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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 7501. |
Prove that |{:(x^2,y^2,z^2),((x+1)^2,(y+1)^2,(z+1)^2),((x-1)^2,(y-1)^2,(z-1)^2):}|=-4(x-y)(y-z)(z-x) |
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Answer» |
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| 7502. |
Provethat the sum of eccentric angles of four concylic points on the ellipse(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is 2npi, wheren in Z |
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Answer» Solution :Let the CIRCLE `x^(2)+y^(2)+2gx+2fy+e-0` CUT the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` in four ponts , P,Q,R and S. Solving circle and ellipse `(x=a cos theta, y=b sin theta)`, we have `a^(2)cos^(2)theta+b^(2)theta+2agcos theta+2bf sin theta+c=0` `rArra^(2)((1-t^(2))/(1+t))+b^(2)((2t)/(1+r^(2)))^(2)+2ag((1-r^(2))/(1+t^(2)))+2bf((2t)/(1+t))+c=0"where"t=tan.(theta)/(2)` `rArra^(2)(1-r^(2))^(2)+4b^(2)t^(2)+2ag(1-t^(2))^(2)(1-t^(2))+4bft(1+f^(2))+c(1+r^(2))^(2)=0` `rArr(a^(2)-2ag+c)t^(4)+4bft^(3)+(-2a^(2)+4b^(2)+2c)t^(2)+4bft+(a^(2)+2ag+c)-0` Roots of equation are `tan.(alpha)/(2),tan.(beta)/(2),tan.(gamma)/(2) and tan. (delta)/(2)`, where `alpha,beta,gamma and delta` ECCENTRIC angles of P,Q,R and S respectively . Now,`tan ((alpha)/(2)+(beta)/(2)+(gamma)/(2)+(delta)/(2))=(s_(1)-s_(2))/(1-s_(2)+s_(4))` Wherem `s_(i)` = sum of products of tangents of HALF angles taken 'i' at a time CLEARLY, from eqution (1), `s_(1)=s_(3)=(-4bf)/(a^(2)-2ag+c)` `tan ((alpha)/(2)+(beta)/(2)+(gamma)/(2)+(delta)/(2))=0` `rArr(alpha+beta+gamma+delta)/(2)=npi,n in Z` `rArr alpha+beta+gamma+delta=2npi,n in Z` |
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| 7503. |
If alpha,beta,gammaare the roots of p x^2+q x^2+r=0,then the value of the determinant |alphabetabetagammagammaalphabetagammagammaalphaalphabetagammaalphaalphabetabetagamma|ispb. qc. 0d. r |
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Answer» pq |
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| 7504. |
The value of the expression 2^(k)((n),(0))((n),(k))-2^(k-1)((n),(1))((n-1),(k-1))+2^(k-2)((n),(2))((n-2),(k-2))….+(-1)^(k)((n),(k))((n-k),(0)) i |
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Answer» `((N),(K))` |
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| 7505. |
If a normal chord at a point t on the parabola y^(2)=4ax subtends a right angle at the vertex, then t equals to |
| Answer» Answer :B | |
| 7506. |
The value of sin^(-1)(cos""(53pi)/(5)) is |
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Answer» a.`(3PI)/(5)` |
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| 7507. |
The standard deviation of 50 values of a vartiable x is 15 is each value of the vartiable is divided by (-3), then the standard deviation of the new set of 50 values of x will be- |
| Answer» ANSWER :C | |
| 7508. |
An insurance company insured 2000 scooter drivers. 4000, car drivers and 6000 truck driver. The probability of an accident involving a scooter, a car and a truck is 1/00,3/100and 3/20 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver ? |
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Answer» SOLUTION :Total number of persons insured `=(2000+4000+6000)=12000`. Let `E_1,E_2andE-3` be the events of choosing a scooter driver, a car driver and truck driver respectively. Then, `P(E_1)=2000/12000=1/6,P(E_2)=4000/12000=1/3andP(E_3)=6000/12000=1/2`. Let E be the event of an insured PERSON meeting with an accident. Then , `P(E//E_1)` = probability that an insured person meets with an accident, given that he is a scooter driver `=1/100` Similarly, `P(E//E_2)=3/100andP(E//E_3)=3/20`. REQUIRED probability `=P(E_1//E)` [by BAYES's theorem] = probability of choosing a scooter driver, given that he meets with an accident `=(P(E//E_1).P(E_1))/(P(E//E_1).P(E_1)+P(E//E_2).P(E_2)+P(E//E_3).P(E_3))` `=((1/100xx1/6))/((1/100xx1/6)+(3/100xx1/3)+(3/20xx1/2))=1/52`. Hence, the required probability is `1/52`. |
