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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. |
C_0+4. C_1 + 7. C_2+…...(n+1) terms = |
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Answer» `(3N +2).2^(n-1)` |
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| 3. |
Assertion (A) : The number of positive integral solutions of x_(1)+x_(2)+x_(3)=10 is 36. Reason (R) : The number positive integral solutions of the equation x_(1)+x_(2)+x_(3)+.......+x_(k)=n is ""^(n-1)C_(k-1), 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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| 4. |
Find the sum of the series .^(84)C_(4)+6xx.^(84)C_(5)+15xx.^(84)C_(6)+20xx.^(84)C_(7)+15xx.^(84)C_(8)+6xx.^(84)C_(9)+.^(84)C_(10). |
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Answer» `= .^(6)C_(6) xx .^(84)C_(4)+ .^(6)C_(5) xx .^(84)C_(5) + .^(6)C_(4) xx .^(84)C_(6)+ .^(6)C_(3) xx .^(84)C_(7) + .^(6)C_(2) xx .^(84)C_(8)+ .^(6)C_(1)xx.^(84)C_(9) + .^(6)C_(0) xx.^(84)C_(10)` `=` Coefficient of `x^(10)` in `(1+x)^(6)(1+x)^(84)` `=` Coefficient of `x^(10)` in `(1+x)^(90)` `= .^(90)C_(10)` |
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| 5. |
Let a = "min" {x^(2)+2x+3: x in R} and b= lim_(theta to 0) (1-cos theta)/(theta^(2)). Then sum_(r=0)^(n) a^(r ) b^(n-r) is |
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Answer» `(2^(N+1)-1)/(3*2^(n))` |
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| 6. |
The ubiquitous AM-GM inequality has many applications. It almost crops up in unlikely situations and the solutions using AM-GM aretruly elegant . Recall that for n positive reals a_(i) I = 1,2 …,n, the AM-GM inequality tells (overset(n) underset(1)suma_i)/n ge ( overset(n)underset(1)proda_i)^((1)/(n)) The special in which the inequality turns into equality help solves many problems where at first we seem to have not informantion to arrive at the answer . If a,b,c are positive integers satisfying (a)/(b+c)+(b)/(c+a) + (c)/(a+b) = (3)/(2), then the value of abc + (1)/(abc) |
| Answer» Answer :3 | |
| 8. |
int_0^1(xdx)/((2x+1)(x+1)) |
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Answer» SOLUTION :`int_0^1(XDX)/((2x+1)(x+1))` int_0^1((1/(x+1)-1/(2x+1))DX ` [In(x+1)-1/2In(2x+1)]_0^1` `In2-1/2(IN3)=In(2/sqrt3)` |
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| 9. |
If f(x), g(x) are primitives of phi (x) then |
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Answer» F(x) = G(x) |
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| 10. |
Let f be a polynormal function defined on [2, 7]. If f(2) = 3 and f'(x) = le 5 for all x in (2, 7), then the maximum possible value attained by f at x = 7 is |
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Answer» 7 |
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| 11. |
If an LPP admits optimal solution at two consecutive vertices of a feasible region, then |
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Answer» the REQUIRED optimal solution is at the midpoint of the line joining the TWO points |
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| 12. |
For vectors bar(a),bar(b),bar( c ),|bar(a)-bar( c )|=|bar(b)-bar( c )| then the value (bar(b)-bar(a)).(bar( c )-(bar(a)+bar(b))/(2))= ……….. |
| Answer» Answer :A | |
| 13. |
The value of 'x' satisfying the equation "log"_2"log"_3(sqrtx+sqrt(x-3))=0 is coprime with : - |
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Answer» 2 |
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| 14. |
A ribbon 40 - inches long is to be cut into three pieces, each of whose lengths is a different integer number of inches {:("Quantity A","Quantity B"),("The least possible length, in",15),("inches, of the longest piece",):} |
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| 15. |
The position vector of a point A is (3, 4, -5) Find, (i) Distance of a point A from XY-plane. (ii) Distance of a point A from X-axis. (iii) Distance of a point A from origin. |
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| 16. |
Evaluate the following integrals. int(1)/(3x^(2)+2x+1)dx |
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| 17. |
Find the vector equation in parametric form and Cartesian equations of a throughthe points (-3, 7, -4) and (13,-5, 2). Find the point where the straight line crosses the xy -plane. |
