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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. |
Let P(x) = 0 be a fifth degree polynomial equation with Integer coefficients that has atleast one Integral root. If P(2) = 13 and P(10) = 5, then find the Integral value of V that must satisfy P(x) = 0. |
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| 2. |
Let S = 0, S' = 0 be two circle intersecting at two distinct points and L_1 be their common chord, L_2 is the line joing their centres and L_3 is the radical axis. Then |
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Answer» `L_1` is not PERPENDICULAR to `L_2` |
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| 4. |
sqrt(1+sinA)=-("sin"(A)/(2)+"cos"(A)/(2))is true if |
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Answer» `(3PI)/(2)ltAlt(5pi)/(2)`only We know ,`1+sinA=("cos"(A)/(2)+"sin"(A)/(2))^(2)` `thereforesqrt(1+sinA)=|"cos"(A)/(2)+"sin"(A)/(2)|` We KOW . `|X|={{:(x ifxge0,),(-x iflt0,):}` `|"sin"(A)/(2)+"cos"(A)/(2)={{:("sin"(A)/(2)+"cos"(A)/(2)",",if2npi-(pi)/(4)le2npi+(3pi)/(4)),(-("sin"(A)/(2)+"cos"(A)/(2))",",if"otherwise"):}` `thereforesqrt(1+sinA)=-("sin"(A)/(2)+"cos"(A)/(2))` when `(3pi)/(4)lt(A)/(2)lt(7pi)/(4)` `RARR(3pi)/(2)ltAlt(7pi)/(2)`. |
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| 5. |
If A=[{:(,1,3,3),(,1,4,3),(,1,3,4):}] then verify that A adj A=|A| I. Also find A^(-1) |
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Answer» Solution :`"we have" |A|=1(16-9)-3(4-3)+3(3-4)=1ne0` `"Now" C_(11)=7, C_(12)=-1, C_(13),C_(21)=-3, C_(22)=1, C_(23)=0, C_(31)=-3, C_(32)=0,C_(23)=1` Therefore ADJ A=`[{:(,7,-3,-3),(,-1,1,0),(,-1,0,1):}]` Now, A(adj A)=`[{:(,1,3,3),(,1,4,3),(,1,3,4):}]` `[{:(,7,-3,-3),(,-1,1,0),(,-1,0,1):}]`=`[{:(,7-3-3,-3+3+0,-3+0+3),(,7-4-3,--3+4+0,-3+0+3),(,7-3-4,-3+3+0,-3+0+4):}]` `[{:(,1,0,0),(,0,1,0),(,0,0,1):}]=(1) [{:(,1,0,0),(,0,1,0),(,0,0,1):}]=|A|I` Also `A^(-1)=(1)/(|A|)=(1)/(1)[{:(,7,-3,-3),(,-1,1,0),(,-1,0,1):}]=[{:(,7,-3,-3),(,-1,1,0),(,-1,0,1):}]` |
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| 7. |
If f : R rarr R , f(x) = [x] , g : R rarr R , g(x) = sinx , h : R rarr R , h(x) = 2x , then ho(gof) = .......... |
| Answer» Solution :N/A | |
| 8. |
Integrate the following functions : int(cos2x)/((sinx+cosx))dx |
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| 9. |
If the value oflim_(n to oo) n^(-n^2) ((n + 1)(n + 1/3)(n + 1/(3^2)) ….(n + 1/(3^(n-1))))^(n) is e^k then k is: |
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Answer» `=lim_(nrarroo)((n+1)/(n))^(n).((n+(1)/(3))/(n))^(n)"............"((n+(1)/(3^(n-1)))/(n))^(n)=lim_(nrarroo)(1+(1)/(n))^(n)(1+(1)/(3n))^((3n)/(3))"...."(1+(1)/(3^(n-1).n))^((3^(n-1).n)/(3^(n-1)))` `e.e^((1)/(3)).e^((1)/(9))"........"e^((1)/(3^(n-1)))=e^(1+(1)/(3)+(1)/(3^(2))"......."oo)=e^((1)/(1-1//3))=e^(3//2)","k=1.50` |
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| 10. |
A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distance 2b from the origin, then the locus of the centre of this circle is |
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Answer» a HYPERBOLA |
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| 11. |
If the line x+y +1=0 intersects the circle x^2+y^2+x+3y=0at two points A and B , then the centre of the circle which passes through the points A,B and the point of intersection of the tangents drawn at A and B to the given circle is |
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Answer» `(5/8,5/8)` |
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| 12. |
Which of the following sentences are propositions and which are not ? Write with reason :2 lt 5 |
| Answer» SOLUTION :`2 GT 5` is a proposition, as it is TRUE. | |
| 14. |
Let X be a B(2, p) and Y be an independent B(4, p). If P(X ge 1)=5//9, then find P(Y ge1). |
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| 15. |
Let f(x) = 4x^(2)-4ax+a^(2)-2a+2 and the golbal minimum value of f(x) for x in [0,2] is equal to 3 The number of values opf a for which the global minimum value equal to 3 for x in [0,2] occurs for the vlue of x lying in (0,2) is |
