Practice paper of mathematics (class 12), Assignments of Mathematics

Download Practice paper of mathematics (class 12) and more Assignments Mathematics in PDF only on Docsity! Q1. 𝐼𝑓  𝑓: 𝑅 − {3 5 } → 𝑅 be defined by 𝑓(𝑥) = 3𝑥+2 5𝑥−3 ,  then A. f−1(x)=f(x) B. f−1(x)=−f(x) C. fof(x)=−x D. 𝑓−1(𝑥) = 1 20 𝑓(𝑥) Answer: A Explanation: Given that, 𝑓(𝑥) = 3𝑥 + 2 5𝑥 − 3 ⇒ 𝑦 = 3𝑥 + 2 5𝑥 − 3 ⇒ 3𝑥 + 2 = 5𝑥𝑦 − 3𝑦 ⇒ 𝑥(3 − 5𝑦) = −3𝑦 − 2 ⇒ 𝑥 = 3𝑦 + 2 5𝑦 − 3 ⇒ 𝑓−1(𝑥) = 3𝑥 + 2 5𝑥 − 3 ⇒ 𝑓−1(𝑥) = 𝑓(𝑥) Q2. 𝐿𝑒𝑡 𝑓: [2, ∞) → 𝑅 𝑏𝑒 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓(𝑥) = 𝑥2 − 4𝑥 + 5, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑓 𝑖𝑠 A. R B. [1, ∞) C. [4, ∞) D. [5, ∞) Answer: B Explanation: 𝑮𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕 𝒇(𝒙) = 𝒙𝟐 − 𝟒𝒙 + 𝟓 𝐋𝐞𝐭 𝒚 = 𝒙𝟐 − 𝟒𝒙 + 𝟓 ⇒ 𝒚 = 𝒙𝟐 − 𝟒𝒙 + 𝟒 − 𝟒 + 𝟓 ⇒ 𝒚 = (𝒙 − 𝟐)𝟐 − 𝟒 + 𝟓 ⇒ 𝒚 = (𝒙 − 𝟐)𝟐 + 𝟏 ⇒ 𝒚 − 𝟏 = (𝒙 − 𝟐)𝟐 ⇒ 𝒙 − 𝟐 = √𝒚 − 𝟏 ⇒ 𝒙 = 𝟐 + √𝒚 − 𝟏 ∴ 𝒚 − 𝟏 ≥ 𝟎, 𝒚 ≥ 𝟏 𝑻𝒉𝒊𝒔 𝒊𝒎𝒑𝒍𝒊𝒆𝒔 𝒓𝒂𝒏𝒈𝒆 𝒊𝒔 [𝟏, ∞) Q3.The set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is A. 720 B. 120 C. 0 D. None of these Answer: C Explanation: We know that if A and B are two non-empty finite sets containing m and n elements respectively , then the number of one-one and onto mapping from A to B is: 𝑛!  𝑖f 𝑚 = 𝑛 0,  𝑖f 𝑚 ≠ 𝑛 Given that 𝑚 = 5 and 𝑛 = 6 ∵ 𝑚 ≠ 𝑛 So,  number of mappings  =  0 Q8.Set A has 3 elements and the set B has 4 elements. Then the number of injective mappings that can be defined from A to B is A. 144 B. 12 C. 24 D. 64 Answer: C Explanation: 𝑻𝒉𝒆 𝒕𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒊𝒏𝒋𝒆𝒄𝒕𝒊𝒗𝒆 𝒎𝒂𝒑𝒑𝒊𝒏𝒈𝒔 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒊𝒏𝒈 𝟑 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒏𝒕𝒐 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒄𝒐𝒏𝒕𝒂𝒊𝒏𝒊𝒏𝒈 𝟒 𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒔 𝒊𝒔𝟒𝑷𝟑 = 𝟒! = 𝟐𝟒. Q9.Let N be the set of natural numbers and the function f : N → N be defined by f(n) = 2n + 3. Then f is A. surjective B. injective C. bijective D. none of these Answer: B Explanation: 𝒇(𝒏𝟏) = 𝒇(𝒏𝟐) ⇒ 𝟐(𝒏𝟏) + 𝟑 = 𝟐(𝒏𝟐) + 𝟑 ⇒ 𝒏𝟏 = 𝒏𝟐 𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝒇 𝒊𝒔 𝒊𝒏𝒋𝒆𝒄𝒕𝒊𝒗𝒆, 𝑩𝒖𝒕 𝒂𝒔 𝒇(𝒏) ≠ 𝟏 𝒐𝒓 𝟐 𝒐𝒓 𝟑 𝒐𝒓 𝟒 𝒘𝒉𝒊𝒄𝒉 𝒐𝒄𝒄𝒖𝒓𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒄𝒐𝒅𝒐𝒎𝒂𝒊𝒏 𝒐𝒇 𝑵, 𝒔𝒐 𝒇(𝒏) = 𝟐𝒏 + 𝟑 𝒊𝒔 𝒏𝒐𝒕 𝒔𝒖𝒓𝒋𝒆𝒄𝒕𝒊𝒗𝒆. Q10. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 tan(1 2 sin−1 3 4 ) 𝑖𝑠: A. 3+√7 4 B. 3−√7 4 C. 4+√7 3 D. 4−√7 3 Answer: D Explanation: Let, tan( 1 2 sin−1 3 4 ) = 𝑦 ⇒ 1 2 sin−1 3 4 = tan−1𝑦 ⇒ sin−1 3 4 = 2tan−1𝑦 ⇒ sin−1 3 4 = tan−1( 2𝑦 1 − 𝑦2 ) ⇒ tan−1 3 √7 = tan−1( 2𝑦 1 − 𝑦2 ) ⇒ 3 √7 = ( 2𝑦 1 − 𝑦2 ) ⇒ 3𝑦2 + 2√7 − 3 = 0 Q11. 𝐼𝑓 sin−1( 2𝑎 1+𝑎2) + cos−1(1−𝑎2 1+𝑎2) = tan−1( 2𝑥 1−𝑥2) 𝑤ℎ𝑒𝑟𝑒 𝑎, 𝑥 ∈]0,1[, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑠: A. 0 B. 