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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. |
(3x+5)(3x+5) |
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Answer» 3x(3x+5) +5(3x+5) 9x²+15x +15x+259x²+30x+25 |
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| 2. |
(3x - 5) (-3x + 5) |
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| 3. |
AD is the median+he Jergtho/DC.△ABCnd BC"6cm , find |
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| 4. |
(ii) (3x 5) (3x 5) |
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Answer» (3x-5)(3x+5)=(3x)²-(5)²=9x²-25 Formulaa²-b²=(a-b) (a+b)Then(3x-5)(3x+5)=(3x)²-(5)²=9x²-25 |
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| 5. |
11. A chord of length 14 em is at a distance of 6 cm from the centre of a circle. The length ofanother chord at a distance of 2 cm from the centre, is:(a) 12 cm(b) 14 cm(c) 16 cm(d) 18 cm. |
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| 6. |
Prove that(1-tanx)²+(1-cotx)²=(secx-cosecx)² |
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| 7. |
[(5x+ 3) (8x + 6)1+ (-9x + 10) (12x2+ 5x -5)] |
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Answer» If you find this solution helpful, Please like it. |
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| 8. |
Example 13. Solve: 3x+-53x-9x5 |
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Answer» 3x + sqrt(9x^2 - 5) = 5(3x - sqrt(9x^2 - 5)) 15x - 3x = 6 sqrt(9x^2 - 5) - sqrt(9x^2 - 5) 12x = 6 sqrt(9x^2 - 5) 2x = sqrt(9x^2 - 5) Squaring both sides, we get4x^2 = 9x^2 - 55x^2 - 5 = 05(x^2 - 1) = 0(x + 1)(x - 1) = 0x = 1, - 1 |
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| 9. |
Solve the equation :1 + secx = cot? |
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Answer» the right answer is x=π/3 |
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| 10. |
3.The coefficient of x in 5x+3is |
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Answer» coefficient of (x×x is 5 the coefficient of x^2 in 5x^2+3 is 5 |
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| 11. |
1913(**)-46*+) = 1117xtil5x-31(7x+1)115x - 3 |
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| 12. |
33+33 |
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Answer» if we add 33 and 33 answer will be 66 if we add 33 and 33 is equal to 66 |
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| 13. |
A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm morethan its breadth. Find the length and the breadth of the rectangle so formed. |
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Answer» thank you: |
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| 14. |
cosec x -1y cosec x+1 secx+ tan x |
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| 15. |
. A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm morethan its breadth. Find the length and the breadth of the rectangle so formed. |
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| 16. |
24. A wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm morethan its breadth. Find the length and the breadth of the rectangle so formed. |
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Answer» thanks |
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| 17. |
X+XtiDifferentiate f(x) = - with respect to x. |
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| 18. |
Find the area of a rectangle ABCD giw(a) AB-12 cm, AD-5 cm |
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Answer» Area of rectangle=AB*AD=12cm*5cm=60 cm² |
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| 19. |
Q2. If AB 12 cm, BC16 cm and AB is perpendicular to BC, then what is the radius of the circlepassing through the points A,B and C? |
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| 20. |
(b) Differentiatend find the value of the2derivative at/f= 0. |
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| 21. |
Page NoDate: 1 |
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| 22. |
\begin{array}{l}{\frac{5}{x+y}-\frac{2}{x-y}=-1} \\ {\frac{15}{x+y}+\frac{7}{x-y}=10}\end{array}where x is not equal to 0and y notequal to 0 |
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Answer» 15/(x+y)-6/(x-y)-15/(x+y)-7/(x-y)=-3-10-13/(x-y)=-131/(x-y)=1x-y=15/(x+y)-2=-15/(x+y)=-1+2=1x+y=5x+y+x-y=1+5=62x=6x=3y=2 |
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| 23. |
secx-1secx+1 |
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| 24. |
Ex飞Reduce the lines 3 x _ 4 y + 4-0and 2 x + 4 y _ 5 = 0 to the normal form and hence(2find which line is nearer to the origin. |
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| 25. |
secx-dr का मान ज्ञात कीजिए।sec x + tan x |
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| 26. |
7. Witeall the integers between the given pairs (write them in the increasing order)(a) 0and-7(b) 4 and 4(c) -8 and 15 (d)30 and-23 |
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| 27. |
14 t the given figure of square ABCD, P. R and S are the midpoints of ABBC, CD and DA respectively. Find thearea of unruled portion.9トーーー14 cm |
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| 28. |
xx xBX X X11IUI 9 '18. A metallic rod X X X X XCD rests on aÑ…thick metallicwire PQRS withx x x xarms PQ andRS parallel tox x x x x xeach other, ata distance 1 = 40 cm, as shown in figure. A uniformmagnetic field B = 0.1 T acts perpendicular to theplane of this paper, pointing inwards (i.e., away fromthe reader). The rod is now made to slide towardsright with a constant velocity v = 5.0 ms.(i) How much emf is induced between the twoends of the rod CD?(ii) What is the direction in which the inducedcurrent flows?(ISC 2013)Hint : V=U BI5 x 0.1 x 40 x 10-2 V = 0.2 VAns. (i) 0.2 V, (ii) direction DCQRD.1DU.ULCDast5 |