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Answer» 1 vertex of this parabola is `((a)/(2),2-2a)` caseI:`0lta//2lt2` In this case f(x) will attain the MINIUM value at `x=a/2` .Thus `f(a/2)^(3)`  or `3=-2a+2 or a =-1/2 (rejected)` CaseII: `(a)/(2)ge2` In this f(x) attains global minimum value at x =2 thus f(2)=3 `3=16-8a+a^(2)-2a+2 or a =5pm sqrt(10)` thus `a=5+sqrt(10)` CaseIII:` (a)/(2)ge0` In this case f(x) attains the global minimum value at x =0 thus f(0)=3 `therefore3=a^(2)-2a+2 or a =1 pm` Thus a =1-`sqrt(2)` Hecne the PERMISSIBEL values of a are `1-sqrt(2) and 2+sqrt(10)` f(x) =`4x^(2)-49x+a^(2)-2a+2`is monotonic in [0,2] Hence the point of minima of FUNCITON should not LIE in [0,2] Now f(x) =0 or 8x-4a=0 or x `=a//2` `(a)/(2)in [0,2] or a in [0,4]` For f(x) to be monotomic [0,2], and[0,4] i/e`ale0` or age4 |
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| 16. |
Let f(x) = 4x^(2)-4ax+a^(2)-2a+2 and the golbal minimum value of f(x) for x in [0,2] is equal to 3 The number of values of a for which the global minimum value equal to 3 for x in [0,2] occurs at the endpoint of interval[0,2] is |
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Answer» 1 vertex of this parabola is `((a)/(2),2-2a)` caseI:`0lta//2lt2` In this case f(x) will attain the minium value at `x=a/2` .THUS `f(a/2)^(3)`  or `3=-2a+2 or a =-1/2 (rejected)` CaseII: `(a)/(2)ge2` In this f(x) ATTAINS global minimum value at x =2 thus f(2)=3 `3=16-8a+a^(2)-2a+2 or a =5 pm sqrt(10)` thus `a=5+sqrt(10)` CaseIII: `(a)/(2)ge0` In this case f(x) attains the global minumum value at x =0 thus f(0)=3 `therefore3=a^(2)-2a+2 or a =1 pm` Thus a =1-`sqrt(2)` HENCE the permissiable values of a are `1-sqrt(2) and 2+sqrt(10)` f(x) =`4x^(2)-49x+a^(2)-2a+2`is monotonic in [0,2] Hence the point of minima of funciton should not lie in [0,2] Now `f(x) =0 or 8x-4a=0 or x =a//2` `(a)/(2)in [0,2] or a in [0,4]` For f(x) to be monotomic [0,2], and[0,4] i/.ea `le 0 or age4` |
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| 17. |
If N is a prime number which divides S=^(39)P_(19)+^(38)P_(19)+^(37)P_(19)+…+"^(20)P_(19), then the largest possible value of N among following is |
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Answer» `41` `=19!{'^(30)C_(19)+^(38)C_(19)+^(37)C_(19)+....+^(20)C_(19)}` `=19!{'^(39)C_(19)+^(38)C_(19)+^(37)C_(19)+....+('^(20)C_(19)+^(20)C_(19))-1}` `=19!{'^(39)C_(19)+^(38)C_(19)+....+('^(21)C_(19)+^(21)C_(19))-1}` `=19!{'^(40)C_(20)-1}` `=((41-1)(41-2)(41-3).....(41-20)-20!)/(20)=(41k)/(20)` `implies` greatest PRIME number is `41` |
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| 18. |
The random variable X has a probability distribution P(X) of the following form, where k is some arbitary real number: P(X=x)={{:(k",","if "x=0), (2k",","if "x=1), (3k",","if "x=2), (0",", otherwise):} (i) Determine the value of k "" (ii) Find P(X lt 2), P(X le 2), P(X ge 2) |
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| 19. |
For positive integers m and n, let gcd (m,n) denote the largest integer that is a factor of both m and n. Find gcd (2015! + 1, 2016! + 1) where nl denotes the factorial of a positive integer n. |
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| 20. |
Equation of the tangent to the curve y=2x^3_6x^2-9 at the point where the curve crosses the y -axis is |
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Answer» only I is TRUE |
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| 22. |
If int e^(x) (x^(2) - 5x+ 8)" dx = " e^(x)f(x) + c then f(x) = |
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Answer» `X^(2) -5x + 12 ` |
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| 24. |
Three numbers are chosen from the set {1, 2, 3, ..., 40} at random without replacement. If p is the probability that three numbers chosen are not consecutive, then 145p - 140.25 is equal to _____ |
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| 25. |
Find the distance of the point(-1,-5,-10) from the point of intersection of line (x-2)/(3)=(y+1)/(4)=(z-2)/(2) and the plane x-y+z=5 measured parallel to the line (x)/(2)=(y)/(3)=(z)/(-6) |
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| 26. |
I: Ifp(q-r)x^2+q(r-p)x+r(p-q) =0has equalrootsthenp,q,rin A.PII :if thesumof therootsof ax^2 + bx+ c=0 hasequalto thesumof thesquaresoftheirreciprocalsthenbc^2 , ca^2, +bx +c=0 is equalto thesum of thesquaresoftheirreciproocals thenbc^2, ca ^2 , ab^2are in A.P |
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Answer» ONLYI is true |
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| 27. |
A vector r of magnitude 3sqrt2 units which makes an angle of (pi)/(4) and (pi)/(2) with y and z-axis, respectively is |
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Answer» `r= pm3hati+3hatj` Therefore, `l^(2)+m^(2)+n^(2)=1` [GIVEN] `l^(2)+1/2 +0=1` `IMPLIES l= pm 1/sqrt(2)` Hence, the required VECTOR `r=3sqrt(2) (l HAT(i)+m hat(j)+n hat(k))` is given by `r=3sqrt(2) (pm 1/sqrt(2) hat(i) +1/sqrt(2) hat(j) + 0hat(k))=r= pm 3 hat(i)+3hat(j)` |
