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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. |
( 4 x ^ 2 - 5 x y %2B 3 ) %2B ( 4 x ^ 2 - 2 x y - 4 ) = |
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Answer» (4x^2 - 5xy + 3) + (4x^2 - 2xy - 4)= 8x^2 - 7xy - 1 |
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| 2. |
rcles touch externally. The sum of their areas is 130T sq. cm and the distance betweentheir centres is 14 cm. Find the radii of the circles. |
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| 3. |
osif cose -2 where x lies in lird quadrant then tinhd the value ot sinco ,thAn2Q6. If tanx +sinxam, tanx-sinx-n, to show that:-- 2-n-dymn.Qz 1f A B, Cand D are the angle of a cyclic quadrilateral then prove that cosA +cosBtors +costcind the maximum and minimum 'slueyi sinxtcos. Ans-V2-2u .T trreeengtesof A, B and Care iaA·P, then prove t -cotBe r-s 15-Q10.To prove that the following identiticssAHB 4-Bsin A +sin Bsin A sin Biv)sin100 sin30 osinsoo sin70%(v)sin200 sinano sin60°sin80%(vi) Sec84+(ix) Sin5xsSsinx-20sin3x +16 sínsx(xi) Cos2x 2sin'y +4cos(x+y) sinxsiny-cos2(x+y) (xii)v3cosec200-sec 20.4(kxili) cosAcos B-2 cos Acos B cos(A B sin AB)iv cos A.cos 2.A.cos 2Q11. Find the principal solution of the following functionsvi cos 1s1+cos(1cos16tan 5 tan 34CoS2Aco4Atai5A tan3 Asec 4 Atan 24never lies between 1/3 and 3fan x2 A cos2 A cos 2 A2 sin 47π 11πsee2 Anstanxe Ans |
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Answer» I dont have enough time i had to submit assignment with in 2 hr Plz hlp me Are working. For it or not Ok.. I'll post the answer Suno serialwise karo 5)since,sin60=√ 3/2 = √ 3/2( sin20sin40sin80) =√ 3/2( sin20sin80sin40) =√ 3/4 [(2sin20sin40)sin80] on applying [cos(A-B)-cos(A+B) = 2sinAsinB] we get, = √ 3/4 (cos20-cos60)sin80 [since,cos(-a)=cosa] = √ 3/4(cos20sin80-cos60sin80) = √ 3/8(2sin80cos20-sin80) = √ 3/8(sin100+sin60-sin80) = √ 3/8( √ 3/2+sin100-sin80 ) = √ 3/8( √ 3/2+sin(180-80)-sin80 ) = √ 3/8( √ 3/2+sin80-sin80 ) [since,sin(180-a)=sina] = √ 3/8( √ 3/2) = 3/16 Saare karo solve Solve the whole questions |
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| 4. |
1. At what points does the line 3x + 4y = 6 cuts the x-axis and thenondinates of any point P in the fourth quadrant wfourth quadrant which is equidistant from thr hy-axis? |
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Answer» To cut at x axis,the value of y must be 0so, putting y=0 in equation3x+4y=63x+4×0=63x+0=63x=6x=2so, the line of equation will cut the x-axis at point (2,0) To cut at y axis,the value of x must be 0so, putting x=0 in the equation3x+4y=63×0+4y=60+4y=64y=6y=6/4y=3/2So,the line of equation will cut y axis at point (0,3/2) |
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| 5. |
Q.9 Find the value of other trigonometric functionswhen Sin xx lies in second quadrant.5' |
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| 6. |
24.1 sin x 1-sin xIfx lies in IInd quadrant, then1n-1-sinxis equal to |
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| 7. |
Point P(2, 3) lies in which quadrant? What will be the co-ordinates of a point Q opposite to itinx-axis ?fourthquadranthavingequaldistancefrom |
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Answer» Both x and y are positive in P. So, it lies in the first quadrant. It's equidistant from x axis. So y coordinate of Q is -3. X coordinate will be constant i.e 2. So Q(2,-3). |
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| 8. |
29 If a+b--2ab- 0, then the family of straight lines ax+ by c0 is concurrent atthe pointsomhus RABC lying entirely in first quadrant or fourth quadrant) of area |
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Answer» if a²+b²-c²-2ab = 0=> a²+b²-2ab = c²=> (a-b)² = c²=> a-b = c => a-c-b = 0, or a-b = -c => a-b+c = 0 so, to make ax+by+c = 0 , look like any of the above results put x = 1 and y = -1 => a(1)+b(-1)+c = 0 => a-b+c = 0✓ so, point is (1,-1) optionB |
