Find the general solution of the differential equation

Question: Find the general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$ Solution: $\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0$ $\Rightarrow \frac{d y}{d x}=-\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}$ $\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=\frac{-d x}{\sqrt{1-x^{2}}}$ Integrating both sides, we get: $\sin ^{-1} y=-\sin ^{-1} x+\mathrm{C}$ $\Rightarrow \sin ^{-1} x+\sin ^{-1} y=\mathrm{C}$...

The areas of two similar triangles are 100 cm2 and 49 cm2 respectively.

Question: The areas of two similar triangles are $100 \mathrm{~cm}^{2}$ and $49 \mathrm{~cm}^{2}$ respectively. If the altitude of the bigger triangle is $5 \mathrm{~cm}$, find the corresponding altitude of the other. Solution: Given: The area of two similar triangles is $100 \mathrm{~cm}^{2}$ and $49 \mathrm{~cm}^{2}$ respectively. If the altitude of bigger triangle is $5 \mathrm{~cm}$ To find: their corresponding altitude of other triangle We know that the ratio of areas of two similar triangl...

In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC.

Question: In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC. Solution: GIVEN: In ΔABC, P divides the side AB such that AP : PB = 1 : 2, Q is a point on AC such that PQ || BC. TO FIND: The ratio of the areas of ΔAPQ and the trapezium BPQC. In ΔAPQ and ΔABC APQ=BCorrespondinganglesPAQ=BACCommonSo,∆APQ~∆ABC (AA Similarity) We know that the ratio of areas of two similar triangles is equal to the rat...

Question: $\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}=$ (a) tan 55 (b) cot 55 (c) tan 35 (d) cot 35 Solution: (a) tan 55 $\frac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 10^{\circ}-\sin 10^{\circ}}$ $=\frac{1+\tan 10^{\circ}}{1-\tan 10^{\circ}} \quad\left[\right.$ Dividing the numerator and denominator by $\left.\cos 10^{\circ}\right]$ $=\frac{\tan 45^{\circ}+\tan 10^{\circ}}{1-\tan 45^{\circ} \times \tan 10^{\circ}}$ $=\tan \left(45^{\circ}+10^{\circ}\right) \quad\left[...

Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Question: Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes. Solution: The equation of a circle in the first quadrant with centre (a,a) and radius (a)which touches the coordinate axes is: $(x-a)^{2}+(y-a)^{2}=a^{2}$ ...(1) Differentiating equation (1) with respect tox, we get: $2(x-a)+2(y-a) \frac{d y}{d x}=0$ $\Rightarrow(x-a)+(y-a) y^{\prime}=0$ $\Rightarrow x-a+y y^{\prime}-a y^{\prime}=0$ $\Rightarrow x+y y^{\prime}-a\left(1+y^{\pri...

ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that :

Question: ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that :(i) $\triangle \mathrm{AOB} \sim \Delta \mathrm{COD}$ (ii) If $\mathrm{OA}=6 \mathrm{~cm}, \mathrm{OC}=8 \mathrm{~cm}$, Find: (a)Area∆AOBArea∆COD(b)Area∆AODArea∆COD Solution: Given: ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. To prove: (i) $\triangle \mathrm{AOB} \sim \Delta \mathrm{COD}$ (ii) If $\mathrm{OA}=6 \mathrm{~cm}, \mathrm{OC}=8 \mathrm{~cm}$ To find: ...

If cot (α + β) = 0,

Question: If cot ( + ) = 0, sin ( + 2) is equal to (a) sin (b) cos 2 (c) cos (d) sin 2 Solution: (a) sin Given: $\cot (\alpha+\beta)=0$ $\Rightarrow \frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)}=0$ $\Rightarrow \cos (\alpha+\beta)=0$ $\Rightarrow \alpha+\beta=\frac{\pi}{2}$ Therefore, $\sin (\alpha+2 \beta)=\sin (\alpha+\alpha+\beta)$ =sin...

