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.
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Date____/_-_1.22862 |
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Answer» ((a+b)(a+b)+(a-b)(a-b))/(a*a+b*b) = (a*a+2ab+b*b+a*a-2ab+b*b)/(a*a+b*b) = (2a*a+2b*b)/(a*a+b*b) = 2(a*a+b*b)/(a*a+b*b) = 2 If you find this answer helpful then like it. |
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| 2. |
18.If the co-ordinates of points A and B are (-2, -2) and (2, 4) respectively, find the co-ordinates of P3such that APAB where P lies on the line segment AB. |
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| 3. |
_1-8uis4 a919207 < ews-ey0) |
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Answer» Like if you find it useful |
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| 4. |
limn^infinite n^_1/n^+1. |
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| 5. |
or cot A-1 cot A2-+sec2 A 1+tanAăź:Sec |
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Answer» cotA - 1 / 2 - sec^2A = cotA / 1 + tanA LHS= cotA - 1 / 2 - sec^2A= (cosA/sinA - 1) / 1 + 1 - sec^2A= [(cosA - sinA) / sinA] /1 - tan^2A= [(cosA - sinA) / sinA] / [1 - sin^2A / cos^2A]= [(cosA - sinA) / sinA] / [(cos^2A - sin^2A) / cos^2A]= (cosA - sinA) / sinA X cos^2A / (cosA + sinA) (cosA - sinA)= cos^2A / sinA (cosA + sinA) RHS= cotA / 1 - tanA= (cosA / sinA) / (1 - sinA/cosA)= (cosA / sinA) / [(cosA - sinA) / cosA]= cosA / sinA X cosA / (cosA - sinA)= cos^2A / sinA (cosA + sinA) LHS = RHSHence proved. |
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| 6. |
9. If A + B = 90°, cot B =Then find TanA |
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Answer» B=90-Acot(90-x) =tanxtherefore, cot(90-A) =tanAtherefore, TanA=3/4 cotB=3/4=>tan(90-B)=3/4=>tanA=3/4. {A+B=90} |
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| 7. |
Show that $ \frac{2 \tan 105^{\circ}}{1-\tan ^{2} 105^{\circ}}=\frac{1}{\sqrt{3}} $ |
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| 8. |
(cosec A-sin A)(sec A-cos A) =tanA+ cot A |
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Answer» (cosecA-sinA)(secA-cosA)=(1/sinA-sinA)(1/cosA-cosA) =(1-sin²A/sinA)(1-cos²A/cosA)=cos²A/sinA *sin²A/cosA=sinAcosA LHS:1/(tanA+CotA)=1/(sinA/cosA +cosA/sinA)=1/(cos²A+Sin²A/cosAsinA)=1/1/(cosAsinA)=cosAsinA=RHS |
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| 9. |
(Cosec A-3sinA) (sec A - cos A ) = 1/tanA +cot A . Prove that |
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| 10. |
\operatorname { tan } A = \sqrt { 2 } - 1 , \text { show that } \frac { \operatorname { tan } A } { 1 + \operatorname { tan } ^ { 2 } A } = \frac { \sqrt { 2 } } { 4 } |
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Answer» now sinAcosA = √2/4 =>( sinA./cosA)*cos A*cosA = √2/4=> tanA.cos²A = √2/4 => tanA/(sec²A) = √2/4=> tanA/(1+tan²A) = √2/4 |
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| 11. |
if tan A = √2-1 then show that :tan A. √2---------- = ---------1+ tan²A. 4 |
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Answer» Given,tan A = root(2) - 1 Then,tan A/ 1 + tan^2 A = root(2) - 1/ 1 + [root(2) - 1]^2 = root(2) - 1/ (1 + 2 +1 - 2root(2)) = (root(2) - 1) / (4 - 2root(2))= root(2)[1 - 1/root(2)]/ 4[1 - 2root(2)/4] = root(2)[1 - 1/root(2)]/ 4[1 - 1/root(2)] = root(2)/4 Hence proved |
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| 12. |
Show that \frac{2 \tan 105^{\circ}}{1-\tan ^{2} 105^{\circ}}=\frac{1}{\sqrt{3}} |
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| 13. |
27. Prove that : हि हि _1-sin®: “{\secO+tan O cos® |
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Answer» √(sec theta- tan theta)²/sec²theta- tan²theta√(sectheta- tantheta)²sec theta- tan thetaas sec²theta-tan²theta=1 RHS = 1/cos theta- sin theeta/costhetasec theta- tan thetaLHS = RHS |
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| 14. |
Prove that;tanf +secfâ1 _1+singtn6-secf+1 cos |
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| 15. |
5 A number when added to its half gives 72. Find the number. |
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| 16. |
3. 5 added to me makes 17. What number am I? |
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Answer» Let the number be x As per given conditionx + 5 = 17x = 17 - 5 = 12 Therefore, Number is 12 17-5=12 is right answer |
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| 17. |
inginthenFrovethat!S A31.2CO-04 |
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Answer» it is right of course it must be. |
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| 18. |
Recall, t is defined as the ratio of the circumference (say c) of a(saya). That is, π- This seems to contradict the fact that πyou resolve this contradiction?3.iti)S14. Represent v9.3 on the number lineRationalise the denominators of the following:งา5 + v27 -2 |
