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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. |
3x+7x=100 then x=? |
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Answer» x=10 is the right answer |
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| 2. |
\begin { equation } \sqrt{1+\tan ^{2} A}= \end { equation } |
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| 3. |
0.Find the remainder when x 100 is divided by x? - 3x + 2. |
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Answer» Letp(x)=x100p(x)=x100andq(x)=x2–3x+2=x2–2x–x+2=x(x–2)–1(x–2)=(x–2)(x–1)q(x)=x2–3x+2=x2–2x–x+2=x(x–2)–1(x–2)=(x–2)(x–1) LetQ(x)Q(x)andr(x)r(x)be the quotient and remainder whenp(x)p(x)is divided byq(x)q(x). Hence by division algorithm, we havep(x)=q(x)⋅Q(x)+r(x)p(x)=q(x)⋅Q(x)+r(x) That is,x100=(x–2)(x–1)Q(x)+r(x)x100=(x–2)(x–1)Q(x)+r(x) Letr(x)=(ax+b)r(x)=(ax+b)where0<degr(x)<20<degr(x)<2 x100=(x–2)(x–1)Q(x)+(ax+b)(1)x100=(x–2)(x–1)Q(x)+(ax+b)(1) Putx=1x=1in Equation(1)(1), we get1=0+(a+b)1=0+(a+b) Hence(a+b)=1(2)(a+b)=1(2) Now put x = 2 in equation (1), we get 2100=0+(2a+b)2100=0+(2a+b) 2a+b=2100(3)2a+b=2100(3) Solving(2)(2)and(3)(3), we geta=(2100–1)a=(2100–1)andb=2–2100b=2–2100 Therefore the remainder=(ax+b)=(2100–1)x+(2–2100) |
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| 4. |
EAMPLE Find the value of k for which the given system of equations hasinfinitely many soliutions: |
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Answer» Like my answer if you find it useful! |
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| 5. |
What should be added to 6-7, to get 12 ?What should be added to 4 to get oz? |
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Answer» 55/12+x=13/2So x=13/2 - 55/12So x=23/1223/12 should be added |
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| 6. |
What number should be subtracted fromto get 12 |
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| 7. |
\begin { equation } \tan \theta=\frac{a}{b} \text { then } \sin \theta= \end { equation } |
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Answer» thank you just as texting app |
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| 8. |
\begin { equation } (\sin x)^{\tan x} \end { equation } |
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| 9. |
what should be divided by(-7)to get 12 |
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| 10. |
9What should be added to 5to get 12? |
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| 11. |
39. What should be added toto get?128 |
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Answer» 19/24 is the number to be added |
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| 12. |
8 Solve the following pairs of equations3-4and 2(3x+y)AIN2(3x-y)8 |
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Answer» 1/3x+y +1/3x-y=3/4....(1) 1/2(3x+y) +1/2(3x-y)=1/81/(3x+y) +1/(3x-y)=1/4....(2) Equation 1 and 2 have same slope, parallel lines No solution |
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| 13. |
1.Find the ratios in which the following straight lines divide the line segments joining the givenpoints. State whether the points lie on the same side or on either side of the straight line.() 3x-4y 7 (2,-7) and (-1,3)(ii) 3x+ 4y 6; (2,-1) and (1, 1)(ii) 2x +3y 5; (0, 0) and (-2,1)Find the point of intersection of the following lines.(i)2.4x+8y-1-0, 2x-y+1:07x ty +3-0, x+y 0(ii)3. Show that the straight lines (a-b)x+ (bc)y c-a, (b-c)x+(c-a)y a-b and(c-a)x + (a-b)y b-c are concurrent.Transform the following equations into the form Li+AL2-0 and find the point of concurrencyof the family of straight lines represented by the equation.a) (2 +5k)x-3(1 +2k) y (2-k)0(ii) (k+ 1)x + (k + 2)y+5 04. |
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| 14. |
Ex.1: Find the length of the ardiameter 10 cm, if the arc is subtending an ansof 36 at the centre.of cirele dfg an angle |
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| 15. |
sin A, cos Asin C, cos C(1) |
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| 16. |
Ex. If log,bc, y-log,ca, z-log,abthen prove that |
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| 17. |
In any AABC, prove that1. a(b cos C - ccos B)3.(b2 -c)cos At cos B+ cos C(a2 + b2 + c2)2abc |
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| 18. |
log 100 + log (0.01+x)t)3x100 |
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| 19. |
(2) Write the following in exponential form.(G) log1, 100 2(i) log,25-2(ii)y log,2-(iii) log 2 11 |
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Answer» (i)100=10² (ii)25=5² (iii)2=2¹ |
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| 20. |
| 1. Find the sum:+19 1+ 0 |
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| 21. |
00 (2)EďŹ= (Q.uo3 + 1)0:5: â900e 328 |
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| 22. |
If log,y 100 and log,x - 10, then the value of y is |
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| 23. |
Solve the equation tan-x + cot?r= 2 |
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| 24. |
rigonomety=ィ E (-T, π) : x 0,+지 . The sum of all distinct solutions of the equationsec x + cosec x + 2(tan x-cot x) = 0 in the set s is equal toル[JEE (Advanced) 2016, Pa2π(C) 0(D)(B)-99 |
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| 25. |
Solve the following equation for xcos(tan-1 x)=sin(cot-14)3 |
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Answer» Cos(tan⁻¹x)=sin(cot⁻¹3/4)Let, cot⁻¹(3/4)=yor, coty=3/4(=perpendicular/base)Using Pythagorus's theorem,hypotenuse=√(p²+b²)=√(3²+4²)=√25=5∴, siny(=p/h)=3/5∴, sin(cot⁻¹3/4)=siny=3/5cos(tan⁻¹x)=3/5or, tan⁻¹x=cos⁻¹(3/5)Again let, cos⁻¹(3/5)=zor, cosz=3/5=base/hypotenuse∴, perpendicular=√5²-3²=√16=4∴, tanz=p/b=4/3∴, tan⁻¹x=zor, x=tanzor, x=4/3 |