If cot (α + β) = 0,

Question: If cot ( + ) = 0, sin ( + 2) is equal to (a) sin (b) cos 2 (c) cos (d) sin 2 Solution: (a) sin Given: $\cot (\alpha+\beta)=0$ $\Rightarrow \frac{\cos (\alpha+\beta)}{\sin (\alpha+\beta)}=0$ $\Rightarrow \cos (\alpha+\beta)=0$ $\Rightarrow \alpha+\beta=\frac{\pi}{2}$ Therefore, $\sin (\alpha+2 \beta)=\sin (\alpha+\alpha+\beta)$ =sin...

In the given figure, ∆ABC and ∆DBC are on the same base BC.

Question: In the given figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect at O, prove thatArea∆ABCArea∆DBC=AODO. Solution: Given: ΔABC and ΔDBC are on the same base BC. AD and BC intersect at O. Prove that: $\frac{\operatorname{Ar}(\triangle \mathrm{ABC})}{\operatorname{Ar}(\triangle \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}$ Construction: Draw $\mathrm{AL} \perp \mathrm{BC}$ and $\mathrm{DM} \perp \mathrm{BC}$. Now, in ΔALO and ΔDMO, we have $\angle \mathrm{ALO}=\angle \...

Prove that

Question: Prove that $x^{2}-y^{2}=c\left(x^{2}+y^{2}\right)^{2}$ is the general solution of differential equation $\left(x^{3}-3 x y^{2}\right) d x=\left(y^{3}-3 x^{2} y\right) d y$, where $c$ is a parameter. Solution: $\left(x^{3}-3 x y^{2}\right) d x=\left(y^{3}-3 x^{2} y\right) d y$ $\Rightarrow \frac{d y}{d x}=\frac{x^{3}-3 x y^{2}}{y^{3}-3 x^{2} y}$ $\ldots(1)$ This is a homogeneous equation. To simplify it, we need to make the substitution as: $y=v x$ $\Rightarrow \frac{d}{d x}(y)=\frac{d}...

If cos P

Question: If $\cos P=\frac{1}{7}$ and $\cos Q=\frac{13}{14}$, where $P$ and $Q$ both are acute angles. Then, the value of $P-Q$ is (a) $\frac{\pi}{6}$ (b) $\frac{\pi}{3}$ (c) $\frac{\pi}{4}$ (d) $\frac{\pi}{12}$ Solution: (b) $60^{\circ}=\frac{\pi}{3}$ $\cos P=\frac{1}{7}, \quad \cos Q=\frac{13}{14}$ Therefore, $\sin P=\sqrt{1-\frac{1}{49}}=\frac{4 \sqrt{3}}{7}$ and $\sin Q=\sqrt{1-\frac{169}{196}}=\frac{3 \sqrt{3}}{14}$ Hence, $\tan P=4 \sqrt{3}, \quad \tan Q=\frac{3 \sqrt{3}}{13}$ $\cos (P-Q)=...

In ∆ABC, D and E are the mid-points of AB and AC respectively.

Question: In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC. Solution: Given: In ΔABC, D and E are the midpoints of AB and AC respectively. To find: Ratio of the areas of ΔADE and ΔABC. Since it is given that D and E are the midpoints of AB and AC, respectively. Therefore, DE || BC (Converse of mid-point theorem)Also,DE=12BC In ΔADE and ΔABC ADE=BCorrespondinganglesDAE=BACCommonSo,∆ADE~∆ABC (AA Similarity) We know that the ratio of areas ...

If A + B + C = π,

Question: If $A+B+C=\pi$, then $\frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}$ is equal to (a) tanAtanBtanC (b) 0 (c) 1 (d) None of these Solution: (c) 1 = 180 Using tan(180 A) = -tanA, we get: $C=\pi-(A+B)$ Now, $\frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}$ $=\frac{\tan A+\tan B+\tan [\pi-(A+B)]}{\tan A \tan B \tan [\pi-(A+B)]}$ $=\frac{\tan A+\tan B-\tan (A+B)}{-\tan A \tan B t \operatorname{an}(A+B)}$ $=\frac{\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A \tan B}}{-\tan A \tan B \times...

