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The JEE questions and answers discussed here will give a much clearer idea of many JEE topics such as gravitation, thermodynamics, atomic structure, chemical bonding, binomial theorem, quadratic equation, etc. The questions have further been answered in an informative and easy to understand manner.

Alternatively, the questions are very important from an examination point of view and can help in quick revision of the different topics. Additionally, students can also clear their doubts and confusion by referring to these questions regarding any concept or topic. Students will find questions that are both conceptual and descriptive including definitions, principles, formulas, examples of, properties of, and types.

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Answer: Solution:Answer: Least k =87, Max k = 89Let total =100n(A)=89, n(B)=98max (n(A),n(B)) ≤ n(AUB) ≤ 10098 ≤ n(A)+n(B)-n(A n B) ≤ 100=> 98 ≤ 98 + 89- k ≤ 100Least k =87Max k = 89...
Answer: Solution:Answer: 5/9 Given: P(A) = p and P(B) = 2p P(A)-P(A∩B)+P(B)-P(A∩B) = 5/93p-2P(A∩B) = 5/90 ≤ P(A∩B) ≤ pFor maximum value: 3p - 2p = 5/9So, p = 5/9...
Answer: Solution:Answer: x = π/6Case 1: tan x ≥ 0cosx/(1+sinx ) = sinx/cosx⇒cos2x = sin2x+sinx⇒1-sin2x = sin2x + sinx⇒2 sin2x + sin x – 1 = 0⇒sin x = -1 ,1/2sin x = 1/2...
Answer: a. Cu2Cl2b. CuCl2c. AgBrd. ZnCl2Solution:Answer: (b)CuCl2 dissolves in water and gives coloured solutions...
Answer: a. (At constant temperature)And as we know, Molar Mass: CO2 > O2 > N2Therefore, Urms : UN2 > UO2 > UCO2Hence, option a) is correct...
Answer: Question: Choose the correct option for the given statements. Statement-I: Titration of a strong acid with a weak base uses methyl orange as an indicator. Statement-II: Titration of a weak acid...
Answer: Solution:Answer: (a)Bromination is highly selective.Reactivity order is: 3o > 2o > 1o...
Answer: Solution:Answer: (2.5)Potash alum = K2(SO4).Al2(SO4)3.24H2OMohr’s salt = (NH4)2Fe(SO4)2.6H2ORatio of water molecule = = 0.25 = 2.5 x 10-1Hence, the correct answer is 2.5...
Answer: a. Br2 / FeBr3, HNO3 / H2SO4, NaCN, H2O / H+b. HNO3 / H2SO4, Br2 / FeBr3, Mg / Ether, CO2, H3O+c. NaCN, H3O+, Br2 / FeBr3, H3O+ / CO2 / H2Od. HNO3 / H2SO4, Br2 / FeBr3 / NaCN/H2OSolution:Answer: (a)...
Answer: a. (1-x)2 f'(x) - 2(f(x)) 2 = 0b. (1-x) 2 f'(x) + 2(f(x)) 2= 0c. (1+x) 2 f'(x) - 2(f(x)) 2= 0d. (1+x) 2 f'(x)+2(f(x)) 2= 0Solution:Answer: (b)f(x) = = cos((2 tan-1(√x)= cos((cos-1 (((1 -...
Answer: a. Ca(HCO3)2,CaOb. CaO,Ca(HCO3)2c. CaCO3,Ca(HCO3)2d. Ca(HCO3)2,CaOSolution:Answer: (c)Ca(OH)2 (aq) + CO2(g) → CaCO3(s) + H2O(l)CaCO3(s) + H2O(l) + CO2(g) → Ca(HCO3)2(aq)Hence, the...
Answer: Solution:Answer: 16(OP) 2 + (AP) 2 + (PC) 2 + (PB) 2=18=> x2 + y2 + x2 + (y - 1) 2 + (x - 1) 2 + y2 + (x - 1) 2 + (y - 1) 2 =18=> 4x2 + 4y2 - 4x - 4y=14=> x2 + y2 - x - y=7/2=>...
Answer: a. ΔH < 0, ΔS < 0b. ΔH < 0, ΔS > 0c. ΔH > 0, ΔS > 0d. ΔH = 0, ΔS > 0Solution:Answer: (a)During adsorption, there is always a...
Answer: Solution:Answer: 40loge (x+y) = 4xyor 1/((x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]or 1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)If x = 0, then y = 1From (i)1 + dy/dx = 4(0 + 1)[0 + 1] = 4Or dy/dx...
Answer: Question: Find the product. Solution:Answer: (a)...
Answer: Solution:Answer: 9/2x2/8 + y2/4 = 1a2 = 8, b2 = 4e = c/a = √((a2 - b2)/a2 ) = √(1-4/8) = √(1/2)(5 - e2 ) = 5 - 1/2 = 9/2...
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