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| 9. |
uocircles touch externally at a point P. From a point Ton the tangent at P,tangents TQand TR areTQ = TR.rawn to the circles with points of contact Q and R respectively. Prove that |
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| 10. |
कीही... TR |
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Answer» Splitting the middle term,7x^2-21x-4x+12=0 7x(x-3)-4(x-3)=0 (7x-4)(x-3)=0 x=4/7 x=3 |
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| 11. |
2 B & TR - |
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Answer» P(n,r) = n! / (n - r)!13p₀₌!13/!13-0=1 |
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| 12. |
(рео) (- 380) + (- 270) |
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Answer» (-380) + (-270) -380-270-650 |
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| 13. |
-16*x - x^2 %2B x^4 %2B 4*x^3 - 12 |
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Answer» x^4 + 4x^3 - x^2 - 16x - 12 = x^4 + x^3 + 3x^3 + 3x^2 - 4x^2 - 4x - 12x - 12 = x^3(x + 1) + 3x^2(x + 1) - 4x(x + 1) - 12(x + 1) = (x + 1)(x^3 + 3x^2 - 4x - 12) = (x + 1)(x^3 + 2x^2 + x^2 + 2x - 6x - 12) = (x + 1)(x^2(x + 2) + x(x + 2) - 6(x + 2)) = (x + 1)(x + 2)(x^2 + x - 6) = (x + 1)(x + 2)(x^2 + 3x - 2x - 6) = (x + 1)(x + 2)(x(x + 3) - 2(x + 3)) = (x + 1)(x + 2)(x + 3)(x - 2) |
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| 14. |
-x^2*(3*y - 6) %2B y^2*(3*x %2B y) %2B 3*x(x %2B 2*y) %2B y(-7*x %2B 2*y^2) |
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| 15. |
2*y^3 %2B 3*(x*y^2) %2B x^3 %2B 3*(x^2*y) |
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| 16. |
Find the product of:3 15 4-X — X-5 16 77(ii)1 42-x2-x-7 922X129X7-15u |
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Answer» (i)9/28(ii)15/9(iii)144/5 1. 9/282. 15/93. 144/5 (1) 9/28 (2) 15/9(3) 144/5 (1) 9/28(2)15/9(3)144/5 9/28 -115/9-2144/5 1)9/28 2)15/93) 144/5 1 ka 9/28, 15/9, 144/5 1.9/28, 2.15/9, 3.144/5 15.42 (a). 9/28,,, (b). 5/3,, (c) 144/5 1.9/282.15/93.144/5 (1) 9/28(2) 5/3(3) 144/5 =3/5×15/16×4/7=9/28 1. 9/282. 5/33. 144/5 The answer is 9/285/3144/5 3/5*15/16*4/7= 3/1*3/4*1/7= 9/28 (1) 3/5*15/16*4/7=9/28(2) 15/7*22/9*7/22=5/3(3) 29/3*30/29*6/15*36/5=144/5 (1) 9/28 | 9.3333....(2) 20/33 | 0.6060....(3) 84/725 | 0.1021 लगभग 1:- 9/82:-5/303:- 48 (1):-(3×3)÷28=9/28 answer hoga (i)-9/28(ii)-5/3(iii)-72/25 I) 3/5×15/16×4/7. 3×3/4×7. 9/28. 1 )0.32142857142 You can search for any mathematical expression, using functions such as: sin, cos, sqrt, etc. You can find a complete list of functions here. ( ) % AC 7 8 9 ÷ 4 5 6 × 1 2 3 − 0 . = + Rad Deg x! Inv sin ln π cos log e tan √ Ans EXP xy = 123 Fx (I)-9/28(ii)-15/9(iii)-432/15 i. 3/5×15/16×4/7=45/140=9/28 1 =9/282=15/93=144/5 1 answer 3 by 7 ,2 answer 15 by 9 ,3 answer 432 by 15 (1)9/28(2)15/9(3)144/5 |
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| 17. |
10008*+270* |
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| 18. |
$\sin x=\frac{3}{5}, x$ lies in second quadrant. |
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Answer» 1 2 3 4 |
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| 19. |
cot x=3/4,x lies in third quadrant |
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Answer» cot x = 3/4 (given) We know that in the third quadrant only tan and cot can remains positive. Therefore tan x = 1/cotx = 1/3/4 =4/3 We know that 1 + tan^2x = sec^2x ⇒ 1 + (4/3)^2= sec^2x ⇒ 1 + 16/9 = sec^2x = 9+16/9 = 25/9 sec^2x = 25/9 or sec x = +- 5/3 since x lies in the third quadrant thereforesec x = -5/3 We know that sec x = 1/cos x ⇒ -5/3 = 1/ cosx ⇒ cosx = 1/-5/3 orcos x =-3/5 We know that sinx = perpendicular / Hypotenuse ⇒sin x = -4/5 Therefore cosec x = 1 / sinx = -5/4 cosec x = -5/4 |
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| 20. |
13secxwhere x lies in fourth quadrant5 |
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| 21. |
x^2-6x-270=0 |
