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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Fsrugdt |
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| 2. |
Cot 7÷8 evaluate:( 1+sin)(1-sin)÷(1+cos)(1-cos) |
| Answer» Given: cot\xa0{tex}\\theta{/tex}\xa0{tex}= \\frac { 7 } { 8 }{/tex}To Evaluate: {tex}\\frac { ( 1 + \\sin \\theta ) ( 1 - \\sin \\theta ) } { ( 1 + \\cos \\theta ) ( 1 - \\cos \\theta ) }{/tex}{tex}= \\frac { 1 - \\sin ^ { 2 } \\theta } { 1 - \\cos ^ { 2 } \\theta } = \\frac { \\cos ^ { 2 } \\theta } { \\sin ^ { 2 } \\theta }{/tex}= cot2{tex}\\theta{/tex}{tex}= \\left( \\frac { 7 } { 8 } \\right) ^ { 2 }{/tex}{tex}= \\frac { 49 } { 64 }{/tex} | |
| 3. |
how many people in tire world immigrated to America and how many specially from europe |
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| 4. |
agar p= 10 hai tho pp kitna hoga or jo bhi answer a gaya usa 300 se divided kar dina |
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| 5. |
9x+3y+12=o |
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| 6. |
Which of the so 5,9,13,17.............is81 |
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Answer» a=5d=4an=81an=a+(n-1)d81=5+(n-1)4n=80/4=20 19 |
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| 7. |
2+2+43332245566+54335676543 |
| Answer» 97667922113 | |
| 8. |
Sin⁴/ a + cos⁴/b= 1/a+bProve that sin8/a³+cos8+b³=1/(a+b)³ |
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| 9. |
Find the value of k for which the points are collinear (72) (51) (3k) |
| Answer» 0 is the answer.Do u want explanation | |
| 10. |
Prove pythagoreas theorem |
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| 11. |
Sn - Sn-1 = ? |
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Answer» -1 Sn- Sn-1=An |
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| 12. |
mx-ny=m^2+n^2x+y=2nSolve by cross- multiplication method |
| Answer» Given equations aremx - ny = m2 + n2 .................(i)x + y = 2m ....................(ii)Herea1 = m, b1 = -n, c1 = -(m2 + n2)a2 = 1, b2 = 1, c2 = -(2m)By cross multiplication method{tex}\\frac{x}{{{{( 2mn + m^2 + n^2)} }}} = \\frac{{ - y}}{{ - {2m^2} + m^2 + n^2 }} = \\frac{1}{{m + n}}{/tex}{tex}\\frac{x}{{{{(m + n)}^2}}} = \\frac{{ - y}}{{ - {m^2} + {n^2}}} = \\frac{1}{{m + n}}{/tex}Now, {tex}\\frac{x}{{{{(m + n)}^2}}} = \\frac{1}{{m + n}} {/tex}{tex}⇒ x = m + n{/tex}And, {tex}\\frac{{ - y}}{{ - {m^2} + {n^2}}} = \\frac{1}{{m + n}}{/tex}{tex} ⇒ y = m - n{/tex}The solutions of the given pair of equations are x = m + n and y = m - n. | |
| 13. |
20x2 |
| Answer» 40 | |
| 14. |
X/a+y/b=a+b |
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| 15. |
X/a+y/b=a+b, x/a×a+y/b×b=2 find the value of X and Y |
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| 16. |
Purn warge banakar hal karne ki vidi se x*x-4x-8=0 ke mul prove kegia? |
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| 17. |
px+qy=p+qpx-qy=p-q |
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| 18. |
Gfuj |
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| 19. |
2x+x-4=0 |
| Answer» It\'s easy | |
| 20. |
1—tanA |
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| 21. |
