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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. |
lim sin x |
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Answer» using L.H rule on differentiation Limx→π cosx/1 = cos(π) = -1 |
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| 2. |
For x >0, lim (sin x)/x+(1/x)sinx ) isx-W(A) 0(C) 1(B) 1(D) 2 |
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Answer» Option c is correct. how |
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| 3. |
limsin gnx → 0 (sin x)m (m<n) |
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Answer» kon se class ka question h ye yrr |
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| 4. |
\lim _{x \rightarrow 0} \frac{\sin 3 x-\sin x}{\sin x} |
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| 5. |
चल की. जिक्र सर1. 2x+3y=93x+4y=35 |
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Answer» 2x + 3y = 9 3x + 4y = 5 multiply 1st equation by 3 and multiply 2nd by 2 and subtract from 1st 6x + 9y - 6x - 8y = 27 - 10 y = 17 put y in 1st equation 2x + 3*17 = 9 2x = 9 - 51 2x = -42 x = -21 If you find this answer helpful then like it. |
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| 6. |
ERCISE 2.1Express each of the following products in the exponential form:333 |
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| 7. |
ANININ(iii) (x-7)(x-9) = 195 \ 7 |
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| 8. |
Solve:2x +y=353x+4y=65 |
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| 9. |
2x+y-35=0; 3x+4y-65=0 |
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| 10. |
if 2x+y-35 &3x+4y-65, find the value of x|y |
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| 11. |
\begin{array}{l}{\text { 2. Find the values using formulae: }} \\ {\text { (i) }(p+4)(p+7)} \\ {\text { (ii) }\left(m+\frac{2}{3}\right)\left(m-\frac{7}{3}\right)}\end{array} |
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Answer» p+4p+7_____p+11 (p+4×p+7=p+11 No it's wrong wrong lakshmi |
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| 12. |
\lim _{x \rightarrow 3} \frac{x+1-\sqrt{x+13}}{x-3} |
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Answer» by taking x=3 we will get 0/0 form so using L'hospital rulelim x->3 (1-1/(2root(x+13)))/1=1-1/2root(3+13)=1-1/2*4=1-1/8=7/8 |
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| 13. |
29.401- cos m.x13. lim |
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Answer» please like my answer if you find it useful |
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| 14. |
af x 1S63. यदि sin डी, cosec ‘(4—) = z—n,mxwmt : |
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Answer» Please like the solution 👍 ✔️ |
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| 15. |
If $ f(x)=8 x^{3} $ and $ g(x)=x^{\frac{1}{3}}, $ find $ f 0 $ |
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| 16. |
The value oflimsin x-cos1sa) 2 |
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| 17. |
(1)16.Lec f(x) be a polynomial of degree '2' and suppose that f(x+1)- S(x)=6x+8 andThen find the value of f(11). |
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Answer» 356 is the correct answer of the given question |
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| 18. |
1. Find the area of a triangle whose sides are 12 cm, 16 cm and 20 cm. |
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Answer» Area of a triangle of sides a , b & c is : √( s(s-a)(s-b)(s-c) ) where s=(a+b+c)/2 ln given trangle , s=(20+12+16)/2=48/2=24 Thus area of given triangle = √(24(24-20)(24-12)(24-16)) =√(24×4×12×8) =√9216 =96 Thus area of triangle of sides 20cm ,12cm and 16cm is 96cm^2 Like my answer if you find it useful! |
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| 19. |
(x - 9)*(x - 7)=195 |
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Answer» (x - 7)(x - 9) = 195 (x^2 - 9x - 7x + 63) = 195 x^2 - 16x - 132 = 0 x^2 - 22x + 6x - 132 = 0 x(x - 22) + 6(x - 22) = 0 (x + 6)(x - 22) = 0 x = - 6, 22 the value of X is -6,22 is the correct answer of the given question. |
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| 20. |
(x-7) (x-9) 195 |
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| 21. |
i) (x~7) (x-9) = 195 |
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| 22. |
1) Find the value of the polynomial 2x 2x+7 using given values for x(ii) x =-11)x =(iii) x0 |
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Answer» thnk u |
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| 23. |
(iii) (x-7) (x-9)-195 |
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| 24. |
2xQ.13 Limx--o ( sin(4x) 1s equal to |
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| 25. |
(P) 1482(v) 165000 (vi)निम्न संख्याओं में से किन संख्याओं का वर्ग सम संसंख्या है?(i) 14 ।(ii) 277(iv) 205(v) 608(vii)1079(viii) 4010 |
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Answer» looking black some is not visible |
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| 26. |
What should be added to 15 to get (-15)? |
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Answer» -30 should be added to 1515+ -30 = -15 |
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| 27. |
110. △ABC is a right triangle right angled at A such that AB=AC andbisector of <C intersects the side AB at D. Prove that AC + AD = BC. |