𝑎 2 C. 2𝑎 1+𝑎2 D. 2𝑎 1−𝑎2 Answer: D Explanation: Let 𝑎 = tan𝜃 Then sin−1( 2𝑎 1 − 𝑎2 ) + cos−1( 1 − 𝑎2 1 + 𝑎2 ) = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ sin−1( 2tan𝜃 1 − tan2𝜃 ) + cos−1( 1 − tan2𝜃 1 + tan2𝜃 ) = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ sin−1sin2𝜃 + cos−1cos2𝜃 = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ 4𝜃 = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ 4tan−1𝑎 = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ 2 ⋅ 2tan−1𝑎 = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ 2 ⋅ tan−1( 2𝑎 1 − 𝑎2 ) = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ tan−1( 2 ⋅ 2𝑎 1 − 𝑎2 1 − ( 2𝑎 1 − 𝑎2)2 ) = tan−1( 2𝑥 1 − 𝑥2 ) ⇒ 4𝑎 1 − 𝑎2 1 − ( 2𝑎 1 − 𝑎2)2 = 2𝑥 1 − 𝑥2 ⇒ 𝑥 = 2𝑎 1 − 𝑎2 Q12. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑓𝑜𝑟 tan−12𝑥 + tan−13𝑥 = 𝜋 4 𝑖𝑠: A. 1 6 B. − 1 6 C. -1 D. 1 Answer: A Explanation: 𝐭𝐚𝐧−𝟏( 𝟏 𝟑 ) + 𝐭𝐚𝐧−𝟏( 𝟏 𝟓 ) + 𝐭𝐚𝐧−𝟏( 𝟏 𝟕 ) + 𝐭𝐚𝐧−𝟏( 𝟏 𝟖 ) = 𝐭𝐚𝐧−𝟏( 𝟏 𝟑 + 𝟏 𝟓 𝟏 − 𝟏 𝟑 ⋅ 𝟏 𝟓 ) + 𝐭𝐚𝐧−𝟏( 𝟏 𝟕 + 𝟏 𝟖 𝟏 − 𝟏 𝟕 ⋅ 𝟏 𝟖 ) = 𝐭𝐚𝐧−𝟏( 𝟖 𝟏𝟒 ) + 𝐭𝐚𝐧−𝟏( 𝟏𝟓 𝟓𝟓 ) = 𝐭𝐚𝐧−𝟏( 𝟒 𝟕 ) + 𝐭𝐚𝐧−𝟏( 𝟑 𝟏𝟏 ) = 𝐭𝐚𝐧−𝟏( 𝟒 𝟕 + 𝟑 𝟏𝟏 𝟏 − 𝟒 𝟕 ⋅ 𝟑 𝟏𝟏 ) = 𝐭𝐚𝐧−𝟏( 𝟒𝟒  +  𝟐𝟏 𝟕𝟕 𝟕𝟕  −  𝟏𝟐 𝟕𝟕 ) = 𝐭𝐚𝐧−𝟏( 𝟒𝟒 + 𝟐𝟏 𝟕𝟕 − 𝟏𝟐 ) = 𝐭𝐚𝐧−𝟏( 𝟔𝟓 𝟔𝟓 ) = 𝐭𝐚𝐧−𝟏(𝟏) = 𝝅 𝟒 Q15. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 tan(2tan−1 1 5 ) 𝑖𝑠: A. 7 12 B. 5 12 C. 9 12 D. 5 9 Answer: B Explanation: 𝑼𝒔𝒊𝒏𝒈 𝒊𝒅𝒆𝒏𝒕𝒊𝒕𝒚 𝟐𝐭𝐚𝐧−𝟏𝒙 = 𝐭𝐚𝐧−𝟏( 𝟐𝒙 𝟏 − 𝒙𝟐 ) 𝒂𝒔, 𝐭𝐚𝐧(𝟐𝐭𝐚𝐧−𝟏 𝟏 𝟓 ) = 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏( 𝟐 × 𝟏 𝟓 𝟏 − ( 𝟏 𝟓 )𝟐 )) = 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏( 𝟐 𝟓 𝟏 − 𝟏 𝟐𝟓 )) = 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏( 𝟐 𝟓 𝟐𝟒 𝟐𝟓 )) = 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏( 𝟏 𝟏𝟐 𝟓 )) = 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏( 𝟓 𝟏𝟐 )) = 𝟓 𝟏𝟐 Q16. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 sin(sec−1 17 15 ) 𝑖𝑠: A. 9 17 B. 8 17 C. 17 8 D. 8 15 Answer: B Explanation: 𝑺𝒊𝒏𝒄𝒆, 𝐬𝐞𝐜−𝟏 𝟏𝟕 𝟏𝟓 = 𝐬𝐞𝐜−𝟏 𝒉 𝒃 , So,In right angle triangle Use Pythagoras theorem to find unknown side as, 𝑝2 + 𝑏2 = ℎ2 ⇒ 𝑝2 = ℎ2 − 𝑏2 ⇒ 𝑝2 = 172 − 152 ⇒ 𝑝2 = 289 − 225 ⇒ 𝑝2 = 64 ⇒ 𝑝 = 8 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 , 𝑢𝑠𝑖𝑛𝑔 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 sin−1 𝑝 ℎ = sec−1 ℎ 𝑏 sin(sec−1 17 15 ) = sin(sin−1 8 17 ) = 8 17 Q17. 𝑇ℎ𝑒 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 cos−1(cos 7𝜋 6 ) 𝑖𝑠: A. 7𝜋 6 B. cos 𝜋 6 C. 5𝜋 6 D. 𝜋 3 Answer: C Explanation: Since 7π/6 does not lie between 0 and π. Therefore cos−1(cos 7𝜋 6 ) ≠ 7𝜋 6 To covert 7π/6 into a value that lies between 0 and π. Let us proceed as, cos−1(cos 7𝜋 6 ) ⇒ cos−1(cos(𝜋 + 𝜋 6 )) (𝑆𝑖𝑛𝑐𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓cos(𝜋 + 𝜃) = cos(𝜋 − 𝜃)) ⇒ cos−1(cos(𝜋 − 𝜋 6 )) ⇒ 5𝜋 6 Answer: B Explanation: This can be solved as, tan−1tan 9𝜋 8 ⇒ tan−1tan(𝜋 + 𝜋 8 ) ⇒ tan−1tan( 𝜋 8 ) ⇒ 𝜋 8 Q22. 