Practice Questions

Q72.If 𝐼𝑥= ∫𝑒sin2𝑥cos𝑥 sin2𝑥- sin𝑥𝑑𝑥 and 𝐼0 = 1, then 𝐼 𝜋 is equal to 3 (1) -1 34 (2) 1 34 2𝑒 2𝑒 3 (3) -𝑒 4 (4) 𝑒 34

Q72.Let p and q be two statements. Then ~(p ∧(p →~q) is equivalent to (1) p ∨(p ∧(~q)) (2) p ∨((~p) ∧q) (3) (~p) ∨q (4) p ∨(p ∧q)

Q72.If α > β > 0 are the roots of the equation ax2 + bx + 1 = 0 , and 1 1−cos(x2+bx+a) 2 1 1 k is equal to lim ( 2(1−αx)2 ) = k ( β −1α ), then x→1α (1) 2β (2) α (3) 2α (4) β

Q72.The equations of two sides of a variable triangle are x = 0 and y = 3 , and its third side is a tangent to the parabola y2 = 6x . The locus of its circumcentre is : (1) 4y2 −18y −3x −18 = 0 (2) 4y2 + 18y + 3x + 18 = 0 (3) 4y2 −18y + 3x + 18 = 0 (4) 4y2 −18y −3x + 18 = 0 JEE Main 2023 (25 Jan Shift 2) JEE Main Previous Year Paper

Q72.The converse of ((~p) ∧q) ⇒r is (1) ((~p) ∨q) ⇒r (2) (~r) ⇒p ∧q (3) (~r) ⇒((~p) ∧q) (4) (p ∨(~q)) ⇒(~r)

Q72.Which of the following statements is a tautology? (1) p →(p ∧(p →q)) (2) (p ∧q) →(~(p) →q) (3) (p ∧(p →q)) →~q (4) p ∨(p ∧q)

Q72.The number of values of r ∈{p, q, ~p, ~q} for which ((p ∧q) ⇒(r ∨q) ∧((p ∧r) ⇒q) is a tautology, is : (1) 1 (2) 2 (3) 4 (4) 3

Q72.The range of 𝑓𝑥= 4sin-1 𝑥2 is 𝑥2 + 1 (1) [0, 2𝜋] (2) [0, 𝜋] (3) [0, 2𝜋) (4) [0, 𝜋) 𝜋 4 𝑒-𝑥tan 50 𝑥𝑑𝑥 Q73. 𝑒-𝜋4 + ∫0 The value of 𝜋 ∫04 𝑒-𝑥(tan49𝑥+ tan51𝑥)𝑑𝑥 (1) 51 (2) 50 (3) 25 (4) 49 JEE Main 2023 (13 Apr Shift 2) JEE Main Previous Year Paper

Q73.Suppose 𝑓: 𝑅→0, ∞ be a differentiable function such that 5𝑓𝑥+ 𝑦= 𝑓𝑥· 𝑓𝑦, ∀ 𝑥, 𝑦∈𝑅, If 𝑓3 = 320, then ∑𝑛=5 0 𝑓𝑛 is equal to: (1) 6875 (2) 6575 (3) 6825 (4) 6528 JEE Main 2023 (30 Jan Shift 1) JEE Main Previous Year Paper

Q73.Negation of (p →q) →(q →p) is (1) (p~) ∨p (2) q ∧(~p) (3) (~q) ∧p (4) p ∨(~q)

Q73.If p, q and r are three propositions, then which of the following combination of truth values of p, q and r makes the logical expression {(p ∨q) ∧((~p) ∨r)} →((~q) ∨r) false ? (1) p = T, q = F, r = T (2) p = T, q = T, r = F (3) p = F, q = T, r = F (4) p = T, q = F, r = F

Q73.Let 𝑓 be a differentiable function such that 𝑥2𝑓𝑥- 𝑥= 4 𝑥𝑡 𝑓𝑡 𝑑𝑡, 𝑓1 = 2 Then 18 𝑓3 is equal to ∫0 3. (1) 210 (2) 160 (3) 150 (4) 180

Q73.Let the positive numbers a1, a2, a3, a4 and a5 be in a G.P. Let their mean and variance be 1031 and mn respectively, where m and n are co-prime. If the mean of their reciprocals is 31 and a3 + a4 + a5 = 14, then 10 m + n is equal to ____________.

