II PUC KCET MATH 2021 DETAILED SOLUTIONS.pdf - CREATIVE PU COLLEGE

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CREATIVE LEARNING CLASSES, KARKALA K-CET DETAILED SOLUTIONS - 2021 DEPARTMENT OF MATHEMATICS SET – C2 1. The cost and revenue functions of a product are given by 𝑐(𝑥) = 20𝑥 + 4000 and 𝑅(𝑥) = 60𝑥 + 2000 respectively where x is the number of items produced and sold. The value of x to earn profit is (A) > 50 (B) > 60 (C) > 80 (D) > 40 ANS: (A) SOL: Profit function = revenue function – cost function P(x) = (60x+2000) – (20x+4000) P(x) = 40x – 2000 Profit exists when P > 0 i.e., 40x – 2000 > 0 ∴ x > 50 2. A student has to answer 10 questions, choosing at least 4 from each of the parts A and B. If there are 6 questions in part A and 7 in part B, then the number of ways can the student choose 10 questions is (A) 256 (B) 352 (C) 266 (D) 426 ANS: (C) SOL: ( 𝟔𝑪𝟒 . 𝟕𝑪𝟔 ) + ( 𝟔𝑪𝟓 . 𝟕𝑪𝟓 ) + ( 𝟔𝑪𝟔 . 𝟕𝑪𝟒 ) = 𝟏𝟎𝟓 + 𝟏𝟐𝟔 + 𝟑𝟓 = 𝟐𝟔𝟔 3. If the middle term of the A.P is 300 then the sum of its first 51 terms is (A) 15300 (B) 14800 (C) 16500 (D) 14300 ANS: (A) SOL: Middle term is 26th term. 𝟑𝟎𝟎 = 𝒂 + (𝒏 − 𝟏)𝒅 = 𝒂 + 𝟐𝟓𝒅 𝒏 𝟓𝟏 𝟓𝟏 𝟓𝟏 𝑺𝟓𝟏 = 𝟐 [𝟐𝒂 + (𝒏 − 𝟏)𝒅] = [𝟐𝒂 + 𝟓𝟎𝒅] = [𝟐{𝒂 + 𝟐𝟓𝒅}] = [(𝟐)(𝟑𝟎𝟎)] = 𝟏𝟓𝟑𝟎𝟎 𝟐 𝟐 𝟐 4. The equation of straight line which passes through the point (𝑎 𝑐𝑜𝑠 3 𝜃 , 𝑎 𝑠𝑖𝑛3 𝜃) and perpendicular to 𝑥 𝑠𝑒𝑐 𝜃 + 𝑦 𝑐𝑜𝑠 𝑒 𝑐𝜃 = 𝑎 is 𝑥 𝑦 (A) 𝑎 + 𝑎 = 𝑎 𝑐𝑜𝑠 𝜃 (B) 𝑥 𝑐𝑜𝑠 𝜃 − 𝑦 𝑠𝑖𝑛 𝜃 = 𝑎 𝑐𝑜𝑠 2 𝜃 (C) 𝑥 𝑐𝑜𝑠 𝜃 + 𝑦 𝑠𝑖𝑛 𝜃 = 𝑎 𝑐𝑜𝑠 2 𝜃 (D) 𝑥 𝑐𝑜𝑠 𝜃 − 𝑦 𝑠𝑖𝑛 𝜃 = −𝑎 𝑐𝑜𝑠 2 𝜃 ANS: (B) SOL: Line perpendicular to 𝒙 𝒔𝒆𝒄 𝜽 + 𝒚 𝒄𝒐𝒔 𝒆 𝒄𝜽 = 𝒂 is given by 𝒙 𝒄𝒐𝒔𝒆𝒄 𝜽 − 𝒚𝒔𝒆𝒄 𝜽 = 𝒌…………(1) (𝟏) passes through (𝒂 𝒄𝒐𝒔𝟑 𝜽 , 𝒂 𝒔𝒊𝒏𝟑 𝜽). Then (𝒂 𝒄𝒐𝒔𝟑 𝜽)𝒄𝒐𝒔𝒆𝒄 𝜽 − (𝒂 𝒔𝒊𝒏𝟑 𝜽)𝒔𝒆𝒄 𝜽 = 𝒌 𝒄𝒐𝒔𝟑 𝜽 𝒔𝒊𝒏𝟑 𝜽 𝒄𝒐𝒔𝟐𝜽 ⇒ 𝒌 = 𝒂( − ) = 𝒂( ) 𝒔𝒊𝒏𝜽 𝒄𝒐𝒔𝜽 𝒄𝒐𝒔𝜽. 𝒔𝒊𝒏𝜽 By substituting in (1) we get 𝒙 𝒄𝒐𝒔 𝜽 − 𝒚 𝒔𝒊𝒏 𝜽 = 𝒂 𝒄𝒐𝒔 𝟐 𝜽 5. The mid points of the sides of a triangle are (1, 5, -1) (0, 4, -2) and (2, 3, 4) then centroid of the triangle 1 1 (A) (1, 4, 3) (B) (1,4, 3) (C) (-1, 4, 3) (D) (3 , 2,4) ANS: (B) 𝟏+𝟎+𝟐 𝟓+𝟒+𝟑 −𝟏−𝟐+𝟒 𝟏 SOL: centroid= ( , , ) = (𝟏, 𝟒, 𝟑) 𝟑 𝟑 𝟑 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 6. Consider the following statements: 𝑎𝑥 2 +𝑏𝑥+𝑐 Statement 1 : 𝑙𝑖𝑚 𝑐𝑥 2 +𝑏𝑥+𝑎 is 1 (where 𝑎 + 𝑏 + 𝑐 = 0) 𝑥→1 1 1 + 1 𝑥 2 Statement 2 : 𝑙𝑖𝑚 is 𝑥→−2 𝑥+2 4 (A) Only statement 2 is true (B) Only statement 1 is true (C) Both statements 1 and 2 are true (D) Both statements 1 and 2 are false ANS: (B) 𝒂𝒙𝟐 +𝒃𝒙+𝒄 SOL: Statement 1: 𝐥𝐢𝐦 𝒄𝒙𝟐 +𝒃𝒙+𝒂 = 𝟏 by putting 𝒙 = 𝟏. It is true 𝒙→𝟏 𝟏 𝟏 + 𝟏 𝟏 𝟏 𝒙 𝟐 Statement 2: 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = − 𝟒 ≠ 𝟒. Thus, it is not true. 