𝑇ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 cos−1(2𝑥 − 1) 𝑖𝑠: A. [0,1] B. [-1,1] C. (-1,1) D. (0, π) Answer: A Explanation: This can be solved as, 𝑓(𝑥) = cos−1(2𝑥 − 1) ⇒ −1 ≤ 2𝑥 − 1 ≤ 1 ⇒ 0 ≤ 2𝑥 ≤ 2 ⇒ 0 ≤ 𝑥 ≤ 1 ⇒ 𝑥 ∈ [0,1] Q23. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 2sec−12 + sin−1(1 2 ) 𝑖𝑠: A. 𝜋 6 B. 5𝜋 6 C. 7𝜋 6 D. 1 Answer: B Explanation: This can be solved as, 2sec−12 + sin−1( 1 2 ) = 2sec−1sec( 𝜋 3 ) + sin−1sin 𝜋 6 = 2( 𝜋 3 ) + 𝜋 6 = 5𝜋 6 Q24. 𝑇ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ [1 𝑥 1][ 1 3 2 2 5 1 15 3 2 ][ 1 2 𝑥 ] = O 𝑖𝑠: A. x=−2, −14 B. x=2,14x C. x=−2, 14 D. x=2, −14 Answer: A Explanation: This can be solved as, [1 𝑥 1][ 1 3 2 2 5 1 15 3 2 ][ 1 2 𝑥 ] = O [16 + 2𝑥 5𝑥 + 6 4 + 𝑥][ 1 2 𝑥 ] = O [1 + 2𝑥 + 15 3 + 5𝑥 + 3 2 + 𝑥 + 2][ 1 2 𝑥 ] = O [16 + 2𝑥 + 10𝑥 + 12 + 𝑥2 + 4𝑥] = O [𝑥2 + 16𝑥 + 28] = O [𝑥2 + 2𝑥 + 14𝑥 + 28] = O (𝑥 + 2)(𝑥 + 14) = O 𝑥 = −2, −14 Q25. 𝐼𝑓 𝐼A = [ 2 0 −1 5 1 0 0 1 3 ], 𝑡ℎ𝑒𝑛 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑚𝑎𝑡𝑟𝑖𝑥 𝑖𝑠: A. [ 3 1 1 5 6 −5 5 −2 15 ] B. [ 3 1 1 3 6 −5 5 −2 15 ] C. [ 3 1 1 15 6 −5 5 −2 2 ] D. [ 3 −1 1 −15 6 −5 5 −2 2 ] Answer: D Explanation: We have given A = [ 2 0 −1 5 1 0 0 1 3 ] Rewrite it as, [ 2 0 −1 5 1 0 0 1 3 ] = [ 1 0 0 0 1 0 0 0 1 ]A ApplyingR1 → R1 2 [ 1 0 − 1 2 5 1 0 0 1 3 ] = [ 1 2 0 0 0 1 0 0 0 1 ]A ApplyingR2 → R2 − 5R1 [ 1 0 − 1 2 0 1 5 2 0 1 3 ] = [ 1 2 0 0 − 5 2 1 0 0 0 1 ]A Q26. 𝐼𝑓 𝐴 𝑖𝑠 𝑠𝑞𝑢𝑎𝑟𝑒 𝑚𝑎𝑡𝑟𝑖𝑥 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝐴2 = 𝐴, 𝑇ℎ𝑒𝑛 (𝐼 + 𝐴) 3 =? A. 7A+I B. 7A−I C. 5A+2I D. 7A+2I Answer: A Q30. 𝑇ℎ𝑒 𝑐𝑜𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑎12 𝑖𝑛 𝑚𝑎𝑡𝑟𝑖𝑥 | 2 −3 5 6 0 4 1 5 −7 | 𝑖𝑠: A. 46 B. -46 C. 12 D. 18 Answer: A Explanation: 𝑪𝒐𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒂𝟏𝟐 𝒊𝒔: 𝑨𝟏𝟐 = (−𝟏)𝟏+𝟐| 𝟔 𝟒 𝟏 −𝟕 | 𝑨𝟏𝟐 = −(−𝟒𝟐 − 𝟒) 𝑨𝟏𝟐 = −(−𝟒𝟔) 𝑨𝟏𝟐 = 𝟒𝟔 Q31.If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m = n, then the order of matrix (5A – 2B) is: A. m × 3 B. 3 × 3 C. m × n D. 3 × n Answer: D Explanation: 𝑰𝒇 𝑨𝟑× 𝒎 𝒂𝒏𝒅 𝑩𝟑 × 𝒏𝒂𝒓𝒆 𝒕𝒘𝒐 𝒎𝒂𝒕𝒓𝒊𝒄𝒆𝒔 𝑰𝒇 𝒎 = 𝒏 , 𝑻𝒉𝒆𝒏 𝑨 𝒂𝒏𝒅 𝑩 𝒉𝒂𝒗𝒆 𝒔𝒂𝒎𝒆 𝒐𝒓𝒅𝒆𝒓 𝒂𝒔 𝟑 × 𝒏 𝒆𝒂𝒄𝒉. 𝑺𝒐 𝒕𝒉𝒆 𝒐𝒓𝒅𝒆𝒓 𝒐𝒇 (𝟓𝑨 – 𝟐𝑩) 𝒔𝒉𝒐𝒖𝒍𝒅 𝒃𝒆 𝒔𝒂𝒎𝒆 𝒂𝒔 𝟑 × 𝒏 Q32.Total number of possible matrices of order 3 × 3 with each entry 2 or 0, is: A. 9 B. 27 C. 81 D. 512 Answer: D Explanation: 𝑻𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒑𝒐𝒔𝒔𝒊𝒃𝒍𝒆 𝒎𝒂𝒕𝒓𝒊𝒄𝒆𝒔 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓 𝟑 × 𝟑 𝒘𝒊𝒕𝒉 𝒆𝒂𝒄𝒉 𝒆𝒏𝒕𝒓𝒚 𝟐 𝒐𝒓 𝟎 𝒊𝒔 𝟐𝟗 = 𝟓𝟏𝟐 Q33.If A and B are symmetric matrices of the same order, then (AB′ – BA′) is a: A. Skew symmetric matrix B. Null matrix C. Symmetric matrix D. None of these Answer: A Explanation: Since (AB′ –BA′)′ = (AB′)′ – (BA′)′ Q34. 