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. |
3+3V2-3+2V21 |
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Answer» it's answer is 5✓2 it's the correct answer 3+3√2-3+2√2=(3-3)+(3√2+2√2)=0+5√2=5√2 5√2 is the correct answer 5√2 at the end it will remain |
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| 2. |
Show that 3V2 is irrational. |
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Answer» Let us assume that 3√2 is rational ie,3√2=a/bwhere a and b are co-prime numbers √2=(a/b)(1/3)=a/3bHere √2 is irrational and a/3b is rational Thus our assumption is wronghence 3√2 is irrational |
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| 3. |
Example 11 : Show that 3V2 is irrational. |
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Answer» let 3 root 2 be rational number.3 root 2 = p/qroot 2 = p/3 q............. (1) from equation 1, we conclude thatp/3 q are integers but root 2 is not a integer.so, our supposition is wrong.hence, 3 root 2 is irrational number. |
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| 4. |
Rationalise the denominator of7 + 3V2 |
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| 5. |
hat (5 +3v2) is an irrational number.1SGiven that 2 is irrational. prove t |
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Answer» Given √2 is irrational Let assume 5+3√2 be rational. 5 + 3√2 = p/q 3√2 = p/q -5 √2 = (p - 5q)/5 So, LHS is irrational and RHS is rational.So, 5+3√2 is irrational |
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| 6. |
5. (cos2 670 - sin2 23) ?What is the value of (cos? 670- sin2 230)?(5+3v2 |
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Answer» We know that,cos²B - sin²A = cos(A + B) * cos ( A - B) Now, cos²67° - sin²23° = cos(67 + 23 ) * cos( 67 - 23 )= cos ( 90 ) * cos45= 0 * 1/√2 .= 0 cos²67 -sin²23 = 0 . Quick Alternative : [ cos²67-sin²23 = cos²(90-23) - sin²23 = sin²23 - sin²23 = 0 ] [ cos²67-sin²23 = cos²67-sin²(90-67) = cos²67-cos²67 = 0 ] Like my answer if you find it useful! |
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| 7. |
find the values of a and b in each of the following- a +b 353+7 3-73v2-2V3NCERT Exemplar Problems] |
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Answer» Thank you so much |
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| 8. |
(sin A+sec A) 2+(cos A+cosec A)2=cot4A +tan 4A |
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Answer» It is wrong please send the question again is it square of (sin A + sec A) ordouble of (sin A + sec A) ⬜ square |
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| 9. |
sum as 6973.which d 7 and 22nd term is 149.difference of the given AP. Then,Hence, there are 38 terms in the AP having their7. Find the sum of first 22 terms of an AP inSol. Let a be the first term and d the commond7 and α22-149a + (22 1)d 149 |
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Answer» Given d = 7,22nd term =149 ,⇒ a+21d = 149⇒a+21(7) = 149⇒a+147 = 149⇒a=149-147=2Sn = (n/2)(2a+(n-1)d)S₂₂ = (22/2)(2×2+(22-1)7)S₂₂ = 11(4+21(7))S₂₂ = 11(4+147)S₂₂ = 11(151)S₂₂ = 1661. |
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| 10. |
. What is meant by latent heat? How willstate of matter transform if latent heais given off?the |
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Answer» Latent heat is thermal energy released or absorbed, by a body or a thermodynamics system, during a constant-temperature process — usually a first-order phrase transition. If we think about substances at a molecular level, gaseous molecules have more vibration than liquid molecules. So when you add heat to a liquid, you are actually causing the molecules to vibrate. The latent heat is the energy required to change the molecular movement. Each substance has a unique latent heat value. state of matter transforms from gas to liquid and from liquid to solid if the latent heat is given off. |
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| 11. |
646-64+164+55-5484 |
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Answer» 646-64+164+55-5484=865-5548= -4619 is correct answer. |