𝒙→−𝟐 𝒙+𝟐 𝒙→−𝟐 𝟐𝒙 𝑎 𝑏 7. If a and b are fixed non-zero constants, then the derivative of 𝑥 4 − 𝑥 2 + 𝑐𝑜𝑠 𝑥 is 𝑚𝑎 + 𝑛𝑏 − 𝑝 where −2 −4 2 (A) 𝑚 = 4𝑥 3 ; 𝑛 = ; 𝑝 = 𝑠𝑖𝑛 𝑥 (B) 𝑚 = ; 𝑛 = 𝑥 3 ; 𝑝 = 𝑠𝑖𝑛 𝑥 𝑥3 𝑥5 −4 −2 2 (C) 𝑚 = ;𝑛 = ; 𝑝 = − 𝑠𝑖𝑛 𝑥 (D) 𝑚 = 4𝑥 3 ; 𝑛 = 𝑥 3 ; 𝑝 = − 𝑠𝑖𝑛 𝑥 𝑥5 𝑥3 ANS: (B) 𝒅𝒚 SOL: 𝒅𝒙 = −𝟒𝒂𝒙−𝟓 + 𝟐𝒃𝒙−𝟑 − 𝒔𝒊𝒏 𝒙. −𝟒 𝟐 Comparing with 𝒎𝒂 + 𝒏𝒃 − 𝒑 we get, 𝒎 = ; 𝒏 = 𝒙𝟑 ; 𝒑 = 𝒔𝒊𝒏 𝒙 𝒙𝟓 8. The Standard Deviation of the numbers 31, 32, 33 …. 46, 47 is 17 472 −1 (A) √ (B) √ (C) 2√6 (D) 4√3 12 12 ANS: (C) SOL: Standard Deviation of the numbers 31, 32, 33 …. 46, 47 is same as standard deviation 𝒏𝟐 −𝟏 𝟏𝟕𝟐 −𝟏 of 1, 2, 3, …, 16, 17 which is given by √ =√ = √𝟐𝟒 = 𝟐√𝟔 𝟏𝟐 𝟏𝟐 9. If P(A) = 0.59, P(B) = 0.30 and P(A ∩ B) = 0.21 then 𝑃(𝐴′ ∩ 𝐵′ ) = (A) 0.11 (B) 0.38 (C) 0.32 (D) 0.35 ANS: (C) SOL: 𝑷(𝑨′ ∩ 𝑩′ ) = 𝑷[(𝑨 ∪ 𝑩)′ ] = 𝟏 − 𝑷(𝑨 ∪ 𝑩) = 𝟏 − 𝑷(𝑨) − 𝑷(𝑩) + 𝑷(𝑨 ∩ 𝑩) = 𝟏 − 𝟎. 𝟓𝟗 − 𝟎. 𝟑 + 𝟎. 𝟐𝟏 = 𝟎. 𝟑𝟐 2𝑥 ; 𝑥 > 3 10. 𝑓: 𝑅 → 𝑅 defined by 𝑓(𝑥) = {𝑥 2 ; 1 < 𝑥 ≤ 3 then f (-2) + f (3) + f (4) is 3𝑥 ; 𝑥 ≤ 1 (A) 14 (B) 9 (C) 5 (D) 11 ANS: (D) SOL: 𝒇(−𝟐) + 𝒇(𝟑) + 𝒇(𝟒) = 𝟑(−𝟐) + 𝟑𝟐 + 𝟐(𝟒) = −𝟔 + 𝟗 + 𝟖 = 𝟏𝟏 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 2𝑥 11. Let A = {x: x ∈ R ; x is not a positive integer}. Define f : A ⟶ R as 𝑓(𝑥) = 𝑥−1, then f is (A) Injective but not surjective (B) surjective but not injective (C) Bijective (D) neither injective nor surjective ANS: (A) 𝟐𝒙 𝟐𝒚 SOL: for 𝒙, 𝒚 ∈ 𝑨, 𝒇(𝒙) = 𝒇(𝒚) ⇒ 𝒙−𝟏 = 𝒚−𝟏 ⇒ 𝟐𝒙𝒚 − 𝟐𝒙 = 𝟐𝒙𝒚 − 𝟐𝒚 ⇒ 𝒙 = 𝒚. Thus, f is injective. 𝟐 ∈ 𝑹 but 2 as no pre image in A. thus, f is not surjective. 12. The function 𝑓(𝑥) = √3 𝑠𝑖𝑛 2 𝑥 − 𝑐𝑜𝑠 2 𝑥 + 4is one-one in the interval −𝜋 𝜋 𝜋 𝜋 −𝜋 𝜋 𝜋 𝜋 (A) [ 6 , 3 ] (B) ( 6 , − 3 ] (C) [ 2 , 2 ] (D) [− 6 , − 3 ) ANS: (A) √𝟑 𝟏 SOL: √𝟑 𝒔𝒊𝒏 𝟐 𝒙 − 𝒄𝒐𝒔 𝟐 𝒙 + 𝟒 = 𝒔𝒊𝒏 𝟐 𝒙 − 𝟐 𝒄𝒐𝒔 𝟐 𝒙 + 𝟒 𝟐 𝝅 𝝅 = 𝒔𝒊𝒏𝟐𝒙. 𝒄𝒐𝒔 𝟔 − 𝒄𝒐𝒔𝟐𝒙. 𝒔𝒊𝒏 𝟔 + 𝟒 𝝅 = 𝒔𝒊𝒏 (𝟐𝒙 − ) + 𝟒 𝟔 𝝅 𝝅 𝝅 𝝅 𝝅 Which will be one-one if − 𝟐 ≤ (𝟐𝒙 − 𝟔 ) ≤ ⇒ −𝟔 ≤ 𝒙 ≤ 𝟐 𝟑 1 13. Domain of the function 𝑓(𝑥) = where [x] is greatest integer ≤ x is √[𝑥]2 −[𝑥]−6 (A) (-∞, -2) ∪ [4, ∞] (B) (-∞, -2) ∪ [3, ∞] (C) [-∞, -2] ∪ [4, ∞] (D) [-∞, -2] ∪ (3, ∞) ANS: (A) SOL: [𝒙]𝟐 − [𝒙] − 𝟔 > 𝟎 ⇒ ([𝒙] + 𝟐). ([𝒙] − 𝟑) > 𝟎 ⇒ [𝒙] < −𝟐 𝒐𝒓 [𝒙] > 𝟑 ⇒ 𝒙 ∈ (−∞, −𝟐) ∪ [𝟒, ∞) 𝜋 14. 𝑐𝑜𝑠 [𝑐𝑜𝑡 −1 ( − √3) + 6 ] = 1 (A) 0 (B) 1 (C) (D) -1 √2 ANS: (D) 𝝅 𝝅 𝝅 SOL: 𝒄𝒐𝒔 [𝝅 − 𝒄𝒐𝒕−𝟏 (√3) + 𝟔 ] = 𝒄𝒐𝒔 [𝝅 − 𝟔 + 𝟔 ] = 𝒄𝒐𝒔 𝝅 = −𝟏 1 5𝜋 √3 15. 