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| 19. |
4 96The area of the front of the woodIn 1,6, find the area of each trapez20 yd2.65 yd40 yd44in14in |
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Answer» The unit is square yards what the answer yes can i see the answer |
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| 20. |
0) A field is in the shape of a trapez ium whose paraliel sides are 35 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. |
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| 21. |
sina ko seca or tana ko cot a m change kro |
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| 22. |
ab*-bc?-ab+c? |
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Answer» ab*b - bc*c - ab + c*c = b(ab - c*c) - 1(ab - c*c) = (b-1)(ab-c*c) |
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| 23. |
40. Which angle in the figure isof measure 80°?ko(11/n2 ko(k +10) |
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Answer» ans in short |
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| 24. |
given slant height is tan 2Show that semi-vertical angle of right circular cone of given surface area andmaximum volume is sin-(1)3. |
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| 25. |
18: Show that vertical angle of a right circular cone of minimum curved surfacethat circumscribes a given sphere is 2sin (2-1) |
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Answer» The formula of curved surface area of right circular cone is : pi*r*l where r is the radius of and l is the slant height. or S = pi*x*(root(a^2 + x^2) Solve for differentiation of S and equate it to zero since curved surface area is given as minimum. that is dS/dx = 0 |
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| 26. |
Show that the right circular cone of least curved surface and given volumehas an altitude equal to v2 time the radius of the base. |
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| 27. |
14 (рез) 111 рди |
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Answer» 111/11=( 100+1)/(10+1)= 10.09Please like the solution 👍 ✔️ |
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| 28. |
radius R isof the Voldmi24. Show that the right circular cone of least curved surface and given volume hasan altitude equal to /2 time the radius of the base.the srmi-ventical angle of the cone of the maximum volume and of |
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| 29. |
(b) Prove that the radius of the right-circular cylinder ofgreatest curved surface, which can be inscribed in a given cone,is half of that of the cone. |
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| 30. |
(vi)Number 5 added to three times the product of numbers m and n. |
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Answer» The number 5 added to three times the product of m and n is written as 3mn + 5. sorry i had to write something else |
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| 31. |
(vi) Five years hence, the age of Jacob will be three times that of his son. Five yeanago, Jacob's age was seven times that of his son. What are their present ages? |
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Answer» Like if you find it useful |
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| 32. |
An Aeroplane is flying at a distance 200 m high from Earth. While flying at this height Aeroplane makes angles of depres-sion at the banks on opposite side of the river are 45 and 60°. Find the width of the river |
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| 33. |
(e) The pictorial representation of data is called a(1) listOii) pictograph(ii) bill |
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Answer» the answer for the above question is pictograph pictograph is the answer pictograph is correct answer. pictograph is the correct answer the pictorial representation of data is called a pictograph the pictorial representation of data is called a pictographs (e) - (ii) pictograph 2)pictographs is the best answer |
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| 34. |
(vi) Five years hence, the age of Jacob will be three times that of his son. Five yago, Jacob's age was seven times that of his son. What are their present agesears |
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Answer» Let us assumeJacob’s current age is x his son’s current age is y Five years hence, Jacob’s age = x + 5 his son’s age = y + 5 As per question;x + 5 = 3(y + 5)x + 5 = 3y + 15x = 3y + 15 – 5 = 3y + 10 Five years ago, Jacob’s age = x – 5 his son’s age = y – 5 As per question;x – 5 = 7(y – 5)x – 5 = 7y – 35x = 7y – 35 + 5 = 7y – 30 Substituting the value of x from first equation in this equation, we get3y + 10 = 7y – 303y + 10 + 30 = 7y7y – 3y = 404y = 40y = 10 Substituting the value of y in first equation, we getx = 3y + 10x = 3 x 10 + 10 = 40 Hence, Jacob’s age = 40 years and son’s age = 10 years |
