100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	In a nuclear power plant, the power discharged to the environment: A. can be made zero by proper design B. must be less than the electrical power generated C. must be greater than the electrical power generated D. can be entirely recycled to produce an equal amount of electrical power E. is not any of the above
Answer» E. is not any of the above

Discussion

2.	High temperatures are required in thermonuclear fusion so that: A. some nuclei are moving fast enough to overcome the barrier to fusion B. there is a high probability some nuclei will strike each other head on C. the atoms are ionized D. thermal expansion gives the nuclei more room E. the uncertainty principle can be circumvented
Answer» A. some nuclei are moving fast enough to overcome the barrier to fusion

Discussion

3.	The binding energy per nucleon: A. increases for all fission events B. increases for some, but not all, fission events C. decreases for all fission events D. decreases for some, but not all, fission events E. remains the same for all fission events
Answer» A. increases for all fission events

Discussion

4.	The function of the control rods in a nuclear reactor is to: A. increase fission by slowing down the neutrons B. decrease the energy of the neutrons without absorbing them C. increase the ability of the neutrons to cause fission D. decrease fission by absorbing neutrons E. provide the critical mass for the fission reaction
Answer» D. decrease fission by absorbing neutrons

Discussion

5.	In the normal operation of a nuclear reactor: A. control rods are adjusted so the reactor is subcritical B. control rods are adjusted so the reactor is critical C. the moderating fluid is drained D. the moderating fluid is continually recycled E. none of the above
Answer» B. control rods are adjusted so the reactor is critical

Discussion

6.	For a controlled nuclear fusion reaction, one needs: A. high number density n and high temperature T B. high number density n and low temperature T C. low number density n and high temperature T D. low number density n and low temperature T E. high number density n and temperature T = 0K
Answer» A. high number density n and high temperature T

Discussion

7.	The energy supplied by a thermal neutron in a fission event is essentially its: A. excitation energy B. binding energy C. kinetic energy D. rest energy E. electric potential energy
Answer» B. binding energy

Discussion

8.	In DeltaABC, sinA, sinB and sinC are in AP, then ex-radii `r_(1),r_(2)` and `r_(3)` will be inA. APB. GPC. HPD. None of these
Answer» Correct Answer - A::B B : Both are biased `P(B) = (1)/(6)" "((1)/(.^(4)C_(2)))` N : Neither is biased `P(N)=(1)/(6)" "((1)/(.^(4)C_(2)))` O : One is biased `P(O) =(2)/(3)` `P(6,6//B)=(1)/(8),(1)/(8) " "P(B//6,6)=((1)/(8.8).(1)/(6))/((1)/(8.8).(1)/(6)+(1)/(6.6).(1)/(6)+(1)/(6.8).(4)/(6))=((1)/(8))/((1)/(8)+(2)/(3.3)+(2)/(3))=(9)/(9+16+48)=(9)/(73)` `P(6,6//N)=(1)/(6).(1)/(6)` `P(6,6//O)=(1)/(6).(1)/(8)`

Discussion

9.	The barrier to fission comes about because the fragments: A. attract each other via the strong nuclear force B. repel each other electrically C. produce magnetic fields D. have large masses E. attract electrons electrically
Answer» A. attract each other via the strong nuclear force

Discussion

10.	Most of the energy produced by the Sun is due to: A. nuclear fission B. nuclear fusion C. chemical reaction D. gravitational collapse E. induced emfs associated with the Sun’s magnetic field
Answer» B. nuclear fusion

Discussion

11.	In a subcritical nuclear reactor: A. the number of fission events per unit time decreases with time B. the number of fission events per unit time increases with time C. each fission event produces fewer neutrons than when the reactor is critical D. each fission event produces more neutrons than when the reactor is critical E. none of the above
Answer» A. the number of fission events per unit time decreases with time

Discussion

12.	In a neutron-induced fission process, delayed neutrons come from: A. the fission products B. the original nucleus just before it absorbs the neutron C. the original nucleus just after it absorbs the neutron D. the moderator material E. the control rods
Answer» A. the fission products

Discussion

13.	The purpose of a moderator in a nuclear reactor is to: A. provide neutrons for the fission process B. slow down fast neutrons to increase the probability of capture by uranium C. absorb dangerous gamma radiation D. shield the reactor operator from dangerous radiation E. none of the above
Answer» B. slow down fast neutrons to increase the probability of capture by uranium