𝐼𝑓 A = [1 3 5 2 7 9 ] 𝑎𝑛𝑑 B = [ 1 5 3 2 0 6 ], 𝑡ℎ𝑒𝑛 A. Only AB is defined B. Only BA is defined C. AB and BA both are defined D. AB and BA both are not defined. Answer: C Explanation: 𝑯𝒆𝒓𝒆 𝑨 = [𝒂𝒊𝒋]𝟐×𝟑𝒂𝒏𝒅 𝑩 = [𝒃𝒊𝒋]𝟑×𝟐 To multiply two marices, the necessary condition is the number of columns in first marix will be equal to number of rows in second matrix. Since both AB and BA are fulfilling this condition., so both AB and BA are defined. Q35.If A and B are square matrices of the same order, then (A + B) (A – B) is equal to: A. A2− B2 B. A2− BA− AB−B2 C. A2− B2+BA−AB D. A2−BA+B2+AB Answer: C Explanation: This can be solved as, (A + B) (A – B) = A (A – B) + B (A – B) = A2− B2+BA−AB Q36.Using properties of determinants, calculate the value of: | 1 𝑥 𝑥2 𝑥2 1 𝑥 𝑥 𝑥2 1 | A. (1−x3)2 B. (1−x3) C. (1−x3)3 D. (1−x2)3 Answer: A Explanation: Explanation: = | 𝒚 + 𝒌 𝒚 𝒚 𝒚 𝒚 + 𝒌 𝒚 𝒚 𝒚 𝒚 + 𝒌 | 𝑹𝟏 → 𝑹𝟏 + 𝑹𝟐 + 𝑹𝟑 = | 𝟑𝒚 + 𝒌 𝟑𝒚 + 𝒌 𝟑𝒚 + 𝒌 𝒚 𝒚 + 𝒌 𝒚 𝒚 𝒚 𝒚 + 𝒌 | = (𝟑𝒚 + 𝒌)| 𝟏 𝟏 𝟏 𝒚 𝒚 + 𝒌 𝒚 𝒚 𝒚 𝒚 + 𝒌 | 𝑪𝟐 → 𝑪𝟐 − 𝑪𝟏 𝑪𝟑 → 𝑪𝟑 − 𝑪𝟏 = (𝟑𝒚 + 𝒌)| 𝟏 𝟎 𝟎 𝒚 𝒌 𝟎 𝒚 𝟎 𝒌 | = (𝟑𝒚 + 𝒌)(𝒌𝟐 − 𝟎) = 𝒌𝟐(𝟑𝒚 + 𝒌) Q39.Using properties of determinants, calculate the value of: | 𝑏 + 𝑐 𝑎 𝑎 𝑏 𝑐 + 𝑎 𝑏 𝑐 𝑐 𝑎 + 𝑏 | A. 2abc B. 4abc C. 3abc D. 5abc Answer: B Explanation: | 𝒃 + 𝒄 𝒂 𝒂 𝒃 𝒄 + 𝒂 𝒃 𝒄 𝒄 𝒂 + 𝒃 | 𝑹𝟏 → 𝑹𝟏 + 𝑹𝟐 + 𝑹𝟑 = | 𝟐(𝒃 + 𝒄) 𝟐(𝒂 + 𝒄) 𝟐(𝒂 + 𝒃) 𝒃 𝒄 + 𝒂 𝒃 𝒄 𝒄 𝒂 + 𝒃 | 𝐓𝐚𝐤𝐢𝐧𝐠 𝟐 𝐚𝐬 𝐜𝐨𝐦𝐦𝐨𝐧 = 𝟐 | 𝒃 + 𝒄 𝒂 + 𝒄 𝒂 + 𝒃 𝒃 𝒄 + 𝒂 𝒃 𝒄 𝒄 𝒂 + 𝒃 | 𝑹𝟏 → 𝑹𝟏 − 𝑹𝟐 = 𝟐 | 𝒄 𝟎 𝒂 𝒃 𝒄 + 𝒂 𝒃 𝒄 𝒄 𝒂 + 𝒃 | = 𝟐[𝒄{(𝒃 + 𝒂)(𝒄 + 𝒂) − 𝒃𝒄} − 𝟎 + 𝒂{𝒃𝒄 − 𝒄(𝒄 + 𝒂)}] = 𝟐(𝒂𝒃𝒄 + 𝒂𝒃𝒄) = 𝟒𝒂𝒃𝒄 Q40.Calculate the area between the given points: A (a,b+c),B (b,c+a),C (c,a+b)A (a,b+c),B (b,c+a),C (c,a+b) A. 0 B. 1 C. 2 D. 3 Answer: A Explanation: Here,  coordinates of points,  𝐴: (𝑎, 𝑏 + 𝑐);  𝐵: (𝑏, 𝑐 + 𝑎);  𝐶: (𝑐, 𝑎 + 𝑏) 𝐴𝑟𝑒𝑎 𝑜𝑓 𝛥𝐴𝐵𝐶 𝑐𝑎𝑛 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑎𝑠 𝛥 = 1 2 | 𝑎 𝑏 + 𝑐 1 𝑏 𝑐 + 𝑎 1 𝑐 𝑎 + 𝑏 1 | 𝑅2 → 𝑅2 − 𝑅1 𝑅3 → 𝑅3 − 𝑅1 = 1 2 | 𝑎 𝑏 + 𝑐 1 𝑏 − 𝑎 𝑎 − 𝑏 0 𝑐 − 𝑎 𝑎 − 𝑐 0 | = 1 2 (𝑎 − 𝑏)(𝑎 − 𝑐)| 𝑎 𝑏 + 𝑐 1 −1 1 0 −1 1 0 | = 1 2 (𝑎 − 𝑏)(𝑎 − 𝑐)[−1 + 1] = 0 Hence, the points A, B, and C are collinear. " Q41. 𝑇ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 | 𝑏2 − 𝑎𝑏 𝑏 − 𝑐 𝑏𝑐 − 𝑎𝑐 𝑎𝑏 − 𝑎2 𝑎 − 𝑏 𝑏2 − 𝑎𝑏 𝑏𝑐 − 𝑎𝑐 𝑐 − 𝑎 𝑎𝑏 − 𝑎2 | 𝑒𝑞𝑢𝑎𝑙𝑠 𝑡𝑜: A. abc (b – c) (c – a) (a – b) B. (b – c) (c – a) (a – b) C. (a + b + c) (b – c) (c – a) (a – b) D. 0 Answer: D Explanation: This can be solved as, A. 1 2 B. √ 3 2 C. √2 D. 2√ 3 4 Answer: A Explanation: 𝐆𝐢𝐯𝐞𝐧,  𝜟 = | 𝟏 𝟏 𝟏 𝟏 𝟏 + 𝐬𝐢𝐧𝜽 𝟏 𝟏 + 𝐜𝐨𝐬𝜽 𝟏 𝟏 | [𝐂𝟏 → 𝐂𝟏 − 𝐂𝟑  𝐚𝐧𝐝 𝐂𝟐 → 𝐂𝟐 − 𝐂𝟑   ] ⇒ | 𝟎 𝟎 𝟏 𝟎 𝐬𝐢𝐧𝜽 𝟏 𝐜𝐨𝐬𝜽 𝟎 𝟏 | = −𝐬𝐢𝐧𝜽𝐜𝐨𝐬𝜽 = −𝐬𝐢𝐧𝟐𝜽 𝟐 𝐍𝐨𝐰,  𝐬𝐢𝐧𝟐𝜽 𝐥𝐢𝐞𝐬 𝐛𝐞𝐭𝐰𝐞𝐞𝐧  − 𝟏 𝐚𝐧𝐝 𝟏,  𝐡𝐞𝐧𝐜𝐞 𝐦𝐚𝐱𝐢𝐦𝐮𝐦 𝐯𝐚𝐥𝐮𝐞  =   𝟏 𝟐 Q45.The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be A. 9 B. 3 C. -9 D. 6 Answer: B Explanation: This can be solved as, Area of triangle  =   1 2 | −3 0 1 3 0 1 0 𝑘 1 | ⇒ 9  =   1 2 (−3(0 − 𝑘) − 0 + 1(3𝑘 − 0)) ⇒ 18  =  3𝑘 + 3𝑘 ⇒ 𝑘 = 3 Q46.If A is a matrix of order 3 × 3, then |3A| = A. 27|A| B. 3|A| C. 9|A| D. |27A| Answer: A Explanation: 𝑭𝒐𝒓 𝒂 𝒔𝒒𝒖𝒂𝒓𝒆 𝒎𝒂𝒕𝒓𝒊𝒙 𝑨 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓𝒏, |𝒌𝑨| = 𝒌𝒏|𝑨| 𝑰𝒇 𝑨 𝒊𝒔 𝒂 𝒎𝒂𝒕𝒓𝒊𝒙 𝒐𝒇 𝒐𝒓𝒅𝒆𝒓 𝟑 × 𝟑 𝑻𝒉𝒆𝒏, |𝟑𝑨| = 𝟑𝟑|𝑨| = 𝟐𝟕 | 𝑨|. Q47. 