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| 7509. |
l=int(sin2x)/(8sin^(2)x+17cos^(2)x)dx is equal to |
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Answer» `-(1)/(9)log_(E)(8sin^(2)x+17cos^(2)x)+c` |
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| 7510. |
If A, B, C are angles of a triangle ABC, then |(sin.(A)/(2),sin.(B)/(2),sin.(C)/(2)),(sin(A+B+C),sin.(B)/(2),sin.(A)/(2)),(cos.((A+B+C))/(2),tan(A+B+C),sin.(C)/(2))| is less than or equal to |
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Answer» `(3sqrt(3))/(8)` |
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| 7511. |
If G(g), H(h) and P(p) are centroid' orthocentre and circumcentre of a triangle and x p+y h+z g=0, then (x, y, z) is equal to |
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Answer» 1, 1, -2  By USING internally division, `(2p+1h)/(2+1)=g implies2p+h-3g=0` But it is given, ` xp+yh+zg=0` `thereforex=2, y=1 and z=-3` |
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| 7512. |
Ifthe angles of a triangle ABC satisfy the equation 81^(sin^(2)x)+81^(cos^(2)x)=30, then the triangle can not be |
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Answer» equilateral |
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| 7513. |
int (1)/(4 + 5 cos 2 theta) d theta = |
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Answer» `(1)/(6) LOG |(3 + tan theta)/(5 - tan theta)| + C ` |
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| 7514. |
Let F(x)= f(x) + f(1/x), where f(x)= int_(1)^(x) (ln t)/(1+t)dt, then F(e)= |
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Answer» 2 |
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| 7515. |
A and B toss a fair coin 50 times each simultaneously. Then find the probability that both of them will not get tails at thesame toss |
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| 7516. |
Let f(x)=x^(2)+ax+b be a quadratic polynomial in which a and b are integers. If for a given integer n, f(n) f(n+1)=f(m) for some integer m, then the value of m is |
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Answer» `N(a+b)+AB` |
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| 7517. |
Prove that (1)/(log_(sqrt(bc))(abc))+(1)/(log_(sqrt(ca))(abc))+(1)/(log_(sqrt(ab))(abc))=1 (ii)(log_(2)10)(log_(2)80)-(log_(2)5)(log_(2)160)=4 (iii) a^(sqrt(log_(a)b))=b^(sqrt(log_(b)a)) (iv) (a^(log_(2^(1//4))2)-3^(log_(27)(a^(2)+1)^(3))-2a)/((7^(4log_(49)a))-a-1)=a^(2)+a+1 |
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| 7518. |
int _(a) ^(b) (dx)/( sqrt(1 + x ^(2))) where a = (e-e^(-1))/( 2 ) &b = (e ^(2) -e ^(-2))/(2 ) |
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Answer» |
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| 7519. |
Iftheta _ 1 and theta _2are the angles of inclination of tangents through a point P to the circlesx^(2) + y^(2)=a^(2),then find the locus of Pwhencot theta _1 + cot theta _1 =k |
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Answer» <P> |
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| 7520. |
Differentiate the following w.r.t.x e^(x^2) |
| Answer» SOLUTION :`d/dx(E^(x^2))=e^(x^2)d/dx(x^3)=e^(x^2)(3x^2)=3x^2e^(x^2)` | |
| 7521. |
If A and B are independent events,show that A^c and B are independent. |
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Answer» SOLUTION :P`(A^c CUP B)`=P(B-A) =P(B)-P(A capB) |
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| 7522. |
If|x|isso small thatx ^ 2andhigherpower ofx maybeneglectedthenan approximatevalue of((1 + (2 )/(3)x) ^(-3) (1 - 15 x ) ^(-1//5))/((2- 3x ) ^ 4)is |