Discussion

14.	Consider all possible fission events. Which of the following statements is true? A. Light initial fragments have more protons than neutrons and heavy initial fragments have fewer protons than neutrons B. Heavy initial fragments have more protons than neutrons and light initial fragments have fewer protons than neutrons C. All initial fragments have more protons than neutrons D. All initial fragments have about the same number of protons and neutronsE. All initial fragments have more neutrons than protons
Answer» E. All initial fragments have more neutrons than protons

Discussion

15.	Fission fragments usually decay by emitting: A. alpha particles B. electrons and neutrinos C. positrons and neutrinos D. only neutrons E. only electrons
Answer» B. electrons and neutrinos

Discussion

16.	`int(x^(2)-1)/((x^(2)+1)sqrt(x^(4)+1))` dx is equal to -A. `sec^(-1)((x^(2)+1)/(sqrt(2)x))+c`B. `(1)/(sqrt(2))sec^(-1)((x^(2+1))/(sqrt(2)x))+c`C. `(1)/(sqrt(2))sec^(-1)((x^(2)+1)/(sqrt(2)))+c`D. None of these
Answer» Correct Answer - B `int(x^(2)(1-(1)/(x^(2))))/(x^(2)(1+(1)/(x^(2)))xsqrt(x^(2)+(1)/(x^(2))))dx` `=int((1-(1)/(x^(2))))/((x+(1)/(x))sqrt((x+(1)/(x))^(2)-2))dx` `x+(1)/(x)=t` `(1-(1)/(x^(2)))dx=dt` `thereforeint(dt)/(tsqrt(t^(2)-2))` `int(dt)/(tsqrt(t^(2)-(sqrt(2))^(2)))` `rArr(1)/(sqrt(2))sec^(-1)""(t)/(sqrt(2))+c`

Discussion

17.	The value of x satisfying the equation `sinx+(1)/(sinx)=(7)/(2sqrt(3))` is given by -A. `10^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)`
Answer» Correct Answer - D Given `((sin^(2)x+1)/(sinx))=(7)/(2sqrt(3))` `rArr2sqrt(3)sin^(2)x-7sinx+2sqrt(3)=0` `sinx=(7pmsqrt(49-48))/(4sqrt(3))` `= (7pm1)/(4sqrt(3))` ltBrgt `=(8)/(4sqrt(3)),(6)/(4sqrt(3))` `(2)/(sqrt(3)),(sqrt(3))/(2)` ltBrgt clearly, x= `60^(@)`

Discussion

18.	Let `A=int_(0)^(1)(e^(x))/(x+1)` dx then answer the following questions in terms of A. Q. `int_(0)^(1)(x^(2)e^(x))/(x+1)dx` equalsA. `A-e`B. `e-2+A`C. `2+A`D. `2-e+A`
Answer» Correct Answer - D `int_(0)^(1)(x^(2)e^(x))/(x+1)dx=int_(0)^(1)((x^(2)-1+1)e^(x))/(x+1)dx` `=int_(0)^(1)(x-1)e^(x)dx+int_(0)^(1)(e^(x))/(x+1)dx` `((x-1)e^(x)dx+int_(0)^(1)(e^(x))/(x+1)dx` `((x-1)e^(x)-e^(x))_(0)^(1)+A=((.a)-(-1-1))+A` `=2-e+A`

Discussion

19.	If `[sinx]+[sqrt(2)cosx]=-3,x in[0,2pi]([.]-GIF), "then" x in`A. `((5pi)/(4),2pi)`B. `[(5pi)/(4),2pi]`C. `(pi,(5pi)/(4))`D. `[pi,(5pi)/(4)]`
Answer» Correct Answer - A::B::C `A^(n+2)-B^(n+2)=(A+B)(A^(n+1)-B^(n+1))-AB(A^(n)-B^(n))` `A^(n)-B^(n)=(A+B)(A^(n)-B^(n))-AB(A^(n)-B^(n))` `impliesI=A+B-AB" "[becauseA^(n)-B^(n)"is invertible"]` `implies (I-A)(I-B)=0` As A, `B ne l`, we get I-A and I-B are singluar matrices.