If A + B + C = π,

Question: If $A+B+C=\pi$, then $\frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}$ is equal to (a) tanAtanBtanC (b) 0 (c) 1 (d) None of these Solution: (c) 1 = 180 Using tan(180 A) = -tanA, we get: $C=\pi-(A+B)$ Now, $\frac{\tan A+\tan B+\tan C}{\tan A \tan B \tan C}$ $=\frac{\tan A+\tan B+\tan [\pi-(A+B)]}{\tan A \tan B \tan [\pi-(A+B)]}$ $=\frac{\tan A+\tan B-\tan (A+B)}{-\tan A \tan B t \operatorname{an}(A+B)}$ $=\frac{\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A \tan B}}{-\tan A \tan B \times...

In the given figure, DE || BC

Question: In the given figure, DE || BC (i) If $\mathrm{DE}=4 \mathrm{~cm}, \mathrm{BC}=6 \mathrm{~cm}$ and Area $(\triangle \mathrm{ADE})=16 \mathrm{~cm}^{2}$, find the area of $\triangle \mathrm{ABC}$. (ii) If $D E=4 \mathrm{~cm}, B C=8 \mathrm{~cm}$ and Area $(\triangle A D E)=25 \mathrm{~cm}^{2}$, find the area of $\triangle A B C$. (iii) If $\mathrm{DE}: \mathrm{BC}=3: 5$. Calculate the ratio of the areas of $\triangle \mathrm{ADE}$ and the trapezium $\mathrm{BCED}$. Solution: In the given ...

tan 3A − tan 2A − tan A =

Question: tan 3A tan 2A tanA= (a) tan 3Atan 2AtanA (b) tan 3Atan 2AtanA (c) tanAtan 2A tan 2Atan 3A tan 3AtanA (d) None of these Solution: (a) tan 3Atan 2Atan A $3 A=2 A+A$ $\Rightarrow \tan 3 A=\tan (2 A+A)$ $=\frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}$ $\Rightarrow \tan 3 A-\tan 3 A \tan 2 A \tan A=\tan 2 A+\tan A$ $\Rightarrow \tan 3 A-\tan 2 A-\tan A=\tan 3 A \tan 2 A \tan A$...

If in ∆ABC, tan A + tan B + tan C = 6,

Question: If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C = (a) 6 (b) 1 (c) $\frac{1}{6}$ (d) None of these Solution: (c) $\frac{1}{6}$ In triangle ABC, $A+B+C=\pi$ We know that $\tan (A+B+C)=\frac{\tan \mathrm{A}+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}$ and $\tan \pi=0$ $\therefore \tan A+\tan B+\tan C-\tan A \tan B \tan C=0$ $\tan A+\tan B+\tan C=\tan A \tan B \tan C$ If tanA+tanB+tanC=6, tanAtanBtanC=6 $\Rightarrow \frac{1}{\tan A \tan B ...

ABC is a triangle in which ∠A = 90°, AN ⊥ BC,

Question: $\mathrm{ABC}$ is a triangle in which $\angle \mathrm{A}=90^{\circ}, \mathrm{AN} \perp \mathrm{BC}, \mathrm{BC}=12 \mathrm{~cm}$ and $\mathrm{AC}=5 \mathrm{~cm}$. Find the ratio of the area of $\triangle \mathrm{ANC}$ and $\triangle \mathrm{ABC}$. Solution: Given: In $\triangle \mathrm{ABC}, \angle A=90^{\circ}, \mathrm{AN} \perp \mathrm{BC}, \mathrm{BC}=12 \mathrm{~cm}$ and $\mathrm{AC}=5 \mathrm{~cm}$. TO FIND: Ratio of the triangles $\triangle A N C$ and $\triangle A B C$. In ∆ANCan...

Form the differential equation representing the family of curves given by

Question: Form the differential equation representing the family of curves given by $(x-a)^{2}+2 y^{2}=a^{2}$ where $a$ is an arbitrary constant. Solution: $(x-a)^{2}+2 y^{2}=a^{2}$ $\Rightarrow x^{2}+a^{2}-2 a x+2 y^{2}=a^{2}$ $\Rightarrow 2 y^{2}=2 a x-x^{2}$ $\ldots(1)$ Differentiating with respect tox, we get: $2 y \frac{d y}{d x}=\frac{2 a-2 x}{2}$ $\Rightarrow \frac{d y}{d x}=\frac{a-x}{2 y}$ $\Rightarrow \frac{d y}{d x}=\frac{2 a x-2 x^{2}}{4 x y}$ $\ldots(2)$ From equation (1), we get: $...

