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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. |
Example 3 Construct a 3 x 2 matrix whose elements are given by ai-3j2 |
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Answer» thanks for helping not coming answers get fast |
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| 2. |
xample 19. The average income of 100 persons is 20. While the average income of all the 150persons is 25, calculate the average income of 50 persons. |
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a 3 x 2 matrix whose elements are given by a[Construct a 3 x 2matrii-3 |
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OME SOLVED PROBLEMS ON MIDPOINT THEOREMXAMPLE 1 If D, E and F are respectively themidpoints of the sides BC, CA and ABof an equilateral triangle ABC, provethat ADEF is also an equilateraltriangle. |
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Answer» Triangle ABC is an equilateral triangle.⇒ AB = BC = ACDE = 1/2 AB EF = 1/2 BC⇒ EF = 1/2 AB [Since AB = BC = AC] DF = 1/2 AC⇒ DF = 1/2 AB [Since AB = BC = AC] DE = EF = DF ∴ ΔDEF is an equilateral triangle. |
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| 5. |
19. Prove thet |
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Answer» (a-b)^2=a^2+b^2-2absimilarily(cosecA-cotA)^2=(cosecA)^2+(cotA)^2-2*cosecA*cotA=1 + (cotA)^2+(cotA)^2-2*1/sinA*cosA/sinA= 1+(cotA)^2+(cotA)^2-2cosA/(sinA)^2=[(sinA)^2 +(cosA)^2+(cosA)^2-2cosA] / (sinA)^2=[1+(cosA)^2-2cosA] /( sinA)^2=[1-cosA]^2 / (sinA)^2=[1-cosA]^2 / (1-cosA)(1+cosA)=(1-cosA)/(1+cosA) |
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| 6. |
19. Prove that:2(sin6e+cos69) - 3(sin40 + cos4e) 1- |
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| 7. |
(i) 963404[NCERT] (ii) 691 [NCERT](iii) 263(ví) 1095 3妆20(iv) 873Τπ 1451 V(v) 1485 3妆43561168 |
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| 8. |
(c) cos 18"(d) 2 sin 18°(NCERT(E))2. If x, y, are positive integers then value ofexpression (x + y) (+) (3 + .x) is(a) = 12 (b) > 811= (c) < 8013 (d) = 4xyz(NCERT(E))9 |
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Answer» Answer: Step-by-step explanation: As x,y are positive integers x +y is also a positive integer |
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| 9. |
angle between a and B>12. find laxbl.[NCERT ENCERT EXemu%Given la l-Wb1-2and a. b |
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Answer» thank you |
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| 10. |
NCERTa and b are the roots of x2-3x + p-0 and c, d are the roots x2-12xa, b, ,d form a G.P. Prove that (+p):(g-p) 17:15+ q0, whereNCERTาะ |
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| 11. |
-5945.If k+ SPk +1e SlL_uk_â[ k+3Pg. find k... &1 find r[C.F |
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Answer» nPk = n!/(n-k)! (k+5)P(k+1) = 11(k-1)/2 (k+3)Pk (k+5)!/(k+5-k-1)! = 11(k-1)/2 * (k+3)!/(k+3-k)! (k+5)(k+4)(k+3)!/4! = 11(k-1)/2 * (k+3)!/3! (k+5)(k+4)/4×3! = 11(k-1)/2 * 1/3! k*k + 9k + 20 = 22k - 22 k*k - 13k + 42 = 0 (k-7)(k-6) = 0 k = 6,7 |
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| 12. |
NCERT) |
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Answer» For one-one function:Let a, bε Domain and f(a) = f(b)=> a³= b³=> a = b so this is one one function |
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| 13. |
7. The cartesian equation of a line is \frac{x-5}{3}=\frac{y+4}{7}=\frac{2-6}{2} Write its vector form |
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| 14. |
The surface area of a sphere is 616 cm2. Find its radius.616 cm |
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Answer» Kya aap mere maths ka question paper of answer bta Doge plz |
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| 15. |
The area of a circle is 616 cm2. Find its circumference. |
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| 16. |
If the surface area of a sphere is 616 cm2, then find its radius |
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| 17. |
5. The area of a circle is 616 cm2. Find its circumference. |
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| 18. |