Discussion

20.	If `f(x) = { 3 + \|x-k\|, x leq k; a^2 -2 + sin(x-k)/(x-k), xgtk}` has minimum at x =k, then show that `\|a\| >2`.A. a`in`RB. `\|a\|lt2`C. `\|a\|gt2`D. `1lt\|a\|lt2`
Answer» Correct Answer - C `1+a^(2)-2gt3` `\|a\|gt2`

Discussion

21.	Q. Two students while solving a quadratic equation in x, one copied the constant term incorrectly and got the roots as 3 and 2. The other copied the constant term and coefficient of `x^2` as `-6` and 1 respectively. The correct roots are :A. 3, -2B. `-3, 2`C. `-6, -1 `D. `6, -1`
Answer» Correct Answer - D Let quandratic equation `x^(2)+bx-6=0` ltBrgt `overset("constant term = wrong")overset(downarrow)("wrong roots=3, 2")` but sum of roots = 3+2 It means, 3+2 = `(-b)/(1)` b = -5 correct `rArrx^(2)-5x-6=0` ltBrgt (x+1)(x-6)=0 x=-1, 6 roots

Discussion

22.	Let `A=int_(0)^(1)(e^(x))/(x+1)` dx then answer the following questions in terms of A. Q. `int_(0)^(1)((x)/(x+1))^(2)e^(x)dx` equalsA. `A-(e)/(2)`B. `(e)/(2)+1-A`C. `(e)/(2)+1-A`D. `(e)/(2)-A`
Answer» Correct Answer - C `int_(0)^(1)((x)/(x+1))^(2)e^(x)dx=int_(0)^(1)(((x+1)^(2)-(2x+1))/((x+1)^(2))).e^(x)dx` `I=int_(0)^(1)e^(x)dx-2int_(0)^(1)(e^(x))/(x+1)dx+int_(0)^(1)(1)/((x+1)^(2))e^(x)dx` `I=(e-1)-2A+(-(e)/(2)+1+A)=(e)/(2)-A`

Discussion

23.	`k=lim_(xtooo)[(sum_(k=1)^(1000)(x+k)^(m))/(x^(m)+10^(1000))]` (mgt101) is -A. 10B. `10^(2)`C. `10^(3)`D. `10^(4)`
Answer» Correct Answer - C `k=lim_(xtooo)[(x^(m)sum_(k=1)^(1000)(1+k/x)^(m))/(x^(m)(1+10^(1000)/(x^(m))))]` `=(1+1+....+1)/(1)=1000`

Discussion

24.	if roots of `ax^(2)+bx+c=0` where `epsiR^(+),` are two positive consecutive even integers, thenA. `\|b\|le6a`B. `\|b\|ge6a`C. `\|b\|=6a`D. None of these
Answer» Correct Answer - B Let roots are `alpha & alpha " " (age2)` `:. alpha + alpha+2=-(b)/(a) ge6, " " i.e.,\|b\|ge6a`

Discussion

25.	`lim_(ntoprop) (n)/(3){((3)/(n)+(9)/(n^(2)))^(2)+((3)/(n)+(18)/(n^(2)))^(2)+((3)/(n)+(27)/(n^(2)))^(2)+….+((3)/(n)+(9)/(n))^(2)}` is less than or equal toA. 62B. 63C. 14D. 21
Answer» Correct Answer - A::B::D `underset(nto prop)(lim)sum_(r-1)^(n)(n)/(3)((3)/(n)+(9r)/(n^(2)))^(2)` `underset(nto prop)(lim)sum_(r=1)^(n)(n)/(3).(9)/(n^(2))(1+(3r)/(n))^(2)` `underset(nto prop)(lim)sum_(r=1)^(n)(3)/(n)(1+(3r)/(n))^(2)` `3int_(0)^(1)(1+3x)^(2)dx` `(cancel(3)(1+3x)^(3))/(cancel(3).3))_(0)^(1)` `(64)/(4)-(1)/(3)=21`