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| 12. |
Q. 9 The square root of (7 + 3V5)(7- 3V5) is |
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Answer» sqrt[(7 + 3(5)^1/2 * (7 - 3(5)^1/2)] Using (a + b)(a - b) = a^2 + b^2 = sqrt(7^2 - 3^2×5) = sqrt(49 - 45) = sqrt(4) = 2 |
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| 13. |
7. If x = 3 - 2V2, find the value of VX + |
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Answer» 2 is the correct answer of the given question |
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| 14. |
032. Find the Inverse Laplace transform of the |
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Answer» hii tum kaha se ho ye sirf chat krne ke liye bnaya gaya hai |
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| 15. |
A:12+3V52-3V5atbu5find the values ofa and b. |
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Answer» the answer therefore is 40 a is 3 root 5 b 8s -3root 5 photo is not clear and it is of icic.board it will be -3 not sure Hence, we gota= -49/41b= -21/41 We need to find the values ofaaandbb, such that the polynomialx3−10x2+ax+bx^3-10x^2+ax+bis exactly divisible byx−1x-1as well asx−2x-2 Assume that :f(x)=x3−10x2+ax+bfx=x3-10x2+ax+b Using remainder theorem, we know that if a polynomialf(x)f(x)is divided byx−cx-c, the remainder isf(c)f(c)x−1=0⇒x=1x-1=0⇒x=1 Applying the factor theorem, we know that: f(x)will be exactly divisible by(x−1)iff(1)=0fxwill be exactly divisible byx-1iff1=0. Hence, we have: f(1)=13−10×12+a×1+b=(1−10+a+b)=−9+a+bf1=13-10×12+a×1+b=1-10+a+b=-9+a+b∴f(1)=0⇒a+b=9...(1)f1=0⇒a+b=9...1 The same method must be applied for the second factor as well,x−2=0⇒x=2x-2=0⇒x=2 Similarly, applying the factor theorem gives us: f(x)will be exactly divisible by(x−2)iff(2)=0fxwill be exactly divisible byx-2iff2=0. Hence, we have: f(2)=23−10×22+a×2+b=(8−40+2a+b)=−32+2a+bf2=23-10×22+a×2+b=8-40+2a+b=-32+2a+b∴f(2)=0⇒2a+b=32...(2)f2=0⇒2a+b=32...2 Solving simultaneously, by subtracting (1)from (2),we have:a=23Solving simultaneously, by subtracting (1)from (2),we have:a=23 Substituting the value ofainto11or22, we get the value ofbwhich is−-14.∴a= 23 andb=−-14 hence,wegot-49/41a -21/41 a= -49/41b= -12/41it is the right answerhope it helps u a=-49/41 b= -21/41 this is the answer a is 3 root 5b 8s-3root 5 a is 3 root 5b 8s -3root 5 first of all 2 + 3√5------------ = a + b√52 - 3√5 then2 + 3√5 2 + 3√5------------ x ---------- = a + b√52 - 3√5 2 + 3√5 4+ 45 +12√5------------------- = a + b√5 4 - 45 49 12√5---- - ----- = a + b√5-41 41 now comparison the both sides a = -49/41b = -12/41 A is 3 root 5 B is 8 -3 root 5 this is very right answer a= -49/45. and b=-12/41 this is write answer value of a is 49and value of b is 12 a=(-49/41)B=(-21/41) |
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| 16. |
Risjie byl 1 €67 2 B2 Lap P |
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Answer» STEPS: 1: On a numberline mark AB = 9.3 unit & BC= 1 unit. 2: Mark O the mid point of AC 3: Draw a semicircle with O as centre & OA as radius 4: At B draw a perpendicular BD. 5: BD = √9.3 unit 6: Now, B as centre, BD as radius, draw an arc, meeting the numberline at E. Now, with BD or BE = √9.3 as radius , with 0 ( origin) of the number line as centre, draw an arc on the the number line, intersecting at point ‘k’. And this point ‘k' lies between integers 3 & 4, & represents √9.3 JUSTIFICATION: BD = √ {(10.3/2)² - (8.3/2)²} => BD = √{10.3²- 8.3²)/4 } => BD = √{(10.3+8.3)(10.3–8.3)/4} => BD = √{18.6*2/4} => BF = √{37.2/4} => BD = √9.3 = BE |
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| 17. |
12 Lap सका कण थी ) pe TR T fld%_ |
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Answer» I help you in english hit like when it help you please like |
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| 18. |
2. Find the square of:3v5 |
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Answer» 1) (3√3/5)^2 = 27/25 |
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| 19. |
Prove that 2-3v5 is an irrational number. |
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| 20. |