Q73.Let 9 = x1 < x2 < … < x7 be in an A.P. with common difference d. If the standard deviation of x1, x2 … , x7 ¯¯is 4 and the mean is x , then x + x6 is equal to : JEE Main 2023 (01 Feb Shift 2) JEE Main Previous Year Paper + 1 ) (2) 34 (1) 18(1 √3 + 8 ) (4) 25 (3) 2(9 √7

Q73.The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12 . If the new mean of the marks is 10. 2. then their new variance is equal to: (1) 4. 04 (2) 4. 08 (3) 3. 96 (4) 3. 92 Q74. ⎡ 1 logx y logx z ⎤ Let x, y, z > 1 and A = logy x 2 logy z . Then adj (adj A2) is equal to ⎣ logz x logz y 3 ⎦ (1) 64 (2) 28 (3) 48 (4) 24

Q73.Let 𝐴= {𝑥∈ℝ: 𝑥+ 3 + 𝑥+ 4 ≤3}, 𝐵= 𝑥∈ℝ: 3𝑥∑𝑟= 1 10𝑟 < 3-3𝑥, where [𝑡] denotes greatest integer function. Then, (1) 𝐵⊂𝐶, 𝐴≠𝐵 (2) 𝐴∩𝐵= 𝜙 (3) 𝐴⊂𝐵, 𝐴≠𝐵 (4) 𝐴= 𝐵

Q73.Let the mean of 6 observations 1, 2, 4, 5, x and y be 5 and their variance be 10 . Then their mean deviation about the mean is equal to (1) 7 (2) 3 3 (3) 8 (4) 10 3 3

Q73.Among the statements (S1) : (p ⇒q) ∨((~p) ∧q) is a tautology (S2) : (q ⇒p) ⇒((~p) ∧q) is a contradiction (1) Neither (S1) and (S2) is True (2) Both (S1) and (S2) are True (3) Only (S2) is True (4) Only (S1) is True

Q73.Let △, ∇∈{∧, ∨} be such that (p →q) △(p∇q) is a tautology. Then (1) △= ∧, ∇= ∨ (2) △= ∨, ∇= ∧ (3) △= ∨, ∇= ∨ (4) △= ∧, ∇= ∧

Q73.The statement B ⇒((~A) ∨B) is not equivalent to : (1) B ⇒(A ⇒B) (2) A ⇒(A ⇔B) (3) A ⇒((~A) ⇒B) (4) B ⇒((~A) ⇒B) ¯¯

Q73.Let the six numbers a1, a2, . . . , a6 be in A. P. and a1 + a3 = 10 .If the mean of these six numbers is 192 and their variance is σ2 , then 8σ2 is equal to (1) 220 (2) 210 (3) 200 (4) 105

Q73.Let 𝑓𝑥= 2𝑥+ tan-1𝑥 and 𝑔𝑥= log𝑒√1 + 𝑥2 + 𝑥, 𝑥∈0, 3. Then (1) There exists 𝑥∈0, 3 such that 𝑓'𝑥< 𝑔'𝑥 (2) max 𝑓𝑥> max 𝑔𝑥 (3) There exist 0 < 𝑥1 < 𝑥2 < 3 such that 𝑓𝑥< 𝑔𝑥, (4) min 𝑓'𝑥= 1 + max 𝑔'𝑥 ∀𝑥∈𝑥1, 𝑥2 Q74. 1 + sin2𝑥 cos2𝑥 sin2𝑥 𝜋 𝜋 Let 𝑓𝑥= sin2𝑥 1 + cos2𝑥 sin2𝑥 , x ∈ 6, 3 . If 𝛼 and 𝛽 respectively are the maximum and the sin2𝑥 cos2𝑥 1 + sin2𝑥 minimum values of 𝑓, then 19 19 (1) 𝛽2 - 2√𝛼= 4 (2) 𝛽2 + 2√𝛼= 4 9 (3) 𝛼2 - 𝛽2 = 4√3 (4) 𝛼2 + 𝛽2 = 2

Q73.Let 𝑦= 𝑓𝑥= sin3𝜋 𝜋 + 5𝑥2 + 1 2. Then, at 𝑥= 1, 3cos 3√2-4𝑥3 (1) 2𝑦' + √3𝜋2𝑦= 0 (2) 2𝑦' + 3𝜋2𝑦= 0 (3) √2𝑦' - 3𝜋2𝑦= 0 (4) 𝑦' + 3𝜋2𝑦= 0

Q73.Let 𝑔𝑥= 𝑓𝑥+ 𝑓1 - 𝑥 and 𝑓"𝑥> 0, 𝑥∈0, 1. If 𝑔 is decreasing in the interval 0, 𝛼 and increasing in the interval 𝛼, 1, then tan-12𝛼+ tan-1 1 tan-1𝛼+ 1 is equal to 𝛼+ 𝛼 5π (1) π (2) 4 (3) 3π (4) 3π 4 2

Q73.The value of the integral ∫-log𝑒2log𝑒2 𝑒𝑥log𝑒𝑒𝑥+ (1) √2 ( 2 + √5 ) 2 √5 (2) ( 2 + √5 ) 2 √5 - log𝑒 √1 + √5 2 log𝑒 √1 + √5 + 2 2 ) 2 ( 2 + ( 3 √5 √2 - √5 √5 ) √5 (3) (4) - + log𝑒 2 log𝑒 + 2 √1 √5 + √1 √5