Explore topic-wise InterviewSolutions in Current Affairs.

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.

b) Given that v2 is a zero of the cubic polynomial of 6x'/2x2-10x-42. Find its othertwo zeroes.

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2.

4X° + X% + 5x ~ 4 3R 2x2 — 6x - 3 को जोड़ें।

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3.

Is log381 rational or irrational ? Justify your answer.

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Log ₃(3⁴)=4 it is not irrationalwe can write it as 4/1 so it is rational

4.

Is \pi a rational number? Justify your answer.

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No, π is an irrational number. Reason: We can not write it is in form of p/q form.

Note: 22/7 is its approximate value used for calculation.

5.

4x3/22. lsx"+a polynomial? Justify your answer.

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yes, as x² + (4x^3/2)/x^1/2 simplified as x² + 4x which is a polynomial.

6.

Find k so thatx2 +2x + k is a factor of2x4 +x3-14x2 +5x+6. Also, find allthe zeroes of the two polynomials.

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7.

Find k so that x + 2x + k is a factor of 2x4 + x - 14x + 5x + 6. Also find all thezeroes of the two polynomials.

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Here is my solution:

Given factor: x2 + 2x + k = 0

Given polynomial: 2x4 + x3 -14x2 + 5x + 6

Divide the polynomial by the factor

x2 + 2x + k ) 2x4 + x3 -14x2 + 5x + 6 ( 2x2 - 3x +(- 8 - 2k)

2x4 + 4x3 +2kx2 ( substract)

------------------------------

- 3x3 +(-14 - 2k)x2 + 5x

- 3x3 - 6x2 - 3kx ( substract)

------------------------------

(- 8 - 2k) x2 +( 5 + 3k)x + 6

(- 8 - 2k) x2 +(-16 - 4k)x + (- 8k - 2k2) ( substract)

-----------------------------------------------------------------

( 21 + 7k)x + (6 + 8k + 2k2)

The remainder is: ( 21 + 7k)x + (6 + 8k + 2k2) = 0

21 + 7k = 0 ⇒ k = -3.

The factors are x2 + 2x - 3 = 0 and 2x2 - 3x - 2 = 0

x2 + 3x - x - 3 = 0 and 2x2 - 4x + x - 2 = 0

x( x + 3 )-1( x + 3) = 0 and 2x (x - 2) + 1(x - 2) = 0

(x - 1)( x + 3) = 0 and (2x + 1)(x - 2) = 0

x = 1 ,3 ,-1 / 2 and 2.

The zeros are 1 ,3 ,-1 / 2 and 2.

1;2;3;-1/2 are zeros of given polynomial.

8.

3T111) sin+ 2 cos π+ 3 . sin3π5 sec π--6 cosec

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9.

7. Find k so that x2 + 2x + k is a factor of 2x4 + x 14x2 +5x + 6. Also find all the zeroes of thetwo polynomials

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( x^2+2x+k )

to be a factor of

(2x^4+ x^3- 14x^2+ 5x + 6)

10.

9. If (x + a) is a factor of two polynomials x++px + q and x2 +mx + n then prove that an-am-p

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(x+a)=.: x=-a; x^2+px+q=(-a)^2+p(-a)+q=a^2-ap+q=0 ; x=-a^2; (-a)^2+m(-a)+n=a^2-ma+n; a^2-ap+q=a^2-ma+n; -ap+q=-ma+n; am-ap=n-q; a=n-q/m-p

11.

(9t-7)(t+6)=(3t-4)(3t+5)

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(9t-7)(t+6)=(3t+5)(3t-4)We move all terms to the left:(9t-7)(t+6)-((3t+5)(3t-4))=0We multiply parentheses ..(+9t^2+54t-7t-42)-((3t+5)(3t-4))=0

We calculate terms in parentheses: -((3t+5)(3t-4)), so:(3t+5)(3t-4)We multiply parentheses ..(+9t^2-12t+15t-20)We get rid of parentheses9t^2-12t+15t-20We add all the numbers together, and all the variables9t^2+3t-20Back to the equation:-(9t^2+3t-20)

We get rid of parentheses9t^2-9t^2+54t-7t-3t-42+20=0We add all the numbers together, and all the variables44t-22=0We move all terms containing t to the left, all other terms to the right44t=22t=22/44t=1/2

12.

5t - 3 = 3t -5

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5t-3=3t-55t-3t=-5+22t=-2t=-1

13.

AWASIYA VIDYALAYA, KURUKSHETRAMathematics AssignmentEvents E and F are such that P (not E or not F) = 0.25. State whetherand F are mutuall exclusive.n that PP(B)

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14.

State Euler's formula for polyhedrons. Using Euler's formula find thenumber of edges if faces are 9 and vertices are 9.

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Euler's formula tells us that

V - E + F = 2;

or, in words: the number of vertices, minus the number of edges, plus the number of faces, is equal to two.

15.

