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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. |
A question paper is divided into 3 sections A, B, C containing 3,4,5 questions respectively. Find the number of ways of attempting 6 questions choosing atleast one from each section. |
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| 3. |
Roots of a quadratic equation x^(2)+5x+3=0 are 4cos^(2)alpha+a, and 4sin^(2)alpha+a. Another quadratic equation is given as x^(2)+px+q=0 where p, q in N and p, q in [1,10]. If roots of second quadratic equation are real then the probability that they are 4cos^(4)alpha + b and 4sin^(4)alpha + b is |
| Answer» Answer :A | |
| 6. |
Let z = x + iy, where x and y are integers. The area of the rectangle whose vertices are the roots of the equation zz^(3)+zbarz^(3) = 350 is |
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Answer» 32 |
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| 7. |
Verify Rolle's theorem for the following functions: f(x)= x^(3) -6x^(2) + 11x-6, x in [2, 3] |
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| 9. |
Translate "If there is a will, there is a way" propositions into symbolic form, stating the prime components |
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Answer» SOLUTION :LET p :There is a will, Q :There is a way. `:.` The given STATEMENT is `prarr q`. |
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| 10. |
Find the number of ways of permuting the letters of the word PICTURE so that no two vowels come together |
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Answer» |
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| 11. |
The range of the function , f (x) = tan^(-1) ((1+x)/(1 - x)) - tan^(-1) xis |
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Answer» `{pi//4}` |
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| 12. |
The set points on the axis of the parabola 2 ((x-1) ^(2) + (y-1) ^(2)) = (x + y) ^(2), from which 3 distinct nomals can be drawn to the parabola, is the set of points (h,k) tying on the axis of the parabola such that |
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Answer» `H GT 3` |
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| 13. |
If alpha is a root of x^(7)=1 and alpha ne 1, then alpha^(101)+alpha^(102)+......+alpha^(205) is : |
| Answer» ANSWER :A | |
| 14. |
Which of the following are true: I : int_(0)^(4) sqrt(x^(2)-4x +4)dx=4 II : int_(0)^(2) sqrt(x^(2)-4x+4)dx=2 III : int_(2)^(4) sqrt(x^(2)-4x+4)dx=2 |
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Answer» Only I, II |
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| 15. |
Solve the following equations : tan^(-1)(x/y)-"tan"^(-1)(x-y)/(x+y) is equal to |
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Answer» `(PI)/2` |
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| 16. |
If A-{a,b,c,d) mention the type of relations on A given below, which of them are equivalence relations? {(a, a),(b,b), (c,c), (d, d)} |
| Answer» Solution :Reflexive, symmetric as WELL as TRANSITIVE .Hence it is an equivalence RELATION. | |
| 17. |
Find the number of all 6-digit natural numbers such that the sum of their digits is 10 and each of the digits 0,1,2,3 occurs at least once in them. |
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| 18. |
If A-{a,b,c,d) mention the type of relations on A given below, which of them are equivalence relations? {(a, a),(b,b)} |
| Answer» SOLUTION :SYMMETRIC and TRANSITIVE but not REFLEXIVE. | |
| 19. |
If a, b, c are non-coplanar vectors and lambda is a real number, then [lambda(A+b)lambda^(2)b lambdac]=[a(b+c)b] for |