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| 28. |
A line makes angles alpha,beta,gamma with the coordinates axes . If alpha+beta=90^@ then gamma is equal to |
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Answer» `0^@` |
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| 29. |
If A_(1),A_(2),A_(3) be the areas of circles x^(2)+y^(2)+4x+6y-19=0, x^(2)+y^(2)=9, x^(2)+y^(2)-4x-6y-12=0 respectively then A_(1):A_(2):A_(3)= |
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Answer» `A_(1)gtA_(2)A_(3)` |
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| 31. |
Equation of one of the tangent passing through (2,8) to the hyperbola 5x^(2)-y^(2)=5 is |
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Answer» 3x+y-14=0 |
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| 32. |
If f(x)= {(mx+1,,x,le,5),(3x-5,,x,>,5):} is continuous, then the value of m is : |
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Answer» (11/4) |
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| 33. |
The symmetric from of the equation of the line x + y -z = 1 and 2x -3y + z = 2 is........... |
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Answer» `(X)/(2) = (y)/(3) = (Z)/(5)` |
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| 34. |
An optically active compound 'X' has molecular formula C_(4)H_(8)O_(3). It evolves CO_(2) with aq. NaHCO_(3). 'X' reacts with LiAlH_(4) to give achiral compound. 'X' is : |
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Answer» `CH_(3)CH_(2)underset(OH)underset(|)(C)HCOOH` |
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| 35. |
Evaluate :int_(0)^(1)a^(2-3x)dx , as limit of sums . |
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| 36. |
I : If |a + b| = |a - b| then |a xx b| = 0 II : (a xx b)^(2) + (a.b)^(2) = |a|^(2) |b|^(2) |
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Answer» only I is ture |
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| 37. |
If f(x) = {(3",",x = 0),(-x^(2) + 3x + k",",0 lt x lt 1),(ax +b",",1 le x le 2):} satisfies the hypothesis of the Lagrange's theorem then (a+b)/k is equal to |
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| 38. |
A random veriate X takes the values 0,1,2,3 and its mean is 1.3 . If P(X =3) = 2P(X =1) and P(X = 2) = 0.3 , then P(X = 0) is equal to |
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Answer» 0.1 |
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| 39. |
Let P(n)= 5^n- 2^n, P(n)is divisibleby3 lamdawherelamdaand nbothare oddpositive integersthentheleastvalueofnandlamdawill be |
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Answer» 13 |
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| 40. |
Assertion (A) : If I_(n)= int tan^(n)xdx then 5 (I_(4) + I_(6)) = tan^(5) x Reason (R) : int tan^(n) " xdxthen " I_(n) = (tan^(n-1)x)/(n) -I_(n- 2)"where n " inN The correct answer is |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 41. |
If the distance of the plane x - y + z + lambda = 0 from the point (1, 1, 1) is d_1 and the distance of this point from the origin is d_2 and d_2d_2 = 5 then find the value of lambda. |
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| 43. |
For each of the differential equations given below, indicate its order and degree(if defined). (i) (d^(2)y)/(dx^(2))+ 5x((dy)/(dx))^(2) - 6y = log x (ii)((dy)/(dx))^(3) - 4((dy)/(dx))^(2) + 7y = sin x (iii)(d^(4)y)/(dx^(4)) - sin ((d^(3)y)/(dx^(3)) = 0 |
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Answer» (II) Order 1; Degree 3 (iii) Order 4; Degree not DEFINED |
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| 44. |
Evaluate the following integrals (i) int_(0)^(1) sin^(-1)((2x)/(1+x^(2)))dx |
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| 46. |
The point of contact of 5x+6y+1=0 to the hyperbola 2x^(2)-3y^(2)=2is |
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Answer» `(5,4)` |
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| 48. |
Show that the following four points in each jof the following are concyclic and find the equation of the circle on which they lie. (1,2),(3,-4),(5,-6),(19,8) |
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| 49. |
int sqrt(x^(2)-25)dx= |
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Answer» `(x)/(2) sqrt(x^(2) - 25) - (25)/(2) cosh^(-1) ((x)/(5)) +C ` |
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