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Answer» ` (1 )/(8) (1+ 7x ) ` `=(1 + (2 ) /(3) x ) ^(-3)( 1- 15x ) ^(-1/5) ( 2 - 3 x ) ^(-4 ) ` `= (1+(2)/(3) x ) ^(-3)( 1-15x ) ^( - 1/5 )2^(- 4 )( 1-(3x ) /(2) ) ^( - 4 ) ` NEGLECTING`x ^ 2`andhigherpowersof` .x.` `= (1)/(16)(1- 3((2)/(3)) x)(1- 15x(-1/5)) ( 1 +(3x ) /(2)(4)) ` `= (1 ) /(16)(1 - 2X ) (1 + 3x )( 1 +6x )` `= (1 ) /(16) (1 + x )(1 + 6x ) ` Neglecting`x ^ 2`andhigherpowersof `.x.` `=(1)/(16)( 1 +7x)` |
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| 7523. |
Consider the polynomialf(x)=1 + 2x + 3x^2 +4x^3for all x in R |
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Answer» `(-1//4,0)` `f(x)=1+2x+3x^2+4x^2` `rArr f(x)=2+6x+12x^2=2(6x^2+3x+1) gt 0` for all `x in R` `rArrf(x)` is striclly increasing on R Now `f(-3//4) lt 0` and `f(-1//2) gt 0` `rArr ` f(x) has a REAL root between `-3/4 " and" -1/2` ALSO f(x) is stricly increasing on R. So f(x) has exactly one real root in the interval `(-3/4, -1/2)` |
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| 7525. |
Corresponding to a triangle ABC, match the items given in List-I with the items given in List II. {:(,"List I",,"List II"),((A),rr_(2)=r_(1)r_(3),(I),angleA=90^(@)),((B),r_(1)+r_(2)=r_(3)-r,(II),b^(2)=c^(2)+a^(2)),((C),r_(1)=r+2R,(III),angleC=90^(@)),(,,(IV),angleB=120^(@)):} The correct match is |
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Answer» `{:(A,B,C),(II,III,I):}` |
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| 7526. |
If a hyperbola has one focus at the origin and its eccentricity is sqrt2 .One of the directrices isx+y+1=0 . Then the centre of the hyperbola is |
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Answer» `(-1,1) ` |
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| 7528. |
Multiply (3sqrt(-7)-5sqrt(-2))(3sqrt(-2)+5sqrt(-2) |
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Answer» SOLUTION : `(3sqrt(-7)-5sqrt(-2))(3sqrt(-2)+5sqrt(-2)` `=3sqrt(7I)-5sqrt(-2i)(3sqrt(2i)+5sqrt(2i)` `=i^2(3sqrt7-5sqrt2)(3sqrt2+5sqrt2)` `=(-1)8sqrt2(3sqrt7-5sqrt2)=-24sqrt(14)+80` |
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| 7529. |
Which of the following options is the only CORRECT combination ? |
| Answer» Answer :C | |
| 7530. |
Which of the following options is the only CORRECT combination ? |
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Answer» <P>(II) (ii) (P) |
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| 7531. |
Using differentiable, find the approximate value of each of the following upto 3 places of decimal.sqrt49.5 |
| Answer» SOLUTION :`sqrt49.5=7.035` | |
| 7532. |
Which of the following options is the only CORRECT combination ? |
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Answer» <P>(III) (II) (S) |
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| 7533. |
If f(x)={{:((sqrt(1+px)-sqrt(1-px))/(x)",",-1lexlt0),((2x+1)/(x-2)",",0lexle1):} continauous on [ -1,1], then p = |
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Answer» `-1/2` |
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| 7534. |
The normal to the circle given by x^2+y^2-6x+8y-144=0 at (8,0) meets the circle again at the point |
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Answer» `(2,-16)` |
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| 7535. |
A plane passes through (2,3,-1) and is perpendicularto the line having direction ratios 3,-4,7 . The perpendicular distance from the origin to this planeis |
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Answer» `3/(SQRT(74))` |
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| 7536. |
For the straightlines 2x - y + 1 = 0 and x - 2y - 2 = 0 find the equation of the (i) bisector of the obtuse angle between them , (ii) bisector of the acute angle between them , |
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Answer» |
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| 7537. |
Show that the points (1,1),(-1,-1) and (-sqrt3,sqrt3) are the vertices of an equilateral triangle. |