Discussion

26.	A polygon has 25 sides, the lengths of which starting from the smallest side are in A.P. If the perimeter of the polygon is 2100 cm and the length of the largest side 20 times that of the smallest, then the length of the smallest side and the common difference of the A.P. is -A. 8 cm, 6 `(1)/(3)` cmB. 6 cm, 6 `(1)/(3)`C. 8 cm, 5 `(1)/(3)` cmD. None of these
Answer» Correct Answer - A Let a be the length of smallest side and d cm the common difference ltBrgt Now `S_(n)=(n)/(2)[2a+(n-1)d]` n = 25, `S_(25)= 2100` ltBrgt `2100=(25)/(2)[2a+24d]" "...(1)` a+12d = 84 `because` The largest side = `25^(th)` side ltBrgt =a+24d = 20a `" "...(2)` solve (1) and (2) a = 8 , d = `6(1)/(3)`

Discussion

27.	x+y+z=15 if 9, x, y, z, a are in A.P. while `(1)/(X)+(1)/(Y)+(1)/(Z)=(5)/(3)if9,X, Y, Z, a` are in H.P., then the value of a will be -A. 1B. 2C. 3D. 9
Answer» Correct Answer - A 9, x, y, z, a are in A.P. x+y+z=15 `rArr3[(9+a)/(2)]=15` ltBrgt a=1 9, X, Y, Z, a are in H.P. `(1)/(9),(1)/(X),(1)/(Y),(1)/(Z),(1)/(a)" are in "A.P.` `(1)/(X)+(1)/(Y)+(1)/(Z)=(5)/(3)rArr3([(1)/(9)+(1)/(a)])/(2)= (5)/(3)rArra=1`

Discussion

28.	If the roots of the cubic `x ^(3) +ax ^(2) + bx +c=0` are three consecutive positive integers, then the value of `(a ^(2))/(b +1) =`A. `(a^(2))/(b+a)=3`B. `(a^(2))/(b+a)=1`C. `(b+1)/(a^(2))=3`D. `(c^(2))/(a^(2))=1`
Answer» Correct Answer - B::C::D `x^(2)+ax^(2)+bx+c=0larroverset(p-1)underset(p+1)p` `a=-3p` `b=p(p-1)+p(p+1)+(p-1)(p+1)` `c=-p(p-1)(p+1)` `(a^(2))/(b+a)=(9p^(2))/(p^(2)-cancel(p)+p^(2)+cancel(p)+p^(2)-cancel(1)+cancel(1))` `=3`

Discussion

29.	If the area bounded by the curve y=f(x), x-axis and the ordinates x=1 and x=b is (b-1) `sin`(3b+4), then-A. f(x)=cos(3x+4)+3(x-1)sin(3x+4)B. f(x)=sin(3x+4)+3(x-1)cos(3x+4)C. f(x)=sin(3x+4)-3(x-1)cos(3x+4)D. None of these
Answer» Correct Answer - B We have `overset(b)underset(1)(int)"f"(x)dx=(b-1)sin(3b+4)` differentiate both side w.r.t. b `"f"(b).1-0=(b-1).3cos(3b+4)+sin(3b+4).1` `"f"(b)=3(b-1)cos(3b+4)+sin(3b+4)` `"f"(x)=3(x-1)cos(3x+4)+sin(3x+4)`

Discussion

30.	If `(1,a), (2,b), (c^2,3)` are vertices of a triangle then its centroid isA. Not be on x axisB. Not be on y axisC. lies at originD. None of these
Answer» Correct Answer - B Centroid `((1+2+c^(2))/(3),(a+b+3)/(3))` It x co-ordinate can not zero.

Discussion

31.	`sum_(n=1)^(prop)tan^(-1)((4n)/(n^(4)+5))=`A. `(pi)/(4)+tan^(-1)2`B. `(3pi)/(4)-tan^(-1)2`C. `tan^(-1)3`D. `(pi)/(4)+cot^(-1)2`
Answer» Correct Answer - B::C::D `sum_(n=1)^(infty)tan^(-1)((4n)/(n^(4)+5))=sum_(n=1)^(infty)tan^(-1)((4n)/(1+(n^(2)+2)^(2)-4n^(2)))` `=sum_(n=1)^(infty)tan^(-1)((4n)/(1+(n^(2)+2n+2)(n^(2)-2n+2)))` `sum_(n=1)^(infty)[tan^(-1)(n^(2)+2n+2)-tan^(-1)(n^(2)-2n+2)]` `=2tan^(-1)(infty)-tan^(-1)1-tan^(-1)2=(3pi)/(4)-tan^(-1)2` `=(pi)/(4)+tan^(-1)(1)/(2)`