𝐼𝑓 |2𝑥 5 8 𝑥 | = |6 −2 7 3 |, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝑖𝑠: A. +3 B. ± 3 C. ± 6 D. +6 Answer: C Explanation: This can be solved as, | 2𝑥 5 8 𝑥 | = | 6 −2 7 3 | 2𝑥2 − 40 = 18 + 14 2𝑥2 = 18 + 14 + 40 2𝑥2 = 72 𝑥2 = 36 𝑥 = ±6 Q48. 𝐼𝑓 𝑥m𝑦n = (𝑥 + 𝑦) m+n , 𝑡ℎ𝑒𝑛 𝑤ℎ𝑎𝑡 𝑖𝑠 d𝑦 d𝑥 ? A. d𝑦 d𝑥 = −𝑦 𝑥 B. d𝑦 d𝑥 = 𝑦 𝑥 C. d𝑦 d𝑥 = 𝑥 𝑦 D. None of these Answer: B Explanation: 𝒙𝒎𝒚𝒏 = (𝒙 + 𝒚)𝒎+𝒏 𝐓𝐚𝐤𝐢𝐧𝐠 𝐥𝐨𝐠 𝐛𝐨𝐭𝐡 𝐬𝐢𝐝𝐞𝐬 𝒎𝐥𝐨𝐠𝒙 + 𝒏𝐥𝐨𝐠𝒚 = (𝒎 + 𝒏)𝐥𝐨𝐠(𝒙 + 𝒚) 𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧 𝒎 𝒙 + 𝒏 𝒚 . 𝒅𝒚 𝒅𝒙 = (𝒎 + 𝒏)[ 𝟏 𝒙+𝒚 . (𝟏 + 𝒅𝒚 𝒅𝒙 )] 𝒎 𝒙 + 𝒏 𝒚 . 𝒅𝒚 𝒅𝒙 = 𝒎+𝒏 𝒙+𝒚 + 𝒎+𝒏 𝒙+𝒚 . 𝒅𝒚 𝒅𝒙 ( 𝒏 𝒚 − 𝒎+𝒏 𝒙+𝒚 ) 𝒅𝒚 𝒅𝒙 = 𝒎+𝒏 𝒙+𝒚 − 𝒎 𝒙 ( 𝒏𝒙+𝒏𝒚−𝒎𝒚−𝒏𝒚 𝒚(𝒙+𝒚) ) 𝒅𝒚 𝒅𝒙 = 𝒎𝒙+𝒏𝒙−𝒎𝒙−𝒎𝒚 𝒙(𝒙+𝒚) ( 𝒏𝒙−𝒎𝒚 𝒚 ) 𝒅𝒚 𝒅𝒙 = 𝒏𝒙−𝒎𝒚 𝒙 𝒅𝒚 𝒅𝒙 = 𝒚 𝒙 Q49. 𝐼𝑓 𝑥 = 𝑎(cos𝑡 + 𝑡sin𝑡), 𝑦 = 𝑎(sin𝑡 − 𝑡cos𝑡), 𝑡ℎ𝑒𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 d 2 𝑦 d𝑡2 A. d2𝑦 d𝑡2 = cos𝑡+sin𝑡 cos𝑡−𝑡sin𝑡 B. d2𝑦 d𝑡2 = 𝑡cos𝑡+sin𝑡 cos𝑡−𝑡sin𝑡 C. d2𝑦 d𝑡2 = 𝑡cos𝑡+sin𝑡 cos𝑡−sin𝑡 D. d2𝑦 d𝑡2 = 𝑡cos𝑡+𝑡sin𝑡 cos𝑡−𝑡sin𝑡 Answer: B Explanation: 𝒙 = 𝒂(𝐜𝐨𝐬𝒕 + 𝒕𝐬𝐢𝐧𝒕), 𝒚 = 𝒂(𝐬𝐢𝐧𝒕 − 𝒕𝐜𝐨𝐬𝒕) 𝒙 = 𝒂(𝐜𝐨𝐬𝒕 + 𝒕𝐬𝐢𝐧𝒕) 𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧 𝒅𝒙 𝒅𝒕 = 𝒂[−𝐬𝐢𝐧𝒕 + 𝒕𝐜𝐨𝐬𝒕 + 𝐬𝐢𝐧𝒕] 𝒅𝒙 𝒅𝒕 = 𝒂(𝒕𝐜𝐨𝐬𝒕) 𝒅𝟐𝒙 𝒅𝒕𝟐 = 𝒂[𝐜𝐨𝐬𝒕 − 𝒕𝐬𝐢𝐧𝒕]    . . . (𝟏) 𝒚 = 𝒂(𝐬𝐢𝐧𝒕 − 𝒕𝐜𝐨𝐬𝒕) 𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧 𝒅𝒚 𝒅𝒕 = 𝒂[𝐜𝐨𝐬𝒕 + 𝒕𝐬𝐢𝐧𝒕 − 𝐜𝐨𝐬𝒕] 𝒅𝒚 𝒅𝒕 = 𝒂(𝒕𝐬𝐢𝐧𝒕) 𝒅𝟐𝒚 𝒅𝒕𝟐 = 𝒂[𝒕𝐜𝐨𝐬𝒕 + 𝐬𝐢𝐧𝒕]       . . . . (𝟐) 𝐎𝐧 𝐝𝐢𝐯𝐢𝐝𝐢𝐧𝐠 . . . (𝟐) 𝐚𝐧𝐝 . . . (𝟏) 𝒅𝟐𝒚 𝒅𝒕𝟐 = 𝒕𝐜𝐨𝐬𝒕+𝐬𝐢𝐧𝒕 𝐜𝐨𝐬𝒕−𝒕𝐬𝐢𝐧𝒕 A. 31 27a B. 32a 27 C. 32 27a D. 32 5a Answer: C Explanation: 𝒙 = 𝒂𝐜𝐨𝐬𝟑𝜽 𝒅𝒙 𝒅𝜽 = 𝟑𝒂𝐜𝐨𝐬𝟐𝜽(−𝐬𝐢𝐧𝜽) 𝒅𝒙 𝒅𝜽 = −𝟑𝒂𝐜𝐨𝐬𝟐𝜽𝐬𝐢𝐧𝜽 𝒚 = 𝒂𝐬𝐢𝐧𝟑𝜽 𝒅𝒚 𝒅𝜽 = 𝟑𝒂𝐬𝐢𝐧𝟐𝜽𝐜𝐨𝐬𝜽 𝒅𝒚 𝒅𝒙 = 𝒅𝒚 𝒅𝜽 𝒅𝒙 𝒅𝜽 = 𝟑𝒂𝐬𝐢𝐧𝟐𝜽𝐜𝐨𝐬𝜽 −𝟑𝒂𝐜𝐨𝐬𝟐𝜽𝐬𝐢𝐧𝜽 𝒅𝒚 𝒅𝒙 = −𝐭𝐚𝐧𝜽 𝒅𝟐𝒚 𝒅𝒙𝟐 = −𝐬𝐞𝐜𝟐𝜽. 𝒅𝜽 𝒅𝒙 𝒅𝟐𝒚 𝒅𝒙𝟐 = 𝐬𝐞𝐜𝟐𝜽 𝟑𝒂𝐜𝐨𝐬𝟐𝜽𝐬𝐢𝐧𝜽 𝒅𝟐𝒚 𝒅𝒙𝟐 = 𝟏 𝟑𝒂𝐜𝐨𝐬𝟒𝜽𝐬𝐢𝐧𝜽 ( 𝒅𝟐𝒚 𝒅𝒙𝟐)𝜽 =  𝝅 𝟔 = 𝟏 𝟑𝒂𝐬𝐢𝐧 𝝅 𝟔 𝐜𝐨𝐬𝟒𝝅 𝟔 𝒅𝟐𝒚 𝒅𝒙𝟐 = 𝟏 𝟑𝒂( 𝟏 𝟐 )( √𝟑 𝟐 )𝟒 𝒅𝟐𝒚 𝒅𝒙𝟐 = 𝟑𝟐 𝟐𝟕𝒂 Q54.If f and g are two continuous functions on their common domain D, then Choose the incorrect or incomplete options from the statements given below. A. f + g is a continuous on D B. f ‒ g is a continuous on D C. f × g is a continuous on D D. f / g is a continuous on D Answer: D Explanation: If f and g are two continuous functions on their common domain D, then f + g is a continuous on D f ‒ g is a continuous on D f × g is a continuous on D f / g is continuous on D ‒ {x: g (x) ≠ 0}. Q55.Which of the following statements is false? A. |x| is continuous at x = 0 B. |x| is differentiable at x = 0 C. |x| is not continuous but not differentiable at x = 0. D. |x| is continuous at x = 1 and ‒1 Answer: C Explanation: The graph of |x| is shown below the graph is continuous at x = 0. But, the graph has a kink at x = 0. Therefore, the function f (x) = |x| is not continuous but not differentiable at x = 0. Q56. 