𝑡𝑎𝑛−1 [ 𝑠𝑖𝑛 ] 𝑠𝑖𝑛−1 [𝑐𝑜𝑠 (𝑠𝑖𝑛−1 )] = √3 2 2 𝜋 𝜋 (A) 0 (B) 6 (C) 3 (D) π *** DATA INSUFFICIENT 2 1 1 −2 1 16. If 𝐴 = [ ] 𝐵 = [3 2]then (𝐴𝐵)′ is equal to 2 1 3 1 1 −3 −2 −3 10 −3 7 −3 7 (A) [ ] (B) [ ] (C) [ ] (D) [ ] 10 7 −2 7 10 2 10 −2 ANS: (B) HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 𝟐 𝟏 𝟏 −𝟐 𝟏 −𝟑 𝟏𝟎 SOL: (𝑨𝑩)′ = [[ ] [𝟑 𝟐]] ′ = [ ] 𝟐 𝟏 𝟑 −𝟐 𝟕 𝟏 𝟏 17. Let M be 2 x 2 symmetric matrix with integer entries, then M is invertible if (A) The first column of M is the transpose of second row of M (B) The second row of M is the transpose of first column of M (C) M is a diagonal matrix with non-zero entries in the principal diagonal (D) The product of entries in the principal diagonal of M is the product of entries in the other diagonal. ANS: (C) SOL: A diagonal matrix with non-zero entries is non-singular. Thus, it is invertible. 18. If A and B are matrices of order 3 and |A| = 5, |B| = 3 then |3AB| is (A) 425 (B) 405 (C) 565 (D) 585 ANS: (B) SOL: |𝟑𝑨𝑩| = 𝟑𝟑 . |𝑨|. |𝑩| = (𝟐𝟕)(𝟓)(𝟑) = 𝟒𝟎𝟓 19. If A and B are invertible matrices then which of the following is not correct? (A) adj A = |A| A-1 (B) det (A-1) = [det (A)]-1 (C) (AB)-1 = B-1A-1 (D) (A + B)-1 = (B-1 + A-1) ANS: (D) SOL: (A + B)-1 = (B-1 + A-1) is false. 𝑐𝑜𝑠 𝑥 1 0 20. If 𝑓(𝑥) = | 0 2 𝑐𝑜𝑠 𝑥 3 | then 𝑙𝑖𝑚𝑓(𝑥) = 𝑥→𝜋 0 1 2 𝑐𝑜𝑠 𝑥 (A) -1 (B) 1 (C) 0 (D) 3 ANS: (A) SOL: 𝒇(𝒙) = (𝒄𝒐𝒔 𝒙). (𝟒𝒄𝒐𝒔𝟐 𝒙 − 𝟑) Then, 𝐥𝐢𝐦 𝒇(𝒙) = 𝐥𝐢𝐦(𝒄𝒐𝒔 𝒙). (𝟒𝒄𝒐𝒔𝟐 𝒙 − 𝟑) 𝒙→𝝅 𝒙→𝝅 = (𝒄𝒐𝒔 𝝅)(𝟒. 𝒄𝒐𝒔𝟐 𝝅 − 𝟑) = −𝟏 1 2 3 3 2 21. If 𝑥 − 2𝑥 − 9𝑥 + 18 = 0 and 𝐴 = |4 𝑥 6| then the maximum value of A is 7 8 9 (A) 96 (B) 36 (C) 24 (D) 120 ANS: (A) SOL: Roots of 𝒙𝟑 − 𝟐𝒙𝟐 − 𝟗𝒙 + 𝟏𝟖 = 𝟎 are 𝟐, 𝟑, −𝟑. 𝟏 𝟐 𝟑 𝑨 = |𝟒 𝒙 𝟔| = −𝟏𝟐𝒙 + 𝟔𝟎…………(1) 𝟕 𝟖 𝟗 (𝟏) attains maximum if 𝒙 = −𝟑. Thus maximum of A is 96. 3 22. At x = 1, the function 𝑓(𝑥) = {𝑥 − 1 1 < 𝑥 < ∞ is 𝑥−1 −∞ < 𝑥 ≤ 1 (A) Continuous and differentiable (B) continuous and non-differentiable (C) Discontinuous and differentiable (D) Discontinuous and non-differentiable ANS: (B) HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA SOL: 𝐥𝐢𝐦+(𝒙𝟐 − 𝟏) = 𝟎 = 𝐥𝐢𝐦− (𝒙 − 𝟏). Thus, f is continuous. 𝒙→𝟏 𝒙→𝟏 ′ (𝒙) 𝟐𝒙, 𝟏 < 𝒙 < ∞ Now, 𝒇 ={ 𝟏, − ∞ < 𝒙 ≤ 𝟏 ⇒ 𝒇′ (𝟏+ ) = 𝟐(𝟏) = 𝟐 and 𝒇′ (𝟏− ) = 𝟏, which are unequal. Thus, f is not differentiable. 𝑑𝑦 23. If 𝑦 = (𝑐𝑜𝑠 𝑥 2 )2 , then 𝑑𝑥 is equal to (A) −4𝑥 𝑠𝑖𝑛 2 𝑥 2 (B) −𝑥 𝑠𝑖𝑛 𝑥 2 (C) −2𝑥 𝑠𝑖𝑛 2 𝑥 2 (D) −𝑥 𝑐𝑜𝑠 2 𝑥 2 ANS: (C) 𝒅𝒚 SOL: 𝒅𝒙 = 𝟐(𝒄𝒐𝒔𝒙𝟐 ). (−𝒔𝒊𝒏𝒙𝟐 ). (𝟐𝒙) = −𝟐𝒙. 𝒔𝒊𝒏 𝟐𝒙𝟐 𝑑 24. For constant a, 𝑑𝑥 (𝑥 𝑥 + 𝑥 𝑎 + 𝑎 𝑥 + 𝑎𝑎 ) is (A) 𝑥 𝑥 (1 + 𝑙𝑜𝑔 𝑥) + 𝑎𝑥 𝑎−1 (B) 𝑥 𝑥 (1 + 𝑙𝑜𝑔 𝑥) + 𝑎𝑥 𝑎−1 + 𝑎 𝑥 𝑙𝑜𝑔 𝑎 (C) 𝑥 𝑥 (1 + 𝑙𝑜𝑔 𝑥) + 𝑎𝑎 (1 + 𝑙𝑜𝑔 𝑥) (D) 𝑥 𝑥 (1 + 𝑙𝑜𝑔 𝑥) + 𝑎𝑎 (1 + 𝑙𝑜𝑔 𝑎) + 𝑎𝑥 𝑎−1 ANS: (B) SOL: Given 𝒂 is a constant. 𝒅𝒚 𝒏 Shortcut formula: 𝒚 = 𝒎𝒏 ⇒ 𝒅𝒙 = 𝒎𝒏 (𝒎 (𝒎′ ) + 𝒏′ (𝒍𝒐𝒈𝒎)) 𝒅(𝒙𝒙 ) 𝒙 ⇒ = (𝒙𝒙 ) (𝒙 (𝟏) + (𝟏. 