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| 26. |
Variance and Standard Deviation |
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| 27. |
\begin { equation } \sqrt{\sec ^{2} \theta+\csc ^{2} \theta}=\tan \theta+\cot \theta \end { equation } |
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Answer» As sin²theta+ cos²theta= 1sectheta=1/costheta RHS |
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| 28. |
\begin { equation } \begin{array}{l}{\text { 39. } \frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0} \\ {\text { 40. } \frac{\tan A+\tan B}{\cot A+\cot B}=\tan A \cdot \tan B}\end{array} \end { equation } |
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| 29. |
\begin { equation } \cot ^{2} 60^{\circ}+\sin ^{2} 45^{\circ}+\sin ^{2} 30^{\circ}+\cos ^{2} 90^{\circ}= \end { equation } |
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Answer» cot^2 60+sin^2 45+sin^2 30+cos^2 90=1/3+1/2+1/4=(4+6+3)/12=13/12 |
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| 30. |
Solve the pair of equations32 |
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| 31. |
Prove that \frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta} )^{2}=\frac{1-\cos \theta}{1+\cos \theta} 1+sin 6 - 0056 ) " 1-cos®1+sin+cos® ) 1+cos® |
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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 ) ] = RHS |
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| 32. |
4lf the 5th term of G P, is 31 and the 2nd term is 24, find the GP |
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| 33. |
Find the value of k if the pair equations given has infinitely many solution:i) 4x - 3y - (k-2) = 0ii) 8x - 6y - k = 0 |
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Answer» Condition for Infintely many solutions a1/a2 = b1/b2 = c1/c2 4/8 = -3/-6 = -(k-2)/-k 1/2 = 1/2 = (k-2)/k k-2/k = 1/2 2k -4 =k k = 4 |
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| 34. |
GIVEN THAT:(1+cos a)(1+cos b)(1+cos c)=(1-cos a)(1-cos b)(1-cos c)TO PROVE:(cos a) - (cos b) - (cos c). |
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| 35. |
\frac{\sin A}{+\cos A}+\frac{\sin A}{1-\cos A}=\sqrt{\frac{1+\cos A}{1-\cos A}}+\sqrt{\frac{1-\cos A}{1+\cos A}} |
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| 36. |
\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}= |
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| 37. |
RE fog for. [1+ cos +) {1+cos (19)[1 + cos(1 + cos2 = 1 |
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Answer» [ 1 + cos( pi/10)][ 1+ cos(3pi/10][1+ cos(7pi/10)]1+ cos(9pi/10]= 1 /2 ; [10][28][64][82]= |
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| 38. |
log - abc logabc + log abc has the value equal tohas the value equal to :100loglogcaabcvab |
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| 39. |
actice TimeDo these divisions. Check your results by multiplicatia) 438 9b) 3480+ 12c) 4507d) 900 10e) 6786f) 2475+11 |
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| 40. |
For some data Z-M 2.5. If the mean of thedata is 20, then Z- |
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Answer» As given M = 20 Z - M = 2.5 Z = 2.5 + 20 = 22.5 |
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| 41. |
If the mean and the s'andard deviation ofx1, x2, , x, are-6 and 4.3 respectively.find the mean and the standard deviationof x, 3, x, +3, .,x, +3. |
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| 42. |
estion9,IfăŤ71-10, then the value of xis |
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Answer» The value of determinant is x(x)-7(x) =-10 x^2-7x+10 =0 x^2-5x-2x+10=0 x(x-5)-2(x-5) =0 (x-2)(x-5) =0 x-2=0 x=2 x-5=0 x=5 So the value of x is 2 and 5. |
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| 43. |
Find the mean deviation about the mean for the data in5.5 10 15 20 25 |
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| 44. |
Ex. 3:Find the general solutions of the equations () 3 cosecx2 |
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| 45. |
1. Find the 20th and nth terms of the GP24 8, |
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Answer» a=5/2r=(5/4)/(5/2)=1/2so 20th tern is ar^19=(5/2)(1/2)^19=5/(2)^20and nth term is (5/2)*(1/2)^(n-1) |
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| 46. |
Find the 20th and nth terms of the GP.5/2,5/4,5/8, |
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| 47. |
entiate W.1.1. d, toesec? x + 3 cos x...of Xis |
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| 48. |
Q-JA student was calculating mean and standard deviation of20 observations 10 and 2 respectively. On rechecking hefound that an observation 8 was incorrect. Calculate thecorrect mean and standard deviation if he replaced wrongitem by 12. 2 |
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Answer» I think this is wrong as 8 would be subtracted and 12 will be added and here 12 is not added |
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| 49. |
0.56789 by 0.4687 correct to 3 decimal places. |
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Answer» 0.56789÷0.4687=1.2116279 Rounding off to three decimal places 1.2116279≈1.212 |
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| 50. |
In an examination 90% of the candidates passed and 80 candidates failed. Find the no, ondcandidates5. |
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