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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.
| 12602. |
The value of k such that the lines 2x-3y+k =0, 3x-4y-13=0 and 8x-11y - 33=0 are concurrent, is |
| Answer» ANSWER :B | |
| 12603. |
If in isa positive inteter D=|(n!,(n+1)!,(n+2)!),((n+1)!,(n+2)!,(n+3)!),((n+2)!,(n+3)!,(n+4)!)| then D/((n!)^(3))-4 is divisible by |
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Answer» `N` |
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| 12604. |
Find the sine of the angle between the vectors hat(i) + 2 hat(j) + 2 hat(k) and 3 hat(i) + 2 hat(j) + 6 hat(k) |
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| 12605. |
If 1^(2)+2^(2)+3^(2)+…..+2003^(2)=(2003)(4007)(334)and(1)(2003)+(2)(2002)+(3)(2001)+…..+(2003)(1)=(2003) (334)(x), then x= |
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Answer» 2005 |
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| 12606. |
Find the angle between the line (x+1)/(2)=y/3=(z-3)/(6) and the plane 10x + 2y – 11z = 3. |
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| 12607. |
Evaluate the integral int _(3) ^(5) (sqrt(x +2 sqrt(2x-4))+ sqrt(x-2 sqrt(2x -4)))dx |
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| 12608. |
The rate of change of area A of a circle of radius r is …………. |
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Answer» `2pir` |
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| 12609. |
If alpha, beta ,gamma are the roots of the equation x^(3)-6x^(2)+11x+6=0, then Sigma alpha^(2)beta+Sigma alpha beta^(2) is equal to, |
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Answer» 80 |
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| 12610. |
If f(x)=ax^(2)+bx+c,a,b,c inR and eqation f(x)-x=0 has imaginary roots alpha,beta,gamma "and " delta be the roots of f(x) -x=0 then |{:(1,alpha,delta),(beta,0,alpha),(gamma,beta,1):}| is |
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Answer» 0 |
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| 12611. |
One kind of cake requires 300 gr of flour and 15 gr of fat and another kind of cake requires 150 gr. of flour and 30 gr of fat. Find the maximum number of cakes which can be made from 7.5 kg of flour and 600 gr. of fat assuming that there is no shortage of the other ingredients used in making the cake. |
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| 12612. |
A and B are events such that P(A)= 0.4, P(B)= 0.3 and P(A cup B)= 0.5.Then P(B' cap A')equal = …….. |
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Answer» `(2)/(3)` |
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| 12613. |
int(sin^3x)/((cos^4 x + 3 cos^2 x+1) tan^(-1) (sec x+ cos x))dx= |
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Answer» `tan^(-1)(sec x+ COS x ) + C` `I=int(SIN^3x)/((cos^4 x + 3 cos^2 x+1) tan^(-1) (sec x+ cos x))dx` `rArr I=int((sin^3x)/(cos^2x))/((cos^2 x + 3+ sec^2 x) tan^(-1) (sec x + cosx ))dx` `rArr I=int1/(1+(sec x + cos x )^2)xx(sin x(1-cos^2x))/(cos^2 x)xx1/(tan^(-1)(sec x + cos x ))dx` `rArr I=int1/(tan^(-1) (sec x + cos x ))xx1/(1+(sec x + cos x )^2) xx (tan x sec x -sinx ) dx` `rArr I=int1/(tan^(-1)(sec x + cos x )) d {tan^(-1) (sec x + cos x)}` `rArr I=log|tan^(-1) (sec x + cos x ) |+C` |
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| 12614. |
If ""^(18)C_(r)=""^(18)C_(r+2) then""^(R )C_(5)= |
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Answer» 23 |
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| 12615. |
If z and omega are two non-zero complex numbers such that |z omega|=1" and "arg(z)-arg(omega)=(pi)/(2), then bar(z)omega is equal to |
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Answer» `-1` |
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| 12616. |
Using differentials find the approximate value of sqrt(49.5) |
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| 12617. |
underset(x to oo)"Lt" ([x]+[2x]+[3x]+....+[nx])/(n^(2))= |
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Answer» x/2 |
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| 12618. |
If the vectors 10hati+3hatj,12hati-15hatj and a hati+11 hatj are collinear a = ………….. |
| Answer» ANSWER :D | |
| 12619. |
If f(x)=(tan[x]pi)/([1+|log(sin^2x+1)|],where [.] denotes the greatest integer function and |.| stands for the modulus of the function,thenf(x) is |
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Answer» DISCONTINUOUS `AA" "X" "epsilon" "I` |
