq \cos \left(\sqrt{3+x+x^{2}}\right) \leq 1$ i. e $-3 \leq 3 \cos \left(\sqrt{3+x+x^{2}}\right) \leq 3$ i. e $3 \cos \left(\sqrt{3+x+x^{2}}\right) \in[-3,3]$...
Read More →Show that
Question: $\frac{d y}{d x}=y \tan x ; y=1$ when $x=0$ Solution: $\frac{d y}{d x}=y \tan x$ $\Rightarrow \frac{d y}{y}=\tan x d x$ Integrating both sides, we get: $\int \frac{d y}{y}=-\int \tan x d x$ $\Rightarrow \log y=\log (\sec x)+\log \mathrm{C}$ $\Rightarrow \log y=\log (\mathrm{C} \sec x)$ $\Rightarrow y=\mathrm{C} \sec x$ ...(1) Now, $y=1$ when $x=0$. $\Rightarrow 1=\mathrm{C} \times \mathrm{sec} 0$ $\Rightarrow 1=\mathrm{C} \times 1$ $\Rightarrow \mathrm{C}=1$ Substituting C = 1 in equat...
Read More →The value of 3 (sin x – cos x)
Question: The value of 3 (sinx cosx)4+ 6 (sinx+ cosx)2+ 4 (sin6x+ cos6x) is ___________ Solution: 3 (sinx cosx)4+ 6 (sinx+ cosx)2+ 4 (sin6x+ cos6x) $=3\left[(\sin x-\cos x)^{2}\right]^{2}+6\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)+4\left[\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\right]$ $=3\left[\sin ^{2} x+\cos ^{2} x-2 \cos x \sin x\right]^{2}+6(1+2 \sin x \cos x)$ $+4\left[\left(\sin ^{2} x+\cos _{1}^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^...
Read More →The value of 3 (sin x – cos x)
Question: The value of 3 (sinx cosx)4+ 6 (sinx+ cosx)2+ 4 (sin6x+ cos6x) is ___________ Solution: 3 (sinx cosx)4+ 6 (sinx+ cosx)2+ 4 (sin6x+ cos6x) $=3\left[(\sin x-\cos x)^{2}\right]^{2}+6\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)+4\left[\left(\sin ^{2} x\right)^{3}+\left(\cos ^{2} x\right)^{3}\right]$ $=3\left[\sin ^{2} x+\cos ^{2} x-2 \cos x \sin x\right]^{2}+6(1+2 \sin x \cos x)$ $+4\left[\left(\sin ^{2} x+\cos _{1}^{2} x\right)^{3}-3 \sin ^{2} x \cos ^{2} x\left(\sin ^{2} x+\cos ^...
Read More →A well with 14 m diameter is dug 8 m deep.
Question: A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment. Solution: Let, r be the radius of well h be the height of well here, h = 8 m 2r = 14 ⟹ r =14/2 = 7m Volume of well $=r^{2} * h$ =22/7 * 7 * 7 * 8 = 22 * 56 $=1232 \mathrm{~m}^{3}$ Let,rebe the radius of embankment hebe the height of embankment Volume of well = Volume of embankment $1232 \mathrm{~m}^{3}=\pi...
Read More →If ABC is a right triangle right-angled at B and M,
Question: If $\mathrm{ABC}$ is a right triangle right-angled at $\mathrm{B}$ and $\mathrm{M}, \mathrm{N}$ are the mid-points of $\mathrm{AB}$ and $\mathrm{BC}$ respectively, then $4\left(\mathrm{AN}^{2}+\mathrm{CM}^{2}\right)=$ (a) $4 \mathrm{AC}^{2}$ (b) $5 \mathrm{AC}^{2}$ (c) $54 \mathrm{AC} 2$ (d) $6 \mathrm{AC}^{2}$ Solution: M is the mid-point of AB. $\therefore B M=A B 2$ N is the mid-point of BC. $\therefore B N=B C 2$ Now, $A N^{2}+C M^{2}=\left(A B^{2}+\left(\frac{1}{2} B C\right)^{2}\...
Read More →The value of tan x+cot
Question: The value of $\tan x+\cot (\pi+x)+\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi-x)$ is____________ . Solution: $\tan x+\cot (\pi+x)+\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi-x)=\tan x+(+\cot x)+\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi-x)$ Since $\cot (\pi+x)=\frac{1}{\tan (\pi+x)}=\frac{1}{+\tan x}=+\cot x$ and $\cot \left(\frac{\pi}{2}+x\right)=\frac{1}{\tan \left(\frac{\pi}{2}+x\right)}=\frac{1}{-\cot x}=-\tan x$ and $\cot (2 \pi-x)=\frac{1}{\tan (2 \pi-x)}=\frac{1}{-\tan x}...
Read More →Show that
Question: $\cos \left(\frac{d y}{d x}\right)=a(a \in R) ; y=1$ when $x=0$ Solution: $\cos \left(\frac{d y}{d x}\right)=a$ $\Rightarrow \frac{d y}{d x}=\cos ^{-1} a$ $\Rightarrow d y=\cos ^{-1} a d x$ Integrating both sides, we get: $\int d y=\cos ^{-1} a \int d x$ $\Rightarrow y=\cos ^{-1} a \cdot x+\mathrm{C}$ $\Rightarrow y=x \cos ^{-1} a+\mathrm{C}$ ...(1) Now, $y=1$ when $x=0$ $\Rightarrow 1=0 \cdot \cos ^{-1} a+C$ $\Rightarrow C=1$ Substituting C = 1 in equation (1), we get: $y=x \cos ^{-1}...
Read More →∆ABC is a right triangle right-angled at A and AD ⊥ BC. Then, BDDC=
Question: ∆ABC is a right triangle right-angled at A and AD BC. Then,BDDC= (a)ABAC2(b)ABAC(c)ABAD2(d)ABAD Solution: Given: In $\triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ}$ and $\mathrm{AD} \perp \mathrm{BC}$. To find: BD: DC $\angle \mathrm{CAD}+\angle \mathrm{BAD}=90^{\circ} \quad \ldots .1 \angle \mathrm{BAD}+\angle \mathrm{ABD}=90^{\circ} \quad \ldots . .2 \quad \angle \mathrm{ADB}=90^{\circ}$ From $(1)$ and $(2), \angle \mathrm{CAD}=\angle \mathrm{ABD}$ In $\triangle A D B$ and $\tr...
Read More →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 $\sin ^{2} x+\operatorname{cosec}^{2} x=2$...
Read More →The trunk of a tree is cylindrical and its circumference is 176 cm.
Question: The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the tree is 3 m. Find the volume of the timber that can be obtained from the trunk. Solution: We know that, circumference = 2r ⟹ 176 = 2r ⟹ r = 176/2 $\Rightarrow r=\frac{176 * 7}{2 * 22}$ ⟹ r = 28 cm Here, height (h) = 3 m = 300 cm Volume of timber $=\mathrm{r}^{2}$ * $\mathrm{h}$ = 22/7 * 28 * 28 * 300 $=44 * 8400=739200 \mathrm{~cm}^{3}$ (or) $0.7392 \mathrm{~m}^{3}$...
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