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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.
| 28501. |
How to prepare? |
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Answer» What to prepare For what?????? |
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| 28502. |
Find the value of for whichthe following equations have real and equalrootsx2+k(2x+k-1)+2=0 |
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| 28503. |
How can I get cbse Guess papers 2019 |
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Answer» Its web not wed It can be on vedantu wed site U can get cbse guess papers from market or internet |
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| 28504. |
Pg 252 of ncert class 10 question number 8 |
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Answer» Please write the question. These questions end on 9 U want solution of this ques?? |
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| 28505. |
2/×÷9 |
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| 28506. |
Volume of hollow cylinder. |
| Answer» Pih(r1sq-r2sq) | |
| 28507. |
Applications of trignometry chp is very tough how to do prepare this chp for board exam |
| Answer» Try to make diagram if diagram will be made u will easily solve the question | |
| 28508. |
I am not able to do coordinate geometry. |
| Answer» Just go through all the formulas which will be used to solve questions. Be thorough with it then u can solve it easily | |
| 28509. |
Find the maximum value of cos theta? |
| Answer» 1 | |
| 28510. |
S.A of hollow cylinder. |
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Answer» pi h(R2+r2) +pi (R2-r2) C.SA of cylinder is SA of Hollow cylinder |
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| 28511. |
How we should find the trigonometric ratioes of 90degaree and 0degaree |
| Answer» let B=base L=altitude and H=hypotenuseAt 900 B =0 and L=H,so sin 90=L/H=1 and cos90=0/H=0 tan 90 = L/B = L/0= infiniteAt 00 L=0 and B=L so sin 0=0 , cos 0 =b/h=1 tan 0=L/B=0/b=0Value of cot,sec and cosec can be found accordingly\xa0 | |
| 28512. |
(1+sinA-cosA/1+sinA+cosA)^2 = 1-cosA/1+cosA |
| Answer» LHS = [(1+sinA-cosA )/(1+sinA+cosA)]²=[ (1+sin²A+cos²A+2sinA-2sinAcosA-2cosA)/(1+sin²A+cos²A+2sinA+2sinAcosA+2cosA) ]²= [( 1+1+2sinA-2sinAcosA-2cosA)/(1+1+2sinA+2sinAcosA+2cosA)]²= { ( 2(1+sinA) -2cosA(sinA+1) / [ 2(1+sinA) + 2cosA(sinA +1 ) ]}²= { (2(1+sinA)[1 - cosA] /[2(1+sinA )(1 + cosA )] }²= [ ( 1 - cosA ) / ( 1 + cosA ) ]²= RHS | |
| 28513. |
X2-4x+a no real roots |
| Answer» No roots | |
| 28514. |
what is the common deference of an ap in which a21-a7=84 |
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Answer» Given.... a 21 - a7 = 84⇒ {a + (21 - 1 ) d} - {a + (7 - 1) d} = 84⇒ a + 20d - a - 6d = 84⇒14 d = 84⇒d = 84/ 14⇒ d = 6Required Common Difference is 6. D =6 |
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| 28515. |
Cosa+cosb |
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Answer» Cosa+1-sina Cos(a+b) |
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| 28516. |
FIND The area of Traingle whose veetices are (2,3) (-1,0) (2,-4) |
| Answer» A(2,3) B(-1,0) C(2,-4)X1=2 X2=-1 X3=2Y1=3 Y2=0 Y3= -4Area of triangle = 1/2 {x1(y2-y3) + x2(y3-y1) + x3(y1-y2)}= 1/2 {2(0-(-4)) + (-1) (-4-3) + 2(3-0)}= 1/2 {8-(-7)+6}= 1/2 {8+7+6}= 1/2 × 21= 21/2\xa0 | |
| 28517. |
(cosp-1)x^2+(cosp)x+sinp is a quadratic equation. What is the value of x ? |