𝒍𝒐𝒈𝒙)) = 𝒙𝒙 (𝟏 + 𝒍𝒐𝒈𝒙) 𝒅𝒙 𝒅(𝒙𝒂 ) 𝒅(𝒂𝒙 ) 𝒅(𝒂𝒂 ) Also, = 𝒂. 𝒙𝒂−𝟏 , = 𝒂𝒙 . 𝒍𝒐𝒈𝒂, =𝟎 𝒅𝒙 𝒅𝒙 𝒅𝒙 𝑑 ∴ (𝑥 𝑥 + 𝑥 𝑎 + 𝑎 𝑥 + 𝑎𝑎 ) = 𝒙𝒙 (𝟏 + 𝒍𝒐𝒈𝒙) + 𝒂. 𝒙 𝒂−𝟏 + 𝒂𝒙 . 𝒍𝒐𝒈𝒂 𝑑𝑥 25. Consider the following statements: 𝑑𝑦 𝑙𝑜𝑔10 𝑒 1 Statement 1: If 𝑦 = 𝑙𝑜𝑔10 𝑥 + 𝑙𝑜𝑔𝑒 𝑥 then 𝑑𝑥 = +𝑥 𝑥 𝑑 𝑙𝑜𝑔 𝑥 𝑑 𝑙𝑜𝑔 𝑥 Statement 2: 𝑑𝑥 (𝑙𝑜𝑔10 𝑥) = 𝑙𝑜𝑔 10 and (𝑙𝑜𝑔𝑒 𝑥) = 𝑙𝑜𝑔 𝑒 𝑑𝑥 (A) Statement 1 is true; statement 2 is false (B) Statement 1 is false; statement 2 is true (C) Both statements 1 and 2 are true (D) Both statements 1 and 2 are false ANS: (A) 𝒅𝒚 𝒅(𝐥𝐨𝐠 𝟏𝟎 𝒙) 𝒅(𝐥𝐨𝐠 𝒆 𝒙) SOL: Statement 1: 𝒅𝒙 = + 𝒅𝒙 𝒅𝒙 𝟏 𝒅(𝐥𝐨𝐠 𝒆 𝒙) 𝟏 = 𝐥𝐨𝐠 . +𝒙 𝒆 𝟏𝟎 𝒅𝒙 𝟏 𝟏 𝟏 = 𝐥𝐨𝐠 .𝒙 + 𝒙 𝒆 𝟏𝟎 𝐥𝐨𝐠 𝟏𝟎 𝒆 𝟏 = + 𝒙 , which is true. 𝒙 By Statement 1, we can observe that the terms in Statement 2 are not the required derivatives, hence, false statement. HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 𝜃 26. If the parametric equation of a curve is given by 𝑥 = 𝑐𝑜𝑠 𝜃 + 𝑙𝑜𝑔 tan 2 and 𝑦 = sin𝜃, then the points 𝑑𝑦 for which = 0 are given by 𝑑𝑥 𝑛𝜋 𝜋 (A) 𝜃 = ,𝑛 ∈ 𝑧 (B) 𝜃 = (2𝑛 + 1) 2 , 𝑛 ∈ 𝑧 2 (C) 𝜃 = (2𝑛 + 1)𝜋, 𝑛 ∈ 𝑧 (D) 𝜃 = 𝑛𝜋, 𝑛 ∈ 𝑧 ANS: (D) 𝒅𝒚 SOL: 𝒚 = 𝒔𝒊𝒏 𝜽 ⇒ 𝒅𝜽 = 𝒄𝒐𝒔 𝜽 𝜽 𝒅𝒙 𝟏 𝜽 𝟏 𝒙 = 𝒄𝒐𝒔 𝜽 + 𝒍𝒐𝒈 tan 𝟐 ⇒ 𝒅𝜽 = (−𝒔𝒊𝒏 𝜽) + 𝜽 . 𝒔𝒆𝒄𝟐 𝟐 . 𝟐 𝒕𝒂𝒏 𝟐 𝟏 = (−𝒔𝒊𝒏 𝜽) + 𝜽 𝜽 𝟐𝒔𝒊𝒏 𝒄𝒐𝒔 𝟐 𝟐 𝟏 = (−𝒔𝒊𝒏 𝜽) + 𝒔𝒊𝒏𝜽 −𝒔𝒊𝒏𝟐 𝜽+𝟏 𝒄𝒐𝒔𝟐 𝜽 = = 𝒔𝒊𝒏𝜽 𝒔𝒊𝒏𝜽 𝒅𝒚 𝒄𝒐𝒔𝜽 Now, 𝒅𝒙 = 𝒄𝒐𝒔𝟐 𝜽 . 𝒔𝒊𝒏𝜽 = 𝒕𝒂𝒏𝜽 𝒅𝒚 = 𝟎 ⟹ 𝒕𝒂𝒏 𝜽 = 𝟎 ⟹ 𝜽 = 𝒏𝝅, 𝒏 ∈ 𝒛 𝒅𝒙 𝑑𝑦 27. If 𝑦 = (𝑥 − 1)2 (𝑥 − 2)3 (𝑥 − 3)5 then 𝑑𝑥 at 𝑥 = 4 is equal to (A) 108 (B) 54 (C) 36 (D) 516 ANS: (D) SOL: Using product rule 𝒅𝒚 = 𝟐(𝒙 − 𝟏)(𝒙 − 𝟐)𝟑 (𝒙 − 𝟑)𝟓 + (𝒙 − 𝟏)𝟐 . 𝟑. (𝒙 − 𝟐)𝟐 (𝒙 − 𝟑)𝟓 𝒅𝒙 +(𝒙 − 𝟏)𝟐 (𝒙 − 𝟐)𝟑 . 𝟓. (𝒙 − 𝟑)𝟒 𝒅𝒚 At 𝒙 = 𝟒, we get, 𝒅𝒙 = [𝟐. 𝟑. 𝟖. 𝟏] + [𝟗. 𝟑. 𝟒. 𝟏] + [𝟗. 𝟖. 𝟓. 𝟏] = 𝟓𝟏𝟔 𝑡2 𝑡 28. A particle starts from rest and its angular displacement (in radians) is given by 𝜃 = 20 + 5. If the angular velocity at the end of t = 4 is k, then the value of 5k is (A) 0.6 (B) 5 (C) 5k (D) 3 ANS: (D) SOL: Differentiating 𝜽 w.r.t t 𝒅𝜽 𝟐𝒕 𝟏 ⇒ = 𝟐𝟎 + 𝟓 𝒅𝒕 𝒅𝜽 𝟑 at 𝒕 = 𝟒 equals 𝟓 = 𝒌 𝒅𝒕 𝟑 𝒕𝒉𝒆𝒏, 𝟓𝒌 = 𝟓 ( ) = 𝟑 𝟓 3 29. If the parabola 𝑦 = 𝛼𝑥 2 − 6𝑥 + 𝛽 passes through the point (0, 2) and has its tangent at 𝑥 = 2 parallel to x axis, then (A) 𝛼 = 2, 𝛽 = −2 (B) 𝛼 = −2, 𝛽 = 2 (C) 𝛼 = 2, 𝛽 = 2 (D) 𝛼 = −2, 𝛽 = −2 ANS: (C) SOL: 𝒚 = 𝜶𝒙𝟐 − 𝟔𝒙 + 𝜷 passes through (𝟎, 𝟐) ⇒ 𝟐 = 𝜷 𝒅𝒚 𝟑 [𝒅𝒙] 𝟑 = 𝟎(𝐩𝐚𝐫𝐚𝐥𝐥𝐞𝐥 𝐭𝐨 𝐱 − 𝐚𝐱𝐢𝐬) ⇒ 𝟐𝜶 (𝟐) − 𝟔 = 𝟎 ⇒ 𝜶 = 𝟐 𝒙= 𝟐 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 30. The function 𝑓(𝑥) = 𝑥 2 − 2𝑥 is strictly decreasing in the interval (A) (−∞, 1) (B) (1, ∞) (C) R (D) (−∞, ∞) ANS: (A) SOL: 𝒇′ (𝒙) < 𝟎 ⇒ 𝟐𝒙 − 𝟐 < 𝟎 ⇒𝒙<𝟏 ⇒ 𝒙 ∈ (−∞, 𝟏) 31. The maximum