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| 22. |
. Explain why 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 iscomposite number |
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Answer» 7× 6 × 5 × 4 × 3 × 2 × 1+ 5 = 5040 +5=5045 = 5045 = 5×1009 We know that a number is called a composite number if it has at least one factor other than 1 and the number itself. Here, 5045 is the product of 2 prime factors 5 and 1009 . Hence,7× 6 × 5 × 4 × 3 × 2 × 1+ 5 is a composite number. |
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| 23. |
76.82x 100 =1482 7.843x 100-7 g ts76.82 x 1,000-6, 82e 7.843 x 1,000-813a) 76.82 x 106826) 7.8303.) Wrire the multipliers of the following.b) 48.792 x1d) 3.808×4メ15.8 xo158c) 15,6931 x4. Find the products.a) 41.7 x 0.25. Solve the following.15693.1b) 0.58 x 0.5c) 34.4 x 0.010.29o. 3a) The length of 1 saree is 5.50 m. Find the length of 14sb) A generator consumes 3.425 L of diesel per hour. Howrequired to run the generator for 8 hours? |
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Answer» 15.8×10=15815.6931×1000=15693.1 |
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| 24. |
6 x^{4}+5 x^{2}+7 x/ x+1 |
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| 25. |
(((x %2B 4)*(x %2B 6))*(x - 7))*(x - 5)=5 |
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| 26. |
\left. \begin array l l ( 3 %2B \sqrt 3 ) ( 2 %2B \sqrt 2 ) & \text (in) ( 3 %2B \sqrt 3 ) ( 3 - \sqrt 3 ) \\ ( \sqrt 5 %2B \sqrt 2 ) ^ 2 & ( \text in ) ( \sqrt 5 - \sqrt 2 ) ( \sqrt 5 %2B \sqrt 2 ) \end array \right. |
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| 27. |
x %2B 2 y ^ 2 %2B 3 x %2B 20 y - 100 = 0 \cdot \text find y \text if x = 10 |
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Answer» x^2+2y^2+3x+20y-100=0X=1010^2+2y^2+3(10)+20y-100=0100+2y^2+30+20y-100=02y^2+20y+30=02(y^2+10y+15)=0y^2+10y+15=0 |
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| 28. |
If the diagonals of a parallelogram are equal, then show that it is a rectan |
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| 29. |
(5ii) In the picture QPlÄŻ ML,nd the value of x. |
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| 30. |
if AB II CD, < APO = 50° and<PRD 127°, find x and y |
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Answer» x = 50° ( Alternate angle)y = 127-50 ( Alternate angle)y = 77° |
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| 31. |
\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\left(\text { when } x \text { lies in } II^{nd \prime} \text { quadrant) }\right. |
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Answer» why u write sinxas sinx/2 sinx=2sinx/2 cosx/2formula sin2A=2sinAcosA that I know but there is any relation b/w 2nd quad |
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| 32. |
lid's division algorithm to find the HCF ofnd 270(Gi) 196 and 38220 (ii) 1651 and 203 |
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| 33. |
3.If sin θ-then find the value of (sin θ + cos θ) + (sin 0-cos θ)2 +(sec θ-tan 0) (sec 0 + tan 0) |
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Answer» wrong |
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| 34. |
) In the given figure AB/CDAPO-50and/PRD- 127, Find x and yA si122C. |
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Answer» x = 50° ( Alternate angle) y = 127 - 50 = 77° ( Alternate angle) |
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| 35. |
If sin 0 - sin (-0) then find 'o'. |
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Answer» when angle is 0 then it could be possible sin¢=sin(-¢)or , sin¢=-sin¢or , 2sin¢=0or , sin¢=0or , ¢=0 |
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| 36. |
le. Find the y13. In each of the figures given below, ABCD is a rectanof χ and y in each case.11035° |
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| 37. |
Explainwhy7x11x13+13and 7 x 6 x 5 x 4 x3exx1 +5 arenumbers. |