2xsquare +x-2 factorisation |
| Answer» x=(-1+√17)/4,(-1-√17)/4[by quadratic formula] | |
| 22. |
Relations between x and y if points are (x,y)(1,2)(7,0) |
| Answer» If the given points are collinear, then the area of the triangle with these points as vertices will be zero.{tex}\\therefore{/tex}{tex}\\frac{1}{2}{/tex}[x (2 – 0) + 1 (0 – y) + 7(y –2)] = 0{tex}\\Rightarrow{/tex}{tex}\\frac{1}{2}{/tex}[2x – y + 7y –14] = 0{tex}\\Rightarrow{/tex}{tex}\\frac{1}{2}{/tex}[2x + 6y –14] = 0{tex}\\Rightarrow{/tex} 2x + 6y – 14 = 0{tex}\\Rightarrow{/tex} x + 3y – 7 = 0 .............Dividing throughout by 2This is the required relation between x and y. | |
| 23. |
If the sum of the zeroes of the cubic polynomial 4x^3-kx^2-8x-12 |
| Answer» – 3 | |
| 24. |
If x3+8x2+kx+18 is completely divisible by x2+6x+9 then find k |
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| 25. |
If discriment is grater than zero then the nature of root |
| Answer» Then the root are real & distinct | |
| 26. |
Sin^2(6x+4)+cos^2(8x+8) |
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| 27. |
3/x+1 +4/x-1 =29/4x-1 |
| Answer» x=-7 or x= 4 | |
| 28. |
Quadratic |
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| 29. |
Prove that 1\\√2 is irrational |
| Answer» Let us assume, to the contrary, that is{tex}\\frac { 1 } { \\sqrt { 2 } }{/tex} rational.So, we can find coprime integers a and b\xa0{tex}( \\neq 0 ){/tex}\xa0such that{tex}\\frac { 1 } { \\sqrt { 2 } } = \\frac { a } { b } \\Rightarrow \\sqrt { 2 } = \\frac { b } { a }{/tex}Since, a and b are integers,\xa0{tex}\\frac { b } { a }{/tex}\xa0is rational, and so is{tex}\\sqrt { 2 }{/tex} rational.But this contradicts the fact that is{tex}\\sqrt { 2 }{/tex} irrational.So, we conclude that is{tex}\\frac { 1 } { \\sqrt { 2 } }{/tex} irrational. | |
| 30. |
72+70+68+ - - - - -+x =952 solve for x |
| Answer» Sn=952 ,a=72, An=x, d=72-70=2 find it now.. | |
| 31. |
121+21 |
| Answer» 142 | |
| 32. |
Define polynomial |
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| 33. |
a^3+b^3=? |
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| 34. |
Q.9 If the LCM of \'a\' and 18 is 36 and the HCF of \'a\' and 18 is 2, then find the value of \'a\'. |
| Answer» a*18=36*2 Therefore,a=4 | |
| 35. |
For what value of k will K + 9, took a -1 and took a + 7 are the consecutive terms of an ap |
| Answer» We know that difference between any two consecutive terms of an AP\xa0is equal{tex}\\therefore{a_2} - {a_1} = {a_3} - {a_2}{/tex}let\xa0{tex}a_1=(k+9), a_2=(2k-1), a_3=(2k+7){/tex}{tex} \\Rightarrow (2k - 1) - (k + 9) = (2k + 7) - (2k - 1){/tex}{tex} \\Rightarrow k - 10 = 8{/tex}{tex} \\Rightarrow k = 18{/tex} | |
| 36. |
What is triangle |
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Answer» triangle is a plane figure bounded by three line segments & having three angles. Triangle is a figure with 3 side and 3 angles. |
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| 37. |
Factorization method |
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| 38. |
what is thales therom |
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| 39. |
Solve quadratic equation for x 4x² - 4a²x + ( a⁴ - b⁴ ) =0 |