驰[9. Find the coordinates of the point which divides the line segment joining the points Af-5, 11)and B(4, -7) in the ratio 7:2. |
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| 19. |
13.If the surface area of a sphere is 616 cm2, find its volume. |
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| 20. |
10. The radius and height of a cylinder are in the ratio 7:2. If the volume of the cylinder is8316 cm3, find the total surface area of the cylinder. |
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| 21. |
14. The radius and height of a cylinder are inthe ratio 7:2. If the volume of thecylinder is 6062364 cm3. Find the totalsurface area of the cylinder? |
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| 22. |
12. Length of a rectangle is 2 less than 3 times its breadth. If the perimeter of the rectangle is 1028 m flength and breadth of the rectangle |
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Answer» suppose breadth= xthen length = 3x-2perimeter= 2(x+3x-2)= 10284x-2= 5144x= 516x= 516/4= 129mbreadth = 3*(129)-2= 385m |
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| 23. |
usaf a right circular cylinder isto that of the original cylindert.the height same fnd the ratio of the volumeanik has a capacity of 6160 cu cm F |
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Answer» Let the radius of or cylinder = rThen radius of reduced cylinder = r/2Height of both cylinder is same = h Volume of original cylinder V1 = pi*r*r*h Volume of reduced cylinder V2 = pi*r/2*r/2*h Then,V2/V1 = (pi*r/2*r/2*h)/(pi*r*r*h) V2:V1 = 1:4 |
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| 24. |
4. A thin bar of length f is placed horizontally along theprincipal axis of a concave mirror of focal length f insuch a way that centre of the bar coincides withcentre of curvature of the mirror. The length of imageof bar will be |
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| 25. |
Practice set 7.1d the volume of a cone if the radius of its base is 1.5 cm and its perpendicularheight is 5 cm. |
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Answer» V=πr^2h/3=22/7*1.5*1.5*5/3=11.78cm^3 |
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| 26. |
1. Find the volume of the cylinder whose radius is0.7cm and height 1.4 cm.2. Find the volume of the cylinder given the areaof its base is 154 cm2 and height 8 cm. |
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Answer» 1) Volume =πr^2 h =(22/7)(0.7)^2 (1.4) cm^3 =2.156 cm^3 2) Volume = base area × height = 154 cm^2 × 8 cm = 1232 cm^3 |
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| 27. |
ífsin 38-cos-60)of e iswhere 3θ and (θ-6") are both acute angles, then the valu(01) |
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Answer» sin3 = cos(-6)sin3 = sin(90- (-6))so comparing both sides we get3 = 90 - +64 = 96 = 96/4 = 24° |
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| 28. |
1. Find the volume of the cylinder whose radius is0.7cm and height1.4 cm.2. Find the volume of the cylinder given the areaof its base is 154 cm2 and height8 cm. |
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Answer» 1) V=πr^2h=π×0.7^2×1.4=2.155cm^3 2)V=A×h=154×8=1232cm^3 |
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| 29. |
8. The radius and height of a cylinder are in ratio 7.: 2. If the volume of the cylinder is 8316 cm3, find the totalsurface area of the cylinder. |
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| 30. |
s. In the adjoining figure, AD is a median ofAABC and DE|| BA. Show that BE is also amedian of AABC. |
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| 31. |
e depreciates at 10% per annum during thefirstyear and at 20% per annum during the second year. What will be its value after 2 years?29, A car is purchased for? 348000. Its valu |
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The radii of the internal and external surfaces of a hollow spherical shell are 3cm and 5cm respectively. If it is melted and recast into a solid cylinder of height 8/3 cm, find the diameter cylinder. |
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| 33. |
Curved surface area of a cylinder is 1980 cm2 and radius of its base is 15cm. Find theheight of the cylinder. (8.Let's learn. |