𝐼𝑡 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓(𝑥) = 𝑥3 + 𝑏𝑥2 + 𝑎𝑥 𝑜𝑛 [1, 3], 𝑅𝑜𝑙𝑙𝑒’𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 ℎ𝑜𝑙𝑑𝑠 𝑤𝑖𝑡ℎ 𝑐 = 2 + 1 √3 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒: A. a = 11, b = − 6 B. a = −11, b = 6 C. a = −11, b = − 6 D. a =11, b = 6 Answer: A Explanation: 𝑮𝒊𝒗𝒆𝒏, 𝒇(𝒙) = 𝒙𝟑 + 𝒃𝒙𝟐 + 𝒂𝒙 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 [𝟏, 𝟑], ⇒ 𝒇′(𝒙) = 𝟑𝒙𝟐 + 𝟐𝒃𝒙 + 𝒂 ⇒ 𝒇′(𝒄) = 𝟑𝒄𝟐 + 𝟐𝒃𝒄 + 𝒂 𝑵𝒐𝒘, 𝑹𝒐𝒍𝒍𝒆’𝒔 𝒕𝒉𝒆𝒐𝒓𝒆𝒎 𝒉𝒐𝒍𝒅 𝒇𝒐𝒓 𝒇(𝒙) 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 [𝟏, 𝟑] 𝒘𝒊𝒕𝒉 𝒄 = 𝟐 + 𝟏 √𝟑 𝐓𝐡𝐞𝐫𝐞𝐟𝐨𝐫𝐞, 𝒇(𝟏) = 𝒇(𝟑) = 𝒇′(𝐜) = 𝟎 𝒇′(𝒄) = 𝟎 ⇒ 𝟑𝒄𝟐 + 𝟐𝒃𝒄 + 𝒂 = 𝟎 . . . (𝒊) 𝑺𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒊𝒏𝒈 𝒄 = 𝟐 + 𝟏 √𝟑  𝐢𝐧 . . . (𝒊),  𝐰𝐞 𝐡𝐚𝐯𝐞 𝟑(𝟐 + 𝟏 √𝟑 )𝟐 + 𝟐𝒃(𝟐 + 𝟏 √𝟑 ) + 𝒂 = 𝟎 ⇒ 𝒂 + 𝟒𝒃 + 𝟏𝟑 + 𝟐 √𝟑 (𝒃 + 𝟔) = 𝟎. . . . (𝒊) 𝑨𝒍𝒔𝒐, 𝒇(𝟏) = 𝒇(𝟑) ⇒ 𝟏 + 𝒃 + 𝒂 = 𝟐𝟕 + 𝟗𝒃 + 𝟑𝒂 ⇒ 𝒂 + 𝟒𝒃 + 𝟏𝟑 = 𝟎 . . . (𝒊𝒊) 𝑺𝒐𝒍𝒗𝒊𝒏𝒈 𝒃𝒐𝒕𝒉 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 … (𝒊) 𝒂𝒏𝒅 … (𝒊𝒊), 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒂 = 𝟏𝟏,  𝒃 = −𝟔. Q57.Find a real number c between (1, 2) such that Lagrange’s mean value theorem holds true for the function f (x) = x (x – 2) on the interval [1, 2]. A. 3/2 B. 2/3 C. ½ D. 5/4 Answer: A Q60. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕𝒔 𝒐𝒏 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒙𝟐 + 𝒚𝟐 − 𝟐𝒙– 𝟑 = 𝟎 𝒂𝒕 𝒘𝒉𝒊𝒄𝒉 𝒕𝒉𝒆 𝒕𝒂𝒏𝒈𝒆𝒏𝒕𝒔 𝒂𝒓𝒆 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝒙 − 𝒂𝒙𝒊𝒔. A. (1, 2) and (1, −2) B. (1, −2) and (1, 2) C. (−1, 2) and (1, −2) D. (−1, 2) and (1, 2) Answer: A Explanation: 𝒙𝟐 + 𝒚𝟐 − 𝟐𝒙 − 𝟑 = 𝟎 𝐎𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐭𝐢𝐨𝐧 𝟐𝒙 + 𝟐𝒚. 𝒅𝒚 𝒅𝒙 − 𝟐 = 𝟎 𝒅𝒚 𝒅𝒙 = 𝟏−𝒙 𝒚 Now, the tangents are parallel to the x-axis if the slope of the tangent = 0 ⇒ 𝑑𝑦 𝑑𝑥 = 0 ⇒ 1 − 𝑥 𝑦 = 0 ⇒ 1 − 𝑥 = 0 ∴ 𝑥 = 1 𝐴𝑙𝑠𝑜, 𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 1 𝑖𝑛 𝑥2 + 𝑦2 − 2𝑥 − 3 = 0 𝑊𝑒 ℎ𝑎𝑣𝑒, 𝑦2 = 4 𝑦 = ±2 Hence, (1, 2) and (1, −2) are the points at which the tangents are parallel to the x-axis. Q61. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕𝒔 𝒐𝒏 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆 𝒚 = 𝒙𝟑 𝒂𝒕 𝒘𝒉𝒊𝒄𝒉 𝒕𝒉𝒆 𝒔𝒍𝒐𝒑𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒊𝒔 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝒕𝒉𝒆 𝒚 − 𝒄𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕. A. (2, 27) B. (3, 27) C. (3, 25) D. (3, 26) Answer: Option B Explanation: y=x3 On differentiation 𝑑𝑦 𝑑𝑥 = 3𝑥2 The slope of the tangent at the point (x, y) is, 𝒅𝒚 𝒅𝒙 ](𝒙,𝒚) = 𝟑(𝒙)𝟐 When the slope of the tangent = equal to the y-coordinate of the point, then 𝒚 = 𝟑𝒙𝟐 𝑨𝒍𝒔𝒐, 𝒚 = 𝒙𝟑 ∴ 𝟑𝒙𝟐 = 𝒙𝟑 𝒙𝟐 (𝒙 – 𝟑) = 𝟎 𝒙 = 𝟎, 𝒙 = 𝟑 𝑾𝒉𝒆𝒏 𝒙 = 𝟎, 𝒕𝒉𝒆𝒏 𝒚 = 𝟎 𝒂𝒏𝒅 𝑾𝒉𝒆𝒏 𝒙 = 𝟑, 𝒕𝒉𝒆𝒏 𝒚 = 𝟑(𝟑)𝟐 = 𝟐𝟕 𝑯𝒆𝒏𝒄𝒆, 𝒕𝒉𝒆 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒑𝒐𝒊𝒏𝒕𝒔 𝒂𝒓𝒆 (𝟎, 𝟎) 𝒂𝒏𝒅 (𝟑, 𝟐𝟕). Q62.A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. The rate at which its area is increasing when radius is 3.2 cm is A. 32.0π cm2/s B. 3.20π cm2/s C. 0.320π cm2/s D. None of these Answer: C Explanation: Let r be the radius of the given disc and A be its area. Then, 𝐴 = 𝜋𝑟2 ⇒ 𝑑𝐴 𝑑𝑡 = 2𝜋𝑟 𝑑𝑟 𝑑𝑡 Now approximate rate of increase of radius 𝑑𝑟 = 𝑑𝑟 𝑑𝑡 𝛥𝑡 = 0.05 cm/s ∴ the approximate rate of increase in area is given by 𝑑A = 𝑑A 𝑑𝑡 (𝛥𝑡) = 2𝜋𝑟( 𝑑𝑟 𝑑𝑡 𝛥𝑡)     = 2𝜋(3.2)(0.05) = 0.320𝜋 𝑐𝑚2/𝑠. Q63. 