slope of the curve 𝑦 = −𝑥 3 + 3𝑥 2 + 2𝑥 − 27 is (A) 1 (B) 23 (C) 5 (D) −23 ANS: (C) 𝒅𝒚 SOL: 𝒅𝒙 = −𝟑𝒙𝟐 + 𝟔𝒙 + 𝟐 𝒅𝟐 𝒚 ⇒ = −𝟔𝒙 + 𝟔 = −𝟔(𝒙 − 𝟏) 𝒅𝒙𝟐 𝒅𝟐 𝒚 Now, 𝒅𝒙𝟐 = 𝟎 ⇒ 𝒙 = 𝟏 𝒅𝟑 𝒚 𝒅𝒚 = −𝟔 < 𝟎, 𝒇𝒐𝒓 𝒙 = 𝟏. Thus, maximum of 𝒅𝒙 is attained at 𝒙 = 𝟏 𝒅𝒚𝟑 𝒅𝒚 ∴ [𝒅𝒙] = −𝟑 + 𝟔 + 𝟐 = 𝟓 𝒙=𝟏 𝑥 3 𝑠𝑖𝑛( 𝑡𝑎𝑛−1 (𝑥 4 )) 32. ∫ dx is equal to 1+𝑥 8 − 𝑐𝑜𝑠( 𝑡𝑎𝑛−1 (𝑥 4 )) 𝑐𝑜𝑠( 𝑡𝑎𝑛−1 (𝑥 4 )) (A) +𝐶 (B) +𝐶 4 4 − 𝑐𝑜𝑠( 𝑡𝑎𝑛−1 (𝑥 3 )) 𝑠𝑖𝑛( 𝑡𝑎𝑛−1 (𝑥 4 )) (C) +𝐶 (D) +𝐶 3 4 ANS: (A) 𝒙𝟑 𝒔𝒊𝒏( 𝒕𝒂𝒏−𝟏 (𝒙𝟒 )) 𝒙𝟑 𝒔𝒊𝒏( 𝒕𝒂𝒏−𝟏 (𝒙𝟒 )) SOL: 𝑰 = ∫ 𝒅𝒙 = ∫ 𝟐 𝒅𝒙 𝟏+𝒙𝟖 𝟏+(𝒙𝟒 ) 𝒙𝟑 𝟏 Taking 𝒕𝒂𝒏−𝟏 𝒙𝟒 = 𝒕 ⇒ 𝟐 𝒅𝒙 = 𝟒 𝒅𝒕 𝟏+(𝒙𝟒 ) 𝟏 ⇒ 𝑰 = 𝟒 ∫ 𝒔𝒊𝒏 𝒕 𝒅𝒕 𝟏 = − 𝟒 𝒄𝒐𝒔 𝒕 + 𝑪 𝟏 = − 𝟒 𝒄𝒐𝒔 (𝒕𝒂𝒏−𝟏 𝒙𝟒 ) + 𝑪 𝑥 2 𝑑𝑥 33. The value of ∫ √𝑥 6 is equal to +𝑎6 (A) 𝑙𝑜𝑔|𝑥 3 + √𝑥 6 + 𝑎6 | + 𝐶 (B) 𝑙𝑜𝑔|𝑥 3 − √𝑥 6 + 𝑎5 | + 𝐶 1 1 (C) 3 𝑙𝑜𝑔|𝑥 3 + √𝑥 6 + 𝑎6 | + 𝐶 (D) 𝑙𝑜𝑔|𝑥 3 − √𝑥 6 + 𝑎6 | + 𝐶 3 ANS: (C) 𝒙𝟐 𝒅𝒙 𝒙𝟐 𝒅𝒙 SOL: I=∫ =∫ √𝒙𝟔 +𝒂𝟔 √(𝒙𝟑 )𝟐 +(𝒂𝟑 )𝟐 𝟏 Taking 𝒙𝟑 = 𝒕 ⇒ 𝒙𝟐 𝒅𝒙 = 𝟑 𝒅𝒕 𝒅𝒕 𝟏 𝟏 ⇒𝑰=∫ = 𝒍𝒐𝒈 |𝒕 + √𝒕𝟐 + 𝒂𝟔 | + 𝑪 = 𝒍𝒐𝒈 |𝒙𝟑 + √𝒙𝟔 + 𝒂𝟔 | + 𝑪 √𝒕𝟐 + (𝒂𝟑 )𝟐 𝟑 𝟑 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 𝑥𝑒 𝑥 𝑑𝑥 34. The value of ∫ (1+𝑥)2 is equal to (A) 𝑒 𝑥 (1 + 𝑥) + 𝑐 (B) 𝑒 𝑥 (1 + 𝑥 2 ) + 𝑐 𝑒𝑥 (C) 𝑒 𝑥 (1 + 𝑥)2 + 𝑐 (D) +𝑐 1+𝑥 ANS: (D) 𝒙𝒆𝒙 𝒅𝒙 (𝒙+𝟏−𝟏)𝒆𝒙 SOL: 𝑰 = ∫ (𝟏+𝒙)𝟐 = ∫ (𝟏+𝒙)𝟐 𝒅𝒙 𝟏+𝒙 𝟏 = ∫ ((𝟏+𝒙)𝟐 − (𝟏+𝒙)𝟐 ) 𝒆𝒙 𝒅𝒙 𝟏 𝟏 = ∫ (𝟏+𝒙 − (𝟏+𝒙)𝟐 ) 𝒆𝒙 𝒅𝒙 𝟏 = 𝒆𝒙 (𝟏+𝒙) + 𝑪 [𝒖𝒔𝒊𝒏𝒈 ∫ 𝒆𝒙 (𝒇(𝒙) + 𝒇′ (𝒙))𝒅𝒙 = 𝒆𝒙 𝒇(𝒙) + 𝑪] 1+𝑠𝑖𝑛 𝑥 35. The value of ∫ 𝑒 𝑥 [1+𝑐𝑜𝑠 𝑥] 𝑑𝑥 is equal to 𝑥 (A) 𝑒 𝑥 𝑡𝑎𝑛 2 + 𝑐 (B) 𝑒 𝑥 𝑡𝑎𝑛 𝑥 + 𝑐 (C) 𝑒 𝑥 (1 + 𝑐𝑜𝑠 𝑥) + 𝑐 (D) 𝑒 𝑥 (1 + 𝑠𝑖𝑛 𝑥) + 𝑐 ANS: (A) 𝟏+𝒔𝒊𝒏 𝒙 𝟏 𝒔𝒊𝒏𝒙 SOL: 𝑰 = ∫ 𝒆𝒙 [𝟏+𝒄𝒐𝒔 𝒙] 𝒅𝒙 = ∫ 𝒆𝒙 (𝟏+𝒄𝒐𝒔𝒙 + 𝟏+𝒄𝒐𝒔𝒙) 𝒅𝒙 𝟏 𝒙 𝒙 = ∫ 𝒆𝒙 (𝟐 𝒔𝒆𝒄𝟐 𝟐 + 𝒕𝒂𝒏 𝟐) 𝒅𝒙 𝒙 = 𝒆𝒙 . 𝒕𝒂𝒏 𝟐 + 𝑪  36. If I n =  tan n 4 x dx where n is positive integer then 𝐼10 + 𝐼8 is equal to 0 1 1 1 (A) 9 (B) 7 (C) 8 (D) 9 ANS: (D)  𝝅/𝟒 SOL: I n =  tan 𝒕𝒂𝒏𝒏−𝟐 𝒙. 𝒕𝒂𝒏𝟐 𝒙𝒅𝒙 n 4 x dx = ∫𝟎 0 𝝅/𝟒 𝑰𝒏 = ∫𝟎 𝒕𝒂𝒏𝒏−𝟐 𝒙. (𝒔𝒆𝒄𝟐 𝒙 − 𝟏)𝒅𝒙 𝝅/𝟒 𝑰𝒏 = ∫𝟎 𝒕𝒂𝒏𝒏−𝟐 𝒙. 𝒔𝒆𝒄𝟐 𝒙. 𝒅𝒙 − 𝑰𝒏−𝟐 𝝅/𝟒 𝑰𝒏 + 𝑰𝒏−𝟐 = ∫𝟎 𝒕𝒂𝒏𝒏−𝟐 𝒙. 𝒔𝒆𝒄𝟐 𝒙. 𝒅𝒙 Taking 𝒕𝒂𝒏𝒙 = 𝒕 ⇒ 𝒔𝒆𝒄𝟐 𝒙𝒅𝒙 = 𝒅𝒕 and 𝒕 = 𝟎 𝒕𝒐 𝟏 𝟏 𝟏 𝒕𝒏−𝟏 𝟏 ⇒ 𝑰𝒏 + 𝑰𝒏−𝟐 = ∫𝟎 𝒕𝒏−𝟐 𝒅𝒕 = [ 𝒏−𝟏 ] = 𝒏−𝟏 𝟎 𝟏 𝟏 Now, 𝑰𝟏𝟎 + 𝑰𝟖 = = 𝟏𝟎−𝟏 𝟗 4042 √𝑥𝑑𝑥 37. The value of ∫0 is equal to √𝑥+√4042−𝑥 (A) 4042 (B) 2021 (C) 8084 (D) 1010 ANS: (B) 𝒃 𝒇(𝒙) 𝒃−𝒂 SOL: ∫𝒂 𝒅𝒙 = 𝒇(𝒙)+𝒇(𝒂+𝒃−𝒙) 𝟐 𝟒𝟎𝟒𝟐 √𝒙𝒅𝒙 𝟒𝟎𝟒𝟐−𝟎 ⇒ ∫𝟎 = = 𝟐𝟎𝟐𝟏 √𝒙+√𝟒𝟎𝟒𝟐−𝒙 𝟐 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 38. The area of the region bounded by 𝑦 = √16 − 𝑥 2 and x- axis is (A) 8 square units (B) 20π square units (C) 16π square units (D) 256 π square units ANS: (A) SOL: 𝒚 = √𝟏𝟔 − 𝒙𝟐 can be expressed as 𝒙𝟐 + 𝒚𝟐 = 𝟏𝟔 whose area Is given by 𝝅𝒓𝟐 . 