Find all zeros ofthe polynomial 3x4-15x3+172+5x-6 i­f two of its zeros are-1/√3 and 1/√3

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16.

3 2. Express the numbers 6825 and 3825 as a product of its factors. Find the L.above numbers by using their product of Prime Factors. Justify answer

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17.

Find all the zeroes of the polynomial 3x4 + 6x -- 2x2 - 10x - 5 if two of its zeroes al

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18.

5)5225A)iD)

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2x + 5 = 02x = -5x = -5/2

2x+5=02x=-5X=-5/2so (b) is ryt option

19.

a) 125Asumla) 3250(b)150360s in 219days at 5%b) 3500per annum. The sum isie) t 3400anounts toid) 3550

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Answer: b)Rs 3500Given:Amount = 3605 rs; rate=5% per annum; time= 219 days=219/365 yrsLet priciple be xFormula:SI=PRT/100=x*5*(219/365)/100=3x/1003x/100+x=3605x=3500

20.

Section: D (4 marks each)23,0btain all other zeros of the polynomial 3x4 + 6x3-2x2-10x-5,if twoof its zeros are.and-li

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p(x)=3x4+6x3-2x2-10x-5

as we have given that two of its zeroes are √(5/3) and -√(5/3). so their factors are;

(x-√(5/3)) and (x+√(5/3)). the product of the factors ;

(x-√5/3) (x+√5/3)=x^2-5/3

=3x^2-5/3=1/3(3x^2-5)

where k=1/3

now g(x)=3x^2-5

dividing p(x) by g(x) we get

q(x)=x^2+2x+1

by division algorithm;

p(x)=g(x)*q(x)+r(x)

=(3x^2-5)(x^2+2x+1)

=(3x^2-5)(x^2+x+x+1)

=(3x^2-5)(x(x+1)+1(x+1)

(3x^2-5)(x+1)(x+1)

the other two zeroes are -1 and-1

21.

s. The product of two polynomials is3x4+5x-21x2 -53x-30. If one of themis x2 -x -6, find the other polynomial.

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22.

(c) OIf θ lies in the second quadrant, then the value of(a) 2sec θ(c) 2cosece(d) of theseasin-sin1 - sin6(b)-2sec θ(d) of these

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23.

express 196 as product of prime factors.

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thanks it is correct

24.

express 96 as a product of prime factors

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96=2 x 48

96=2 x 2 x 24

96=2 x 2 x 2 x 12

96=2 x 2 x 2 x 2 x 6

96=2 x 2 x 2 x 2 x 2 x 3

96= 2^5 x 3

96=2×48 2×2×24 2×2×2×12 2×2×2×2×6 2×2×2×2×2×3

96=2x2x24=2x2x2x12=2x2x2x2x6=2x2x2x2x2x3Answer

25.

etOR1-sineProve that V 1+sinesec- tanednn drm rm and รกcm and then:

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LHS = √(1 + sin∅)/√(1 - sin∅)= √(1 + sin∅) × √(1 + sin∅)/√(1 -sin∅)×√(1 +sin∅)= √(1 + sin∅)²/√(1 -sin²∅)= (1 + sin∅)/√cos²∅= (1 + sin∅)/cos∅= 1/cos∅ + sin∅/cos∅= sec∅ + tan∅ = RHS

Like my answer if you find it useful!

26.

4. Find the distance between the points A(Cos0,0) and B(0,sin6)

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27.

11. Express 1947 into product of prime factors.

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Prime factorisation of 1947:1947=3 x 11 x 59

28.

Express as a product of prime factors 7429

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7429 as a product of its prime factors...17×19×23=7429

29.

rdului1S.LTOUTLO.By using the suitable identities, solve (1001) - (999)

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1000008000 this is the answer

30.

A shuttle cock used for playing badmintonhastheshape of a frustum of a cone mounted on ahemisphere the external diameters of the frustumare 5 cm andexternal surface area of the cock.2 cm and the height is 7 cm. Find

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Solution

Referrence

31.

the remainder when x^15 is divided by x-2 is

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this is wrong solution try again

32.

5id the remainder when x17 + 15 is divided by x + 1

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33.

{(sin +cos )‘7 +(sin6 - cosd Y=2

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34.

dby A shuttle cock used for playing badminton has theshape of a frustum of a cone mounted on ahemisphere the external diameters of the frustumare 5 cm and 2 cm and the height is 7 cm. Findexternal surface area of the cock.

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r1= 1 cm,r2= 2.5 cm,h= 6 cml = √h² + (r2-r1)²= √(6)² + (2.5-1)² cm= √36 + 2.25 cm= √38.25 cm= 6.18cmSo,External Surface area = (curved surface area of frustum) + (curved surface area of hemisphere)= 22/7(r1+r2)l +2.22/7.r1= 22/7 [(1+2.5)6.18 + 2×(1)²] cm²= 22/7 (3.5 × 6.18 +2) cm²= 22/7 (21.63+2) cm²= 22/7 × 23.63 cm²= 519.86/7 cm²= 74.26 cm²

35.