Discussion

32.	The equations of the perpendicular bisector of the sides AB and perpendicular bisector of the sides AB and AC of a `triangle ABC` are x-y + 5 = 0 and x + 2y = 0 respectively, if the point Ais (1,-2), then the equation ofthe line BC isA. 14x+23y=40B. 14x-23y=40C. 23x+14y=40D. 23x-14y=40
Answer» Correct Answer - A A is (1, -2) B is image of a w.r.t. x-y+5=0 i.e. (-7, 6) C is image of A w.r.t. x + 2y =0 `" "i.e.((11)/(5),(2)/(5))` Now equation of BC is `y-6= ((2//5)-6)/((11//5)+7)(x+7)` `rArr14x+23y=40`

Discussion

33.	is The function `f(x)=(x^2-1)\|x^2-3x+2\|+cos(\|x\|)` is differentiable not differentiable at (a)-1 (b)0 (c)1 (d)2A. `-1`B. `0`C. `1`D. `2`
Answer» Correct Answer - A::B::C `cos(\|x\|)=cosx` `(x^(2)-1)\|x^(2)-3x+2\|=(x-1)\|x-1\|.(x+1)\|x-2\|` The only point where f is not differentiable is `x=2`

Discussion

34.	If f(x) = `e^(x)`, then `lim_(xto0) (f(x))^((1)/({f(x)}))` (where { } denotes the fractional part of x) is equal to -A. f (1)B. f (0)C. f (-`oo`)D. Does not exist
Answer» Correct Answer - D `{c^(x)}=[{:(e^(x)-1",",xgt0^(+)),(e^(x)",",xlt0^(-)):}` `underset(xto0^(-))(lim)(e^(x))^((1)/(e^(x)))=1` `underset(xto0^(+))(lim)(e^(x))^((1)/(e^(x)-1))=e` `rArr` limit does not exist at x = 0

Discussion

35.	Find the number of positive integers `n` such that 105 is a divisor of `n^(2)+n+1`
Answer» The digit in the unit place of `n(n+1)+1` is 1,3, or 7 `implies` 5 does not divides `n^(2)+n+1` `implies` 105 is not a divisor of `n(n+1)+1`,

Discussion

36.	If `f(x)={{:(,[x]+[-x],,,xne2 ),(," "lamda,,,x=2):}` then `f` is continuous at x=2, provided `lamda` is equal to -A. 1B. 0C. -1D. 2
Answer» Correct Answer - B `becausef(x)` is continuous at x = 2 `thereforef(2)=underset(xto2)(lim)f(x)` `rArr lamda=underset(xto2)(lim)[x]+[-x]` `lamda`=-1 `{{:(because" we know "[x]+[-x]=0: x in1),(" "-1:x notin1):} `

Discussion

37.	if `x=1^1/1+2^2/3+...1001^2/2001, y=1^1/3+2^2/5+...1001^2/2003` then `([x-y])/10` isA. 500B. 450C. 510D. 555
Answer» Correct Answer - A::B `x-y=1^(2)((1)/(1)-(1)/(3))+2^(2)((1)/(3)-(1)/(5))….1001^(2)((1)/(2001)-(1)/(2003))` `=(1^(2).2)/(1.3)+(2^(2)2)/(3.5)…+(2.1001^(2))/(2001.2003)` `(x-y)/(2)=((1^(2))/(1.3)+(2^(2))/(3.5)…(1001^(2))/(2001.2003))` `T_(r)=(r^(2))/((2r-1)(2r+1))` `=(1)/(4)((4r^(2)+1)/((2r-1))(-1)/((2r+1)))` `=(1)/(4)(1+(1)/((2r-1)(2r+1)))` `=(1)/(4)(1)+((1)/((2r-1)-(1)/((2r+1)))` `sum_(r=1)^(1001)T_(r)=(1)/(4)(1001)+(1)/(8)(1-(1)/(2003))` `(x-y)/(2)=(1001)/(2)+(2002)/(2003.8)` `x-y=(1001)/(2)+(1001)/(2.2003)` `[x-y]=500`