Answer» Solution : `THEREFORE absbar(AB) = sqrt((-1-1)^2 + (-1-1)^2) = 2sqrt2` Slope of `absbar(BC) sqrt((-1+sqrt3)^2 + (-1-sqrt3)^2)` = `sqrt(2)(1+3) = 2sqrt2` `absbar(AC) = sqrt((1+sqrt3)^2 + (1-sqrt3)^2)` = sqrt(2)(1+3) = 2sqrt2` `therefore BARAB = barBC = barAC` `therefore` ABC is an equilateral TRIANGLE. |
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| 7538. |
Two dice are thrown simultaneously. The probability of obtaining a total score of 5 is |
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Answer» `(1)/(18)` |
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| 7539. |
Statement-1: int_(0)^(pi//2) x cot x dx=(pi)/(2)log2 Statement-2: int_(0)^(pi//2) log sin x dx=-(pi)/(2)log2 |
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Answer» Statement-1 is TRUE, Statement-2 is True,Statement-2 is a correct explanation for Statement-1. Now, `underset(0)overset(pi//2)int X cot x dx=[x log sin x]_(0)^(x//2)-underset(0)overset(pi//2)int log sin x dx` `rArr underset(0)overset(pi//2)int x cot x dx=0underset(x to 0)LIM x logsin x-(-(pi)/(2)log2)`[Using Statement-2] `rArr underset(0)overset(pi//2)int x cot x dx=(pi)/(2)log 2` So, statement-1 is true. Also,Statement-2 is a correct explanation for statement. |
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| 7540. |
If A and B are matrices of order 3 and |A| = 5 , |B| = 3, then |3AB| = 27xx5xx3=405. |
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| 7543. |
Positive Integer just greater than (1+0.0001)^(10000) is : |
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Answer» 2 now we know that `2le (1+(1)/(n))^(n)lt 3, n GE 1, n in N` HENCE POSITIVE Integer just greater that `(1+0.0001)^(1000)` is 3. |
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| 7544. |
Ifveca=hati-hatk,vecb=xhati+hatj+(1-x)hatk,vecc=yhati+xhatj+(1+x-y)hatkshow that[veca,vecb,vecc]depends on neither .x nor y. |
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Answer» onlyx |
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| 7545. |
Let A be the event of having 53 sundays and B be the event of having 53 Mondays in a leap year. Decide whether A, B are independent or not. |
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| 7546. |
For the reaction CH_(3)COCH_(3)+Br_(2)overset(H^(+))toCH_(3)COCH_(2)Br+H^(+)+Br^(-) the following data was collected {:(["Acctone"],[Br_(2)],[H^(+)],"Rate of reaction "(Ms^(-1)),),(" "0.15,0.025,0.025," "6xx10^(-4),),(" "0.15,0.050,0.025," "6xx10^(-4),),(" "0.15,0.025,0.050," "12xx10^(-4),),(" "0.20,0.025,0.025," "8.0xx10^(-4),):} Calculate order of the reaction w.r.t. CH_(3)COCH_(3)andBr_(2) |
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Answer» Solution :`"RATE"=k."[ACETONE]"^(x).[Br_(2)]^(y).[H]^(z)` <BR> `6xx10^(-4)=k.(0.15)^(x)(0.025)^(y)(0.025)^(z)`. . . .(i) `6xx10^(-4)=k.(0.15)^(x)(0.050)^(y)(0.025)^(z)`. . . .(ii) `12xx10^(-4)=k.(0.15)^(x)(0.025)^(y)(0.050)^(z)`. . . .(iii) `8xx10^(-4)=k.(0.20)^(x)(0.025)^(y)(0.025)^(z)`. . . .(iv) (i) `divide` (ii), `1((0.025)/(0.050))^(y)` y=0, hance rate does not depend upon the CONCENTRATION of `Br_(2)`. (i) `divide` (iii), `(6)/(12)=((0.025)/(0.050))^(z),(1)/(2)=((1)/(2))^(z)` `:.z=1` (i) `divide` (iv), `(3)/(4)=((3)/(4))^(x)` `:.x=1""]` |
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| 7547. |
Integrate the following intdx/(sqrt(1-(x-1)^2) |
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Answer» SOLUTION :`(int(DX)/(sqrt(1-(x-1)^2))` put x-1=t then dx=dt `dt/(sqrt(1-t^2)=sin^(-1)t+C` |
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| 7548. |
LetA ={ 1,2,3,4,5}and Rbe therelationon AdefinedbyR= {(a,b}: b=a^2} writeRin rosterform |
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| 7549. |
Write the sum of the order and degree of the differential equation : d/(dx){((dy)/(dx))^4}=0 |
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| 7550. |
bar(a),bar(b) and bar( c ) are three unit vectors. bar(a)_|_bar(b) and bar(a)||bar( c ) then bar(a)xx(bar(b)xx bar( c )) = ……………. |
| Answer» Answer :B | |