Add 3v2 + 5v3 2V5 and 3V5 - 2-2V3 |
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Answer» 3√2 + 5√3 -2√5 +3√5 -√2 -2√3=2√2 +3√3 +√5 hit like if you find it useful |
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| 21. |
EXERCISE 2.1)State the signs of trigonometric functionscos((i) sin (159°(ii)(111) tan (250°)(iv)3πsec|- |
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Answer» Thank |
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| 22. |
ed Examples:: If tan theta= -2 and theta lies in the second quadrant,find the values of other trigonometricfunctions. |
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| 23. |
TRIGONOMETRIC FUNCTIONS8. Prove that Sec8A 1 tan 8Asec 4A-1 tan 2A |
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Answer» L.H.S. = (sec 8A -1) / (sec 4A -1)=> [(1 - cos 8A)/ cos 8A] / [(1 - cos 4A)/ cos 4A]=> [2 sin² 4A / 2 sin² 2A] * [cos 4A / cos 8A]=> [(2 sin 4A * cos 4A) * sin 4A / cos 8A] / [2 sin² 2A] => [(sin 8A / cos 8A) * sin 4A] / [2 sin² 2A]=> tan 8A * (sin 4A / 2 sin² 2A)=> tan 8A * (2sin 2A * cos 2A / 2 sin² 2A)=> tan 8A * ( cos 2A / sin 2A)=> tan 8A * cot 2A => tan 8A / tan 2A ==> R.H.S.
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| 24. |
TRIGONOMETRIC FUNCTIONSEXERCISE 3.3t that:in cos6-an 422. 2sincosec70Find the value ofa (sin 75(11) tan 15° |
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| 25. |
caplace transform of to 45 cost. |
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| 26. |
Solutions of the equation\sqrt{5} x^{2}-8 x+3 \sqrt{5}=0 are(A) 5, 3(c) 5, 3V5(B) 5, 3V5(D) 4, 2 |
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Answer» root(5)x^2-5x-3x+3root(5)=0root(5)x(x-root(5))-3(x-root(5))=0(root(5)x-3)(x-root(5))=0so x=root(5) or 3/root(5) |
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| 27. |
2V2V5 = 2.236)Find:646ii) 325iv)162v) 2435 |
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Answer» 64^1/6now2^6=64 64^1/6=232^1/5now2^5=32 hence 32^1/5=2 16^3/2=(16)^1/2)^3=4^3=64 |
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| 28. |
wrlte uieTauoIIf x = 3 + 2V2, then find the value of VX |
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Answer» l |
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| 29. |
14. Find the value of sin(2tan-14)-costan-126)14. Find the value of sin 2 tan+cos(tan 2V2) |
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| 30. |
)Nametheproductobtairedinchlor-alkalĂprocess.) Name the gases liberated at anode and at cathode respectively. |
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Answer» The products obtained in chlor - alkali process aresodium hydroxide,chlorinegas, andhydrogengas. # the gas liberated at anode ischlorineand gas liberated at cathode ishydrogen |
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| 31. |
3+2V22 + 1ă2-1&. Show that= |
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| 32. |
7. The angles of a quadrilateral are in the ratio 1:2:3:4. Find the smallest and the largest angle.8. Show that23+2V2 |
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Answer» 7. Let the angles of quadrilateral be x, 2x, 3x , 4x sum of all angles of quadrilateral = 360° x + 2x + 3x + 4x = 360° 10x = 360° x = 360/10 = 36° Ans. the four angles are :- x = 36°2x = 2 × 36 = 72°3x = 3 × 36 = 108°4x = 4 × 36 = 144° 8. √2+1/√2-1 = √2+1/√2-1 * √2+1/√2+1 = (√2+1)²/(√2²-1²) = √2² + 2√2 + 1/2-1 = 2+1+2√2 = 3+2√2 Like my answer if you find it useful! |
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| 33. |
2.(26 +2 v762 2v2is equal to(4) 0(B) 2/2(D) 217 |
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| 34. |
Find the inverse laplace transform of 1/(s-a)^5 |
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| 35. |
Laplace theorem |