𝟏 𝟏 Required area = 𝟐 (𝝅𝒓𝟐 ) = 𝟐 (𝟏𝟔𝝅) = 𝟖𝝅 sq. units (Given as 8 sq. units) 𝑥2 𝑦2 39. If the area of the Ellipse is 25 + 𝜆2 = 1 𝑖𝑠 20𝜋 square units, then 𝜆 is (A) ±4 (B) ±3 (C) ±2 (D) ±1 ANS: (A) SOL: Area of ellipse is given by 𝝅𝒂𝒃. Here, 𝒂 = 𝟓, 𝒃 = 𝝀 ⇒ 𝝅(𝟓)(𝝀) = 𝟐𝟎𝝅 ⇒ 𝐩𝐨𝐬𝐬𝐢𝐛𝐥𝐞 𝐯𝐚𝐥𝐮𝐞𝐬 𝐨𝐟 𝛌 𝐚𝐫𝐞 ± 𝟒 40. Solution of Differential Equation xdy − ydx = 0 represents (A) A rectangular Hyperbola (B) Parabola whose vertex is at origin (C) Straight line passing through origin (D) A circle whose centre is origin ANS: (C) SOL: xdy − ydx = 𝟎 ⇒ 𝒙𝒅𝒚 = 𝒚𝒅𝒙 𝒅𝒚 𝒅𝒙 ⇒∫ =∫ 𝒚 𝒙 ⇒ 𝒍𝒐𝒈 𝒚 = 𝒍𝒐𝒈 𝒙 + 𝒍𝒐𝒈 𝑪 ⇒ 𝒍𝒐𝒈 𝒚 = 𝒍𝒐𝒈 𝒙𝑪 ⇒ 𝒚 = 𝑪𝒙, a straight line passing through the origin. 𝑑𝑦 𝑦 +1 41. The number of solutions of 𝑑𝑥 = when 𝑦(1) = 2 is 𝑥−1 (A) three (B) one (C) infinite (D) two ANS: (C) 𝒅𝒚 𝒚 +𝟏 𝒅𝒚 𝒅𝒙 SOL: 𝒅𝒙 = ⇒ ∫ 𝒚+𝟏 = ∫ 𝒙−𝟏 𝒙−𝟏 ⇒ 𝒍𝒐𝒈 (𝒚 + 𝟏) = 𝒍𝒐𝒈 (𝒙 − 𝟏) − 𝒍𝒐𝒈 𝑪 𝒙−𝟏 𝒙−𝟏 ⇒ 𝒍𝒐𝒈 𝑪 = 𝒍𝒐𝒈 (𝒚+𝟏) ⇒ 𝑪 = 𝒚+𝟏……(1) Substituting 𝒙 = 𝟏, 𝒚 = 𝟐 in (1), we get 𝑪 = 𝟎 ⇒ 𝒙 = 𝟏, only one solution. 42. A vector 𝑎⃗ makes equal acute angles on the coordinate axis. Then the projection of vector 𝑏⃗⃗ = 5𝑖̂ + 7𝑗̂ − 𝑘̂ on a⃗ is 11 11 4 3 (A) 15 (B) (C) 5 (D) 5√3 √3 ANS: (B) SOL: Equal acute angles can be 𝜶 = 𝜷 = 𝜸 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 𝟏 𝟏 𝟏 𝟏 ⇒ 𝟑𝒄𝒐𝒔𝟐 𝜶 = 𝟏 ⇒ 𝒄𝒐𝒔𝜶 = ⃗⃗ = . thus, 𝒂 𝒊̂ + 𝒋̂ + ̂ 𝒌 √𝟑 √𝟑 √𝟑 √𝟑 𝟓 𝟕 𝟏 + − 𝟏𝟏 Projection of ⃗𝒃⃗ on 𝒂 ⃗⃗ = ⃗𝒃⃗. 𝒂 ̂ = √𝟑 √𝟑 √𝟑 = 𝟏 √𝟑 43. The diagonals of a parallelogram are the vectors 3𝑖̂ + 6𝑗̂ − 2𝑘̂ and − 𝑖̂ − 2𝑗̂ − 8𝑘̂ then the length of the shorter side of parallelogram is (A)2√3 (B)√14 (C) 3√5 (D)4√3 *** DATA INSUFFICIENT 44. If 𝑎⃗. b⃗⃗ = 0 and a⃗ + b⃗⃗ makes an angle 600 with 𝑎⃗ then ⃗⃗⃗⃗⃗⃗ = 2|𝑏| (A)|𝑎| ⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗ (B) 2|𝑎| |𝑏| (C) ⃗⃗⃗⃗⃗⃗ |𝑎| = √3 ⃗⃗⃗⃗⃗⃗ |𝑏| (D) √3 ⃗⃗⃗⃗⃗⃗ |𝑎| = ⃗⃗⃗⃗⃗⃗ |𝑏| ANS: (D) SOL: ⃗⃗| |𝒃 ⃗⃗| |𝒃 𝒕𝒂𝒏𝟔𝟎° = |𝒂⃗⃗| ⇒ √𝟑 = |𝒂⃗⃗| ⃗𝒃⃗ 60° ⇒ √𝟑 ⃗⃗⃗⃗⃗⃗ |𝒂| = ⃗⃗⃗⃗⃗⃗ |𝒃| ⃗⃗ 𝒂 45. If the area of the parallelogram with 𝑎⃗ and 𝑏⃗⃗ as two adjacent sides is 15 sq. units then the area of the parallelogram having 3𝑎⃗ + 2𝑏⃗⃗ and 𝑎⃗ + 3𝑏⃗⃗ as two adjacent sides in sq. units is (A) 45 (B)75 (C)105 (D)120 ANS: (C) ⃗⃗ × ⃗𝒃⃗| = 𝟏𝟓. Now, |(𝟑𝒂 SOL: |𝒂 ⃗⃗ + 𝟐𝒃 ⃗⃗) × (𝒂 ⃗⃗ + 𝟑𝒃 ⃗⃗)| = |𝟗(𝒂 ⃗⃗) + 𝟐(𝒃 ⃗⃗ × 𝒃 ⃗⃗ × 𝒂 ⃗⃗)| = |𝟗(𝒂⃗⃗ × 𝒃⃗⃗) − 𝟐(𝒂 ⃗⃗)| ⃗⃗ × 𝒃 ⃗⃗ × 𝒃 = 𝟕|𝒂 ⃗⃗| = 𝟕 ∗ 𝟏𝟓 = 𝟏𝟎𝟓 46. The equation of the line joining the points (−3,4,11) and (1, −2,7) is 𝑥+3 𝑦−4 𝑧−11 𝑥+3 𝑦−4 𝑧−11 (A) = = (B) = = 2 3 4 −2 3 2 𝑥+3 𝑦+4 𝑧+11 𝑥+3 𝑦+4 𝑧+11 (C) = = (D) = = −2 3 4 2 −3 2 ANS: (B) 𝒙+𝟑 𝒚−𝟒 𝒛−𝟏𝟏 𝒙+𝟑 𝒚−𝟒 𝒛−𝟏𝟏 SOL: 𝟏+𝟑 = −𝟐−𝟒 = 𝟕−𝟏𝟏 ⇒ = = 𝟒 −𝟔 −𝟒 𝒙+𝟑 𝒚−𝟒 𝒛−𝟏𝟏 ⇒ = = 𝟐 −𝟑 −𝟐 𝒙+𝟑 𝒚−𝟒 𝒛−𝟏𝟏 ⇒ = = −𝟐 𝟑 𝟐 √3 1 √3 √3 1 −√3 47. The angle between the lines whose direction cosines are ( 4 , 4 , ) and ( 4 , 4 , ) is 2 2 𝜋 𝜋 𝜋 (A)𝜋 (B)2 (C) 3 (D) 4 ANS: (C) √3 √3 𝟏 𝟏 √3 √3 𝟏 SOL: If direction cosines are given, then 𝒄𝒐𝒔 𝜽 = |( 𝟒 . ) + (𝟒 . 