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Answer» exactly TWO VALUES of `lambda` `|{:(lambda(a_(1)+b_(1)),lambda(a_(2)+b_(2)),lambda(a_(3) + b_(3))),(lambda^(2)b_(1),lambda^(2)b_(2),lambda^(2)b_(3)),(lambdac_(1),lambdac_(2),lambdac_(3)):}|=|{:(a_(1),b_(1)+c_(1),b_(1)),(a_(2),b_(2)+c_(2) ,b_(2)),(a_(3),b_(3)+c_(3),b_(3)):}| = |{:(a_(1),a_(2),a_(3)),(b_(1)+c_(1),b_(2)+c_(2),b_(3)+c_(3)),(b_(1),b_(2),b_(3)):}|` `RARR lambda^(4)|{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|= - |{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|` `rArr lambda^(4) = - 1` So, no value of `lambda` exists. |
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| 20. |
Identify the function from the above graph |
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Answer» `tan^-1x` |
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| 21. |
Let S,T ,Ube three non voidsetsand f: S rarrTg: T rarr U so thatgof : S rarrUis surjective then |
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Answer» G and f are both surjective |
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| 22. |
If the events A and B are mutually exclusive events such that P(A) = (1)/(3)(3x+1) and P(B) = (1)/(4) (1-x) , then the set of possible values of x lies in the interval |
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Answer» [0,1] |
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| 23. |
Show that (i) sin^(-1)(2xsqrt(1-x^(2)))=2sin^(-1)x,-1/(sqrt(2))lexle1/(sqrt(2)) (ii) sin^(-1)(2xsqrt(1-x^(2)))=2cos^(-1)x,1/(sqrt(2))lexle1 |
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| 24. |
Find the set of values of p for which line 2px-4y+2pi-9=0 intersect the curve y=cos^(-1)at three distinct points. |
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Answer» Solution :We have curve `y =f(x)=cos^(-1)(2x-1)` and line 2px-4y+2pi-p=0 Domain of y =`f(x)=cos^(-1)(2x-1) is [0,1)` `f(0)=cos^(-1)=pi and f(1)=cos^(-1)=0` Graph of the function =f(x) is as shown in the following figure.  CLEARLY curve has point of inflection at `(1//2,pi//2)` Examining the line , we find that it PASSES through `(1//2,pi//2)`  This line must INTERSECT the curve atthree distinct points. Now `(dy)/(DX)=(-2)/sqrt(1-(2x-1))^(2)=(-1)/sqrt(x-x^(2))=-(x-x^(2))^(-1//2)` `therefore ((dy)/(dx))_(x=0.5)=-2` =slope of tangent to the curve at `(1//2,pi//2)` It can be verified that that points `(0,pi)` and `(1//2,pi//2)and (1,0)` are collinear slope of line joining these points is `(pi-(pi)/(2))/(0-(1)/(2))=-pi` Hence given line intersect the curve at three distinct point if its slope is less than '-2' but more than or equal, to `-pi` `therefore (p)/(2)in[(-pi,-2) rarr p in [-2pi,-4)` |
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| 25. |
The solution set of the constraints x+2y ge 11, 3x+4y le 30, 2x+5y le 30, x ge 0, y ge 0 includes the point. |
| Answer» ANSWER :C | |
| 27. |
Repeated roots : If equation f(x) = 0, where f(x) is a polyno- mial function, has roots alpha,alpha,beta,… or alpha root is repreated root, then f(x) = 0 is equivalent to (x-alpha)^(2)(x-beta)…=0, from which we can conclude that f(x)=0 or 2(x-alpha)[(x-beta)...]+(x-alpha)^(2)[(x-beta)...]'=0 or (x-alpha) [2 {(x-beta)...}+(x-alpha){(x-beta)...}']=0 has root alpha. Thus, if alpha root occurs twice in the, equation, then it is common in equations f(x) = 0 and f'(x) = 0. Similarly, if alpha root occurs thrice in equation, then it is common in the equations f(x)=0, f'(x)=0, and f'''(x)=0. If a_(1)x^(3)+b_(1)x^(2)+c_(1)x+d_(1)=0 and a_(2)x^(3)+b_(2)x^(2)+c_(2)x+d_(2)=0 have a pair of repeated roots common, then |{:(3a_(1),2b_(1),c_(1)),(3a_(2),2b_(2),c_(2)),(a_(2)b_(1)-a_(1)b_(2),c_(1)a_(2)-c_(2)a_(1),d_(1)a_(2)-d_(2)a_(1)):}|= |
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Answer» 0 `"Then "g(x)=a_(2)x^(3)+b_(2)x^(2)+c_(2)x+d_(2)=0" MUST have roots "alpha,alpha, gamma.` Then `a_(1)alpha^(3)+b_(1)alpha^(2)+c_(1)alpha+d_(1)=0"(1)"` `"and "a_(2)alpha^(3)+b_(2)alpha^(2)+c_(2)alpha+d_(2)=0"(2)"` `alpha" is alos a root of equations "f'(x)=3a_(1)x^(2)+2b_(1)x+c_(1)=0 and ` `g'(x)=3a_(2)x^(2)+2b_(2)x+c_(2)=0.` Then `3a_(1)alpha^(2)+2b_(1)alpha+c_(1)=0"(3)"` `"and "3a_(2)alpha^(2)+2b_(2)alpha+c_(2)=0"(4)"` `"Also, from "a_(2)(1)-a_(1)(2),"we have"` `(a_(2)b_(1)-a_(1)b_(2))alpha^(2)+(c_(1)a_(2)-c_(2)a_(1))alpha+d_(1)a_(2)-d_(2)a_(1)=0"(5)"` Eliminating `alpha" from "(3),(4), and (5)` we have `|{:(3a_(1),2b_(1),c_(1)),(3a_(2),2b_(2),c_(2)),(a_(2)b_(1)-a_(1)b_(2),c_(1)a_(2)-c_(2)a_(1),d_(1)a_(2)-d_(2)a_(1)):}|=0.` |
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| 28. |
Repeated roots : If equation f(x) = 0, where f(x) is a polyno- mial function, has roots alpha,alpha,beta,… or alpha root is repreated root, then f(x) = 0 is equivalent to (x-alpha)^(2)(x-beta)…=0, from which we can conclude that f(x)=0 or 2(x-alpha)[(x-beta)...]+(x-alpha)^(2)[(x-beta)...]'=0 or (x-alpha) [2 {(x-beta)...}+(x-alpha){(x-beta)...}']=0 has root alpha. Thus, if alpha root occurs twice in the, equation, then it is common in equations f(x) = 0 and f'(x) = 0. Similarly, if alpha root occurs thrice in equation, then it is common in the equations f(x)=0, f'(x)=0, and f'''(x)=0. If x-c is a factor of order m of the polynomial f(x) of degree n (1ltmltn), then x=c is a root of the polynomial [where f^(r)(x) represent rth derivative of f(x) w.r.t. x] |
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Answer» `F^(m)(x)` g(x) is polynomial of degree n-m. Then x=c is common root for the equations `f(x)=0,f^(1)(x)=0, f^(2)(x)=0,…,f^(m-1)(x)=0` where f'(x) represent rth DERIVATIVE of f(x) w.r.t. x,. |
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| 29. |
If log(1+x)-(2x)/(2+x) is increasing, then |
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Answer» `0 LT X ltinfty` |
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| 30. |
If sqrt(x+y)-sqrt(x-y)=c," then "(d^(2)y)/(dx^(2)) |
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Answer» `(1)/(y)((DY)/(DX))^(2)` |
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| 31. |
Prove by mathematical induction that n^(5)and n have the same unit digit for any natural numbern. |
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Answer» Solution :We have to prove that `n^(5) - n` is divisible by 10. For `n = 1, 1^(5) - 1 = 0` is divisible by 10. Also, `n = 2 , 2^(5) - 2 = 30` is divisible by 10. Thus, `P(1)` and `P(2)` are true. `K^(5) - k = 10m"........"(1)` Now, `(k + 1)^(5) - (k +1)` `= k^(5) + 5K^(4) + 10k^(3) + 10 k^(2) + 5 k + 1 -k - 1` `= (k^(5) - k) + 5 k(k^(3) + 1) + 10 k^(3) + 10 k^(2)` `= 10 m + 10 k^(3) + 10 k^(2) + 5k(k^(3) + 1)` [Using (1)] Clearly, `k(k^(3)+1)` is EVEN for `k in N`. Thus, `P(k+1)` is true WHENEVER`P(k)` is true, So, by the principleof mathematical induction,`P(n)` is true for any natural number n. |
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| 33. |
Which of the followingstatements is/are correct with respect to surface phenomenon ? |
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Answer» POTASSIUM ferrocyanide can cause greater coagulation in a basic dye as compared to `Na_(2)HPO_(3)`. `Na_(2)HPO_(3)rarr2Na^(+)+HPO_(3)^(2-)` higher negative charge over`[Fe(CN)_(6)]^(4-)` so it can cause more coagulation in a basic dye. (B) Starch aqua-sol is a lyophilic sol so it can be used as protective colloid. (C) Freundlichadsorptionisotherm fails at high pressure |
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| 34. |
If the straight line joining the origin and the points of intersection of y=mx+1 and x^2+y^2=1be perpendicular to each other , then find the value of m. |
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| 35. |
Shwo that the difference of the squares of the tangents to two coplanar circles from any point P in the plane of the circles varies as the perpendicular from P on their radical axis. Also, prove that the locus of a point such that the difference of the squares of the tangents from it to two given circles is constant is a line parallel to their radical axis. |
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| 36. |
intcos^5xsin^3xdx |
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Answer» SOLUTION :`intcos^5xsin^3xdx` =`intcos^5x(1-cos^2x)sinxdx` [PUT cosx=t Then sinxdx=-DT] =`intt^5(1-t^2).(-dt)=INT(t^7-t^5)dt` =`1/8t^8-1/6cos^6x+C` |
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| 37. |
Evaluate int(1)/(2sin^(2)+3sinxcosx-2cos^(2)x)dx |
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| 38. |
Find the asymptodes of the hyperbola 2x^(2)+5xy+2y^(2)-11x-7y-4=0 |
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| 39. |
For the equation (e ^(-x))/(1+x)= lamda which of the following statement (s) is/are correct ? |
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Answer» when `lamda in (0,oo)` EQUATION has 2 real and DISTINCT ROOTS |
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| 40. |
If the two adjacent sides of a parallelogram are given by hat(i)+2hat(j)+3hat(k) and -3hat(i)-2hat(j)+hat(k), then the lengths of the diagonals are |
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Answer» `2, 6sqrt(5)` |
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| 42. |
Integrate the following functions tan^2 (2x-3) |
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Answer» SOLUTION :`INT tan^292x-3) DX` =`int (sec^2(2x-3)-1)dx` =`tan(2x-3)2 - x+c` |
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| 43. |
inte^x((2+sin2x)/(1+cos2x))dx |
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Answer» SOLUTION :`I= inte^x((2+sin2x)/(1+cos2x))DX` =`inte^x((2+2sinxcosx)/(2cos^2x))dx` =`inte^x(sec^2x+tanx)dx` =`inte^xsec^2xdx+inte^xtanxdx` =`inte^xsec^2xdx+e^xtanx-inte^xsec^2xdx+c` =`e^xtanx+c` |
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| 44. |
Choose the correct answer Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is |
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Answer» `(37)/(221)` |
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| 45. |
The sides of an equilateral triangle are increasing at the rate of sqrt 3 cm/sec. find the rate at which the area of the triangle is increasing when the side is 4 cm long. |
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Answer» Solution :Let x be the LENGTH of each SIDE of an equilateral TRIANGLE : Given thatdx/dt =`SQRT 3` cm/sec. if A is the area of the triangle then `A = sqrt 3/4 x^2 rArr dA/dt = sqrt 3/4 cdot 2x dx/dt = sqrt 3/2 x cdot dx/dt` Now dA/dt](x=4) = sqrt 3/2 xx 4xx sqrt 3=6` therefore Area of the triangle is INCREASING at the rate of 6 `cm^2/sec`. |
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| 47. |
For a moderately skewed distribution |
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Answer» M.D.=Q.D. |
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| 48. |
Find the points on the curve x^(2) + y^2 – 2x – 3 = 0 at which the tangents are parallel to the x-axis. |
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| 49. |
What is the probability that 6 is obtained on one of the dice in a throw of two dice, given that the sum is 7. |
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