eSaral - JEE | NEET | Class 8-10 Exam Prep App

A point P is 26 cm away from the centre O of a circle and the length

Question: ApointPis 26 cm away from the centreOof a circle and the lengthPTof the tangent drawn fromPto the circle is 10 cm. Find the radius of the circle. Solution: Let us put the given data in the form of a diagram. We have to find OT. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Therefore $\angle O T P$ is a right angle and triangle $O T P$ is a right triangle. We can find the length of $T P$ using P...

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Question: If $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$, then $x=$ Solution: $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$ $\Rightarrow \tan ^{-1} x=\tan ^{-1} 1-\tan ^{-1} \frac{1}{3}$ $\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{1-\frac{1}{3}}{1+1 \times \frac{1}{3}}\right)$ $\left[\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right), x y-1\right]$ $\Rightarrow \tan ^{-1} x=\tan ^{-1}\left(\frac{\frac{2}{3}}{\frac{4}{3}}\right)$ $\Rightarrow \tan ^{-1} x=\tan ^{...

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term,

Question: If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero. Solution: Given: $10 a_{10}=15 a_{15}$ $\Rightarrow 10[a+(10-1) d]=15[a+(15-1) d]$ $\Rightarrow 10(a+9 d)=15(a+14 d)$ $\Rightarrow 10 a+90 d=15 a+210 d$ $\Rightarrow 0=5 a+120 d$ $\Rightarrow 0$ $\Rightarrow 0=a+24 d$ $\Rightarrow a=-24 d \ldots(\mathrm{i})$ To show: $a_{25}=0$ $\Rightarrow$ LHS : $a_{25}=a+(25-1) d$ $=a+24 d$ $=-24 d+24 d \quad(\operatorname{From}(\mathrm{...

Find the length of a tangent drawn to a circle with radius 5 cm,

Question: Find the length of a tangent drawn to a circle with radius 5 cm, from a point 13 cm from the centre of the circle. Solution: Let us first put the given data in the form of a diagram. We have to find TP. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Thereforeis a right angle and triangleOTPis a right triangle. We can find the length ofTPusing Pythagoras theorem. We have, $T P^{2}=O P^{2}-O T^{2}...

If O is a point within ∆ABC, show that:

Question: If $O$ is a point within $\triangle A B C$, show that: (i) $A B+A CO B+O C$ (ii) $A B+B C+C AO A+O B+O C$ (iii) $O A+O B+O C\frac{1}{2}(A B+B C+C A)$ Solution: Given:In triangle ABC, O is any interior point.We know that any segment from a point O inside a triangle to any vertex of the triangle cannot be longer than the two sides adjacent to the vertex.Thus, OA cannot be longer than both AB and CA (if this is possible, then O is outside the triangle).(i) OA cannot be longer than both AB...

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Question: If $0x\frac{\pi}{2}$, then $\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)=$___________________. Solution: $\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)$ $=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-x\right)\right]+\cos ^{-1}\left[\cos \left(\frac{\pi}{2}-x\right)\right]$ $=\frac{\pi}{2}-x+\frac{\pi}{2}-x$ $=\pi-2 x$ If $0x\frac{\pi}{2}$, then $\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)=$ $\pi-2 x$...

If PT is a tangent at T to a circle whose centre is O and OP = 17 cm,

Question: IfPTis a tangent at T to a circle whose centre isOandOP= 17 cm,OT= 8 cm, Find the length of the tangent segmentPT. Solution: Let us put the given data in the form of a diagram. We have to find TP. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Thereforeis a right angle and triangleOTPis a right triangle. We can find the length ofTPusing Pythagoras theorem. We have, $T P^{2}=O P^{2}-O T^{2}$ $T P...

If $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$, then $\alpha-\beta=$

[question] Question. If $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$, then $\alpha-\beta=$ (a) $\frac{\pi}{6}$ (b) $\frac{\pi}{3}$ (C) $\frac{\pi}{2}$ (d) $-\frac{\pi}{3}$ [/question] [solution] Solution: (a) $\frac{\pi}{6}$ We have $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$ Now, $\alpha-\beta=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right)-\tan ^{-1} \frac{2 x-...