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| 28518. |
Find area of rhombus? |
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Answer» Area of rhombus = 1/2 products of its diagonals A=pq/2, |
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| 28519. |
What are the first 5 terms of an ap |
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Answer» No need to worry about sst,sst is always week for maths good studentsAnswer all question write something what you rememberSome time topper in mathsgets minimum marks in sst. so not to worry.before exam read small and MCQ aur rat loRead like stories,write like stories.All the best\xa0 Same here I am in little tension only about ssst Yr plz ques toh complete kroo |
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| 28520. |
How to find LCM of 3 numbers by prime factorisation method |
| Answer» Take the given three no suppose -4,6,8 and factories them seprately 4-2^2,6-2x3,8-2^3 and then take their LCM -2^3x3. That\'s it | |
| 28521. |
Prove that tan0/1-cot0+cot0/1-tan0 |
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Answer» Full question is tan0/1-cot0+cot0/1-tan0= 1+ sec.cosec Complete yur question first........? Kiske equal prove krna hai |
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| 28522. |
Find the nature of the quadratic equation 3xroot 2-4root 3 x+4=0 |
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Answer» If we use D then no real root exist Use D =b^2 - 4ac |
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| 28523. |
If 2x,x+10,3x+2 are in AP. Find the value of x |
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Answer» In this ap , d= a2-a1----(1)d=a3-a2--------(2)From eq. 1and 2a2-a1=a3-a2x+10-(2x)=3x+2-(x+10)x+10-2x=3x+2-x- 10-x+10=2x-810+8=2x+x18=3x x=6 Put formula 2b=a+c X=6 |
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| 28524. |
3x/2 - 5y/2 = -2x/3 + y/2 = 13/6 |
| Answer» Put them both equal to 36 you will get two equation then compare them you will get the answer | |
| 28525. |
What is the probability that a non leap year has 53 Mondays |
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Answer» 53/365 Ryt 1/7 |
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| 28526. |
Find the discriminant of quadratic equation 2x^2-4x+3=0 and hence find the nature of its root |
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Answer» B^2-4ac=(4)^2-4(2)(3)=16-24=-8.Therefore -8<0 Hence it have no real roots D=-8.if D is less than zero than no real root exist D=16-24D=-8 D=<0 |
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| 28527. |
Find the sum of n terms of the series (4-1/n)+(4-2/n)+.... |
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Answer» (4-1/n)+(4-2/n)+.......=4n-1/n(1+2+3........n){tex}\\begin{array}{l}\\text{=4n-}\\frac1{\\mathrm n}\\times\\frac{\\mathrm n(\\mathrm n+1)}2\\\\\\text{=4n-}\\frac{n+1}2\\\\\\text{=}\\frac{8n-n-1}2\\\\\\text{=}\\frac{7n-1}2\\end{array}{/tex} Sn = (7n-1)/2 |
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| 28528. |
Cot theta + cosec theta +1÷cosec theta - cot theta + 1=cosec theta - cot theta +1 |
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| 28529. |
Cbse 2014 question paper |
| Answer» ???? | |
| 28530. |
Show that any positive odd integer is in the form 4q+1 or 4q+3 when q is some integer |
