Practice Questions

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.If the domain of the function f(x) = loge(4x2 + 11x + 6) + sin−1(4x + 3) + cos−1( 10x+63 ) is (α, β] , then 36|α + β| is equal to (1) 54 (2) 72 (3) 63 (4) 45

Q72. n→∞{(2 1 1 1 1 1 1 (1) 1 (2) 0 (3) √2 (4) 1 √2

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.Let 𝑆 be the set of all solutions of the equation cos-12𝑥- 2cos-1√1 - 𝑥2 = 𝜋, 𝑥∈-1 2, 12. Then ∑𝑥∈𝑆2sin-1𝑥2 is equal to -2𝜋 (1) 0 (2) 3 (3) 𝜋- sin-1√3 (4) 𝜋- 2sin-1√3 4 4

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.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.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

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

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 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.Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and α(> 0), and the mean and standard deviation of marks of class B of n students be respectively 55 and 30 −α. If the mean and variance of the marks of the combined class of 100 + n students are respectively 50 and 350 , then the sum of variances of classes A and B is (1) 500 (2) 450 (3) 650 (4) 900

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 S be the set of all values of a1 for which the mean deviation about the mean of 100 consecutive positive integers a1, a2, a3, … . , a100 is 25 . Then S is (1) ϕ (2) {99} (3) N (4) {9}

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.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.The negation of (p ∧(−q)) ∨(−p) is equivalent to (1) p ∧(−q) (2) p ∧q (3) p ∨(q ∨(−p)) (4) p ∧(q ∧(−p))

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

Q73.Let [x] denote the greatest integer function and f(x) = max{1 + x + [x], 2 + x, x + 2[x]}, 0 ≤x ≤2 , where f is not continuous and n be the number of points in (0, 2), where f is not differentiable. Then (m + n)2 + 2 is equal to (1) 2 (2) 11 (3) 6 (4) 3 α, β > 0 , then α4 −β4 is equal to dx = α1 loge( α+1β ),

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 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 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.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 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

Q74.Let A, B, C be 3 × 3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements (S1) A13 B26 −B26 A13 is symmetric (S2) A26C 13 −C 13 A26 is symmetric Then, (1) Only S2 is true (2) Only S1 is true (3) Both S1 and S2 are false (4) Both S1 and S2 are true Q75. 1 3 √10 √10 1 −i Let A = ⎡ ⎤ and B = , where i = √−1. If M = AT BA , then the inverse of the matrix −3 1 [0 1 ] ⎣ √10 √10 ⎦ AM2023 AT is (1) [10 −2023i1 ] (2) [1−2023i 01 ] (3) [12023i 10 ] (4) [10 2023i1 ] m, such that x −cos x) + m