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. |
Q17. Differentiate sin (cosx²) w.r.t. x |
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| 2. |
Differentiate (sin 2x)" + sin-1 V3x with respect to x. |
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Answer» wrong answer |
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| 3. |
Differentiate sin 2x w.r. to x by Ist Principle. |
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| 4. |
35. Differentiate sin x w.r.t x form first principles |
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| 5. |
Differentiate wrt x\sqrt{\frac{(1+\sin 2 x)}{(1-\sin 2 x)}} |
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Answer» thank you very much |
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| 6. |
integrate wrt x9xe^3x |
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| 7. |
36. The sum of the roots of 3x2 +5x 2 is |
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Answer» thanx |
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| 8. |
22. Find the domain of the following functions :yo li) log(4x – 3) Elielog(9 - x2) |
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| 9. |
Find the storm AP the sum is 36product is 1620 |
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| 10. |
Different sinx wrt x from 1 principle. |
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| 11. |
Different sinx wrt x from 1 principle.1. |
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| 12. |
.Q-19 Prove that : 3 cos 68º.cosec22°43º. tan 47º. tan 12°tan 600 tan 780-6-V3 |
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Answer» 3 cos 68. cosec 22 - 1/2 tan 43 . tan 47. tan 12. tan 60 . tan 78 ⇒3 sin 22 . cosec 22 - 1/2 cot 47 .tan 47 .cot 78 .√3 . tan 78 (since cos 68 = sin 22, tan 43 = cot 47,tan 12 = cot 78 and tan 60 =√3) ⇒3 * 1/cosec 22 * cosec 22 - 1/2 cot 47 * 1/cot 47 . cot 78 . 1/cot 78 *√3 ⇒3 -√3/2 ⇒(6 -√3)/ is the answer 2 |
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| 13. |
\operatorname { tan } \frac { \pi } { 12 } + \operatorname { tan } \frac { \pi } { 6 } + \operatorname { tan } \frac { \pi } { 12 } \operatorname { tan } \frac { \pi } { 6 } = 1 |
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| 14. |
65. The angle between the pair of tangents drawn tothe ellipse 3x2 +2y 5 from the point12(a) tan(b) tan (6/5)ortan"(12)(d) tan-1(12.5) |
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Answer» x^2/5/3+y^2/5/2=1y=mx+-√5m^2/3+5/2passing through (1,2)4m^2+24m-9=0M1+m2=-6and m1m2=-9/4tan^2A=(m1+m2)^2-4m1m2/(1+m1m2)^2=144/5tanA=12/√5A=tan^-1(12/√5) |
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| 15. |
If(cosx) -(cos y) find ay |
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| 16. |
9 x \sqrt { 5 } + \operatorname { log } _ { x } 5 x = \frac { 9 } { 4 } + \operatorname { log } _ { x } ^ { 2 } \sqrt { 5 } |
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| 17. |
( 0 \operatorname { log } 5 + \operatorname { log } ( 5 x + 1 ) = \operatorname { log } ( x + 5 ) + 1 |
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| 18. |
Differentiate the followi. (i) sin 2x |
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| 19. |
Differentiate the following from the firstprinciple:-11. sin-1(2x)2. tan2 |
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| 20. |
If x = acose and y = bsinθ then find ay whereθ is a parameter.d2dx2 |
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Answer» Use parametric differentiation to do it first calculate dy/d(theta) =bcos() now dx/d(theta)=-asin() now y`=-bcos/asin and then dy`/d(theta)= -b/a×(-sin^2-cos^2)/sin^2=bcosec^2/a now d^2y/dx^2=(bcosec^2/a)-asin=-b/a^2 cosec^3 |
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| 21. |
First term of an arithmetic progress is 1, last term is 11 and sum 36. Find thenumber ofterms and common difference. |
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Answer» please find the common difference |
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| 22. |
Doind the sum ofindicated number ofterms in each ofthe following A.Ps(i) 16, 11.6 , 23 termsTHIs(ii) 0.5,-1.0,-1.5, ...., 10 terms4' 2 |
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| 23. |
3.following are zeroes of the poly(oi p(vii)期 H-1, -73,-34. Find the zeno of the polynomial in each of the follapx)x-5(v) p(x)= 3xa+d,c#0,c,dare realnumbers.mi) p(1)Remainder Theorems consider two numbers 15 and 6. You know that vtient 2 and remainder 3. Do you remember h153(6 x 2)+3e that the remainder 3 is less than the di12 (6x 2)+0t is the remainder here? Here the remaindeof 12 or 12 is a multiple of 6.ow, the question is: can we divide one polynomi |