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| 18. |
Mean of 10 numbers is 6. It was later observed that one number was misread as 9. When the correct means was 7, then the correct value of that number is |
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Answer» 19 |
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| 20. |
Evaluate the following inegrals int(sinx)/(sin(a+x))dx |
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| 21. |
The value of .^(15)C_(0)^(2)-.^(15)C_(1)^(2)+.^(15)C_(2)^(2)-"...."-.^(15)C_(15)^(2) is |
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Answer» 15 (if n is odd) and in the question `n = 15` (odd). Hence, sum of given SERIES is 0. |
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| 22. |
A 2.5 gm impure samplecontainingweakmonoacidic base (Mol. Wt = 45) is dissolved in 100ml waterand titrated with 0.5 M HCL when ((1)/(5))^(th) of the base was neutralised the pH was foundto be 9 andatequivalent pointpH of solution is 4.5. Given : All data at 25^(@)C & log2 = 0.3. Select correct statement(s). |
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Answer» `k_(b)` of base is less that `10^(6)` |
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| 23. |
If a=sum_(n=1)^(oo)((2n))/(2n-1),b sum_(n=1)^(oo) (2n)/(2n+1!) then ab = |
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Answer» 1 |
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| 24. |
The value of sin^(-1) ((2 sqrt2)/(3)) + sin^(-1) ((1)/(3)) is equal to |
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Answer» `(2pi)/(3)` |
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| 26. |
Using elementary transformations, find the inverseof the matrices [(3,10),(2,7)] |
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| 27. |
Prove that the curves 2x^(2)+4y^(2)=1 and 6x^(2)-12y^(2)=1 cut each other at right angles . |
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| 28. |
Using differentials, find the approximate value of each of the up to 3 places of decimal. (82)^((1)/(2)) |
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| 29. |
The sum of the series 1^(2) + 3^(2) + 5^(2) + ......+n' |
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Answer» `(n(n + 1)(2N + 1))/2` |
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| 30. |
The probability distribution of a discrete random variable X is given as under : Calculate : (i) The value of A if E(X) = 2.94 (ii) Variance of X |
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| 31. |
Match the following. I. int sec^(2)x "cosec"^(2)dx "" a) x+c II. int tan^(2)x cot^(2) xdx= "" b) tan x-x+c III. Int sin^(2)x sec^(2) x dx= "" c)-cot x-x+c IV. Int cos^(2) x "cosec"^(2) x dx= "" d) tan x-cot x+c |
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Answer» a,B,c,d |
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| 32. |
The point on the line 3y - 4x + 11 = 0 equidistant from (3, 2) and (-2,3) is |
| Answer» Answer :D | |
| 33. |
ABCD is a rectangle. If A = (2, 3), C = (8, 11) and BD is parallelto y-axis then B and D are |
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Answer» (5, 12), (5, 2) |
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| 35. |
Match the following {:(I.(1+sin 2x)-cos x + sin x,(a)sin x=1//2),(II.(2sin x -cos x)(1+cos x)=sin^(2)x,(b)tan x =-1),(III.tan theta + cot theta =2,(c)theta = pi//6),(,(d)theta=pi//4):} |
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Answer» a, B, c |
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| 36. |
Two cards drawn from a well shuffled deck of 52 cards with replacement. The probability that both cards are queens is |
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Answer» `(1)/(13) xx (1)/(13)` |
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| 37. |
Consider the parabola y^2 =4x and the ellipse 2x^2 + y^2 =6, intersecting at P and Q.(a) Find the area enclosed by the parabola and the commonchord of the ellipse and parabola.(b) If tangent and normal at the point P on the ellipse intersect the x-axis at T and G respectively then find the area of the triangle PTG. |
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| 38. |
{:("Column A","in the figure, P and Q are center of the two circle of radii 3 and 4 respectively. A and B are the points at which a common tangent touches each circle","Column B"),(AB,,PQ):} |