| Answer» Let a=bq+r(where r=0,1,2,3)But in this we have only to take odd numbersSo r=1,3Put the value of r you will get the answer hope it helps you..... | |
| 28531. |
If sin(thita) = cos(thita), then find the value of 2tan(thita)+cos(square thita) |
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Answer» But it\'s answer is 5/2? √2 + 2/2 2+sin^2thetha |
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| 28532. |
Find the coordinates of the point on y_axis which is nearest to the point(-2,5) |
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| 28533. |
Mid point theorem |
| Answer» Refer 9 class ncert | |
| 28534. |
743,22×4,333 |
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Answer» 322,037,226 322037226 Use calculator |
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| 28535. |
Write the exponent of 3 in the prime factorisation of 162 |
| Answer» 162=18 x 9=2 x 9 x 9= 2 X 3^4hence exponent =4 | |
| 28536. |
Show that 21 power n cannot end with the digit 0,2,4,6 and 8 for any natural number n |
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Answer» {tex}\\begin{array}{l}21^{\\mathrm n}=3^{\\mathrm n\\;}\\times7^{\\mathrm n}\\\\\\mathrm{Now}\\;3^{\\mathrm n\\;}\\;\\mathrm{and}\\;7^{\\mathrm n\\;}\\;\\mathrm{are}\\;\\mathrm{always}\\;\\mathrm{odd}\\;\\mathrm{for}\\;\\mathrm{all}\\;\\mathrm{values}\\;\\mathrm{of}\\;\\mathrm n\\\\\\mathrm{odd}\\;\\times\\;\\mathrm{odd}\\;=\\mathrm{odd}\\\\\\mathrm{Hence}\\;\\;3^{\\mathrm n\\;}\\times7^{\\mathrm n}\\;\\mathrm{can}\\;\\mathrm{not}\\;\\mathrm{end}\\;\\mathrm{with}\\;2,4,6,8\\end{array}{/tex} It should have an even number as factor in it .but it is not there so it cannot end with digit 0,2,4,6,8 |
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| 28537. |
The 6th term of an AP is zero. Prove that it\'s 31st term is five times the 11th term. |
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Answer» 31st term is a + (31-1)da + 30d..........(1)....... 11 th term is a +(11-1)da + 10dso 5 times 11 term is5a + 50d.........(2)...... so 5 times 11 term is5a + 50d.........(2)......on comparing with 1 a + 4a + 50das a = -5da + 4(-5d) + 50da - 20d + 50da + 30d31st term.... pehle vala answer mt dekho.... ye apka solution hai Ek kam kijiye put a6 = 0, ---> a+5d=0..., (a=5d), ab find a31 and a11...... ▪a31= a+30d ....... ▪a31=5d +30d = 35 d........ and a11=a+10d ▪a11=5d +10d =15d...... Pehle 31st trm nikalo phir 11th trm. Nikalo aur uske baad a=-5d put ksr dena Hey, its proved bt yahan solition thoda bada hai, apko kaise dun mai |
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| 28538. |
2asquarexsquare+b (6asquare+1)x+3bsquare=0 |
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Answer» Sry, 6 a square b x hona chahiye Sir, second step me, 6 abx square {tex}\\begin{array}{l}2\\mathrm a^2\\mathrm x^2+\\mathrm b(6\\mathrm a^2+1)\\mathrm x+3\\mathrm b^2=0\\\\2\\mathrm a^2\\mathrm x^2+6\\mathrm a^2\\mathrm x+\\mathrm{bx}+3\\mathrm b^2=0\\\\2\\mathrm a^2\\mathrm x(\\mathrm x+3\\mathrm b)+\\mathrm b(\\mathrm x+3\\mathrm b)=0\\\\(\\mathrm x+3\\mathrm b)(2\\mathrm a^2\\mathrm x+\\mathrm b)=0\\\\\\mathrm x=-3\\mathrm b,-\\frac{\\mathrm b}{2\\mathrm a^2}\\end{array}{/tex} Use quadratic formula easy answer would come out Yur question is not clear......, write it properly ? |
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| 28539. |
Find the point on x-axis which is equidistant from (2,5) & (-2,9) |
