Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 18601. |
If asin30 + bcos30=c then prove that acos30-bsin30=√a×a+b×b-c×c |
| Answer» | |
| 18602. |
Jdi |
| Answer» | |
| 18603. |
Find the nth term and sum of n terms of the following 1+5+13+29 |
|
Answer» Question galat hai It is not in A.P |
|
| 18604. |
In equilateral triangle ABC AD is perpendicular to BC prove tht 3BC×BC =4AD×AD |
| Answer» Use Pythagoras thm. U will prove it | |
| 18605. |
For some a and b,if HCF of 55 and 210 is 210a+55b,then find the value of a and b |
| Answer» a= 5 and b=19 | |
| 18606. |
sab ignore kar rhae hai mujhe !!!!! iss liye pic dali\xa0 |
|
Answer» kk Hmm koi ni bhai ab Mt dalna ok chal sorry yrr @aru That jagah pic dal rhe ho..Batana nhi hai toh pic Mt dala karo..Aur specially 9 mein bilkul nhi Bss ab pic Mt dalna? Mat roo yar chup ho jaamein ignore nahi krta but yrr koi rply hi nhai kar rha thamai bore ho rha tha vse toh socha chalo or kisi se trick puchne ki request kara the hai lol heheh Pic mat dalo So sorry harsh Harsh yaar pic dalne ko mna kiya tha na bhai Awwww no one is ignoring u bhai abb toh kisi ko nahi batane wala mai !!! ek ko bhi nhai Bhai tere pair pakad raha hooo ye bata ye picture kaise daaali............ |
|
| 18607. |
3√5 is irrational find |
|
Answer» We know √5 is an irrational so 3√5 is also an irrational . You can prove this in 1st unit Yes |
|
| 18608. |
I a givrn triangle pqr we have given ps=3cm and sr ab apni |
| Answer» | |
| 18609. |
hey.... Ishan...\xa0\xa0 |
|
Answer» Aur btao.. Nhi... Yaar.. Ignore kr rhi ho.?? Aa gaya me yaad tumhe me yaar.... Naraz ho kya??\xa0 |
|
| 18610. |
Kisko lgta h ki vo 90% above la skta h? |
|
Answer» Kismat ke sath aapka hard work bhi hai kshitij Delhi Correct I got 1st position in my class since 7 years. Ise main apni kismat manta hu Vese tum kaha se ho Vo unhi ki madad krte h Jo hardworking hote h Kismat kuch cheez hoti nhi hai bhai... kismat to kutti cheez hoti hai??? True Kshitiz agar tum preparation nhi kroge to bhagwan bi Kuch nhi kr skte Toh aajayenge aapko Kismat bhi kch hoti h Pdh to rhe hi h Haa Padhai karo toh lagega Sb bhagwaan bhrose Kuch keh nhi skte bro |
|
| 18611. |
Show that exactly one of the number n,n+2 or n+4 is divisible by 3 |
| Answer» Let n =3kthen n + 2 = 3k + 2and n + 4 = 3k + 4Case 1: When n=3k ,n is divisible by 3 ............(1)n + 2 = 3k + 2or, n + 2 is not divisible by 3n + 4 = 3k + 4= 3(k + 1) + 1or, n + 4 is not divisible by 3Case 2:When n=3k+1, n is not divisible by 3\xa0n + 2 = (3k + 1) + 2=3k + 3 = 3(k + 1){tex} \\Rightarrow{/tex}\xa0n+ 2 is clearly divisible by 3..........................(2)n + 4 = (3k + 1) + 4= 3k + 5= 3(k + 1) + 2{tex}\\Rightarrow{/tex}\xa0n + 4 is not divisible by 3Case 3:When n=3k+2,n is not divisible by 3\xa0n + 2 = (3k + 2) + 2= 3k + 4(n + 2) is not divisible by 3x + 4 = 3k + 6 = 3(k + 2){tex}\\Rightarrow{/tex}\xa0n + 4 is divisible by 3........................(3)Hence, from (1),(2) and (3) it is clear that\xa0exactly one of the numbers n, n + 2, n + 4, is divisible by 3. | |
| 18612. |
Prove that sinA-cosA +1/sinA+cosA-1=1/secA-tanA |
|
Answer» I know but samaj me Ni aaa Rha in ncert Niii ho Rha Rationalise kro isse |
|
| 18613. |
hii... Frnds... ☺☺ |
|
Answer» Mujhe bhi aa gya... how r u all??? ?? \xa0 Hii Hii sinthi?? Hi hyyy. ??dear Hello |