Discussion

38.	The function `f(x)=(2x-1)/(x-2)(xne2)` is such thatA. it is inverse of itselfB. decreases in `(-infty,2)` and `(2,infty)`C. it has a graph entirely above x-axisD. `underset(xto0^(+))(lim)f(e^(x^(-1))=2`
Answer» Correct Answer - A::B::D `f(x)=2+(3)/(x-2)` is bijective `f^(-1)(x)=2+(3)/(x-2)` `underset(xto0)(lim)f(e^((1)/(x)))=2`

Discussion

39.	From the given options complete the series: 4, 9, 16, 25, _______.1. 282. 303. 264. 36
Answer» Correct Answer - Option 4 : 36 Calculation: The series follows following pattern 2² = 4 3² = 9 4² = 16 5² = 25 So, 62 = 36 ∴ The required number is 36

39.

From the given options complete the series: 4, 9, 16, 25, _______.1. 282. 303. 264. 36

Answer» Correct Answer - Option 4 : 36

Calculation:

The series follows following pattern

2² = 4

3² = 9

4² = 16

5² = 25

So, 62 = 36

∴ The required number is 36

Discussion

40.	If u : v = 6 : 5, then what is (4u + 3v)/(v/5)?1. 232. 243. 344. 39
Answer» Correct Answer - Option 4 : 39 Given u : v = 6 : 5 Calculation u/v = 6/5 u = 6v/5 [4(6v/5) + 3v]/(v/5) [(24v/5) + 3v]/(v/5) [(24v + 15v)/5]/(v/5) (39v/5)/(v/5) = 39 ∴ 39

40.

If u : v = 6 : 5, then what is (4u + 3v)/(v/5)?1. 232. 243. 344. 39

Answer» Correct Answer - Option 4 : 39

Given

u : v = 6 : 5

Calculation

u/v = 6/5

u = 6v/5

[4(6v/5) + 3v]/(v/5)

[(24v/5) + 3v]/(v/5)

[(24v + 15v)/5]/(v/5)

(39v/5)/(v/5) = 39

∴ 39

Discussion

41.	A and B can do a work in 12 days, B and C can do it in 15 days and C and A can do it in 20 days. If A, B and C work together, then they will complete the same work in:1. 14 days2. 12 days3. 5 days4. 10 days
Answer» Correct Answer - Option 4 : 10 days Given: A and B can do a work in 12 days. B and C can do it in 15 days C and A can do it in 20 days Calculation: Work done by A and B in 1 day = 1/12 ….(i) Work done by B and C in 1 day = 1/15 ….(ii) Work done by C and A in 1 day = 1/20 ….(iii) Adding (i), (ii) and (iii) we get Work done by 2(A + B + C ) = 1/12 + 1/15 + 1/20 ⇒ (5 + 4 + 3)/60 = 12/60 = 1/5 ⇒ Work done by 1 × (A + B + C ) = (1/5) × 1/2 ⇒ Work done by A, B & C in 1 day = 1/10. ∴ A, B and C together can complete the work in 10 days

41.

A and B can do a work in 12 days, B and C can do it in 15 days and C and A can do it in 20 days. If A, B and C work together, then they will complete the same work in:1. 14 days2. 12 days3. 5 days4. 10 days

Answer» Correct Answer - Option 4 : 10 days

Given:

A and B can do a work in 12 days.

B and C can do it in 15 days

C and A can do it in 20 days

Calculation:

Work done by A and B in 1 day = 1/12 ….(i)

Work done by B and C in 1 day = 1/15 ….(ii)

Work done by C and A in 1 day = 1/20 ….(iii)

Adding (i), (ii) and (iii) we get

Work done by 2(A + B + C ) = 1/12 + 1/15 + 1/20

⇒ (5 + 4 + 3)/60 = 12/60 = 1/5

⇒ Work done by 1 × (A + B + C ) = (1/5) × 1/2

⇒ Work done by A, B & C in 1 day = 1/10.

∴ A, B and C together can complete the work in 10 days

Discussion

42.	1, 3, 7, 15, 31, What is the sixth number in the series of numbers given above?1. 532. 633. 554. 65
Answer» Correct Answer - Option 2 : 63 1 + 2¹ = 3 3 + 2² = 7 7 + 2³ = 15 15 + 2⁴ = 31 31 + 2⁵ = 31 + 32 = 63 ∴ Sixth number in the series is 63

42.