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Answer» on x axis y= 0let the coordinates are O (x,0),A(2,5),B=(-2,9){tex}\\begin{array}{l}\\mathrm{OA}=\\mathrm{OB}\\\\\\mathrm{OA}^2=\\mathrm{OB}^2\\\\(\\mathrm x-2)^2+(0-5)^2=(\\mathrm x+2)^2+(0-9)^2\\\\\\mathrm x^2-4\\mathrm x+4+25=\\mathrm x^2+4\\mathrm x+4+81\\\\-8\\mathrm x=81-25=56\\\\\\mathrm x=-7\\\\\\mathrm{hence}\\;\\mathrm{the}\\;\\mathrm{point}\\;\\mathrm O\\;(-7,0)\\end{array}{/tex} Check exercise 7.1 of ncert same aisa question hai ncert me... apko proper solution isi app me mil jayega |
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| 28540. |
If the quadratic equation px2- 2√5px +15=0 has two equal roots then find the value of p. |
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Answer» 1/3 b2=4ac( 2√5 )^2=4*15*p20=60pp=1/3 P=3 |
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| 28541. |
Explain why 1323334356715 is a composite number? |
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Answer» Fine Bcz may b it has more than 2 factors....so it cn b considered as composite no. |
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| 28542. |
If sec theta+ tan theta= p so find the value of cosec theta. |
| Answer» Secθ+tanθ=p ----------------------(1)∵, sec²θ-tan²θ=1or, (secθ+tanθ)(secθ-tanθ)=1or, secθ-tanθ=1/p ----------------(2)Adding (1) and (2) we get,2secθ=p+1/por, secθ=(p²+1)/2p∴, cosθ=1/secθ=2p/(p²+1)∴, sinθ=√(1-cos²θ)=√[1-{2p/(p²+1)}²]=√[1-4p²/(p²+1)²]=√[{(p²+1)²-4p²}/(p²+1)²]=√[(p⁴+2p²+1-4p²)/(p²+1)²]=√(p⁴-2p²+1)/(p²+1)=√(p²-1)²/(p²+1)=(p²-1)/(p²+1)∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1)\xa0 | |
| 28543. |
2+root3 and 2-root3 please solve this and give eq with proper explanation |
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Answer» Multiply both numerator and denominator of the fraction by the conjugate of the denominator to clear the square root from the denominator. 2+√3and2-√32+√3+2-√3(2)²-(√3)²4-31 |
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| 28544. |
Will this year papers contain only textual questios |
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Answer» Firstly tell me by textual questions what do you mean.. Its serious bro I know the a answerThe answer is that you will get your answer on 7 March |
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| 28545. |
X square+9x+40=0 , xsquare-10x+24=0 , xsquare-2x-15=0 , xsquare+3x-10=0 |
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| 28546. |
If (a+b)=0,than sin (a_b) can be reduced to |
| Answer» It should be cos(a+b)=0then, cos(a+b)=cos90°=>a+b=90°=>a=90°-bNowsin(a-b)=sin(90-b-b)=sin(90-2b)=cos2b [as sin(90-x)=cosx] | |
| 28547. |
4x_5Y+16=0 and 2x+y_6=0 solve graphically the system of linear equations |
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Answer» X. Y. 1. 46. 811. 12First equations answerX. Y3. 02. 21. 4Second equations answerNow put the value in graphYou will get your ans. 4x-5y=-16. [ X. Y. 4x=-16+5y. 1. 4X=-16+5y÷4. 6. 8 11. 12]---put the value in graph2x+y=6. [ X. Y. 2x=6-y. 3. 0x=6-y÷2. 2. 2 1. 4]----put the value in graph |
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| 28548. |
(x-2)/(x+2)+3(x+2)/(x-2)-4=0 solve this equation |
| Answer» X-2/x+2+3(x+2)/x-2=4(X-2)²+3(x+2)(x+2)/(x+2)(x-2)=4(X-2)²+3(x+2)²/(x+2)(x-2)=4X²+4-4x+3(x²+4+4x)/x²-4=4X²+4-4x+3x²+12+12x/x²-4=44x²+8x+16/x²-4=44x²+8x+16=4x²-168x=-16-16x=-4 | |
| 28549. |
/5 |
| Answer» | |
| 28550. |
sin square 30 degree |
| Answer» (1/2)² = 1/4 | |