|
| 18614. |
Blow out /but /used to /they we /candles /light a lamp |
| Answer» Blow out candle but used to light a lamp I think !!! | |
| 18615. |
If ax+by=a2+b2and bx+ay=0find the value of x and y |
| Answer» No one knows | |
| 18616. |
Exescise 11.1 |
|
Answer» Wo to constructions h ,????? Achhe questions hai. 11.1 me. |
|
| 18617. |
P(x)=x³-3x²+5x-3, g(x)=x²-2 |
|
Answer» Right Quotient is x-3Remainder is 7x -9 |
|
| 18618. |
If p times pth term is q times qth term. Show that ( p+q)is term is 0 |
| Answer» p * ap = q * aqp{ a + (p-1) d } = q { a + ( q - 1 ) d }ap + p2d - pd = aq + q2d - qdap - aq = - p2d + q2d - qd + pda (p - q ) = d ( q2 - p2 + p - q )a ( p - q ) = d { ( q - p ) ( q + p) + p - q }a ( p - q ) = d ( p - q ) { -1 ( p + q) + 1 }a = d ( - p - q + 1 ) .... ( 1) ( p + q ) th term = a + (n - 1 ) dhere , n = no. of terms= a + ( p + q - 1 ) d = o .. (2)substituting a = d ( - p - q + 1 ) in ( 2 )= d ( - q - p + 1 ) + ( p + q - 1 ) d= -dp - dq + d + pd + qd - d= 0 | |
| 18619. |
yaar bahut bore ho ra hu??? |
|
Answer» Haa bhai ho gyi corrupt bhai but teri pic curroupt hori hai Acha!! |
|
| 18620. |
27,23,19,.....,-65 find the eleventh term from the last term of the A.P. |
|
Answer» On reversing a.p.=-65,-61.........,27a=-65 d=+4,an=27a+(n-1)d=an-65+(n-1)4=27(n-1)4=92n-1=23n=24A11=a+10d=-65+10×4=-65+40=-25 Omid ha sahi ha -65+40 =-25 -25 - 25 T11 from last -13 T11=-25 |
|
| 18621. |
Given ab/pq=ac/pr=ad/pm, prove that triangle abc- triangle pqr |
| Answer» | |
| 18622. |
If the mean of N natural numbers is 5N/9.Find the value of N. |
| Answer» I think its 9 bt not sure | |
| 18623. |
Evaluate √20+√20+√20+.....+infinity |
|
Answer» easy see in foundation book in quadratic equation Proved Its 5 |
|
| 18624. |
If the rootsh of equation (a2+b2)-2(ac+bd) +(c2+d2)=0 are equal prove that a/b=c/d |
| Answer» According to the question,the given equation is,(a2 + b2)x2 - 2(ac + bd)x + (c2 + d2) = 0The discriminant of the given equation is given byD = [-2(ac + bd)]2 - 4 {tex}\\times{/tex} (a2 + b2) {tex}\\times{/tex} (c2 + d2){tex}\\Rightarrow{/tex} D = 4(ac + bd)2 - 4(a2c2 + a2d2 + b2c2 + b2d2){tex}\\Rightarrow{/tex} D = 4(a2c2 + b2d2 + 2abcd) - 4(a2c2 + a2d2 + b2c2 + b2d2){tex}{/tex}{tex}{/tex}D= 4(2abcd - a2d2 - b2c2){tex}\\Rightarrow{/tex} D = -4[(ad)2 + (bc)2 - 2(ad)(bc)]{tex}\\Rightarrow{/tex} D = -4(ad - bc)2Since the roots of the given equation are given to be equal, therefore,Discriminant, D = 0{tex}\\Rightarrow{/tex} -4(ad - bc)2 = 0{tex}\\Rightarrow{/tex} (ad - bc)2 = 0 [as -4\xa0{tex}\\ne{/tex}\xa00]{tex}\\Rightarrow{/tex} ad - bc = 0{tex}\\Rightarrow{/tex} ad = bc{tex} \\Rightarrow \\frac{a}{b} = \\frac{c}{d}{/tex} | |
| 18625. |
this is the show guys koi dekta hai?? |
|
Answer» Yupp kanishka dekha hai?? Harsh Ye tumne kaise kiya Yaar.... Mujhe ye samajh nhi aa rha.... Kya iske lye fb se connect krte h.. Hey harsh tumne kaise kiya ye magic bolo na yaar. Mai pics nahi dalungi pakka bas bol do Of course I have seen this show ten times all the seasons. I\'m a huge fan ohh.yrr its my fav Yaar... Iske lye fb id honi chhaiye kya Nahi dekhte Noo |
|
| 18626. |
FOR U ANSHUL\xa0 |
| Answer» Kaise kiya | |
| 18627. |
Write the rational number between √2 and √3 |
| Answer» √3×√2 | |
| 18628. |
sin18°÷cos72° |
|