1, 3, 7, 15, 31, What is the sixth number in the series of numbers given above?1. 532. 633. 554. 65

Answer» Correct Answer - Option 2 : 63

1 + 2¹ = 3

3 + 2² = 7

7 + 2³ = 15

15 + 2⁴ = 31

31 + 2⁵ = 31 + 32 = 63

∴ Sixth number in the series is 63

Discussion

43.	What is the value of \( 2^{55}-2^{54}-2^{53} \)
Answer» 2⁵⁵ - 2⁵⁴ - 2⁵³ = 2⁵³(2² - 2 - 1) = 2⁵³(4 - 3) = 2⁵³

43.

What is the value of \( 2^{55}-2^{54}-2^{53} \)

Answer»

2⁵⁵ - 2⁵⁴ - 2⁵³

= 2⁵³(2² - 2 - 1)

= 2⁵³(4 - 3)

= 2⁵³

Discussion

44.	if `r_(1)` and `r_(2)` are distances of points on the ellipse `5x^(2)+5y^(2)+6xy-8=0` which are at maximum and minimum distance from the origin thenA. `r_(1)+r_(2)=3`B. `\|r_(1)-r_(2)\|=1`C. `\|r_(1)-2r_(2)\|=0`D. `r_(1)+2r_(2)=4`
Answer» Correct Answer - A::B::C::D Let any point `(rcostheta,rsintheta)` in xy plane We have to maximize & minimize r `5r^(2)cos^(2)theta+5r^(2)sin^(2)theta+6r^(2)sinthetacostheta-8=0` `5r^(2)+3r^(2)sin2theta-8=0` `r^(2)=(8)/(5+3sin2theta)` `r_(max)=2` `r_(min)=1` `r_(1)+r_(2)=3`

Discussion

45.	\( \left(x^{2}-7 x+1\right)(3 x-4) \)
Answer» ans is 4 ok bye (x² - 7x + 1) (3x - 4) = 3x³ - 21x² + 3x - 4x² + 28x -4 = 3x³ - 25x² + 31x - 4

45.

\( \left(x^{2}-7 x+1\right)(3 x-4) \)

Answer»

ans is 4 ok bye

(x² - 7x + 1) (3x - 4)

= 3x³ - 21x² + 3x - 4x² + 28x -4

= 3x³ - 25x² + 31x - 4

Discussion

46.	`int(dx)/(ax^(2)+bx+c)=k_(1)tan^(-1)(x+A)/(B)+C` ifA. `agt0,b^(2)-4acgt0`B. `alt0,b^(2)-4acgt0`C. `alt0,b^(2)-4aclt0`D. `alt0,b^(2)-4aclt0`
Answer» Correct Answer - C::D `int(dx)/(ax^(2)+bx+c)=(1)/(a)int(dx)/((x+(b)/(2a))^(2)-(b^(2)-4ac)/(4a^(2)))` `k_(1)=tan^(-1)(x+A)/(B)+C` if `(b^(2)-4ac)/(4a^(2)lt0`

Discussion

47.	Consider the circle `x^(2)+y^(2)=1` and thhe parabola `y=ax^(2)b(agt0)`. This circle and parabola intersect atA. four distinct points is `agtbgt1`B. no point if `blt-1`C. two distinct points if `-1ltblt1`D. one point if `b=1`
Answer» Correct Answer - A::B::C `x^(2)+(ax^(2)-b)^(2)=1` `impliesa^(2)x^(4)+(1-2ab)x^(2)+(b^(2)-1)=0` `impliesa^(2)t^(2)+(1-2ab)t+(b^(2)-1)=0` `impliesf(t)=0` `D=4a^(2)-4ab+1` `agtbgt1impliesDgt0`, `f(0)gt0` and `(2ab-1)/(2a^(2))gt0impliest_(1)gt0,t_(2)gt0` `implies` four distinct real values of x `blt-1impliesDgt0,f(0)gt0` and `(2ab-1)/(2a^(2))lt0` `impliest_(1)lt0,t_(2)lt0implies` no real value of x `-1ltblt1impliesf(0)lt0impliest_(1)gt0,t_(2)lt0` `implies` two distinct real values of x

