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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. |
\log _{2} \log _{2} \log _{4} 256+2 \log _{2} 2 |
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| 2. |
( i ) ^ { x + 2 } = 256 |
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Answer» 2^(x+2) = 256=> 2^(x+2) = 2^8 => x+2 = 8=> x = 8-2 = 6 |
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| 3. |
() 270(0) (2x3)8,39 (2(i) (4),佳)1(i) ( ).(i) [訂ずー3 |
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Answer» (2x3)^(-2)6^(-2)1/6^21/36 |
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| 4. |
49 y ^ { 2 } + 84 y z + 36 z ^ { 2 } |
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| 5. |
-49*b^2 %2B 4*a^2 - 12*a %2B 9 |
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| 6. |
Show that 11 - V2 is an irrational, if given that v2 is an irrational. |
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Answer» We know √2 is irrational. Let 11 - √2 be rational then 11 - √2 = r , where r is rational 11 - r = √2 This is an absurd result as LHS is rational and RHS is irrational . So, a contradiction to our assumption hence 11 - √2 is irrational |
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| 7. |
27. यदि 4 x b = |a-b | तो 2x3-5x7 का मान होगा : |(a) -1 (b) 1(c) 3(d) -2 |
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Answer» a) -1 is the correct answer -1 correct answer hai bhai |
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| 8. |
Show that V2 is irrational. |
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Answer» Let us assume that √2 is rational Thus √2= a / b where a,b are integers b ≠ 0we also suppose that a / b is written in the simplest form Now √2 = a / b 2 = a2/ b2 2b2= a2 ∴ 2b2is divisible by 2 a2is divisible by 2 a is divisible by 2 ∴ let a = 2ca2= 4c2 2b2= 4c2⇒b2= 2c2 Thus 2c2 is divisible by 2 Thus b2 is divisible by 2 Thus b is divisible by 2 Thus a are b are divisible by 2 . This contradicts our supposition that a/b is written in the simplest form Hence our supposition is wrong Hence √2 is irrational number. |
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| 9. |
4. Obtain all other zeroes of the polynomial x-3/2x3+ 3x2 +3,2x-4, if two of its zeroes are2 and 22 |
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Answer» P(x) = x⁴ - 3 √2 x³ + 3 x² + 3√2 x - 4 given that (x - √2) , (x - 2√2) are factors of P(x) as √2 and 2√2 are two zeroes of P(x) = 0. (x -√2) (x - 2√2) = x² -3√2 x + 4 is a factor of P(x). let A be a constant. We can write the constant term in the second factor by : -4/4 = -1... dividing the constant terms. let (x² - 3√2 x + 4) (x² + A x -1 ) = P(x) Now compare the coefficients of x³ : A - 3√2 = -3√2 => A = 0 coefficient of x : 4A + 3√2 = 3√2 => A = 0 so the other factors are : x² - 1 = 0 so x = 1 and -1 are the other factors. This method of multiplying the factors and comparing coefficients is simple. |
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| 10. |
\int_{-1}^{1} x^{17} \cos ^{4} x d x=0 |
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| 11. |
b^2-68b+256 |
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| 12. |
(256)^{-\left(4^{-3 / 2}\right)} |
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| 13. |
174x+31-x |
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| 14. |
w that the quadrilateral formed bying the mid-points of the pairs ofjacent sides of a square is a square. |
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| 15. |
(a)256รท16รท2รท 18รท9ร2 |
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Answer» 6204÷94 |
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| 16. |
when 2^256 is divided by 17 the remainder will be?? |
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| 17. |
[32 x V2 x I3 x «ठि] बराबर है |
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Answer» (2)^1/3 * (2)^1/2 * (3)^1/3 * (3)^1/2 = (2)^[1/3 + 1/2] * (3)^[1/3 + 1/2] = (2)^5/6 * (3)^5/6 |
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| 18. |
arg 49 x ^ 2 %2B 4 x %2B 8 |
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Answer» sum of zeros = -b/a = -4/1 = -4 |
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| 19. |
( 4 x %2B 3 y ) ^ 2 - 49 z ^ 2 |
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Answer» (4x + 3y)^2 - (49z)^2 Using a^2 - b^2 = (a + b)(a - b) = (4x + 3y + 7z)(4x + 3y - 7z) |
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| 20. |
Q2. Find the roots of 6x2 – V2 x - 2 = 0. |
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| 21. |
a^2 %2B 14*a %2B 49 |
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Answer» a^2+14a+49a^2+7a+7a+49a(a+7)+7(a+7)(a+7)(a+7)a= -7,-7 |
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| 22. |
sec r(a)fcs sccxrdr.dlxINCERTIre |
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Answer» ∫(sec2x/cosec2x)dx∫(1/cos2x)/(1/sin2x)dx ∫(1/cos2sin) . (sin2x/1)dx ∫tan2x dx ∫(sec2x-1) dx tanx - x + c tisri len समझाओ tisri len samjhao |
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| 23. |
Is V2 x V 8 irrational ? Justify your answer ? |
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| 24. |
Find the number of positive integral solutions of abc= 45 |
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| 25. |
Let y = V2 + V2 , z-J2-12 and x = y + z then the value of x is(A) 2 y(C) 2z(B) 2y(D) 4 |
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| 26. |
Q 17: Two zeroes of the polynomial p x)remaining zeroes of P(x), if any2x , - 3x33x2 + 6x2 are v2 andV2. Find the |
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Answer» your handwriting is nice.keep up👍 |
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| 27. |
13. Solve the equation using the method of factorization2r+3 2x3 172x32x 3 432OR |