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| 24. |
Find ay, whendx1. x = at?, y = 2at |
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Answer» x = at² y = 2at dy/dx d(2at)/ d(at²) divide by dt d(2at)dt/dt d(at²) 2a/ 2at 1/t |
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| 25. |
llcos:sin-1:2,thenfindthevalueof4cos θ + 2 sin θ10.From the given figure, prove that θ +0-90Also prove that there are two other rightangled triangles. Find sina, cos β andtan .1520129E16 |
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| 26. |
Product of its zero respectively1/4,1 |
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Answer» Any quadratic equation having a and b as zeroes can be written as x² - ( a + b)x + ab = 0 x² - ( 1/4 + 1) x + 1/4 × 1 = 0 x² - 5x/4 + 1/4 = 0 4x² - 5x + 1 = 0 |
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| 27. |
ayFind-, if x = a( θ + sin θ ), y = a(1 + cos θ ).ax |
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| 28. |
it sina 0 - 3 sin a caso then plonethat tan olar 12 |
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Answer» 1+sin^2 theta=3 sintheta cos theta (we know that sin^2 theta + cos^2 theta =1) = ( sin^2 theta + cos^2 theta ) + sin ^2 theta = 3 sin theta cos theta = sin^2 theta + cos^2 theta + sin ^2 theta = 3 sin theta cos theta = cos^2 theta + 2 sin^2 theta = 3 sin theta cos theta On dividing by cos^2 theta, we get = 1 + 2 tan^2 theta = 3 tan theta Let tan theta = b 2b^2 - 3b + 1 = 0 = (2b-1)(b-1) = 0 b = 1 or 1/2 So, tan theta = 1 or 1/2. |
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| 29. |
\frac { 1 - \operatorname { cos } \theta + \operatorname { sin } \theta } { 1 + \operatorname { cos } \theta + \operatorname { sin } \theta } \text { interms of } \operatorname { tan } \frac { \theta } { 2 } |
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Answer» (1-cosx+sinx)/(1+cosx+sinx)=(2sin^2x/2+2sinx/2cosx/2)/(2cos^2x/2+2sinx/2cosx/2)=2sinx/2(sinx/2+cosx/2)/2cosx/2(sinx/2+cosx/2)=tanx/2 |
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| 30. |
wwwma1. Find ten rational numbers between andles |
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Answer» it's shortcut. hope u like this answer |
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| 31. |
log x5.dxx2 |
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| 32. |
y=\frac{x^{2}-1}{x^{2}+1} wrt x |
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| 33. |
(vii) Differentiate sin-1 (3x 4x3) with respectto х. |
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| 34. |
3 Differentiate sin (cos(x)) with respect to x.ition for |
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| 35. |
le 9.6. Differentiate sin 2x w.r. to x by Ist Principle. |
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| 36. |
2. What are the different ways in which infectiousdiseases can spread ? |
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| 37. |
Prove that if the number ofterms of an A. P. is odd then the middle term is the A M.between the first and last terms |
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Answer» i) Let the 1st term of the AP be 'a' and its common difference be 'd' ii) As the number of terms are odd, the nth term is (2n + 1) iii) Hence nth term (last tern) is: a + (2n + 1 - 1)d = a + 2nd iv) So Arithmetic mean between 1st & last term is:(a + a + 2nd)/2 = 2(a + nd)/2 = a + nd v) The middle term of (2n + 1)th term is: (2n + 1 + 1)/2 = (n + 1)Hence middle term is (n + 1)th term = a + (n + 1 - 1)d = a + nd Thus from steps (iv) & (v), Middle term = AM of 1st and last terms. |
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| 38. |
AT WHICH SUM WILL THE SIMPLEINTEREST AT THE RATE OF 3 3/4, PERANNUM BE 210 IN 2 1/3 |
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Answer» Sum = (100 * 210)/(15/4 * 7/3) = 21000 * 4/35 = 2400. |
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| 39. |
If the divisor, gootient and remainder are.10o, 28 and zero respectively then thedividend is |
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Answer» use dividend=quotient*divisor+remainder Answer is 2800 quotient*divisor=dividend28*100=2800 use dividend=quotient*divisor+remainder answer is 2800 quotient*divisor= divident28*100=2800 the answer is 2800 use this answer dividend is 2800 becoz it is due to dividend=divisor*quotient quotient*divisor=dividend 28*100=2800 100×28=2800divisor× quotient=divident use formula : Dividend= divisor*quotient + remainder then answer would be 100*28+0=2800 dividend=divisor*quotient+remainder= 100*28+0=2800+0=2800 Divident=quotient*divisor +remainder Formula is a=bq+r And answer is 2800 Let divident be xas we know, x÷100=28now, we transposed 100 in to R. H. S. then we get, x=28*100x=2800 2800 is the answer of this question twenty eight hundred 28 X 100 + 0= 2800 dividend quotient*divisor=divident 28*100=2800 quotient*fivisor=divident 28*100=2800 2800dividend=divisor*quotient+remainder dividend =answer is 2800 dividend=quotient*divisor+remainder answer is 2800 dividend =divisor × qutent + remainderdividend=100×28+0=2800 dividend=quotienti division+remainder answer is 2800 the correct Answer is 2800 is. question answer zero if the divisor qootient and remainder are 100,28 and zero respectively then the dividend is 100*28+0 = 2800 so the dividend is 2800 use divide =quentent divisor+reminder and answer is 2800 use dividend=quotient*divisor+ remainderanswer is 2800 2800 is the right answer quotient ×divisor =divident 28×100=2800 use dividend = quotient*divisor+remainderanswer is 2800 use dividend=question*divisor+remainder Answer is 2800 2800 is. the answer By using Dividend= Divisor* Quotient + Remainder , dividend = 100*28+0= 2800 2800 is the right answer........ the dividend is 2800 28*100=2800 is right answer Dividend_is_2800 Becoz_it_DueDividend=Divisor*Quotientcorrect_answer divident=divisor*quotient+remainder2800 this answer is 1000%sure 2800 2800 is the answer of the question 2800 is the answer i am correct 28×100=2800quotient*divison=dividend the answer is 2800 it's correct answer if the divisor qootien and remainder are 100,28 and zero respectively then the dividend is dividend is 3.5714285714 The divine r is 3.5714285714 dividend=divisor * quotient +remainderthen we get dividend= 100*28+0 =2800+0 =2800 quotient ❌ divisor= dividend 28❌100= 2800 28×100dividend =divisor*quotient+remainder 2800 as dividend= diviser* quotient+ reminder 2800 is the correct answer 28*100dividend =divisor *quotient +reminder use dividend=divisor×quotient+remainder therefore answer is 2800 my answer is 2800 bro am I right use divided = question*divisor+remainder answer is 2800 Dividend=divisor×quotient+remainder =100×28+0dividend=2800 ✓ I don't know about it right answer is 2800 2800 is the answer quotient *divisor +reminder 100*28=2800this is right answer. use dividend=quotient*dividor+remainder answer is 2800 dividend=divisor*quotient+remainder,so,100*28+0=2800 28*100=2800 use is Answers Dividend=Queficient *divisor+remainder 2800 if the divisor qootint and remainder are 100,28 and answer is 2800 and so on Dividend=quotient*divisor+remainder answer is 2800 2800 divided answer this question on my mind |
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| 40. |
कि के मे. न हे कर से केक,(x —2)(x + 3)(x - 3)(x + 4) =40 AN e की प्रकृति ज्ञातकीजिए। (2013, 16, 17)o e TFTR=TTTIT |
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| 41. |
if tan A=5/12 find sinA+cosA_tanA/cosA+cosecA_cotA |
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| 42. |
find the value of sina interms of cosa |
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| 43. |
Zero Marks: 0 In all other cases.unita digit of the lent positive integer which when divided by 6, 7, 8, 9 and 12 lesremainder 1 is |
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| 44. |
Find ay , when x-e2t . cos t , y = e2t, sin t .dx |
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| 45. |
A test인42. Studenta-H 6 ad-there studenagiver E ClasA2 S tudenn서o manY |
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Answer» Number of stidents that passed= 42 * 6/7 = 36 students Number of students that failed= 42 - 36 = 6. |
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| 46. |
)If y = log cos x2, then x =at ay has the value :dx |
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| 47. |
o) Find the average of the first 10 multiples of 15. |
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Answer» 15,30,45,60,75,90,105,130,135,150so average=(15+30+45+60+75+90+105+120+135+150)/10=825/10=82.5 |
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| 48. |
EXAMPLE 4.37Differentiate (sin)+ wrt |
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| 49. |
Find the sum of first 10 multiples of 6. |
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Answer» Herea= 6d= 6n= 10so Sn= n/2(2a+(n-1)d) = 5(12+54)66*5= 330 |
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| 50. |
EXAMPLE 3.15Differentiate sin'z by first principle. |
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