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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. |
3. Find the distance between the points (a cos 65°, 0) and (0, a cos 259). |
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| 2. |
7. Solve for x and y by cross-multiplication method: 4 +5y 7 and 3 +4y. x |
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| 3. |
By cross multiplication method\frac{x+y}{x y}=2, \quad \frac{x-y}{x y}=6 |
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| 4. |
an Ď34. lim 1-tan x)IC |
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Answer» by taking t= x-3.14/4 |
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| 5. |
11.12 2980 एक न्यून कोण त्रिमुज है, और 42) L BC तAC’=AB*+ BC*-2BC x BD |
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Answer» inΔADC-AC²=AD²+DC²(pythagoras पाइथागोरस प्रमेय theorem)...................1inΔADB-AB²=AD²+BD²(same पाइथागोरस प्रमेय reason).............................2AC²=AD²+(BC-BD)² (BD+DC=BC) from 1AC²=AD²+BC²+BD²-2·BC·BD [(a-b)²=a²+b²-2ab]AC²=(AD²+BD²)+BC²-2·BC·BDfrom 2AC²=AB²+BC²-2.BC.BDhence proved इति सिद्धम्म |
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| 6. |
all the parts:Determine P, so that the following equationcoincident roots: t2 + p2 2 (p + 1) t.(10and 6,-7) are coln |
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Answer» For coincident roots,b^2=4ac(2(p+1))^2=4(p^2)4(p^2+1+2p)=4p^28p=-8p=-1 |
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| 7. |
\operatorname { lim } _ { x \rightarrow \frac { \pi } { 6 } } \frac { \operatorname { cot } ^ { 2 } x - 3 } { \operatorname { cosec } x - 2 } |
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Answer» by putting x=pie/6 we will get 0/0 form so using L'hospital rulelim x->pie/6 -2cotxcosec^2x/-cosecxcotx=lim x->pie/6 2cosecx=2(2)=4 |
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| 8. |
In âłABC, DEll BC, Find the value of x.x +3E. |
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Answer» x/x+1= x+3/x+5x(x+5)= (x+1)(x+3)x^2+5x= x^2+4x+35x-4x= 3x= 3 |
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| 9. |
24. In a AABC, it is given thatD.LA: LB:4C 3:2:1 and CDLAC.Find LECD |
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| 10. |
1. Show that the statementp: "If x is a real number such that x' +4x(i) direct method,0, then x is 0" is true by(ii) method of contradiction, (ii) method of contrapositive |
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| 11. |
If from an external point B of a circle with centerer, two tangents BC and BD aredrawn such thatDBC-120, prove that BO= 2BC. |
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| 12. |
10. In the adjoining figure, ABCD is aquadrilateral in which diag. BD 14 cm. IfALL BD and CM丄BD such that AL = 8 cmand CM- 6 cm, find the area of quad. ABCD. |
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| 13. |
method:() x+y=5and 2x-3y = 4 |
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| 14. |
In Fig. 6.53, ABD is a triangle right angled at Aand AC丄BD. Show that'll obi(i) AB BC. BD(i) AC2 -BC. DC(iii) AD-BD, CD |
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| 15. |
by multiplication method x-3y-3=03x-9y-2=0 |
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Answer» These lines are parallel to each other asa1/a2= b1/b2 |
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| 16. |
се! 20. Show that:a the points (0,-2), (3,1). (0, 4), (-3, 1) arevertices of a squaretheA'a the points (o -1), (2, 3), (6, 7), (8, 3) are thevertices of a rectangle |
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Answer» 1)PR²= (0-0)²+(-2-4)²= 36SQ²=(-3-3)²+(1-1)²= 36the above two diagonals are equal further, PQ²= 18QR²= 18so the proof is obvious |
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| 17. |
Which of the following is false ?(A) Gopi says the distance of a point P (x, y,) from origin0, 01 is(B) Centroid of triangle whose vertices are (3, -5), (-7,4) (10, -2) is (2, -1)(C) Slope of A (0, 4), B (4, 0) is 1.(D) Mid-point of the line segment joining points (2, 7) and (12,-7) is (7,0) |
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Answer» (c) The slope of A(0,4) B(4,0) is 0-4/4-0 = -4/4 =-1. supet |
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| 18. |
For what value of k, (k > 0), is the area of the trianglewith vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to53 sq. units? |
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| 19. |