𝑰𝒕 𝒊𝒔 𝒈𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕 𝒂𝒕 𝒙 = 𝟏, 𝒕𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒙𝟒 − 𝟔𝟐𝒙𝟐 + 𝒂𝒙 + 𝟗 𝒂𝒕𝒕𝒂𝒊𝒏𝒔 𝒊𝒕𝒔 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆, 𝒐𝒏 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 [𝟎, 𝟐]. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒂. A. 120 B. 110 C. -110 D. -120 Answer: A Explanation: 𝒇(𝒙) = 𝒙𝟒 − 𝟔𝟐𝒙𝟐 + 𝒂𝒙 + 𝟗 𝒇′(𝒙) = 𝟒𝒙𝟑 − 𝟏𝟐𝟒𝒙 + 𝒂 𝒇′(𝟏) = 𝟒 − 𝟏𝟐𝟒 + 𝒂 = −𝟏𝟐𝟎 + 𝒂 𝒇′(𝟏) = 𝟎 −𝟏𝟐𝟎 + 𝒂 = 𝟎 𝒂 = 𝟏𝟐𝟎 Q64.Find the rate of change of the area of a circle with respect to its radius r when r = 4 cm A. 5π B. 4π C. 8π D. 2π Answer: C Explanation: The area of a circle, A=πr2 On differentiating both sides On differentiating both sides 𝑑𝐴 𝑑𝑟 = 𝑑 𝑑𝑟 (𝜋𝑟2) = 2𝜋𝑟 𝐻𝑒𝑟𝑒 𝑑𝐴 𝑑𝑟 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑟 = 4 𝑐𝑚 𝑑𝐴 𝑑𝑟 = 2𝜋𝑟 = 2𝜋(4) = 8𝜋 Explanation: ∫ 𝒅𝒙 𝟏+𝒙𝟐 √𝟑 𝟏 𝑳𝒆𝒕 𝑰 = ∫ 𝒅𝒙 𝟏+𝒙𝟐 √𝟑 𝟏 = [𝐭𝐚𝐧−𝟏𝒙]𝟏 √𝟑 = 𝐭𝐚𝐧−𝟏(√𝟑) − 𝐭𝐚𝐧−𝟏(𝟏) = 𝝅 𝟑 − 𝝅 𝟒 = 𝝅 𝟏𝟐 Q69. 𝑬𝒗𝒂𝒍𝒖𝒂𝒕𝒆: ∫ (𝟏+𝐥𝐨𝐠𝒙)𝟐 𝒙 𝒅𝒙 A. (1+log𝑥)3 3 + 𝐶 B. (1+log𝑥)3 2 + 𝐶 C. (1+log𝑥)2 2 + 𝐶 D. (1−log𝑥)3 3 + 𝐶 Answer: A Explanation: ∫ (𝟏+𝐥𝐨𝐠𝒙)𝟐 𝒙 𝒅𝒙 𝑳𝒆𝒕 𝟏 + 𝐥𝐨𝐠𝒙 = 𝒕 𝟏 𝒙 𝒅𝒙 = 𝒅𝒕 𝑷𝒖𝒕 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝟏 𝒙 𝒅𝒙 𝒊𝒏 𝒕𝒉𝒆 𝐞𝐱𝐩𝒓𝒆𝒔𝒔𝒊𝒐𝒏 = ∫ 𝒕𝟐𝒅𝒕 = 𝒕𝟑 𝟑 + 𝒄 = (𝟏+𝐥𝐨𝐠𝒙)𝟑 𝟑 + 𝒄 Q70. 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 ‘𝒂’ 𝒊𝒇 ∫ 𝟑𝒙𝟐𝒅𝒙 = 𝟖 𝒂 𝟎 A. 1 B. 3 C. 2 D. 4 Answer: C Explanation: ∫ 𝟑𝒙𝟐𝒅𝒙 = 𝟖 𝒂 𝟎 𝟑. [ 𝒙𝟑 𝟑 ]𝟎 𝒂 = 𝟖 [𝒙𝟑]𝟎 𝒂 = 𝟖 𝒂𝟑 − 𝟎 = 𝟖 𝒂 = 𝟐 Q71. ∫𝐥𝐨𝐠(𝟏 + 𝒙𝟐)𝒅𝒙 =? A. xlog(1+x2)−2x+2tan−1x+C B. −xlog(1+x2)−2x−2tan−1x+C C. xlog(1+x2)+2x+2tan−1x+C D. xlog(1+x2)−2x+2tan−1(1+x2)+C Answer: A Explanation: 𝐋𝐞𝐭,  𝐈 = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)𝒅𝒙 ⇒ 𝐈 = ∫ 𝟏 ⋅ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)𝒅𝒙 ⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − ∫ (𝒙 ⋅ 𝟏 𝟏+𝒙𝟐 ⋅ 𝟐𝒙)𝒅𝒙 ⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝟐∫ ( 𝒙𝟐 𝟏+𝒙𝟐)𝒅𝒙 ⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝟐∫ ( 𝒙𝟐+𝟏−𝟏 𝟏+𝒙𝟐 )𝒅𝒙 ⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝟐∫ 𝒅𝒙 + 𝟐∫ 𝟏 𝟏+𝒙𝟐 𝒅𝒙 ⇒ 𝐈 = 𝒙𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝟐𝒙 + 𝟐𝐭𝐚𝐧−𝟏𝒙 + 𝐂. Q72. 𝑰𝒇 ∫ 𝒆𝒕 𝟏+𝒕 𝒅𝒕 𝟏 𝟎   = 𝒂, 𝒕𝒉𝒆𝒏 ∫ 𝒆𝒕 (𝟏+𝒕)𝟐 𝒅𝒕 𝟏 𝟎   =? A. 𝑎 − 1 + 𝑒 2 B. 