𝟒) −(𝟐 . )| =𝟐 𝟒 𝟐 −𝟏 𝟏 𝝅 ⇒ 𝜽 = 𝒄𝒐𝒔 (𝟐) = 𝟑 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 48. If a plane meets the coordinate axes at A, B and C in such a way that the centroid of triangle ABC is at the point (1, 2, 3) then the equation of the plane is 𝑥 𝑦 𝑧 𝑥 𝑦 𝑧 (A) 1 + 2 + 3 = 1 (B) 3 + 6 + 9 = 1 𝑥 𝑦 𝑧 1 𝑥 𝑦 𝑧 (C) 1 + 2 + 3 = 3 (D) 1 − 2 + 3 = −1 ANS: (B) 𝒂 𝒃 𝒄 SOL: Centroid (𝟏, 𝟐, 𝟑) = (𝟑 , 𝟑 , 𝟑) ⇒ 𝒂 = 𝟑, 𝒃 = 𝟔, 𝒄 = 𝟗 𝐱 𝐲 𝐳 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐩𝐥𝐚𝐧𝐞 𝐢𝐧 𝐢𝐧𝐭𝐞𝐫𝐜𝐞𝐩𝐭 𝐟𝐨𝐫𝐦 𝐢𝐬 𝐚 + 𝐛 + 𝐜 = 𝟏 𝒙 𝒚 𝒛 ⇒𝟑+𝟔+𝟗 =𝟏 49. The area of the quadrilateral ABCD, when A (0, 4, 1), B (2, 3, -1), C (4, 5, 0) and D (2, 6, 2) is equal to (A) 9 sq. units (B) 18 sq. units (C) 27 sq. units (D) 81 sq. units ANS: (A) SOL: 𝑨𝑩 ̂ and 𝑩𝑪 ⃗⃗⃗⃗⃗⃗⃗ = 𝟐𝒊̂ − 𝒋̂ − 𝟐𝒌 ̂ ⃗⃗⃗⃗⃗⃗⃗ = 𝟐𝒊̂ + 𝟐𝒋̂ + 𝒌 𝒊̂ 𝒋̂ 𝒌̂ ⃗⃗⃗⃗⃗⃗⃗ × 𝑩𝑪 Required area = |𝑨𝑩 ⃗⃗⃗⃗⃗⃗⃗| = ||𝟐 −𝟏 ̂ | = √𝟖𝟏 = 𝟗 −𝟐|| = |𝟑𝒊̂ − 𝟔𝒋̂ + 𝟔𝒌 𝟐 𝟐 𝟏 50. The shaded region is the solution set of the inequalities (A) 5𝑥 + 4𝑦 ≥ 20, 𝑥 ≤ 6, 𝑦 ≥ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 (B) 5𝑥 + 4𝑦 ≤ 20, 𝑥 ≤ 6, 𝑦 ≥ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 (C) 5𝑥 + 4𝑦 ≥ 20, 𝑥 ≤ 6, 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 (D) 5𝑥 + 4𝑦 ≥ 20, 𝑥 ≥ 6, 𝑦 ≤ 3, 𝑥 ≥ 0, 𝑦 ≥ 0 ANS: (C) SOL: Shaded region is (i) above the line 𝟓𝒙 + 𝟒𝒚 = 𝟐𝟎 (ii) to the left of the line 𝒙 = 𝟔 (iii) below the line 𝒚 = 𝟑 these conditions are satisfied by 𝟓𝒙 + 𝟒𝒚 ≥ 𝟐𝟎, 𝒙 ≤ 𝟔, 𝒚 ≤ 𝟑, 𝒙 ≥ 𝟎, 𝒚 ≥ 𝟎 3 1 4 51. Given that A and B are two events such that 𝑃(𝐵) = 5 P(A/B) = 2 and P(𝐴 ∪ 𝐵) = 5 then P(A) = 3 1 1 3 (A) (B) 2 (C) 5 (D) 5 10 ANS: (B) SOL: 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩) 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨|𝑩). 𝑷(𝑩) 𝑷(𝑨) = 𝑷(𝑨 ∪ 𝑩) + 𝑷(𝑨|𝑩). 𝑷(𝑩) − 𝑷(𝑩) 𝟒 𝟏 𝟑 𝟑 𝟏 𝑷(𝑨) = 𝟓 + (𝟐 . 𝟓) − 𝟓 = 𝟐 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 52. If A, B and C are three independent events such that P(A)=P(B)=P(C)=P then P (at least two of A, B, C occur) = (A) 𝑃3 − 3𝑃 (B) 3𝑃 − 2𝑃2 (C) 3𝑃2 − 2𝑃3 (D) 3𝑃2 ANS: (C) SOL: 𝐏 (𝐚𝐭 𝐥𝐞𝐚𝐬𝐭 𝐭𝐰𝐨 𝐨𝐟 𝐀, 𝐁, 𝐂 𝐨𝐜𝐜𝐮𝐫) = 𝑷(𝑨𝑩𝑪′ ) + 𝑷(𝑨𝑩′ 𝑪) + 𝑷(𝑨′ 𝑩𝑪) + 𝑷(𝑨𝑩𝑪) = 𝑷. 𝑷. (𝟏 − 𝑷) + 𝑷. (𝟏 − 𝑷). 𝑷 + (𝟏 − 𝑷). 𝑷. 𝑷 + 𝑷. 𝑷. 