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Answer» 4x2-4a2x+(a4-b4)=0a = 4, b = -4a{tex}^2{/tex}, c =\xa0{tex}a^4-b^4{/tex}D =\xa0{tex}b^{2}-4 a c{/tex}=\xa0{tex}\\left(-4 a^{2}\\right)^{2}-4 \\times 4 \\times\\left(a^{4}-b^{4}\\right){/tex}{tex}=16 a^{4}-16 a^{4}+16 b^{4}{/tex}D = 16 b4Therefore,by quadratic formula,we have,{tex}x=\\frac{-b \\pm \\sqrt{D} }{ 2 a}{/tex}{tex}=\\frac{-\\left(-4 a^{2}\\right) \\pm 4 b^{2}}{2 \\times 4}{/tex}{tex}=\\frac{4 a^{2} \\pm 4 b^{2}} {8}{/tex}x =\xa0{tex}\\frac{ a^{2}\\pm b^2} {2}{/tex}x =\xa0{tex}\\frac{a^2+b^2}{2}{/tex} or x={tex}\\frac{a^2-b^2}{2}{/tex}\xa0\xa0{tex}{/tex}\xa0{tex}{/tex}Hence,the roots of given quadratic equation is ,{tex}\\frac{a^2+b^2}{2} ,\\frac{a^2-b^2}{2}{/tex} yeah you are right |
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| 40. |
What is sin formula |
| Answer» Sin =perpendicular/hypotenuse | |
| 41. |
Two variables in one equation |
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| 42. |
Prove that-1/alpha^4 + 1/beta^4 = alpha^4 +beta^4 / (alpha -beta)^4 |
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| 43. |
The 4th trem of an Ap is zeors .prove that 25th term of the APis three times its 11th term |
| Answer» a4=0a+(4-1)d=0a+3d=0 ....(1)We need to prove thata25=3×a11i.e, a+24d = 3(a+10d)Proof:a+24d. =3(a+10d)a+3d+21d =3(a+3d+7d)(a+3d)+21d =3{(a+3d)+7d}0+21d =3{0+7d} ....Using(1)21d =21d L.H.S = R.H.S | |
| 44. |
To prove (1+tanAtanB)²+(tanAtanB)²=sec²Asec²B |
| Answer» We have,L.H.S = (1 + tanA tanB)2 + (tan A - tanB)2{tex}\\Rightarrow{/tex}\xa0L.H.S = (1 + 2tanA.tanB + tan2A.tan2B) + (tan2A + tan2B-2tanA.tanB) {tex}\\left[\\because\\left(a+b\\right)^2=a^2+2ab+b^2,\\;\\left(a-b\\right)^2=a^2-2ab+b^2\\right]{/tex}{tex}\\Rightarrow{/tex}\xa0L.H.S = 1 + tan2A.tan2B + tan2A + tan2B{tex}\\Rightarrow{/tex}\xa0L.H.S = (1 + tan2A) + (tan2B + tan2A.tan2B){tex}\\Rightarrow{/tex}\xa0L.H.S = (1 + tan2A) + tan2B(1 + tan2A) {tex}\\left[\\;taking\\;\\tan^2B\\;as\\;a\\;common\\;\\right]{/tex}{tex}\\Rightarrow{/tex} L.H.S = (1 + tan2A)(1 + tan2B) = sec2A sec2B = R.H.S {tex}\\left[\\because1+\\tan^2\\theta\\;=\\;sec^2\\theta\\right]{/tex}therefore,(1 + tanA.tanB)2 + (tan A - tanB)2\xa0{tex}={/tex}\xa0sec2A.sec2BHence proved. | |
| 45. |
How to do mode in statistics |
| Answer» Mode=l + (f1-f0/2f1-f0-f2) * h | |
| 46. |
Prove that (5+3root2)is irrational |
| Answer» 5+3√2=p/q.p and Q are co-primes 3√2=p-5q/Q. √2=p-5q/3q.Which is rational but √2 is is irrational.Therefore this contradicts that our assumption is wrong.5+3√2 is irrational. | |
| 47. |
1/x+1 -1/x-7 =11/30 |
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| 48. |
Is there any Vkvstudent? |
| Answer» Yes I am also a KV student. | |
| 49. |
Show that :Sin²45°+Cos²45°=1 |
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Answer» Sin²+Cos²=1. (identity) NowSin²45+cos²45 =1Proved Second method÷since sin^2A+cos^2A=1. therefore, sin^245+cos^245=1 (1/√2)^2+(1/√2)^2=1/2+1/2=1 |
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| 50. |
I got 90 marks in maths if any body have doubt you can ask in comment |
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