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Answer» the Laplace transform is an integral transform named after its inventor Pierre-Simon Laplace. It takes a function of a real variable t (often time) to a function of a complex variable s (complex frequency). The transform has many applications in science and engineering. The Laplace transform is similar to the Fourier transform. While the Fourier transform of a function is a complex function of a real variable (frequency), the Laplace transform of a function is a complex function of a complex variable. Laplace transforms are usually restricted to functions of t with t ≥ 0. A consequence of this restriction is that the Laplace transform of a function is a holomorphic function of the variable s. Unlike the Fourier transform, the Laplace transform of a distribution is generally a well-behaved function. Techniques of complex variables can also be used to directly study Laplace transforms. As a holomorphic function, the Laplace transform has a power series representation. This power series expresses a function as a linear superposition of moments of the function. This perspective has applications in probability theory. |
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| 36. |
write ĂĽll tIUI 2HanaA v2-1, then proveA cos A =2v2 |
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Answer» TanA=√2-1or, tanA=(√2-1)(√2+1)/(√2+1)or, tanA=(2-1)/(√2-1)or, tanA=1/√2-1∴, tanA+1/tanA=1/√2-1+√2-1or, (tan²A+1)/tanA=[1+(√2-1)²]/(√2-1)or, sec²A/tanA=(1+2-2√2+1)/(√2-1)or, (1/cos²A)/(sinA/cosA)=(4-2√2)/(√2-1)or, 1/sinAcosA=(4-2√2)(√2+1)/(√2-1)(√2+1)or, 1/sinAcosA=(4√2-4+4-2√2)/(2-1)or, 1/sinAcosA=2√2or, sinAcosA=1/2√2 |
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| 37. |
V2 + 1-3+2V22-18. Show that |
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| 38. |
(A) 2 12/22-12V21(B)(C)(D)V2+12/2 |
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| 39. |
Find the final value of the function whose Laplace transformF(s)=\frac{s+6}{s(s+3)} |
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| 40. |
Inverse Trigonometric Functionstan-1(1-r)=itan-1 x, (x > 0).SolteIOSOLUTION We have |
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| 41. |
If V2=1.414,V3-1.732,find the value of 3V3+2V2 + 3V3-2V2 |
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Answer» hit like if you find it useful |
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| 42. |
3V5+2V2 |
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Answer» Thank you very nice |
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| 43. |
Add 2V2 + 5v3 and v2 -3V3 |
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| 44. |
KOLI PUN BOOK4. Heights and distances :From a point on the grhovering at some height abeint P on the ground, the angles of elevation of the top of a 20m tall building and of a helicopter,at some height above the top of the building are 30 and 60° respectively. Find the height atwhich the helicopter is hovering. (above the ground) |
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| 45. |
2+3+2V2V2- 18. Show that |
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| 46. |
23. If x = 3-2V2 find (xx) |
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Answer» not understand hello |
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| 47. |
v2+ 3+2v22- 18. Show that |
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| 48. |
3+2V2-/2 + 112-18. Show that= |
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Answer» (√2 + 1)/(√2 - 1) = (√2 + 1)*(√2 + 1)/(√2 - 1)*(√2 + 1) = (2+1+2√2) / (2-1) = 3+2√2 tq |
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| 49. |
3 An aeroplane flying at a height of 4000 m from the ground passes verticallabove another aeroplane at an instant when the angle of elevation of twoplanes from the same point on the ground are 60° and 45 respectivelyFind the vertical distance between the aeroplanes at that instant. (take V31.732)and 45° respectively |
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| 50. |
15From a point P on the ground, the angle of elevation of the top of a 10 m tall buildingand a helicopter, hovering over the top of the building, are 309 and 60° respectivelyFind the height of the helicopter above the ground. |
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Answer» hit like if you find it useful |
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