Answer» 1 1 1 1 1 1 》Cos (90-72)/cos 72》cos 72 /cos 72》 1 |
|
| 18629. |
Guys give me some idea about paper presentation in boards exam...plz give your active response |
| Answer» Youtube pe "lestute \'s paper presentation" dekh lo. | |
| 18630. |
In triangle ABC , write cos (B+C/2) in terms of A. |
| Answer» Sum of three angles of triangle is\xa0{tex}180^{\\circ}{/tex}A + B + C = 180°or, B + C = 180° - ADivide both sides by 2{tex}\\frac{B+C}{2}=\\frac{180^{\\circ}}{2}-\\frac{A}{2}{/tex}{tex}\\frac{B+C}{2}=90-\\frac{A}{2}{/tex}{tex}\\therefore \\quad \\cos \\left( \\frac { B + C } { 2 } \\right) {/tex}{tex}= \\cos \\left( 90 - \\frac { A } { 2 } \\right){/tex}{tex}= \\sin \\frac { A } { 2 }{/tex} | |
| 18631. |
SinA+cosA÷sinA -cosA=root 1+sin2A÷root 1-sinA prove that |
| Answer» | |
| 18632. |
Why there are numbers |
| Answer» | |
| 18633. |
Sum of all 3 digit no. Which leaves remainder 2 when divided by 5 |
| Answer» | |
| 18634. |
hello friends????❤️❤️❤️ |
|
Answer» Hey Ok I am also fine Hi Fine and u How are u Hi |
|
| 18635. |
Are you all fine |
|
Answer» Mai jodhpur se hu Yes and what about you Kha se ho aap Hi hii |
|
| 18636. |
X-1/X=3 |
|
Answer» x=-1/2 X=-1/2 |
|
| 18637. |
Perimeter of a quadrant of a circle of radius r |
|
Answer» Semi circle perimeter= πr+2rPerimeter of a quadrant= πr+2r/2 Sorry I think it\'s πr/2+2r πr/2+2r r/2(π+4) πr+2r. I. Think Pie r +2r/2 Perimeter =1/4*2 pie r =1/2 pie r |
|
| 18638. |
Divide f x is equal to x ki power 4 minus 5 x + 6 x is equal to 2 - x square |
| Answer» Pphala questions sahi sa type karo Like this 2x^2-5x+8=0It is example not a question | |
| 18639. |
Pichle ten years paper must h? |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 18640. |
What is parallogram |
| Answer» A parallelogram is a quadrilateral with pairs of opposite sides parallel and equal. | |
| 18641. |
How will you find q (x) and r (x).If p (x) and g (x) are any two |
| Answer» p(x)/g(x) | |
| 18642. |
Blue print of 2018 |
| Answer» Check the pattern in the syllabus :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 18643. |
FIND THE RATIO IN WHICH LINE SEGMENT JOINING A(1,-5),(-4,5) IS DIVIDED BY THE xAxis |
| Answer» Let the point of division be P. Let the ratio be K : 1.Then{tex} P \\to \\left\\{ {\\frac{{(K)( - 4) + (1)(1)}}{{K + 1}},\\frac{{(K)(5) + (1)( - 5)}}{{K + 1}}} \\right\\}{/tex}{tex}P \\to \\left\\{ {\\frac{{ - 4K + 1}}{{K + 1}},\\frac{{5K - 5}}{{K + 1}}} \\right\\}{/tex}{tex}\\because{/tex} P lies on the x-axis and we know that on the x-axis the ordinate is 0.{tex}\\therefore \\;\\frac{{5K - 5}}{{K + 1}} = 0{/tex}{tex}\\Rightarrow{/tex} 5K - 5 = 0{tex}\\Rightarrow{/tex} 5K = 5{tex}\\Rightarrow K = \\frac{5}{5} = 1{/tex}Hence, the required ratio is 1 : 1.Putting K = 1, we get{tex}P \\to \\left\\{ {\\frac{{ - 4(1) + 1}}{{1 + 1}},\\frac{{5(1) - 5}}{{1 + 1}}} \\right\\}{/tex}{tex}P \\to \\left\\{ { - \\frac{3}{2},0} \\right\\}{/tex} | |
| 18644. |
Prove that sum of squares of diagonals of a parellogram is equal to the sum of squares of its sides. |