If 3 sin x + 4 cos x = 5,

Question: If 3 sinx+ 4 cosx= 5, then 4 sinx 3 cosx= (a) 0 (b) 5 (c) 1 (d) None of these Solution: (a) 0 $3 \sin x+4 \cos x=5$ $\frac{3}{5} \sin x+\frac{4}{5} \cos x=1$ Let $\cos \alpha=\frac{3}{5}$ and $\sin \alpha=\frac{4}{5}$. $\therefore \cos \alpha \sin x+\sin \alpha \cos x=1$ $\Rightarrow \alpha+x=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{2}-\alpha$ ....(1) We have to find the value of $4 \sin x-3 \cos x$. $4 \sin \left(\frac{\pi}{2}-\alpha\right)-3 \cos \left(\frac{\pi}{2}-\alpha\right) \qu...

If 3 sin x + 4 cos x = 5,

Question: If 3 sinx+ 4 cosx= 5, then 4 sinx 3 cosx= (a) 0 (b) 5 (c) 1 (d) None of these Solution: (a) 0 $3 \sin x+4 \cos x=5$ $\frac{3}{5} \sin x+\frac{4}{5} \cos x=1$ Let $\cos \alpha=\frac{3}{5}$ and $\sin \alpha=\frac{4}{5}$. $\therefore \cos \alpha \sin x+\sin \alpha \cos x=1$ $\Rightarrow \alpha+x=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{2}-\alpha$ ....(1) We have to find the value of $4 \sin x-3 \cos x$. $4 \sin \left(\frac{\pi}{2}-\alpha\right)-3 \cos \left(\frac{\pi}{2}-\alpha\right) \qu...

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively.

Question: The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas. Solution: Given: The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. To find: Ratio of areas of triangle. We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes. $\frac{\operatorname{ar}(\text { triangle } 1)}{\operatorname{ar}(\text { triangle } 2)}=\left(\frac{\text { al...

If tan A

Question: If $\tan A=\frac{a}{a+1}$ and $\tan B=\frac{1}{2 a+1}$, then the value of $A+B$ is (a) 0 (b) $\frac{\pi}{2}$ (c) $\frac{\pi}{3}$ (d) $\frac{\pi}{4}$ Solution: (d) $\frac{\pi}{4}$ $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$ $=\frac{\frac{a}{a+1}+\frac{1}{2 a+1}}{1-\frac{a}{(a+1)(2 a+1)}}$ $=\frac{2 a^{2}+a+a+1}{2 a^{2}+3 a+1-a}$ $=\frac{2 a^{2}+2 a+1}{2 a^{2}+2 a+1}$ $=1$ Therefore, $A+B=\tan ^{-1}(1)=\frac{\pi}{4}$....

The areas of two similar triangles are 25 cm2 and 36 cm2 respectively.

Question: The areas of two similar triangles are $25 \mathrm{~cm}^{2}$ and $36 \mathrm{~cm}^{2}$ respectively. If the altitude of the first triangle is $2.4 \mathrm{~cm}$, find the corresponding altitude of the other. Solution: Given: The area of two similar triangles is 25cm2and 36cm2respectively. If the altitude of first triangle is 2.4cm To find: The altitude of the other triangle We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding ...

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

Question: For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation. Solution: (i) $y=a e^{x}+b e^{-x}+x^{2}$ Differentiating both sides with respect tox, we get: $\frac{d y}{d x}=a \frac{d}{d x}\left(e^{x}\right)+b \frac{d}{d x}\left(e^{-x}\right)+\frac{d}{d x}\left(x^{2}\right)$ $\Rightarrow \frac{d y}{d x}=a e^{x}-b e^{-x}+2 x$ Again, differentiating both sides with respect tox, we get: $\frac{d^{2} y}...