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Answer» If column A is LARGER |
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| 39. |
Draw the graph of f(x) = xcosx-sinx, x in [-3pi, 3pi] |
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Answer» SOLUTION :`F(x) = xcosx-sinx` Clearly, domain is R. ALSO `f(x)` is non-periodic. `f^(')(x)=-xsinx` `f^(')(x) =0 rArr x=-3pi, -2pi, -pi, 0, pi, 2pi, 3pi` `f^(')(0^(-)) = (-)(-)(-) lt 0` and `f^(')(0^(+)) = (-)(+)(+) lt 0` So `x=0` is the POINT of inflection (as the derivative does not CHANGE sign in the neighbourhood of `x=0` Sign scheme of `f^(')(x)` is as follows:  Clearly, f is decreasing at `x=0` and has the point of minima at `x=pi, -2pi` and point of maxima at `x=-pi, 2pi` Since `f(x)` is an odd function, we check the graph for `x int [0, 3pi]` `f(0)=0, f(pi)=-pi` Thus, `f(x)` decreases from `0` to `-pi` in the interval `(0, pi)` Thus, `f(x)` increases from `-pi` to `2pi` in the interval `(pi, 2pi)` `f(3pi) =-3pi` Thus, `f(x)` decreases from `2pi` to `-3pi` in the interval `(2pi, 3pi)` From the above discussion the graph of `y=f(x)` is as shown in the following figure.
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| 40. |
Let A be set of first ten natural numbers and R be a relation on A, defined by (x,y)in Rimpliesx+2y=10, then domain of R is |
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Answer» `{1,2,3,…,10}` |
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| 41. |
If the focus is (1,-1) and the directrix is the line x+2y-9=0, the vertex of the parabola is |
| Answer» ANSWER :B | |
| 42. |
Integrate (2x)/( ( x^(2) + 1) ( x^(2) + 3) ) with respect to x. |
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| 43. |
Find adjoint of each of the matrices [{:(5,8,1),(0,2,1),(4,3,-1):}] |
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| 44. |
Solving problems in co-ordinate geometry with given set of conditions causes difficulties at times One such case is equation of family of lines. The equation y -1 = m (x -2) represent a family of lines passing through (2,1). But this does not include the line x -2 =0. Since the slope of the line x -2 =0 is infinity. Therefore in case there are two solutions to a problem and only one solution is being obtained aigebralcally we have to re-examine. A line y = mx through origin cuts the parallel lines x + y =3 and y =5at A and B respectively. The distance between the two points of intersection is d. If AB=2 then the quadratic equation formed is |
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Answer» `(4-d^(2)) m ^(2) + 2md ^(2) + 4-d ^(2) =0` |
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| 45. |
The minimum positive integral value of m such that ( 1073)^(71)-m may be divisible by 10, is |
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Answer» 1 |
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| 46. |
Solving problems in co-ordinate geometry with given set of conditions causes difficulties at times One such case is equation of family of lines. The equation y -1 = m (x -2) represent a family of lines passing through (2,1). But this does not include the line x -2 =0. Since the slope of the line x -2 =0 is infinity. Therefore in case there are two solutions to a problem and only one solution is being obtained aigebralcally we have to re-examine. In previous question if d=2 then |
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Answer» There is only one LINE (x-axis) OUTING intercept 2 between the parallel lines. |
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| 48. |
int (1)/((1 + x^(2)) sqrt(1-x^(2))) dx= |
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| 50. |
A rectangle is inscribed in a circle with a perimeter lying along the line 3y = x+7. If the two adjacement verticles of the rectangle are (-8, 5) and (6,5), then the area of the rectangle ( in sq units) is |
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Answer» 72  Let `(-8, beta) and (6, beta)` are the coordinates of the other vertices of rectangle as shown in the figure. Since, the mid-point of line JOINING points (-8, 5) and `(6, beta)` lies on the line 2y = x + 7. `therefore 3((5+beta)/(2))=(-8+6)/(2)+7` `rArr 15 + beta=-2+14` `rArr 3beta=-3rArrbeta = -1` Now, area of rectangle `=|-8-6|xx|beta-5|` `=14+6=84` |
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