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| 38. |
-x %2B 30*(text*((2*x^2)*(b*y))) - 13*x %2B 6*x^3 - 13*x^2 - 6 |
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| 39. |
\left. \begin{array} { l } { 13 ( x - y ) = 29 - 3 x } \\ { 13 x + 4 y = 9 } \end{array} \right. |
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| 40. |
llnDEr Which is divisible by 4, 6, 8 and 12. Also find its square root.. The students of class VII of a school donated 6,625 for Drought Relief Fund. Each student éupees as the number of students in the class. Find the number of students in the class[Hint: Find the square root of 5,625.]Simnlift |
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Answer» Let there be x students.Each student donates ₹x.Total amount = x² = 5625∴ Number of students in the class, x = √5625 = 75 |
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| 41. |
mu the area of the remaining paperD. A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. Iftdiameter of the wheel is 60 cm calculate the speed per hour with which the boy is cyclign.Answers |
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Answer» thanks |
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| 42. |
133. A boy is cycling in such a way that the wheels of his bicycle are making140 revolutions per minute. If the diameter of a wheel is 60 cm, calculatethe speed (in km/h) at which the boy is cycling. |
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| 43. |
सत्यापित करो कि 08 80+ sin60° V31+ cos60 + sin 30 2 |
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Answer» cos30=√3/2 sin60=√3/2 cos60=1/2 sin30=1/2hence√3/2+√3/2/1+1=√3/2 |
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| 44. |
(3) 2009 (4) 201034. 35 is a surd of the order of:12m (2) 160 cmm (4) 72 cmI divides his herd ofmong his four sons soson gets one-half ofthe second son(3),(4)(1) 1(3) 343. If x = sin A cos B,25 |
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Answer» yes the given answer is correct We have to find the order of the surd ³√5 A root of a positive real quantity is called a surd. A surd with index of root 3 is called a third order surd. If x is a positive integer with nth root, then n is called the surd when the value is irrational. Hence, we can say In expression n is the order of surd and x is known as radicand. ∴ In the given question, the order of the surd ³√5 is 3 (3) is correct option |
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| 45. |
7. In the given figure, if AB II CD, ZAPO = 50 and Z PRD = 127",find x and yRD |
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Answer» Use property of Alternate interior angles?APR = ?PRD50° + y = 127°y = 127 - 50y = 77°use same property of Alternate interior anglesAPQ = PQR50° = x x = 50° and y = 77° |
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| 46. |
(yani. Find the sum of 51 terms of A.P. in which II ndand III rd terms are 14 and 18 respectively. |
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| 47. |
sin 0 =find the value of all T-ratios of 0.1. If sin 0 =V3 |
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Answer» Sin theta =(3^1/2)/2=sin60°so,theta=60° |
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| 48. |
हि -536 + 63"’/',)&52),3»" |
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Answer» 36.8 - (-53.6+63.4)×4.8 = 36.8 - 9.8×4.8 = 36.8 - 47.04 = -10.24 If you find this answer helpful then like it. wrong answer |
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| 49. |
(at) 2(R1) I(a)(a) ant |
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Answer» wrong ans koi triangle nhi bnega |
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| 50. |
is right angled.ORTwo angles of triangle are equal and the third angle is greater than each of these angles by 30°Find all the angles of the triangle. |
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Answer» thanks |
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