If 9th term of an A.P. is zero,

Question: If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term. Solution: Given: $a_{9}=0$ $\Rightarrow a+(9-1) d=0 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+8 d=0$ $\Rightarrow a=-8 d \ldots(\mathrm{i})$ To prove. $a_{29}=2 a_{19}$ Proof: LHS : $a_{29}=a+(29-1) d$ $=a+28 d$ $=-8 d+28 d \quad($ From $(\mathrm{i}))$ $=20 d$ RHS : $2 a_{19}=2[a+(19-1) d]$ $=2(a+18 d)$ $=2 a+36 d$ $=2(-8 d)+36 d \quad($ From $(\mathrm{i}))$ $=-16 d+36 d$ $=20 d$ LHS = RHS Hence, ...

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Question: If $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$, then the value of $x$ is______________. Solution: Let $\sin ^{-1} x=\theta \Rightarrow \sin \theta=x$. $\therefore \cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$ $\Rightarrow \cos 2 \theta=\frac{1}{9}$ $\Rightarrow 1-2 \sin ^{2} \theta=\frac{1}{9}$ $\Rightarrow 2 \sin ^{2} \theta=1-\frac{1}{9}=\frac{8}{9}$ $\Rightarrow \sin ^{2} \theta=\frac{4}{9}$ $\Rightarrow \sin \theta=\pm \frac{2}{3}$ $\Rightarrow x=\pm \frac{2}{3}$ $(\sin \theta=x...

The 6th and 17th terms of an A.P. are 19 and 41 respectively,

Question: The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term. Solution: Given: $a_{6}=19$ $\Rightarrow a+(6-1) d=19 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+5 d=19 \quad \ldots(1)$ And, $a_{17}=41$ $\Rightarrow a+(17-1) d=41 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+16 d=41 \quad \ldots(2)$ Solving the two equations, we get, $16 d-5 d=41-19$ $\Rightarrow 11 d=22$ $\Rightarrow d=2$ Putting $d=2$ in the eqn (1), we get: $a+5 \times 2=19$ $\Rightarrow ...

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Question: If $\tan ^{-1} \frac{a}{x}+\tan ^{-1} \frac{b}{x}=\frac{\pi}{2}$, then $x=$_________________. Solution: We know $\tan ^{-1} x=\cot ^{-1} \frac{1}{x}$ $\therefore \tan ^{-1} \frac{b}{x}=\cot ^{-1} \frac{1}{\left(\frac{b}{x}\right)}=\cot ^{-1} \frac{x}{b}$ ....(1) So, $\tan ^{-1} \frac{a}{x}+\tan ^{-1} \frac{b}{x}=\frac{\pi}{2}$ (Given) $\Rightarrow \tan ^{-1} \frac{a}{x}+\cot ^{-1} \frac{x}{b}=\frac{\pi}{2}$ [Using (1)] $\Rightarrow \frac{a}{x}=\frac{x}{b}$ $\left(\tan ^{-1} y+\cot ^{-1...

Find the next five terms of each of the following sequences given by:

Question: Find the next five terms of each of the following sequences given by: (i) $a_{1}=1, a_{n}=a_{n-1}+2, n \geq 2$ (ii) $a_{1}=a_{2}=2, a_{n}=a_{n-1}-3, n2$ (iii) $a_{1}=-1, a_{n}=\frac{a_{n}-1}{n}, n \geq 2$ (iv) $a_{1}=4, a_{n}=4 a_{n-1}+3, n1$. Solution: In the given problem, we are given the first, second term andthenthterm of an A.P. We need to find its next five terms (i) $a_{1}=1, a_{n}=a_{n-1}+2, n \geq 2$ Here, we are given that $n \geq 2$ So, the next five terms of this A.P would...