15 Divide the polynomial 3x4-4x3 3x(x - 1)and find its quotient and remainder.

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36.

A shuttle cock used for playing badminton has the shape of frustum of a cone mounted on a hemi-sphereThe external diameters of the frustum are 5 cm and 2 cm, the height of entire shuttle cock is 7 cm. Find the, external surface area

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Given:r1= 1 cm,r2= 2.5 cm,h= 6 cml = √h² + (r2-r1)²= √(6)² + (2.5-1)² cm= √36 + 2.25 cm= √38.25 cm= 6.18cmSo,External Surface area = (curved surface area of frustum) + (curved surface area of hemisphere)= 22/7(r1+r2)l +2.22/7.r1= 22/7 [(1+2.5)6.18 + 2×(1)²] cm²= 22/7 (3.5 × 6.18 +2) cm²= 22/7 (21.63+2) cm²= 22/7 × 23.63 cm²= 519.86/7 cm²

37.

6.1sin6, find the values of cost and tanf

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sint = 7/25 cost = √(1-49/625) = √(576/625) = 24/25 tant = sint/cost = (7/25)/(24/25) = 7/24

38.

Prove that: \lsece._tan9+ _1l+sin6tan 0 -secg4q - cos 0

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39.

OR1-sineProve that

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40.

the lawn.b. A shuttle cock used for playing badminton has the shape of a frustum of a conemounted on a hemisphere. The external radii of the frustum are 5cm and 2cnmThe height of the entire shuttle cock is 7cm. Find its external surface area.OR

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= 1 cm,r2= 2.5 cm,h= 6 cml = √h² + (r2-r1)²= √(6)² + (2.5-1)² cm= √36 + 2.25 cm= √38.25 cm= 6.18cmSo,External Surface area = (curved surface area of frustum) + (curved surface area of hemisphere)= 22/7(r1+r2)l +2.22/7.r1= 22/7 [(1+2.5)6.18 + 2×(1)²] cm²= 22/7 (3.5 × 6.18 +2) cm²= 22/7 (21.63+2) cm²= 22/7 × 23.63 cm²= 519.86/7 cm²= 74.26 cm²

41.

hollow cone is cut by a plane parallel to the base and the upperportion is removed. If the curved surface of the remainder is of thecurved surface of the whole cone, find the ratio of the line segmentinto which the cone's altitude is divided by the plane.89

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hit like if you find it useful

42.

! cos-sin8+127, Drove i = cosech +cotनल c0s8-+sin6-1)

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LHS = (cos∅ - sin∅ + 1)/(cos∅ + sin∅-1) dividing sin∅ both Numerator and denominator. = (cos∅/sin∅ - sin∅/sin∅ + 1/sin∅)/(cos∅/sin∅ + sin∅/sin∅ - 1/sin∅)= (cot∅ - 1 + cosec∅)/(cot∅ + 1 - cosec∅)

now, put 1 = cosec²∅ - cot²∅ in numerator

= {cot∅ + cosec∅ - (cosec²∅-cot²∅)}/(cot∅-cosec∅ +1)= (cosec∅+cot∅)(1 - cosec∅ + cot∅)/(cot∅-cosec∅+1)= cosec∅ + cot∅ = RHS

Hence proved

43.

यदि ८७6 - जे 5 (1+sin6)(1—sin®)8 (1+cos0)(1—cos®)मान ज्ञात कीजिए |

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44.

divided by 32 quotient is 791 remainder 15 dividend is 25227

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45.

When a number is divided by 32, the quotient is 791 and remainder 15What is the number?(1) 25237 (2) 25327 (3) 25227 (4) 25247

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46.

Prove that Sin6(1 + Cost) has maximum value at θ = 3

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Other than this value all are having a lower value as sin (60)= √3/2and cos(60)= 1/2so√3/2(1+1/2)√3/2(3/2)3√3/4

47.

फिर कम जात प घ्कगे प लत प ८० उजब्छप कब्ब्ठ कच्चाProve that sin6(1 + (करा + ८०560 + ८०६8) न उह८8 + ८०5९८8

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sinθ(1+tanθ)+cosθ(1+cotθ)=secθ+cosecθ

soling LHS

sinθ+sin²θ/cosθ + cosθ + cos²θ/sinθ=(sinθ+cos²θ/sinθ)+(cosθ+sin²θ/cosθ)=1/sinθ + 1/cosθ=cosecθ + secθ

48.

e I+sin6+‘ ८059हा ८0590 1+ 5090

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49.

यंदि sec0 = 13 तो (1 +5in0) (1 - sin6) का मन: 12 . {1+ cos0)} (1 —cos®) -

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50.

0. Prove that:SIn6

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2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1= 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1=2-6sin²θcos²θ-3+6sin²θcos²θ+1=-1+1=0 (Proved)