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| 35. |
A circular flower bed ithe area of the path? Dosurrounded by a path 5 mwide. The diameter of flower bed is 65 m. What is |
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Answer» A circular flower bed is surrounding by a path 5m wide the diameter of flower bed is 65mso, radius of circular flower bed = diameter/2 = 65/2 = 32.5m area of circular bed = π × (32.5)² = 1056.25π m² radius of circular bed + path = 32.5 + 5 = 37.5 m Area of circular bed + path = π × (37.5)² = 1406.25π m² Area of path = Area of (circular bed + path) + area of circular bed = 1406.25π - 1056.25π = 350 m² |
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| 36. |
35. An aeroplane is flying at a height of 300 m above the ground. Flyingat this height the angles of depression from the aeroplane of two pointson both banks of a river in opposite directions are 45 and 609spectively. Find the width of the river. [Use 3 -1.732 (cBSE 2017LLeSE 20005evel, theges fromro1nd te anles of elevation of the hottomn0 0 |
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| 37. |
(a+b)-(a - b)?- is equal toa'bab? |
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Answer» answer=4/(a-b)right ans like this 4/(a-b). is very very right answer 4/(a-b) is correct answer. |
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| 38. |
cpcsPKPAPLES BCDegm AB c t BAB Proe ths BC |
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| 39. |
|(C) रुधनापुङD)_y = 4bab7. समीकरण 2 -2 और 4करने वाले x और y के मान है।(A) x = a, y = b ।(B) x = 2a, y=-2b(C) x = a, y=-b(D) x = 2a, y = 2b |
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Answer» 2(x/a) + y/b = 2.....(1)x/a - y/b = 4...........(2) Add eq(1) and eq(2) 3x/a = 6x = 6a/3 = 2a Put value of x in eq(2)2a/a - y/b = 4y/b = 2 - 4y = - 2b (B) is correct option |
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| 40. |
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flowerbed is 66 m. What is the area of this path? (π= 3.14)13. |
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Answer» so much thank you sir it mean me alot |
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| 41. |
3. A circular flower bed is surrounded by a path 4 m wide. The diameter of the flowerbed is 66 m. What is the area of this path? (π = 3.14)Cor |
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| 42. |
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flowerbed is 66 m. What is the area of this path? (t 3.14)13.66m |
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| 43. |
Prove that the radius of the right circular cylinderof greatest curved surface area which can beinscribed in a given cone is half of that of thecone. |
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Answer» H = height of the cone , R = radius of the coneh = height of the cylinder, r = radius of the cylinder,S = lateral surface area of the cylinder S = 2π r hr = R (1 - h/H) . S = 2π (R (1 - h/H)) hdS/dh = 2π (R/H) (H - 2h)d²S/dh² = -4π (R/H)H - 2h = 0 .......... set dS/dh = 0 to find the stationary pointsh = H/2 .............. as d²S/dh² < 0 r = R (1 - (H/2)/H) ....... plug h=H/2 into r = R (1 - h/H)r = R/2 |
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| 44. |
15+14 |
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Answer» 15+14=29 . . |
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| 45. |
(20+2x)(14+2x)-(20)(14)=111 |
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| 46. |
ength of the shadow of a tree is Vi times the height of the tree at a particular pointof time. What is the angle of elevation of the sun rays with the ground at that time |
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| 47. |
F=eet 0=fâx () |
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Answer» From second equation,y=-3x=y x=-3 |
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| 48. |
UR BABKNOW?EET YOUPAGE NODATEReviston3)Solve :-Stara- 7XB3 |
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| 49. |
14) 2x += 6 and 2x-1--1,then 4x2-_?ĐĐŁ3y29y2 |
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Answer» (a+b)(a-b)=a^2-b^2(2x+1/3y)(2x-1/3y)=4x^2-1/9y^26(-1/2)=4x^2-1/9y^24x^2-1/9y^2=-3 |
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| 50. |
The circumference of a circular flower bed is 77 n find its area |
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Answer» Circumference = 2 × pi × r 77 = 2 × 22/7 × r r = 7×77/44 Area = pi × r × r = 22/7 × 7×77/44 × 7×77/44 = 41503/88 If you find this answer helpful then like it. |
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