In the given figure, prove that

Question: In the given figure, prove that(i)CD + DA + ABBC(ii)CD + DA + AB + BC2AC. Solution: Given: QuadrilateralABCDTo prove:(i)CD + DA + ABBC(ii)CD + DA + AB + BC2ACProof:(i) In $\Delta A C D$, $C D+D AC A \quad \ldots(1)$ In $\Delta A B C$, $A B+C AB C \quad \ldots(2)$ Adding $(1)$ and $(2)$, we get $C D+D A+A B+C AC A+B C$ $\therefore A B+C D+D AB C$ (ii) In $\Delta C D A$ $C D+D AC A \quad \ldots(3)$ In $\Delta B C A$, $B C+A BC A \quad \ldots(4)$ Adding $(3)$ and $(4)$, we get $C D+A D+B ...

The first term of an A.P. is 5, the common difference is 3 and the last term is 80;

Question: The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms. Solution: Here,a= 5,d= 3,an= 80 Let the number of terms ben. Then, we have: $a_{n}=a+(n-1) d$ $\Rightarrow 80=5+(n-1) 3$ $\Rightarrow 75=(n-1) 3$ $\Rightarrow 25=(n-1)$ $\Rightarrow 26=n$ Thus, there are 26 terms in the given A.P....

The first term of an A.P. is 5, the common difference is 3 and the last term is 80;

Question: The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms. Solution: Here,a= 5,d= 3,an= 80 Let the number of terms ben. Then, we have: $a_{n}=a+(n-1) d$ $\Rightarrow 80=5+(n-1) 3$ $\Rightarrow 75=(n-1) 3$ $\Rightarrow 25=(n-1)$ $\Rightarrow 26=n$ $\Rightarrow 26=n$ Thus, there are 26 terms in the given A.P....

How many terms are there in the A.P.

Question: (i) How many terms are there in the A.P. 7, 10, 13, ... 43? (ii) How many terms are there in the A.P. $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2}, \ldots, \frac{10}{3} ?$ Solution: (i) 7, 10, 13...43 Here, we have: a= 7 $d=(10-7)=3$ $a_{n}=43$ Let there benterms in the given A.P. Also, $a_{n}=a+(n-1) d$ $\Rightarrow 43=7+(n-1) 3$ $\Rightarrow 36=(n-1) 3$ $\Rightarrow 12=(n-1)$ $\Rightarrow 13=n$ Thus, there are 13 terms in the given A.P. (ii) $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2}, ...

If tan-12, tan-13 are measures of two angles of triangle, then the measure of its third angle is

Question: If $\tan ^{-1} 2, \tan ^{-1} 3$ are measures of two angles of triangle, then the measure of its third angle is________________. Solution: Let the measure of third angle of the triangle bex. Now, $\tan ^{-1} 2+\tan ^{-1} 3+x=\pi$ (Angle sum property of triangle) $\Rightarrow \pi+\tan ^{-1}\left(\frac{2+3}{1-2 \times 3}\right)+x=\pi$ $\left[\tan ^{-1} x+\tan ^{-1} y=\pi+\tan \left(\frac{x+y}{1-x y}\right)\right.$, if $\left.x y1\right]$ $\Rightarrow \tan ^{-1}(-1)+x=0$ $\Rightarrow-\frac...

If tan are measures of two angles of triangle, then the measure of its third angle is

Question: If $\tan ^{-1} 2, \tan ^{-1} 3$ are measures of two angles of triangle, then the measure of its third angle is________________. Solution: Let the measure of third angle of the triangle bex. Now, $\tan ^{-1} 2+\tan ^{-1} 3+x=\pi$ (Angle sum property of triangle) $\Rightarrow \pi+\tan ^{-1}\left(\frac{2+3}{1-2 \times 3}\right)+x=\pi$ $\left[\tan ^{-1} x+\tan ^{-1} y=\pi+\tan \left(\frac{x+y}{1-x y}\right)\right.$, if $\left.x y1\right]$ $\Rightarrow \tan ^{-1}(-1)+x=0$ $\Rightarrow-\frac...

