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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. |
sin ax + bx20. 0 ax+ sin bx,a,b, a +b |
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Answer» lim x->0sinax+bx/ax+sinbxlim x->0a(sinax/ax)+b/a+b(sinbx/bx)a*1+b/a+b*1( as lim x->0 sinx/X=1)hencea+b/a+b=1 |
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| 2. |
(ax^2 + bx + c)(x + 1)-(a³x²-bx + c)(x-1) |
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| 3. |
10. Ganesh walks 8 km in 2 hours. Find the distance he can cover |
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Answer» Let to be Distance x then, we know that 1 hour=60 minutes then, 2 hour=2×60 =) 120 minutes Now, \frac{8}{120} \times 45x \frac{45}{15} = 3km thanks |
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| 4. |
Set of even numbers |
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Answer» set of even number-(2,4,6,8)2n |
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| 5. |
Define pre image of an element under a function f:A → B |
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| 6. |
In Fig. 8.42, X is any point within a squlABCD. On AX a square AXYZ is described.Prove that BX= DZ.Fig. 8.42 |
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| 7. |
8. ABCD is a rectangle. X and Y are points on side AD andBC respectively such that AY-BX. Show that BY = AXand LBAY = <ABX.Fig.9 |
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| 8. |
sin ax +bxlimx→0 ax + sin bxa, b, a + b0 |
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| 9. |
君faぁPM.L QRTママ职fa, PM2= QM.MIR |
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Answer» InΔQMP andΔQPR <QMP =<QPR (EACH = 90 degree) <Q =<Q (COMMON)⇒ ΔQMP SIMILARΔQPR ....(1) (AA similarity) Again inΔPMR andΔQPR, <PMR = <QPR (EACH = 90 degree) <R = < R (COMMON)⇒ ΔRMP similarΔQPR .......(2) (AA similarity)From (1) and (2) we get ΔQMP similarΔPMRTherefore, The corresponding sides are proportional QM/PM = PM/RM⇒ QM.RM= PM. PM (BY CROSS-MULTIPLICATION)⇒ PM² = QM.RM |
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| 10. |
(i)Set of even numbers |
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Answer» 2,4,6,8,10,12,14,16,18,20 2,4,6,8,10,12,14,16,18,20 |
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| 11. |
Define pre image of an element under a function fA -B |
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| 12. |
lim┬(x→1) ( ax^2+bx+c/ cx^2 + bx + a ), a+b+c ≠0 |
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Answer» put in the limitshencea*1+b*1+c/c*1+b*1+a=a+b+c/a+b+c=1 |
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| 13. |
917.IntheFigBX =AB, BY = -BC and AB = BC. Show thatBX = BY. |
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| 14. |
In ABCD, 1(AB) = 13 cm, 1 (DC)=9 cm,(AD) = 8 cm, find the area of ABCD. |
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| 15. |
InABCD, 1(AB)13 cm,I(DC) 9 cm, l(AD) 8 cm,find the area of ABCD. |
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Answer» Thank u bro |
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| 16. |
Find integ tadion2 |
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| 17. |
Set of even numbersSet of negative integ |
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Answer» set of even numbers = { 2n | n belongs to Z} set of negitive numbers = { z | z belongs to Z} |
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| 18. |
Let A [1, 2, 3. 4) and Z be the set of integDefine fA-z by fx)3+7. Show that f isa function from A to Z. Also find the range of f |
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Answer» f(x)=3x+7domain of f is Aso we have to take values from Aso f(1)=10f(2)=13f(3)=16f(4)=19so range is {10,13,16,19} |
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| 19. |
Find the value of K so that points A (5, 5), B (K. 1) and C (11, 7) are collinear.Differentiate : sin-×4x w.r.t.x |
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| 20. |
Blailasin? x dx |
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| 21. |
xpress each number as a product of its prime factors:(ii) 156 (iii) 3825 (iv) 5005(i) 140nirs nf integ |
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Answer» 140 = 2×2×5×7 156 = 2×2×3×13 3825 = 3×3×5×5×17 5005 = 5×7×11×13 |
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| 22. |
Differentiate1. sin 4x |
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| 23. |
L , 2 1700Evaluate: 4 (sin’30° + cos’60°) - 3 (5in’60° - cos*45°) + 7 tan’6( |
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Answer» hit like if you find it useful |
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| 24. |
1. Find area of trapezium with parallel sides 10 cm and12 cm and aktitude 8 cm.Find the area of a quadrilateral with diagonal 10 m andoffsets 8 m and 5 m.2. |
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Answer» 1)area of trapezium= =1/2 (sum of parallel side)(altitude) =1/2(10+12)(8)cm^2 =88 cm^2 |
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| 25. |
In ABCD, (AB)13 cm,(DC)-9 cm, /(AD)8 cm,find the area of ABCD |
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Answer» area of trapezium=1/2×sum of parallel sides ×height=1/2×(13+9)×8=84 |
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| 26. |
\begin{array}{l}{6(a x+b y)=3 a+2 b} \\ {6(b x-a y)=3 b-2 a}\end{array} |
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| 27. |
45. If 20% of A = 30% ofof Cthen A B Cis:,, at A: B : C=? |
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Answer» Given, A x 20/100 = B x 30/100 = 1/6 of C Then, A/5 = 3 B/10 = C/6A: B = 3:2 & B: C = 5:9 Therefore, the ratio of A:B:C = 3 x 5:2 x 9 = 15:10:18 |
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| 28. |
If a + B = 900B = 90° and a.: B = 2:1, then thee of sina : sin B is O (SSC FCI 2012(6) 2:101:1(d) 2:1a) 3:1 |
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| 29. |