L calculatemea uca.-d LA-1 LB & LC |
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| 20. |
Prove that: $\left|\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}\right|=-\frac{2}{\cos x},$ where $\frac{\pi}{2}<x<\pi$ |
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| 21. |
2 T B, T 1.oSRtLB |
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| 22. |
If sum of first 7 terms of an A.P is 49 and sum of its first 17 terms is 289find the sum of the first 'n' terms.OR |
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| 23. |
If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum offirst n terms of the A.P3. |
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| 24. |
(1) Ä°f:|-8a dl-:|-- 8 and12. From the given determinants, form the twosimultaneous equations in x and y |
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Answer» 2x - 3y = 8 -x - 3y = 12 2x - 3y - (-x -3y) = 8 - 12 2x - 3y + x + 3y = -4 3x = -4 x = -4/3 -8/3 - 3y = 8 3y = -8/3-8 = -32/3 y = -32/9 |
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| 25. |
\begin{array} { l } { \text { Form two simultaneous equations from the } } \\ { \text { given determinants and solve. (3 marks each) } } \\ { \text { (i) } \left| \begin{array} { c c } { 3 } & { y } \\ { 4 } & { x } \end{array} \right| = 18 \text { and } \left| \begin{array} { c c } { y } & { x } \\ { 2 } & { - 1 } \end{array} \right| = - 1 } \end{array} |
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Answer» 3x-4y=18-y-2x=-12x+y=13x-4y+8x+4y=18+4=2211x=22so x=24+y=1so y=-3 |
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| 26. |
) A AR & ८निन८ 8 प्भ७* सशसर 3 F IP LA gy भले भरे A0 |
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| 27. |
(1) ADD(ii) AB + BC + CD>DA(iii) AB + BC + CD + DA> AC + BD10. Prove that the sum of all the angles of a quadrilateral is 364 |
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Answer» By drawing a diagonal to a quadrilateral it forms 2 equal triangles according to the triangle theory 'sum of angle of 3 angles of triangle is 180 'as there ar 2 triangle with their angle 180 and 180 therefore the sum of the angle of the 2 triangle is 180+180=360 |
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| 28. |
5. ABCD is quadrilateral. 1AB +BC+ CD +DA <2 (AC+ BD)? |
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Answer» thanks for helping me in my home work. |
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| 29. |
1.Show that in a quadrilateral ABCD, AB + BC+ CD+ DA > AC + BD |
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| 30. |
30. If COS A = m and COSA = n, then prove that(m2 + n2) cos²B = n2 |
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Answer» Given:cosA/cos B=m; cos A/sin B=n To prove:(m²+n²)cos²B=n²now (m²+n²)cos²B={(cos²A/cos² B)+(cos ²A/sin ²B}cos²B=cos²A{ (1/cos²B) + (1/sin ²B) }cos²B=cos ²A{(sin²B+cos ²B)/(sin²Bcos²B)cos²B =(cos ²A cos ²B)/(sin²Bcos²B)=(cos²A/sin²B)=n²=RHS |
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| 31. |
k/find the orthocenter of the triangle with the following vertices(-2,-1), (6,-1) and (2,5) |
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Answer» By inspection, we see that the base of the triangle formed by the vertices (-2,-1) and (6,-1) is symmetric about the line x=2,so this altitude goes through the point (2,-1), and we know the x coordinate of the orthocentre is 2.To find the y coordinate, we use the fact that an altitude will be perpendicularto the line formed by any two vertices, and will pass through the 3rd vertex.Using (6,-1) and (2,5), the line isy + 1 = ((5+1)/(2-6)(x - 6) -> y = -3/2x + 8And, since perpendicular lines have negative reciprocal slope, the line for the altitude isy + 1 = 2/3(x + 2) -> y = 2/3x + 1/3The intersection point of x = 2 and y = 2/3x + 1/3 gives the orthocentrey = 2/3*2 + 1/3 = 5/3Ans: (2,5/3) |
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| 32. |
MAPLE 14 Using slopes, show that the vertices (-2,-1),(4,0), (3, 3) and (-3, 2) arethe vertices of a parallelogram.ration of the driven |
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| 33. |
प्णज्त्फ - l ]Y1y h ) £. कर 191 फ्श pace - flfi7 R ८०५ iy! Cand <A d)( e “tmhd Tyb\h—l\ हैं. लि a0 व g 5 2 (ARCova baxed al o |
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Answer» 1)When we toss three coins simultaneouslythen the possible of outcomes are: (HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT) respectively; where H is denoted for head and T is denoted for tail. |