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Answer» x=32 jit gur har sar sar sae |
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| 28. |
(x + 3) (1-x) = 17,x:W, 2 .(x-2) x 4'23,Solve for x: |
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| 29. |
3. Using the properties of proportion, solve for4x"+117x, given%2[2013] |
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| 30. |
as co-ordinate axes, prove that theABCD is a square of side a. AB and AD are takenequation of the circle circumscribing the square isa(r) |
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| 31. |
3y +4=5y-4 |
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Answer» PLEASE LIKE THE SOLUTION |
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| 32. |
(2x-3y)+(5y-3x) |
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Answer» 2x-3y+5y-3x= -x+2y is the correct answer |
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| 33. |
how✓2is irrational? |
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Answer» Let √2 = a / b where a,b are integers b ≠ 0 We also suppose that a / b is written in the simplest form.Now √2 = a / b ⇒ 2 = a2/ b2 ⇒ 2b2= a2∴ 2b2is divisible by 2 ⇒ a2is divisible by 2 ⇒ a is divisible by 2 ∴ let a = 2ca2= 4c2⇒ 2b2= 4c2 ⇒b2= 2c2 ∴ 2c2 is divisible by 2 ∴ b2 is divisible by 2 ∴ b is divisible by 2 ∴ a and b are divisible by 2 This contradicts our supposition that a/b is written in the simplest formHence our supposition is wrong ∴ √2 is irrational number. |
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| 34. |
5y+10=3y+4 |
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Answer» 5y+10=3y+45y-3y=4-102y= -6y= -6/2= -3y= -3 is answer 5y+10=3y+45y-3y=4-102y=-6y=-6/2y=-3 5y+10=3y+45y-3y=4-102y=-6y=-3 Answer 5y + 10 = 3y + 45y - 3y = 4 - 102y = -6y = -3 is right answer |
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| 35. |
x2-1/2 | = a, show that dy=x(1-tana)x+y2 ,2EXAMPLE 12 If tan-dxy(1+ tan a) |
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Answer» If you like the solution, Please give it a 👍 |
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| 36. |
52. The equation of a linewhich is equidistant from the linesy -8andy--2is(d) none of these(a) y 6(b) y 4(c)y 3 |
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Answer» required line is equidistant from y=-2 and y=8now total distance between both the line is 10so half of it is 5so resultant line is y=3 |
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| 37. |
2. Which is the smallest fraction that must be add to the square root of 5 4 9 so that theresulting number is a positive whole number?2 |
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Answer» Question says that which is the smallest fraction that must be subtracted from the square root of 5 4 9 so that the resulting number is a positive whole number. square root of 549 is approx 23.4307490277 That means nearest positive square root value is 23. Square of 23 is 529. Difference between 549 and 529 is 549-529=20 Hence required smallest fraction is 20 or you can also write 20/1 in fraction form. thank a lot.. wish I could also help but I'm in a situation like this.. So sorry |
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| 38. |
Integration (tan^4 x sec^2 x dx |
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| 39. |
5. The smallest positive integral valie of n for whichis purelyimaginary with positive imaginary part, is(a) 1(c) 5(b) 3(d) none of these |
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| 40. |
ind the area of a rhombus if its vertices are (3, 0), (4, 5), (-1,4) and (-2,-1) taken inorder. Hint : Area of a rhombus 5 (product of its diagonals) |
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| 41. |
The perimeter of the ends of a frustum of a cone are 44 cm and 8.4 π cm. If the depthis 14 cm., then find its volume. |
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Answer» Given, Perimeter, P1= 2πr1= 44 cm Then, r1=44/2π=22×7/22= 7 cm Perimeter, P2= 2πr2= 8.4π cm Then, r2=8.4π/2π= 4.2 cm Given: h = 14 cm Volume =1/3πh(r12+ r22+ r1r2) =1/3x π x 14 [(7)2+ (4.2)2+ (7 x 4.2)] =1/3x22/7x 14(49 + 17.64 + 29.4) = 1408.58 cm3 |
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| 42. |
Find the area of a rhombus if its vertices are (3,0), (4, 5), (-1,4) and (2,-1) taken inorder. [Hint : Area of a rhombus(product of its diagonals)] |
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Answer» thanki |
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| 43. |
Find the area of a rhombus if its vertices are (3,0), (4, 5),(-1,4) and (-2,-1) taken inorder. [Hint: Area of a rhombus (product of its diagonals)] |
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| 44. |
make a squareAB=4BC=5CD=7DA=8 |
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Answer» A square has all the sides.. equal , so it is not possible to make a square.. with the given dimensions it is not possible because it has all sides equal |
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| 45. |
9 ^ { n } - 8 n - 1 \text { is divisible by } 64 |
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| 46. |
11. ज्ञात कीजिए: () 642 |
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| 47. |
\text { add }\4 y \left( 3 y ^ { 2 } + 5 y - 7 \right) \text { and } 2 \left( y ^ { 3 } - 4 y ^ { 2 } + 5 \right) |
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| 48. |
1. Find :(642(1) 325(iii) 1253 |
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| 49. |
1. ज्ञात कीजिए: (i) 642 |
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| 50. |
1. Find:(i) 642 |
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Answer» √64√2*2*2*2*2*22*2*2= 8 thanks |
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