Discussion

48.	If the sum of the 100 terms of an A.P. is `-1` and the sum of the even numbered terms lying in first 100 terms is 1, thenA. common difference of the sequence is `(-3)/(50)`B. first term of the sequence is `(-149)/(50)`C. `100^(th)` term of the sequence is`(74)/(25)`D. arithmetic mean of the odd numbered terms is `(1)/(25)`
Answer» Correct Answer - B::C `(100)/(2)(2a+99d)=-1,(50)/(2)(2(a+d)+49xx2d)=1` solving we get `d=((3)/(50)),a=(-149)/(50)` `t_(100)=(74)/(25),AM=-(1)/(25)`

Discussion

49.	if `sqrt(alpha_(1)-1)+2sqrt(alpha_(2)-4)+3sqrt(alpha_(3)-9)+4sqrt(alpha_(4)-16)=(alpha_(1)+alpha_(2)+alpha_(3)+alpha_(4))/(2)` where `alpha_(1),alpha_(2),alpha_(3),alpha_(4)` are all real. ThenA. `alpha_(1)+alpha_(2)=10`B. `alpha_(2)+alpha_(3)=26`C. `alpha_(4)-alpha_(3)=12`D. `alpha_(4)-alpha_(2)-alpha_(1)=22`
Answer» Correct Answer - A::B::D if `2sqrt(alpha_(1)-1)+4sqrt(alpha_(2)-4)+6sqrt(alpha_(3)-9)+8sqrt(alpha_(4)-16)` `=(alpha_(1)-1)+1+(alpha_(2)-4)+4+(alpha_(3)-9)+9+(alpha_(4)-16)+16` `(sqrt(alpha_(1)-1)-1)^(2)+(sqrt(alpha_(2)-4)-2)^(2)+(sqrt(alpha_(3)-9)-3)^(2)+(sqrt(alpha_(4)-16)-4)^(2)=0` `=sqrt(alpha_(1)-1)=1,sqrt(alpha_(2)-4)=2,sqrt(alpha_(3)-9)=3`. `sqrt(alpha_(4)-16)=4` `impliesalpha_(1)=2,alpha_(2)=8,alpha_(3)=18,alpha_(4)=32`

Discussion

50.	Prove thatCot18+Cot30=Cosec12
Answer» LHS = Cot18+Cot30 = \({cos18^o\over sin18^o}+ {\sqrt3}\) =\({cos18^o+\sqrt3sin18^o \over sin18^o}\) =\({2(sin30^ocos18^o+\cos30^osin18^o )\over sin18^o}\) =\({2sin(30^o+18^o)\over sin18^o}\) =\({2sin48^o\over sin18^o}\) =\({2sin48^osin12^o\over sin18^osin12^o}\) =\({cos36^o -cos60^o\over sin18^osin12^o}\) =\({{{\sqrt{5} +1\over 4}-{1\over2} }\over sin18^osin12^o}\) =\({{{\sqrt{5} -1\over 4}}\over sin18^osin12^o}\) = \({{sin18^o}\over sin18^osin12^o}\) = \({cosec12^o}\) =RHS proved

50.

Prove thatCot18+Cot30=Cosec12

Answer»

LHS = Cot18+Cot30

= \({cos18^o\over sin18^o}+ {\sqrt3}\)

=\({cos18^o+\sqrt3sin18^o \over sin18^o}\)

=\({2(sin30^ocos18^o+\cos30^osin18^o )\over sin18^o}\)

=\({2sin(30^o+18^o)\over sin18^o}\)

=\({2sin48^o\over sin18^o}\)

=\({2sin48^osin12^o\over sin18^osin12^o}\)

=\({cos36^o -cos60^o\over sin18^osin12^o}\)

=\({{{\sqrt{5} +1\over 4}-{1\over2} }\over sin18^osin12^o}\)

=\({{{\sqrt{5} -1\over 4}}\over sin18^osin12^o}\)

= \({{sin18^o}\over sin18^osin12^o}\)

= \({cosec12^o}\)

=RHS proved

Discussion

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Prove thatCot18+Cot30=Cosec12