\left. \begin{array} { l } { 5 x + 6 y \text { and } - 8 x - 3 y \quad B .15 x y \text { and } - 18 x y } \\ { - 6 p ^ { 2 } + 8 q ^ { 2 } - 2 p q - 5 \text { and } 2 p ^ { 2 } - 3 q ^ { 2 } - 2 p q + 3 } \\ { - a ^ { 2 } - 2 b ^ { 2 } - 3 a b - 9 \text { and } 2 a ^ { 2 } - 3 b ^ { 2 } - 6 } \end{array} \right. |
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| 30. |
\begin{array} { l } { \text { Evaluate the following integrals: } } \\ { \text { (i) } \int \sin ^ { 5 } x d x } \end{array} |
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| 31. |
\begin{array}{l}{\text { Evaluate the following integ}} \\ {\text { 1. (i) } \int \sin ^{2} x d x}\end{array} |
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Answer» Thank you |
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| 32. |
\begin{array} { l } { \text { Evaluate the following limits: } } \\ { \text { 1. } \lim _ { x \rightarrow 0 } \frac { \sin 4 x } { 6 x } } \end{array} |
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| 33. |
in a square ABCD AC is equal to 8 cm find the area of ABCD |
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Answer» 64 is a correct answer diagonal of square=diagonal ×diagonal÷2 64 is the right answer sorry 8×8=64÷2 =32 I posted the incomplete answer 64 cm² will be the area of square . 64 is the correct answer of this question 64 cm sq is the correct answer of this question |
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| 34. |
3. Construct a square ABCD in which(i) each side is 5 cm. |
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Answer» 1]draw base 5 cm2] draw angles from both end points of 90 degrees3] from first end point mark an arc of 5 cm same from other4] join the points of intersection of two arcs to the segments that is the 90 degrees lines . |
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| 35. |
In square ABCD, diagonal AC is 10\sqrt{2} cm. Findthe length of its side. |
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| 36. |
Pageだ5ba, s、s& more then(he numbers!は、ho the class9 |
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Answer» Let the no. of girls be 5xNo. of boys=7xAccording to the question5x+8=7x7x-5x=82x=8x=8/2x=4No. of girls=5*4=20no. of boys=7*4 or 5*4+8=28Hence the total strength of the class is 20+28=48 |
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| 37. |
Constructa AABCin which BC8.8 cm, B C 30. Measure AB and AC, What do you observe? |
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| 38. |
28. A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If thecurved surface area of the remainder isof the curved surface area of the wholecone, find the ratio ofline segment into which cone's altitude is divided by the plane. |
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Answer» Like if you find it useful |
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| 39. |
.Diameterofthehaseofacei1ndiscurvedsurface area. |
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| 40. |
3:4, find the value of 2+yORe vertices of AABC are A(6,-2), B(O, -6) and C(4, 8). Find the co-ordinates of mid-points of AB, BCand AC.either side of a 75 m high building and in line with base of building observe the angles1 60 Find the distance between the two men. (Use |
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Answer» Given thatA(6,-2), B(0,-6),C(4,8)mid point formal = x1 + x2 /2 ,y1+y2 /2mid point for A and B= x1 + x2/2 , y1 +y2 / 2= 6+0/2 ,-2-6 /2= (3,-4)mid point for the B and C= o+4 /2 , -6+8 /2= (2,1)mid point for A and C= 4+6 /2 , 8-2 /2= (5,3). |
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| 41. |
classmateDatePageHomecookequation andD) solve the followingIcheck your answert.2) 5x-2 212 |
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| 42. |
11. A 2-m tall man walks at a uniform speed of 5 km per hour away from a6-metre-high lamp post. Find the rate at which the length of his shadovwincreases |
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| 43. |
mple 7. Evaluate l sin' x cos" x dx |
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| 44. |
ce. How much did they paint togeteQ. 4. Prabha was3s given 7 of the basket of apples. What fraction of a25% 416 3twere left in the basket?SolutionSolution:xmple: Find t |
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Answer» 3/7 is givenso left in the basket are 1-3/7=4/7 |
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| 45. |
A circular car racetrack has innercircumference of 272 m and outercircumference of 304 m. Find the iatof the track. |
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Answer» thanks |
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| 46. |
Ogermple 9. What should be added to -15 |
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Answer» -3/10 + 4/15-9 + 8/30 1/30 -3/10 + 4/15-9 + 8/30 1/30 17/30 is right ans check all my dear friend 1 upon 5 is your amswer 17/30 is answer of your questions |
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| 47. |
Thove+hat area e antdlescr2oles esti bed on oyne et eb.diagewel.创agonal |
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| 48. |
5.Calculate the area ofthe shaded rector region as shown in Fig.π =60° |
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Answer» area of sector = theta/ 360° × pi ×r² so area of shaded region 60°/360 pi × 9² - 60°/360 pi × 4²60°/360 pi (9²-4²)(1/60)× 22/7 × 65 = 3.405 |
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| 49. |
s dataS.In centimetresIn degreesclAssmateDatePage6 0to |
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Answer» why dis the lord look puzzle? draw a rhombus abcd with ab 4cm and acute angle equals 60 degree |
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| 50. |
Q.Find the area ofthe shaded region where ABCD is a square. |
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Answer» area of shaded region is 42 cm.square. |
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