| Answer» Given: ABCD is a parallelogram whose diagonals are AC and BD.To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2Construction: Draw AM {tex} \\bot {/tex} DC and BN{tex} \\bot {/tex} D(Produced)Proof: In right triangle AMD and BNC.AD = BC ..............Opp.sides of a ||gmAM = BN ............Both are altitudes of the same parallelogram to the same base{tex}\\therefore {/tex}\u200b{tex}\\triangle {/tex}\u200b AMD {tex}\\cong{/tex}\u200b{tex}\\triangle {/tex}\u200bBNC ................RHS congruence criterion{tex}\\therefore {/tex} MD = NC .........(1).........CPCTIn right triangle BND,{tex}\\because {/tex}{tex}\\angle{/tex} N=90°{tex}\\therefore {/tex} BD2 = BN2 + DN2 .............By Pythagoras theorem= BN2 + (DC + CN)2= BN2 + DC2 + CN2 + 2DC.CN= (BN2 + CN2) + CN2 + 2DC.CN= BC2 + DC2 + 2DC.CN ..........(2)In right triangle BNC with {tex}\\angle{/tex}N = 90oBN2+CN2 = BC2 ......By Pythagoras theoremIn right triangle AMC{tex}\\because {/tex}{tex}\\angle{/tex} M=90o{tex}\\therefore {/tex} AC2 = AM2 + MC2= AM2 = (DC - DM)2= AM2 + DC2 + DM2 - 2DC.DM= (AM2 + DC2) + DC2 - 2DC.DM= AD2 + DC2 - 2DC.DM{tex}\\because {/tex} In right triangle AMD with {tex}\\angle{/tex} M=90°AD2 = AM2 + DM2 ..........[By Pythagoras theorem]= AD2 + AB2 - DC.CN .......From(1)Adding (3) and (2) ,we getAC2 + BD2 = (AD2 + AB) + (BC2 + DC) = AB2 + BC2 + BC2 + CD2 + DA2 | |
| 18645. |
Any friend here |
| Answer» Yes | |
| 18646. |
Prove that square root of 6 is irrational number? |
| Answer» If possible, let {tex}\\sqrt { 6 }{/tex} be rational and let its simplest form be {tex}\\frac { a } { b }{/tex} then, a and b are integers having no common factor other than 1, and {tex}b \\neq 0{/tex}.Now, {tex}\\sqrt { 6 } = \\frac { a } { b } {/tex}{tex}\\Rightarrow 6 = \\frac { a ^ { 2 } } { b ^ { 2 } }{/tex} [on squaring both sides]{tex}\\Rightarrow 6b^2 = a^2{/tex} .................(i){tex}\\Rightarrow{/tex} 6 divides {tex}a^2{/tex} [{tex}\\because{/tex} 6 divides {tex}6b^2{/tex}]{tex}\\Rightarrow{/tex} 6 divides {tex}a{/tex}Let {tex}a = 6c{/tex} for some integer {tex}c{/tex}putting {tex} a = 6c{/tex} in (i), we get{tex}a^2 = 36c^2{/tex}{tex}6b^2 = 36c^2 \\;\\;\\;[6b^2 = a^2] {/tex}{tex}\\Rightarrow b^2 = 6c^2{/tex}{tex}\\Rightarrow{/tex} 6 divides {tex}b^2{/tex} [{tex}\\because{/tex} 6 divides {tex}6c^2{/tex}]{tex}\\Rightarrow{/tex} 6 divides {tex}b{/tex} [{tex}\\because{/tex} 6 divides {tex}b^2 = 6{/tex} divides {tex}b{/tex}]Thus, 6 is a common factors of {tex}a{/tex} and {tex}b{/tex}But, this contradicts the fact that {tex}a{/tex} and {tex}b{/tex} have no common factor other than 1The contradiction arises by assuming that {tex}\\sqrt { 6 }{/tex} is rational.Hence {tex}\\sqrt { 6 }{/tex} is irrational. | |
| 18647. |
2x+9 + x = 13 (2x+9 is in under root√) solve for x |
|
Answer» Solve this quadratic equation than you will get value of x √2x+9=13-x[by during on both side]2x+9=169-x²+26x=2x=169-9-x²+26x=x²=160+26x-2x=x²=160+24xX²-24x-160X² 20 and 8 |
|
| 18648. |
Is this is an low level question |
| Answer» | |
| 18649. |
Oswal book questions is very important? |
|
Answer» Vidhya question bank is very important I m solving this only |
|
| 18650. |
How 2/√3 multipli 3/4 become √3/2 |
|
Answer» I know.. When we multiply two by root three with three by four it becomes six byfour root three. Then rationalise denominator by multiplying root three on both numerator and denominator thnwe will get. The answr.. 2/√3 × 3/4 = 6/4√3 by cancelling 6 nd 4 by 2 remain 3/2√3 now cancel √3by 3 it will be √3/2 I know Please tell fairly Please tdll |
|