Prove that in a triangle, other than an equilateral triangle,

Question: Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than $\frac{2}{3}$ of a right angle. Solution: Given: InΔ∆ABC,BCis the longest side. To prove: $\angle B A C\frac{2}{3}$ of a right angle, i.e., $\angle B A C60^{\circ}$ Construct: Mark a pointDon sideACsuch thatAD=AB=BD.Proof:InΔ∆ABD,∵∵AD=AB=BD (By construction) $\therefore \angle 1=\angle 3=\angle 4=60^{\circ}$ Now $\angle B A C=\angle 1+\angle 2=60^{\circ}+\angle 2$ but $60^{\cir...

If $\cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$, then $4 x^{2}-12 x y \cos \frac{\theta}{2}+9 y^{2}=$

[question] Question. If $\cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$, then $4 x^{2}-12 x y \cos \frac{\theta}{2}+9 y^{2}=$ (a) 36 (b) $36-36 \cos \theta$ (c) $18-18 \cos \theta$ (d) $18+18 \cos \theta$ [/question] [solution] Solution: (c) 18 − 18 cosθ We know $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)$ $\therefore \cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$ $\Rightarrow \cos ^{-1}\left(\frac{x}{3} \frac{y}{2}-\sqrt{...

Solve the following

Question: (i) Which term of the sequence $24,231 / 4,221 / 2,213 / 4 \ldots$ is the first negative term? (ii) Which term of the sequence $12+8 i, 11+6 i, 10+4 i, \ldots$ is (a) purely real (b) purely imaginary? Solution: (i) $24,231 / 4,221 / 2,213 / 4 \ldots$ This is an A.P. Here, we have: a = 24 $d=\left(23 \frac{1}{4}-24\right)=\left(-\frac{3}{4}\right)$ Let the first negative term be $a_{n}$. Then, we have: $a_{n}0$ $\Rightarrow a+(n-1) d0$ $\Rightarrow 24+(n-1) \quad(-3 / 4)0$ $\Rightarrow ...

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Question: If $3 \tan ^{-1} x+\cot ^{-1} x=\pi$, then $x=$ Solution: $3 \tan ^{-1} x+\cot ^{-1} x=\pi$ (Given) $\Rightarrow 2 \tan ^{-1} x+\frac{\pi}{2}=\pi$ $\Rightarrow 2 \tan ^{-1} x=\pi-\frac{\pi}{2}=\frac{\pi}{2}$ $\left(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$ $\Rightarrow \tan ^{-1} x=\frac{\pi}{4}$ $\Rightarrow x=\tan \frac{\pi}{4}=1$ If $3 \tan ^{-1} x+\cot ^{-1} x=\pi$, then $x=$___1___...

D is any point on the side AC of ΔABC with AB = AC. Show that CD < BD.

Question: Dis any point on the sideACof ΔABCwithAB=AC. Show thatCDBD. Solution: Given: InΔ∆ABC,AB=ACTo prove:CDBDProof:InΔ∆ABC,Since,AB=AC (Given) So, $\angle A B C=\angle A C B$ In $\Delta A B C$ and $\Delta D B C$, $\angle A B C\angle D B C$ $\Rightarrow \angle A C B\angle D B C \quad[$ From (i) $]$ $\Rightarrow B DC D \quad$ (Side opposite to greater angle is longer.) $\therefore C DB D$...

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Question: If $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$, then $\cot ^{-1} x+\cot ^{-1} y=$_________________. Solution: We know $\tan ^{-1} a+\cot ^{-1} a=\frac{\pi}{2}$, for all $a \in \mathrm{R}$ .....(1) Now, $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$ (Given) $\Rightarrow \frac{\pi}{2}-\cot ^{-1} x+\frac{\pi}{2}-\cot ^{-1} y=\frac{4 \pi}{5}$ [Using (1)] $\Rightarrow \cot ^{-1} x+\cot ^{-1} y=\pi-\frac{4 \pi}{5}$ $\Rightarrow \cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{5}$ If $\tan ^{-1} x+\tan ^{-1...

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