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| 28. |
If x =and ythen which of the following is true, if A> 11(A) f is greater than y(B) x is equal to y(D) of themA)/X 1s greater than v(C) r is less than y |
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Answer» A > 1 x = A/(A - 1) y = 1/(A-1) So, x > y Reason : at 1/(A-1) both are equal and A > 1 when multiplied them x become greater . A) option is correct |
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| 29. |
9 29, ABCDE is apensagon A lime heooghD ssbuced at F Shiow8 paraillel to ACthat) ar (ACB) LACFar AEDF) ar (ABCDEFig. 9.25 |
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Answer» tnx bro u r genies |
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| 30. |
The figure shows a cylinder, from which 2hemispheres are removed from both ends. Theheight of the cylinder is 20 cm and base diameteris 12 em.20 cm12 cma. Find the radius of the hemisphere removed.b. Find the surface area of the solid obtained.c. Find the volume of the solid obtained.619 |
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| 31. |
Find the equation of the right bisector of the line segment joining the points (3,4) and (-1, 2). |
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| 32. |
10 AABC is a right triangle right angled at A such that AB AC andbisector of C intersects the side AB at D. Prove that AC+ AD- BC. |
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| 33. |
(ii) 196 ŕ¤ŕ¤° 38220 |
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Answer» 38220 = 196 x 195 + 0 H.C.F = 196 hit like if you find it useful |
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| 34. |
(i) 196 31R 38220 |
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| 35. |
38220 =196 X 195+ 0 |
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Answer» RHS 196*195 + 0 Applying Bodmas we will first multiply then add196*195 = 38220then 38220+0 =38220so 196*198 + 0 = 38220 |
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| 36. |
eun |
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Answer» Rational numbers between 3/5 and 7/8 => 3/5 = 24/40 and 7/8 = 35/40 so, 3 rational numbers are = 28/40 = 7/10= 30/40 = 3/4= 32/40 = 4/5 |
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| 37. |
2(2+2)+2=2 |
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| 38. |
1 Prave that1 6 |
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Answer» Cos20°cos40°cos60°cos80°=(1/2)(2cos20°cos40°)(1/2)cos80° [∵,cos60°=1/2]=(1/4)[cos(20°+40°)+cos(20°-40°)]cos80°=(1/4)(cos60°+cos20°)cos80°=(1/4)(cos60°cos80°+cos20°cos80°)=(1/4)(1/2)cos80°+(1/4)cos20°cos80°=(1/8)cos80°+(1/4)(1/2)(2cos20°cos80°)=(1/8)cos80°+(1/8)[cos(20°+80°)+cos(20°-80°)]=(1/8)cos80°+(1/8)(cos100°+cos60°)=(1/8)cos80°+(1/8)cos100°+(1/8)cos60°=(1/8)(cos80°+cos100°)+(1/8)×(1/2)=(1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)=(1/8)(2cos90°cos10°)+(1/16)=0+(1/16) [cos90°=0]=1/16 (proved) |
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| 39. |
154Mathematics - VII14. In the adjoining figure, find the measure of 4BCDC.1150 |
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| 40. |
15. An employee works in a company on a contract of 30 days on the condition that hewill receive 200 for each day he works and he will be fined 20 for each day he |
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| 41. |
Mathematics for Class VI20.2815. Calculate the area of the pentagon ABCDE, where AB= AE and with dimensions as shown inFig. 36.m20 cm12 cmC10mD |
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Answer» The area of the pentagon is 160 the area of pentagon is 60 the area of pentagon is 160 |
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| 42. |
Q 21: Find the value of p for which the points (-5, 1). (1.p) and (4,-2) are collinear22:)ABCDE is polygon whose vertices are A (-1,0), B (4,0). C (4,4). D (0,7) and E (-6, 2) Findthe area of the polygon |
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Answer» please like the solution 👍 ✔️👍 |
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| 43. |
prove that the right bisector of a chord of a circle bisects the corresponding arc |
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| 44. |
fined value of afy |
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| 45. |
Find the area of the minor segment of a circle of radius 42cm, if length of the correspondingarc is 44cm. |
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| 46. |
(ii) 196 38220 |
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| 47. |
(ii) 196 और 38220 |
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Answer» 196*38220= 7491120 is the correct answer 195 is the right answer 144 5.514345 is a right answer 195 is the correct answer 195 is the right answer 195 right answer for this question |
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| 48. |
(ii) 196 and 38220 |
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| 49. |
(ii)196 and 38220 |
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Answer» 38220 = 196 x 195 + 0 H.C.F = 196 |
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| 50. |
Find five rational numbers between and55 |
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