𝑷 = 𝑷𝟐 [𝟑 − 𝟐𝑷] = 𝟑𝑷𝟐 − 𝟐𝑷𝟑 53. Two dice are thrown. If it is known that the sum of numbers on the dice was less than 6 the probability of getting a sum as 3 is 1 5 1 2 (A) (B) 18 (C) 5 (D) 5 18 ANS: (C) SOL: Favourable outcomes are (𝟏, 𝟐) and (𝟐, 𝟏). Total outcomes are (𝟏, 𝟏), (𝟏, 𝟐), (𝟏, 𝟑), (𝟏, 𝟒), (𝟐, 𝟏), (𝟐, 𝟐), (𝟐, 𝟑), (𝟑, 𝟏), (𝟑, 𝟐), (𝟒, 𝟏) 𝟐 𝟏 Required probability = 𝟏𝟎 = 𝟓 54. A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of cars and plant Y manufactures 30% of cars, 80% of cars at plant X and 90% of cars at plant Y are rated as standard quality. A car is chosen at random and is found to be of standard quality. The probability that it has come from plant X is 56 56 56 56 (A) (B) 84 (C) 83 (D) 79 73 ANS: (C) 𝟕 𝟑 𝟖 𝟗 SOL: 𝑷(𝑬𝟏 ) = 𝟏𝟎 , 𝑷(𝑬𝟐 ) = 𝟏𝟎 , 𝑷(𝑨|𝑬𝟏 ) = 𝟏𝟎 and 𝑷(𝑨|𝑬𝟐 ) = 𝟏𝟎 𝑷(𝑬𝟏 ).𝑷(𝑨|𝑬𝟏 ) Required 𝑷(𝑬𝟏 |𝑨) = 𝑷(𝑬 𝟏 ).𝑷(𝑨|𝑬𝟏 )+𝑷(𝑬𝟐 ).𝑷(𝑨|𝑬𝟐 ) 𝟕 𝟖 . 𝟓𝟔 𝟏𝟎 𝟏𝟎 𝑷(𝑬𝟏 |𝑨) = 𝟕 𝟖 𝟑 𝟗 = 𝟖𝟑 . + . 𝟏𝟎 𝟏𝟎 𝟏𝟎 𝟏𝟎 55. In a certain town 65% families own cell phones, 15000 families own scooter and 15% families own both. Taking into consideration the families own at least one of the two, the total number of families in the town is (A) 20000 (B) 30000 (C) 40000 (D) 50000 ANS: (B) SOL: Let 𝒙 be the total population. 𝟔𝟓 𝟏𝟓 Given, 𝒏(𝑨) = 𝟏𝟎𝟎 𝒙 , 𝒏(𝑩) = 𝟏𝟓𝟎𝟎𝟎, 𝒏(𝑨 ∩ 𝑩) = 𝟏𝟎𝟎 𝒙 and 𝒏(𝑼) = 𝒏(𝑨 ∪ 𝑩) 𝟔𝟓 𝟏𝟓 ⇒ 𝒙 = 𝟏𝟎𝟎 𝒙 + 𝟏𝟓𝟎𝟎𝟎 − 𝟏𝟎𝟎 𝒙 ⇒ 𝒙 = 𝟑𝟎𝟎𝟎𝟎 𝑛(𝐴) 56. A and B are non -singleton sets and n(A × B) = 35. If 𝐵 ⊂ 𝐴 then 𝐶𝑛(𝐵) = (A) 28 (B) 35 (C) 42 (D) 21 ANS: (D) SOL: Since A and B are non-singleton sets and B⊂A we get; 𝒏(𝑨) = 𝟕 and 𝒏(𝑩) = 𝟓 ⇒ 𝟕𝑪𝟓 = 𝟐𝟏 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 𝑥 57. Domain of f(𝑥) = 1−|𝑥|is (A) 𝑅 − [−1,1] (B) (−∞, 1) (C) (−∞, 1) ∪ (0,1) (D) 𝑅 − {−1, 1} ANS: (D) SOL: 𝟏 − |𝒙| ≠ 𝟎 ⇒ |𝒙| ≠ 𝟏 ⇒ 𝒙 ≠ ±𝟏 ⇒ 𝒙 ∈ 𝑹 − {−𝟏, 𝟏} 58. The value of 𝑐𝑜𝑠 1 200° + 𝑡𝑎𝑛 1 485° is 1 3 3 1 (A) (B) 2 (C) − 2 (D) − 2 2 ANS: (A) SOL: 𝒄𝒐𝒔 𝟏 𝟐𝟎𝟎° + 𝒕𝒂𝒏 𝟏 𝟒𝟖𝟓° = 𝐜𝐨𝐬(𝟏𝟎𝟖𝟎° + 𝟏𝟐𝟎°) + 𝐭𝐚𝐧(𝟏𝟒𝟒𝟎° + 𝟒𝟓°) = 𝒄𝒐𝒔𝟏𝟐𝟎° + 𝒕𝒂𝒏𝟒𝟓° 𝟏 𝟏 = −𝟐+𝟏 = 𝟐 59. The value of 𝑡𝑎𝑛 1° 𝑡𝑎𝑛 2° 𝑡𝑎𝑛 3° . . . . . . . . . 𝑡𝑎𝑛 8 9° is 1 (A) 0 (B) 1 (C) 2 (D) -1 ANS: (B) SOL: 𝒕𝒂𝒏 𝟏° 𝒕𝒂𝒏 𝟐° 𝒕𝒂𝒏 𝟑° … … … 𝒕𝒂𝒏 𝟖 𝟗° = 𝒕𝒂𝒏𝟏°𝒄𝒐𝒕𝟏°. 𝒕𝒂𝒏𝟐°𝒄𝒐𝒕𝟐° … 𝒕𝒂𝒏𝟒𝟔°𝒄𝒐𝒕𝟒𝟒°. 𝒕𝒂𝒏𝟒𝟓° = 𝟏. 𝟏. 𝟏 … 𝟏. 𝟏 =𝟏 1+𝑖 𝑥 60. If (1−𝑖) = 1 then (A) 𝑥 = 4𝑛 + 1; 𝑛 ∈ 𝑁 (B) 𝑥 = 2𝑛 + 1; 𝑛 ∈ 𝑁 (C) 𝑥 = 2𝑛; 𝑛 ∈ 𝑁 (D) 𝑥 = 4𝑛; 𝑛 ∈ 𝑁 ANS: (D) 𝟏+𝒊 𝒙 SOL: (𝟏−𝒊) = 𝟏 (𝟏+𝒊)(𝟏+𝒊) 𝒙 ⇒ [(𝟏−𝒊)(𝟏+𝒊)] = 𝟏 𝒙 𝟏+𝒊𝟐 +𝟐𝒊 ⇒[ ] =𝟏 𝟏+𝟏 𝒙 ⇒ 𝒊 = 𝟏 ⇒ 𝒙 = 𝟒𝒏, 𝒏 ∈ 𝑵 Department of Mathematics ❖ Mr. Ashwath S L ❖ Mr. Ganapathi K S ❖ Mr. Aditya Vati K ❖ MR. Gopal Reddy ❖ Mr. Rakshith B S CREATIVE EDUCATION FOUNDATION MOODBIDRI (R) Website : www.creativeedu.in Phone No. : 9019844492 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA