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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.
| 2. |
A shopkeeper manufactures gold rings and chains.The combined number of rings and chains manufactured per day is at most:(i) 16 (ii) 24 (iii)24 It takes (i) one hour (ii) half an hour (iii) one hour yo make a ring and(i) half an hour (ii) one hour (iii) half an hour for a chain.The maximum number of hours available is (i) 12(ii)16 (iii)16. If the profit on a ring is : (i)₹ 300 (ii) ₹ 100(iii) ₹ 300 and on a chain is : (i)₹ 200 (ii)₹ 300 (iii) ₹ 190. How many of each should be manufactured daily, so as to maximize profti?Form an LPP and solve it graphically. |
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Answer» (ii) 16 CHAINS ; Max.Profit = ₹ 4,800 (iii) 8 Gold Rings and 16 Chains; Max.Profit = ₹ 5,440. |
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| 3. |
Prove that the sequence 1/2, 1/2, 3/4, 1/4, 7/8, 1/8...... with the general term x_(n)={{:(,1-(1)/(2^((n+1)//2)),"if n is odd"),(,(1)/(2^(pi//2)),"if n is odd"):} has no limit. |
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| 5. |
IFsin^(16)alpha= 1/5then thevalue of(1) /(cos ^2 alpha) +(1)/( 1 + sin ^2alpha) +(2)/(1+ sin^4alpha) + ( 4 )/( 1+ sin ^8 alpha) isequalto |
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Answer» 2 |
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| 6. |
If x[(2),(3)]+y[(-1),(1)]=[(10),(5)], find the value of x and y. |
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| 7. |
I=int sec^(-1)((1-x)/(1+x))dx |
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| 9. |
9^11 + 11^9 is divisible by |
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Answer» 7 |
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| 10. |
On a diagram mark the area bounded by the parabola y^(2) =4xand the circle x-4 =4cos theta ,y =4sin thetaabove the x-axis and obtain the area by integration |
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| 11. |
Evaluate: int_(-log2)^(log2) e^(-|x|) dx. |
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| 12. |
Differentiatex^2(1+x)(2-x) |
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Answer» Solution :`y=x^2(1+x)(2-x)` `dy/dx=2x(1+x)(2-x)+x^2(2-x)+x^2(1+x)(-1)` `[therefore/dx(uvw)=(du)/dx CDOT V cdot w+u(DV)/dx cdotw+u cdotv(dw)/dx` `4x+2x^2-2x^3-2x^2-x^3-x^2-x^3` `=4x+3x^2-4x^3` |
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| 14. |
If the circle x^(2) + y^(2) + 2x + 3y + 1 = 0 cuts another circle x^(2) + y^(2) + 4x + 3y + 2 = 0 in A and B, then the equation of the circle with AB as a diamter is |
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Answer» `x^(2) + y^(2) + x + 3y + 3 = 0` `S = x^(2) + y^(2) + 2x + 3y + 1 = 0` and `S^(1) -= x^(2) + y^(2) + 4x + 3y + 2 = 0` `implies` EQUATION of radical axis is `S - S^(1) = 0` -2x - 1 = 0 2x + 1 = 0 Equation of required circle is `implies (x^(2) + y^(2) + 2x + 3y + 1) + lambda (2x + 1) = 0` Centre, `C = (-1 - lambda, (-3)/(2))` LIES on equation (1) then `implies - 2 - 2 lambda + 1 = 0` `implies 2 lambda = - 1` `implies lambda = (-1)/(2)` Equation (2) `implies x^(2) + y^(2) + 2x + 3y + 1 - (1)/(2) (2x + 1) = 0` `implies 2x^(2) + 2y^(2) + 4x + 6y + 2 - 2x - 1 = 0` `:. 2x^(2) + 2y^(2) + 2x + 6y + 1 = `, is required circle |
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| 15. |
If 2% of a given lot of manufactured parts are defective then find the probability that a sample of 100 items has no defective item. |
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| 16. |
If the k^(th) term is the middle term in (x^2 - (1)/(2x))^20 find T_k and T_(k+3) |
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| 17. |
The maximum value of f(x)=((1)/(x))^(x) is ………… |
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Answer» E |
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| 18. |
If n is odd then C_0^2 - C_1^2+C_2^2-…....+(-1)^nC_n^2= |
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Answer» 0 |
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| 19. |
If [alpha]+[beta]+[gamma]=7 then the values of c, where [.] represent the greates integer function are, |
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Answer» (-20,-16) `therefore g(X)=3(x-2)(x-4)` Singn scheme of g(X)  for three real roots of f(X)=`x^(3)-9x^(2)+24+c=0` in the INTERVAL (-20,-16) `f(x)=clt0` `f(1)=1-9+24+c=c+16lt0 for forall c in (-20,-1)` `f(2)=8-36+48+c=c+20gt0` `alphain (1,2) or [alpha]=1` `thereforebeta in (3,4)if c in (-18,-16)` now `f(4)=64-144+96+c=16+clt0 if CIN (-18,-16)` or `betan in (3,4) if c in (-18,-16)` Now `f(4)=64-144+96+c=16+clt0 forall c in (-20,16)` or ` gamma(4,5) or [gamma]=4` Thus `[alpha]+[beta]+[gamma]={{:(1+2+4-20ltclt-18),(1+3+4,-18l,tclt-16):}` Now if `c in (-20,-18)` `alphain (1,2) beta (2,3)gamma in (4,5)` or `[alpha]+[beta]+[gamma]=7` If `c in (-18,-16),alpha in (1,2) , beta (3,4) gamma in (4,5) then [alpha+[beta]+[gamma]=8` |
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| 20. |
If f : R rarr Rbe given by f(x) = tan x , then f^(-1)(1) is ....... |
| Answer» Solution :N/A | |
| 21. |
If the probability that a fluorescent light has a useful life of at least 600 hours is 0.9, find the probabilities that among 12 such lights exactly 10 will have a useful life of at least 600 hours, |
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| 22. |
If the probability that a fluorescent light has a useful life of at least 600 hours is 0.9, find the probabilities that among 12 such lights at least 11 will have a useful life of at least 600 hours, |
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| 23. |
A and B are two such thatP(A) != 0 , P(B//A) if : (i) A is subset of B (ii) A cap B = phi are respectivey : |
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| 24. |
Let a be anon zero vector . If X= hati xx(axxhati),y =hatjxx(axx hatj) and Z= hatk xx(axxhatk)then [x y z] = |
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Answer» |a| |
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| 25. |
Letf(x) = a_(0) + a_(1)x + a_(2)x^(2) + …+ a_(2n) x^(2n) and g(x) = b_(0) + b_(1) x + b_(2)x^(2) + …+ b_(n-1) x^(x-1) + x^(n) + x^(n+1) + …+ x^(2n) . Iff(x) = =g (x + 1) , finda_(n) in terms of n. |
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| 26. |
The ratio of the number of girls to the number of boys in a certain class is 3:5. If there is a total of 32 students in the class, how many girls are in the class? |
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| 27. |
By the method of integration find the area of the triangle formed by the straight linesx+ 3y -8 =0, 5x -y -8=0andx-y+ 4=0 |
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| 28. |
Six X's have to be placed in the squares of the figure below, such that each row contains atleast one X, this can be done in |
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Answer» 24 ways |
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| 29. |
Express the following as trigonometric ratios of some acute angles. tan 458^@ |
| Answer» Solution :`tan 458^@ = tan (5(pi)/2 + 8^@) = -COT 8^@` | |
| 30. |
The sum of values of 'x' satisfying the equation ||x-1|-2|=1 is :- |
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Answer» 4 |
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| 31. |
Evaluate the following integrals. int(1)/(sin(x-a)sin(x-b))dx |
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| 32. |
Using integration, find the area of the triangular region whose sides have the equations y = 2x + 1, y = 3x + 1 and x = 4. |
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| 33. |
If locus of point P(z) in complex plane is |z+z_(1)|+|z+z_(2)|=4 where A represents z_(1) as (1,0) and B represents z_(2) as (-1, 0) and Q(omega) is moving point inside the locus of P(z) such that all internal angle bisectors of triangle /_\PAB concurrent at Q. Then, answer the following questions if |omega-omega_(1)|+|omega-omega_(2)|=2 If minimum value fo |omega-z_(1)|+|omega-z_(2)| is equal to m, then [m] is (where [.] denotes greatest integer part) |
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Answer» 1 Now, `2ae=2xxsqrt(2/3)` `|z_(1)-z_(2)|=2` `:.[m]=2( :' m LT |z_(1)-z_(2)|)` |
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| 34. |
Check the injectivity and surjectivity of the following function . f: [-1,1] rarr [-1,1] , f(x) = x |x| |
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| 35. |
Integrate the function is exercise. sqrt(1+(x^(2))/(9)) |
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| 36. |
Ify=x^((lnx)^ln(lnx)) , then (dy)/(dx) is equal to |
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Answer» `(y)/(X)(LNX^(lnx-1)+2ln+(lnx))` |
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| 37. |
If P (A) = 0.4,P (B | A)= 0.3 and P (B^c | A^c)=0.2. find P(A cup B) |
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Answer» <P> SOLUTION :`P(A CUP B)=P(A)+P(B)-P(ACAPB)` |
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| 38. |
Find the sine of the angle between the vectors bar(a)=3hati+hatj+2hatk and bar(b)=2hati-2hatj+4hatk. |
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| 39. |
Solve the inequation log_(x^(2)+2x-3)((|x+4|-|x|)/(x-1)|gt0 |
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Answer» The given inequation is valid for `(|x+4|-|x|)/((x-1))gt0` and `x^(2)+2x-3gt0,!=1`…….i Now consider the following cases: CASE I If `0ltx^(2)+2x-3lt1` `implies4lt(x+1)^(2)lt5` `implies-sqrt(5)lt(x+1)lt-2` or `2ltx+1ltsqrt(5)` `implies-sqrt(5)-1ltxle-3` or `1ltxltsqrt(5)-1` `:.x epsilon (-sqrt(5)-1,-3)uu(1,sqrt(5)-1)`...ii Ten `(|x+4|-|x|)/((x-1))lt1` Now `xlt-4`, ten `(-(x+4)+x)/((x-1))lt1` `implies 1+4/(x-1)gt0` `implies((x+3))/((x-1))gt0` `:.x epsilon (-oo,-3)uu(1,oo)` `impliesx epsilon (-oo,-4)[:' x lt -4]`.iii `-4lexlt0`, then `(x+4+x)/((x-1))-1lt0` `implies((x+5))/((x-1))LT0` `:.x epsilon (-5,1)` `impliesx epsilon [-4,0)[:'-4lexlt0]` ..iv and `xge0` then `((x+4-x)/((x-1))lt1` `implies1=4/(x-1)gt0` `implies((x-5))/((x-1))gt0` `:.x epsilon (-oo,1)uu(5,oo)` `impliesx epsilon [0,1)uu(5,oo)[ :' x ge0]`.......v From eqs (iii), (iv) and (v) we get `x epsilon (-oo,1)uu(5,oo)` .vi Now common valuesin Eq. (ii) and (iv) is `x epsilon (-sqrt(5)-1,-3)`..vii Case II If `x^(2)+2x-3gt1` `impliesx^(2)+2x+1gt5implies(x+1)^(2)gt5` `impliesx+1lt-sqrt(5)` or `x+1gtsqrt(5)` `:.x epsilon(-oo,-1-sqrt(5))uu(sqrt(5)-1,oo)`........viii Then `(|x+4|-|x|)/((x-1))gt1` Now `xlt-4`, then `(-4)/(x-1)gt1` `implies1+4/(x-1)lt0` `implies (x+3)/(x-1)lt0` `:. x epsilon (-3,1)` Which is false `[:' x lt-4]` `-4lexlt0` then `(2x+4)/((x-1))-1gt0` ltbRgt `implies((x+5))/((x-1))gt0` `:.x epsilonn (-oo,-5)uu(1,oo)` Which is false `[:'-4lexlt0]` and `xge0` then `4/(x-1)gt1` `implies1-4/(x-1)lt0` `implies(x-5)/(x-1)lt0` `:.x epsilon (1,5)`....ix which is false `[:'xge0]` Now comon values in Eq (viii) and (ix) is `:. x epsilon (sqrt(5)-1,5)` .......x Combinin Eqs (viii) and (x) we get `x epsilon (-sqrt(5)-1,-3)uu(sqrt(5)-1,5)` |
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| 40. |
Coefficient of x^n in 1+(a+bx)/(1!) + ((a+bx)^2)/(2!)+ ....oo = |
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Answer» `e^a/(N!)` |
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| 41. |
Equation of line passing through a point (1, 4) and the sum of the intercept on the positive axes is minimum is ………….. |
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Answer» `2x+y-6=0` |
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| 42. |
Matrices X and Y are inverse of each other then........ |
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Answer» `XY=I, YX=-I` |
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| 43. |
Find the equation of normal to the curve x^((2)/(3))+y^((2)/(3))=a^((2)/(3)) at ((a)/(2sqrt(2)),(a)/(2sqrt(2))). |
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| 44. |
Which of the following compound statements are false? p : 17 is odd and prime number q :23 is even and prime number r : Volume of cube isa^3and total surface areais 4a^2where a is the length of side of a cube |
| Answer» Answer :B | |
| 45. |
If A = ((2,52,152),(4,1-6,358),(6,162,620)) then the determinant of adj (2A) is equal to |
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Answer» 64 |
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| 46. |
How many numbers between 10 and 10,000 can be formed by using the digits 1, 2, 3, 4, 5 if Digits can be repeated |
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| 47. |
Find x such that the four points A(3, 2, 1), B(4, x, 5), C(4, 2, -2) and D(6, 5, -1) are coplanar. |
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| 48. |
Let A and B .............. |
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Answer» `implies P(AUUB)'=0.4` `implies P(A UU B)=0.6` `P(A)+P(B)-P (A nn B)=0.6` `implies P(A nn B)=0.1` ltbgt `P(A). P(B) NE P(A nn B) implies A & B` are not INDEPENDENT events `P(A/bar(B))=((A nn bar(B)))/(P(bar(B)))=(P(A)-P(A nn B))/(1-P(B))=0.2/0.6=1/3` `P(B/bar(A))=(P(B nn bar(A)))/(P(bar(A)))=(P(B)-P(A nn B))/(1-P(A))=0.3/0.7=3/7` |
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