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| 12620. |
Integrate the function is exercise. sqrt(x^(2)+3x) |
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| 12621. |
If omega is a complex cube root of unity, then (a+b omega+c omega^(2))/(c+a omega+b omega^(2))+(c+a omega+b omega^(2))/(a +b omega+c omega^(2))+(b+c omega+a omega^(2))/(b+c omega+a omega^(5))= |
| Answer» ANSWER :D | |
| 12622. |
ifcosalpha + sinalpha= 3/4, thensin^6alpha+ cos ^6 alpha= |
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Answer» `(877)/(1024` |
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| 12623. |
Internal bisectors of DeltaABC meet the circumcircle at point D, E, and F Ratio of area of triangle ABC and triangle DEF is |
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Answer» `GE 1` `angleADE = angleABE = (B)/(2)` Similarly, `angleFDA = angleFCA = (C)/(2)` `rArr angleFDE = (B+C)/(2)` and `ANGLEDEF = (A+C)/(2) and angleDFE = (A +B)/(2)`  In `DeltaDEF`, by sine rule, `(EF)/(sin (angleFDE)) = 2R` `rArr EF = 2R COS ((A)/(2))` Then, area of `DeltaDEF` `= 2R^(2) sin ((A+B)/(2)) sin ((B+C)/(2)) sin ((A+C)/(2))` `=2R^(2) cos ((A)/(2)) cos ((B)/(2)) cos ((C)/(2))` Now `(Delta_(ABC))/(Delta_(DEF)) = (2R^(2) sin A sin B sin C)/(2R^(2) cos ((A)/(2)) cos((B)/(2)) cos((C)/(2)))` `= 8sin ((A)/(2)) sin ((B)/(2)) sin ((C)/(2)) le 1` `rArr Delta_(ABC) le Delta_(DEF)` |
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| 12624. |
A: The angle of rotation to remove the xy-term in the equation 2x^(2) + sqrt(3)xy + 3y^(2) =9 is pi//6. R: The angle of rotation of the axes to eliminate xy term in the equation. ax^(2) + 2hxy + by^(2) + 2gx + 2fy +c=0 is 1/2 tan^(-1)((2h)//(a-b)) |
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Answer» Both A and R are TRUE and R is the CORRECT explanation of A. |
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| 12625. |
Three integers are selected from the integers 1,2,…….., 1000. The number of ways in which these integers can be selected such that their sum is divisible by 4 are__________________. |
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| 12626. |
A random variable X has the following probability distribution : then P(X le 1) = …....... |
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Answer» 0.55 |
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| 12627. |
(d)/(dx)y=tan^(-1)[(sqrt(1+sinx)-sqrt(1-sinx))/(sqrt(1+sinx)+(sqrt(1-sinx))]= |
| Answer» Answer :C | |
| 12628. |
If the extremities of diagonal of a squareare (1,-2,3),(2,-3,5) then the length of its side is |
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Answer» `sqrt(6)` |
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| 12629. |
Which of the following options is the only CORRECT combination ? |
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Answer» <P>(I) (II) (P) |
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| 12630. |
ABC is an isosceles triangle inscribed in a circle of radius r. If AB=AC and h is the altitude form A to BC. If P is perimeter and A is the area of the triangle then find the valueof lim_(hto0)(A)/(P^(3)). |
Answer» Solution :In `DeltaABC, AB=AC, AD_|_BC(D" is midpoint of "BC)`  Let r= radius of CIRCUMCIRCLE `:.""OA=OB=OC=r` Now `BD=sqrt(BO^(2)-OD^(2))=sqrt(r^(2)-(h-r)^(2))=sqrt(2rh-h^(2))` `:.""BC=2sqrt(2rh-h^(2))` Also` AB^(2)=BD^(2)+AD^(2)=2hr-h^(2)+h^(2)=2hr`. `:.""AB=AC=sqrt(2hr)` `:."""PERIMETER",P=2sqrt(2rh-h^(2))+2sqrt(2hr)` `:."""Area of "DeltaABC=(1)/(2)xxBCxxAD=hsqrt(2rh-h^(2))` So,`""underset(hto0)lim(A)/(P^(3))=(hsqrt(2rh-h^(2)))/(8(sqrt(2rh-h^(2))+sqrt(2hr))^(3))` `=underset(hto0)lim(h^(3//2)sqrt(2r-h))/(8h^(3//2(sqrt(2r-h)+sqrt(2r))^(3)))` `=underset(hto0)lim(sqrt(2r-h))/(8[sqrt(2r-h)+sqrt(2r)]^(3))` `=(sqrt(2r))/(8(sqrt(2r)+sqrt(2r))^(3))=(1)/(128r)` |
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| 12631. |
Find the centre and radius of the circle 3x^(2)+ 3y^(2) - 6x + 4y - 4 = 0 |
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| 12632. |
Statement-1: The equation x log x= 2-x is satisfied by at least one value of x lying between 1 and 2 Statement-2: The function f(x) = x log x is an increasing function in [1,2] and g(x) = 2-x is a decreasing function in [1, 2] and the graphs represented by these functions intersect at a point in [1,2] |
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| 12633. |
If f(x) = x + (x^3)/(3!) +(x^5)/(5!) +(x^7)/(7!)+ .... then f'(x) = |