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| 29. |
19/ Prove that :--=-secx-tan x cosr cosx secx+tan x |
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| 30. |
(b)RectangleIn the rectangle, x: y = 2:7.Find x and y. |
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Answer» 20 and 70 degree is right ans |
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| 31. |
ABCD is a rectangle. Find the values of x and y. |
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Answer» X-y =14.....(i)x+y=30.......(ii)so, x=14+y....(iii)in equ ii,14+y+y=3014+2y=302y=16y=8so x =14+y=14+8=22 |
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| 32. |
Find the area of a shaded Aregion in the figure, whereABCD is a square of side14 cm. |
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| 33. |
20. In the given figure, ABCD is a square of side 14 cm.Semicircles are drawn with each side of square as diameter.Find the area of the shaded region. |
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| 34. |
7. In the given figure, ABCD is a llgm in which Ddiagonals AC and BD intersect at O. Ifar(llgm ABCD) is 52 cm2, then the ar(AOAB)?(a) 26 cm(b) 18.5 cm2 |
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| 35. |
Ifar(llgm ABCD) 96 cm2 and AB 12 cm. Find(i) the area of ADEC9a.(ii) the length of the altitude of the ADEC. |
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| 36. |
StarlinDate:-1Page:fndmethodl of completiag the squaxethe xocks of the equatton sa2 |
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Answer» 5x^2 -6x -2 = 0 x^2 -(6/5)x -(2/5) = 0 (x)^2 - [2.(3/5).x] + (3/5)^2 - (3/5)^2 -(2/5) = 0 [x-(3/5)]^2 = (2/5)+(9/25) [x-(3/5)]^2 = (19/25) x-(3/5) = √19/5 x = (3/5) ± √19/5 Roots of the equation are x = [3+(√19)]/5 & x = [3-(√19)]/5_______________ |
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| 37. |
coso," show that (m2 + m2) cos2 β,,sinv)is any point on the line joining the point A( |
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| 38. |
Differentiate1.Ď+1 |
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Answer» d(x²+1) d(x+1)(x+1)× --------- - ---------- × (x²+1) dx dx--------------------------------------------------- (x+1)²= (x+1)2x - (x²+1) ----------------- (x+1)² 2x²+ 2x-x²-1= ------------------- (x+1)² = x²+2x-1 ------------- (x+1)² |
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| 39. |
7. Find theequation of the line joining the point (3, 5) to the point of intersection of the line41+y-1-0and 7 x-3y-35 = 0. |
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| 40. |
If p(x, y) is any point on the line joining thepoint A(a,0) and B(0,b) then show that-1 |
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| 41. |
The mid point of the line joining the points (4,0) and(-6, 0) is : |
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Answer» (4,0) and (-6,0)andmidpoint will be 4-6/2,0+0/2=-2/2,0=-1,0 |
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| 42. |
Differentiate xsn (sin xw.r.t. xcos x |
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| 43. |
sin{cos(x^2)} differentiate w.r.t x |
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| 44. |
1 - X, X-VI-laat r =XXV |
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| 45. |
x+y=20x-y=10 abcd is a rectangle find tha x and y |
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Answer» Adding both will get2x= 30x= 15cmthen y= 20-15= 5cm |
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| 46. |
Differentiate(secx 1)/(secx+1) |
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| 47. |
Find the area of the shaded region iwhere ABCD is a square of side 14 cArea of square ABCD14 x 14 cm 19 |
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| 48. |
. Find the coordinates of the points of trisection of theline joining the points (2, 3) and (6, 5) |
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| 49. |
. AB 11 CD in trapezium ABCD. If <A=y +60.ZB-x+6rLC 3x -40° and D 3y-80°, then find all the angesof ABCD |
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| 50. |
ABCD is a parallelogram. X and Y are the midpoints of BC and CD. Find the ratio of ar (AAXY)to ar (llgm ABCD).22. |
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Answer» Given that ABCD ia a parallelogram. X and Y are the mid points of BC and CD Construction: Join BD Since X and Y are the mid points of sides BC and CD respectively, therefore in triangle BCD, XY//BD and XY=1/2 BD implies area of triangle CYX= 1/4 area of triangle DBC { In triangle BCD, if X is the mid point of BC and Y is the mid pt of CD then area triangle CYX=1/4 area triangle DBC} IMPLIES AREA TRIANGLE CYX= 1/8 parallelogram ABCD [Area parallelogram is twice the area of triangle made by the diagonal] Since parallelogram ABCD and triangle ABX are between same // lines AB and BC and BX=1/2BC Therefore, area triangle ABX= 1/4 area //gm ABCD Similarly, area triangle AYD= 1/4 area parallelogram ABCD Now, area triangle AXY= area/parallelogram ABCD- {ar triangleABX + ar AYD + ar CYX} = parallelogramABCD - {1/4 + 1/4 + 1/8} area of parallelogram =area of parallelogram- 5/8 area parallelogram ABCD =3/8 area /parallelogram ABCD. |
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