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Answer» 45467ghhhugftfffcyvubctcv |
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2. Find the coordinates of the point on y-axis which is nearest to the point (-2, 5) |
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Q2Find the point on the curve y2 = 4x which is nearest to the point (2,-8). |
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| 36. |
Independent Practice3.3 Ozfraction tiles or number lines.compare. Use >,<, or=. Check your answer using6. O}1. 50 OżMIT10.308 |
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Answer» 2/6=1/33/5<5/62/3 > 1/34/10 < 1/23/4 >1/32/3 =6/9 |
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Prove that the point (2,-2), (-2,1) and (5,2) are vertices of a right angled triangle. |
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| 38. |
ग 3 1+ [1+{524 -14 (1343-143)}] को हल कीजिए। |
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Answer» 1 is the correct answer of the given question |
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| 39. |
Parallel and Perpendicular LnieIndependent Practice1. Indicate whether the lines are parallel, perpendicular. or neither. Justify youranswer.a, y = 4x-1 and 12x3y+ 7b. x-7y-10 and 2x +14y 21c. 5x +6y 18 and 18x -15y 36d. x-1 and y-1 |
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Answer» For perpendicular lines product of slopes should be -1 and parallel lines have same slope so a) product of slopes is 4*(1/4) = 1 , neither.b) product of slopes is (-1/7)*(1/7) = -1/49 , neither.c) product of slopes is (6/5)*(-5/6) = -1 , perpendicular.d) these are also perpendicular. |
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| 40. |
DAILY PRACTICEFind the range of the function f(x) = x2+ 6x-7 when(a) x el-8o)-24(c |
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| 41. |
Find the term independent of x in (x -1/x2)3n |
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Answer» please hit on like button if u like the solution |
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| 42. |
35°, then angle ATQ wouldIn the given figure O is the centre of the arte with ATOMbe equal to :88.A) 55(T) 56"(C) 3ID) 54 |
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Answer» OQ=OT(radius of circle)So,angle OQT=angle OTQ=35angle ATQ=90-35=55Option A is correct |
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| 43. |
If 15tan” © + 4sec” 0 = 23, then find the value of(sec © + cosec 0)%—sin’ § |
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| 44. |
4 (132)-21. Simplify: V143 |
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Answer» 4√(132/143)^-24√(143/132)^24√(13/12)^2=2√13/124 is within over root2 is also within over rootor(13/12)^1/2 |
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13. Show thate median of a trierigle divides it into two triangles of equal arte |
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| 46. |
If 3r-cosec 0 andcot 0, find the value of3r-.t |
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| 47. |
If cosec 0 =10, find the values of all T-ratios of e. |
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Answer» Given,cosec theta = H/P = root(10) Using pythagoras theoram H^2 = P^2 + B^2 10 = 1^2 + B^2 B^2 = 10 - 1 = 9B = 3 Therefore, Sin theta = P/H = 1/root(10)Cos theta = B/H = 3/root(10)Tan theta = P/B = 1/3Sec theta = H/B = root(10)/3Cot theta = B/P = 3/1 = 3 |
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| 48. |
ramateăă¤tone |
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| 49. |
6) In AABC, prove that s2(ii) If 00 θ 90° , find the least value of4 cosec-0 + 9 sin,0. (iii) Ifx tan 30°+y co 60 0 and 2x - y tan 45 1, then find the value of x and y.22 |
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| 50. |
Q. 10. If tanecosec?e - sec? ofind the value ofcosec e + sec? 0sin AQ. 11. If cosec A=2, find the value of COLA+,1+cos A |
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Answer» tanx=1/V7; cosec^2x - sec^2x/ cosec^2x+ sec^2x=(V7/3)^2-(V7/4)^2/ (V7/3)^2+(V7/4)^2=7/9-7/16/ 7/9+7/16=7-7/16(9)/14/16(9)=14/1 |
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