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Answer» `SIN HX ` |
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| 12634. |
Let R be a relation on X, If R is symmetric then xRy impliesyRx. If it is also transitive then xRy and yRx implies xRx.So whenver a relation is symmetric and transitive then it is also reflexive. What is wrong in this argument ? |
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Answer» Solution :Let R is a relation on X.` If R is SYMMETRIC the `xRy implies yRx` If R is ALSO transitive then xRy and yRx `impliesxRx` ` implies` Whenever a relation is symmetric and transitive , then it is reflexive. This argument is wrong because the symmetry of R does not IMPLY dom R does not imply domR =Xand for reflexive `XRX AA x in X`. |
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| 12635. |
The value of the infinite series (1^(2)+2^(2))/(lfloor1)+(x^(4))/(lfloor2)+(x^(6))/(lfloor3).... Then the value of "log"_(e) y is |
| Answer» Answer :C | |
| 12636. |
Find the angle between the vectors""oversetrarra=overset^^i+overset^^j-overset^^k "and"oversetrarrb=overset^^i-overset^^j+overset^^k |
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Answer» SOLUTION :`cos THETA=(veca.vecb)/(AB)= (1-1-1)/(SQRT3 sqrt3)=(-1)/3` `theta=cos^(-1)(-1)/3)` |
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| 12638. |
The value of ""^(n)C_(3)+""^(n)C_(7)+""^(n)C_(11)+…. is : |
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Answer» `1/2{2^(n-1)-2^(n//2)"SIN"(NPI)/(4)}` |
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| 12639. |
Evaluate the following integrals inte^(x)sinxdx |
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| 12640. |
Find inverse , by elementary row operations (if possible), of the following matrices (i)[{:(1,3),(-5,7):}] (ii)[{:(1,-3),(-2,6):}] |
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Answer» (II) `thereforeA^(-1)` |
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| 12641. |
Solve the system of equations {:(2x+5y=1),(3x+2y=7):}, |
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| 12642. |
If a circle passes through the point (a,b) and cuts the circlex^(2) + y^(2) = k^(2) orthogonally, the equation of the locus of its centre is |
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Answer» ` 2ax + 2BY = a^(2) + B^(2) + K^(2)` |
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| 12643. |
Find c of the R olle’s theorem for the functions f(x)=log(x^(2)+3)/(4x) in [1,3] |
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| 12644. |
Showthat thelines vec( r)=(2hat(i) -3hat(k)) + lambda(hat(i) +2hat(j)+3hat(k)) " and" vec(r )=(3hat(i) +6hat(j)+3hat(k)) + mu(2hat(i) +3hat(j) +4hat(k)) intersect. Also findtheirpoint ofintersection. |
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| 12645. |
(1+x^2/(2!)+x^4/(4!)+....oo)^2 = |
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Answer» `1+(2X^2)/(2!)+(2^3x^4)/(4!)+(2^5x^6)/(6!)+....OO` |
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| 12646. |
int(5 cot x + 1)/((cot x - 1)(cot x - 2) sin^(2) x) dx = 6log|f(x)| + 11log|g(x)| + c, then (f(x), g(x)) = |
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Answer» `(COT, X - 1), (cot x - 2)^(-1))` |
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| 12647. |
Any material entering the trachea causes :- |
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Answer» vimiting |
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| 12648. |
Let f(t)=|{:(cost,t,1),(2sint,t,2t),(sint,t,t):}|" then "underset(trarr0)lim(f(t))/t^2 equal to".........." |
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Answer» 0 |
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| 12649. |
For the series21,22,23 ,….. K - 1 , kthe G,Mof the first and last numberexistin the given series.If 'k' is a three digit number, then 'k' can attain |
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Answer» 5 VALUES |
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| 12650. |
A 3 xx 3 determinant has its entries 1 or-1 . The number of such determinant in 2^9 =512. We will call a 3 xx 3 determinant with entries 1 or -1 as minus special if product of elements of any rows or any columns is -1 for example |{:(1,-1,1),(1,1,-1),(-1,1,1):}| is a minus special determinant. The number of nxxn minus special determinates must be |
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Answer» `2^(n-1)` |
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