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| 34. |
n2-3n -40 |
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Answer» n^2-3n-40=n^2-8n+5n-40=n(n-8)+5(n-8)=(n+5)(n-8) therefore,n= 8,-5 |
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| 35. |
Try These aForm as manyshapes as youcan bycombining twoor more setsquares. Drawthem on squaredpaper and notetheir lines ofsymmetry |
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| 36. |
Find the sum of the first 17 terms of the AP whose terms is given by tn=7-4n |
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| 37. |
0.12 Find the sum of first 24 terms of the list of numbers whose nt'term is given by a-3 +2n. |
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Answer» part 1 thanku very much |
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| 38. |
1. 35 added to a number gives 217. Find the number. |
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Answer» 2 17+35= 2 +(1+3)+(7+5)=2+(4)+(12=252 Let the unknown number be x X+35=217 X=217-35 X=182plz like 182 is the correct answer of the given question 182 is right answer of this question. please like my answer 182 is right answer I think I am right |
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| 39. |
4.One fifth of a number is 9. find the number |
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| 40. |
1.Five added to twice a number gives 13, Find the number, |
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Answer» thanks |
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| 41. |
3. By using slopes, show that the points A(0,4), B(2,10), C(3,13), are collinear. |
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| 42. |
Convert the following percentages todecimals\begin{array}{ll}{\text { (i) } 50 \%} & {\text { (ii) } 33 \frac{1}{3} \%}\end{array} |
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Answer» 50/100 = 1/2 33.33/100 = 1/3 66.67/100 = 2/3 |
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| 43. |
34. If 3 tan 20 = 3 and 0°< 0 S 90°. Find the value of 3V3 cos e + 2Sine - 6tan²0OR |
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Answer» Given,root(3)tan 2theta = 3 tan 2theta = 3/root(3) = root(3)tan 2theta = tan 60theta = 60/2 = 30 Therefore,3root(3) cos theta + 2sin theta - 6 tan^2 theta = 3root(3) cos 30 + 2sin 30 - 6 tan^2 30 = 3root(3)*root(3) + 2*1/root(2) - 6*(1/root(3))^2 = 3*3 + root(2) - 6/3 = 9 - 2 + root(2) = 7 + root(2) |
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| 44. |
0 30. 51 हो, तो 34-28 का मान ज्ञात कीजिए7 0\ |
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Answer» 3A-2B = -1 0 3 -6 15 11 8 10. 27 |
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| 45. |
JUWrite each of the following as percentages.19(11) 20(iii) |
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Answer» To find percentage multiply by 100hence3/5*100=3*20=60%and19/20*100=19*5=95% 95 % (I)=3/5*100=60%19/20*100=95% |
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| 46. |
14 Using slopes, show that the vertices (-2,-1)4,0),(3,3)and (-3,2the vertices of a parallelogram. |
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| 47. |
48. यदि 5 sin-0 + cos20 = 2, तो sin 0 का मूल्य ज्ञात कीजिए।34।। |
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Answer» 5sinA^2=2- cosA^2=2(1-cosA^2)=2(SinA^2); 5 sinA^2-2sinA^2=3 SinA=1; sinA=1/3 |
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| 48. |
The slopes of the lines, which make an angle 45° with the line 3x-y+5=0 are |
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| 49. |
i n ^ { 2 } 34 ^ { 0 } + \operatorname { sin } ^ { 2 } 56 ^ { \circ } + 2 \operatorname { tan } 18 ^ { \circ } \operatorname { tan } 72 ^ { \circ } - \operatorname { cot } ^ { 2 } 30 ^ { \circ } |
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Answer» go and ask this question to your teacher 👌 sin²34°+ sin²(90°- 56°)+ 2tan18°tan(90°− 72°)− cot²30°. =sin²34°+cos²34°+2tan18°cot18°-cot²30 = 1+2-(√3)² = 1+2-3 = 0 But? 😐😐gggggg |
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| 50. |
e a squared line paper to draw each to the following.All possible rectangles of 30 square units and find the factors of 30All possible rectangles of 30 square units and find the factors of 49.ich of the following number are divisible by 2? What kind ofe they.69085815Maths-5(b) 5820(e) 1204(c) 1896(f) 2503 |
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