𝑎 + 1 − 𝑒 2 C. a D. a2 Answer: B Explanation: According to question ∫ 𝑒𝑡 1 + 𝑡 𝑑𝑡 1 0   = 𝑎 ⇒ | 1 1 + 𝑡 𝑒𝑡|0 1 − ∫[− 𝑒𝑡 (1 + 𝑡)2 ]𝑑𝑡 1 0   = 𝑎 ⇒ 𝑒 1 + 1 − 1 + ∫ 𝑒𝑡 (1 + 𝑡)2 𝑑𝑡 1 0   = 𝑎 ⇒ 𝑒 2 − 1 + ∫ 𝑒𝑡 (1 + 𝑡)2 𝑑𝑡 1 0   = 𝑎 ⇒ ∫ 𝑒𝑡 (1 + 𝑡)2 𝑑𝑡 1 0   = 𝑎 + 1 − 𝑒 2 Q73. ∫ 𝟏 𝐬𝐢𝐧𝟐𝒙𝐜𝐨𝐬𝟐𝒙 𝒅𝒙 =? A. tanx + cotx + C B. (tanx + cotx)2 + C C. tanx − cotx + C D. (tanx − cotx )2+ C Answer: C Explanation: This can be solved as, Explanation: The given equations are 2𝑦 = 𝑥 . . . . . . . . . . . . . . . . . . . . . (1) 𝑎𝑛𝑑 𝑥2 + 𝑦2 = 32 . . . . . . . . . . . (2) Solving (1) and (2) as we find that the line and the circle meet at B(4, 4) in the first quadrant (Fig). Draw perpendicular BM to the x-axis. Area of the region bounded by a circle and a line can be drawn as, Since general equation of the circle passing through origin is : 𝑥2 + 𝑦2 = 𝑟2 𝐶𝑜𝑚𝑝𝑎𝑖𝑟𝑖𝑛𝑔 𝑥2 + 𝑦2 = 32 𝑔𝑖𝑣𝑒𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑥2 + 𝑦2 = 𝑟2, 𝑤𝑒 𝑔𝑒𝑡 𝑟2 = 32 ⇒ 𝑟 = 4√2 For region OMBO, limits will be from 0 to intersecting point i.e 4 and for the region BMAB, limits will be 4 to 4√2 Therefore, the required area = area of the region OBMO + area of the region BMAB. = ∫ 𝒚𝟏 𝟒 𝟎 𝒅𝒙 + ∫ 𝒚𝟐 𝟒√𝟐 𝟒 𝒅𝒙 = ∫ 𝟒 𝟎 𝒙𝒅𝒙 + ∫ √𝟑𝟐 − 𝒙𝟐 𝟒√𝟐 𝟒 𝒅𝒙 = [ 𝒙𝟐 𝟐 ]𝟎 𝟒 + [ 𝟏 𝟐 𝒙√𝟑𝟐 − 𝒙𝟐 + 𝟑𝟐 𝟐 𝐬𝐢𝐧−𝟏 𝒙 √𝟑𝟐 ]𝟒 𝟒√𝟐 = [ 𝟏𝟔 𝟐 − 𝟎] + [ 𝟏 𝟐 𝟒√𝟐√𝟑𝟐 − 𝟑𝟐 + 𝟏𝟔𝐬𝐢𝐧−𝟏 𝟒√𝟐 𝟒√𝟐 ] − [ 𝟏 𝟐 𝟒√𝟑𝟐 − 𝟏𝟔 + 𝟏𝟔𝐬𝐢𝐧−𝟏 𝟒 𝟒√𝟐 ] = 𝟖 + [𝟎 + 𝟏𝟔𝐬𝐢𝐧−𝟏𝟏] − [𝟐 ⋅ 𝟒 + 𝟏𝟔𝐬𝐢𝐧−𝟏 𝟏 √𝟐 ] = 𝟖 + 𝟏𝟔𝐬𝐢𝐧−𝟏𝟏 − 𝟖 − 𝟏𝟔𝐬𝐢𝐧−𝟏 𝟏 √𝟐 = 𝟏𝟔( 𝝅 𝟐 − 𝝅 𝟒 ) = 𝟏𝟔( 𝟐𝝅 − 𝝅 𝟒 ) = 𝟒𝝅 Hence the required area is 4π Square units. Q76. 𝑻𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒆𝒈𝒊𝒐𝒏 {(𝒙, 𝒚): 𝟎 ≤ 𝒚 ≤ 𝒙𝟐 + 𝟏, 𝟎 ≤ 𝒚 ≤ 𝒙 + 𝟏; 𝟎 ≤ 𝒙 ≤ 𝟐} 𝒊𝒔: A. 16/3 sq units B. 13/3 sq units C. 11/3 sq units D. 7/3 sq units Answer: C Explanation: 𝑻𝒉𝒆 𝒄𝒖𝒓𝒗𝒆𝒔 𝒂𝒓𝒆𝒙 = 𝟐, 𝒚 = 𝒙 + 𝟏, 𝒚 = 𝒙𝟐 + 𝟏 𝑻𝒉𝒆 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒄𝒖𝒓𝒗𝒆𝒔 𝒚 = 𝒙 + 𝟏, 𝒚 = 𝒙𝟐 + 𝟏 𝒙𝟐 + 𝟏 = 𝒙 + 𝟏 𝒙(𝒙 − 𝟏) = 𝟎 𝒙 = 𝟎, 𝟏 The shaded area is the required area. ∫ 𝑦2 1 0 𝑑𝑥 − ∫ 𝑦1 1 0 𝑑𝑥 + ∫ 𝑦2 2 1 𝑑𝑥 = ∫(𝑥 + 1) 1 0 𝑑𝑥 − ∫(𝑥2 + 1) 1 0 𝑑𝑥 + ∫(𝑥 + 1) 2 1 𝑑𝑥 = [ 𝑥2 2 + 𝑥]0 1 − [ 𝑥3 3 + 𝑥]0 1 + [ 𝑥2 2 + 𝑥]1 2 = [ 1 2 + 1] − [ 1 3 + 1] + [ 4 2 + 2 − 1 2 − 1] = 1 2 − 1 3 + 4 − 1 2 = − 1 3 + 4 = −1 + 12 6 = 11 3 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠 Q80.Calculate the area under the curve y = 2√x included between the lines x = 0 and x = 1. A. 2/3 sq units B. 1 sq units C. 4/3 sq units D. 5/3 sq units Answer: C Explanation: Since 𝐴𝑟𝑒𝑎 = ∫ 2√𝑥𝑑𝑥 1 0 ⇒ 𝐴𝑟𝑒𝑎 = 2[ 𝑥3/2 3 ⋅ 2]0 1 ⇒ 𝐴𝑟𝑒𝑎 = 2[ 2 3 ⋅ 1 − 0] = 4 3 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠 Q81. 𝑾𝒉𝒂𝒕 𝒊𝒔 𝒕𝒉𝒆 𝒑𝒂𝒓𝒕𝒊𝒄𝒖𝒍𝒂𝒓 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 (𝒕𝒂𝒏−𝟏𝒚 − 𝒙) 𝒅𝒚 = (𝟏 + 𝒚𝟐) 𝒅𝒙? (𝑮𝒊𝒗𝒆𝒏 𝒕𝒉𝒂𝒕, 𝒂𝒕 𝒙 = 𝟎 𝒘𝒆 𝒉𝒂𝒗𝒆 𝒚 = 𝟎) A. 𝑥 = tan−1𝑦 + 𝑒−tan−1𝑦 B. 𝑥 = tan−1𝑦 − 1 + 𝑒tan−1𝑦 C. 𝑥 = tan−1𝑦 − 5 + 𝑒−tan−1𝑦 D. 𝑥 = tan−1𝑦 − 1 + 𝑒−tan−1𝑦 Answer: D Explanation: (tan−1𝑦 − 𝑥)𝑑𝑦 = (1 + 𝑦2)𝑑𝑥 𝑑𝑥 𝑑𝑦 = (tan−1𝑦 − 𝑥) (1 + 𝑦2) 𝑑𝑥 𝑑𝑦 = tan−1𝑦 1 + 𝑦2 − 𝑥 1 + 𝑦2 𝑑𝑥 𝑑𝑦 + 𝑥 1 + 𝑦2 = tan−1𝑦 1 + 𝑦2 Integrating factor = 𝑒 ∫ 𝑑𝑦 1+𝑦2 = 𝑒tan−1𝑦 𝑁𝑜𝑤, 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑓𝑎𝑐𝑡𝑜𝑟 𝑒tan−1𝑦 𝑑𝑥 𝑑𝑦 + 𝑒tan−1𝑦. 𝑥 1 + 𝑦2 = 𝑒tan−1𝑦. tan−1𝑦 1 + 𝑦2 𝑥. 𝑒tan−1𝑦 = ∫ 𝑒tan−1𝑦. tan−1𝑦 1 + 𝑦2 . 𝑑𝑦 Let 𝑡 = tan−1𝑦 𝑑𝑡 = 𝑑𝑦 1 + 𝑦2 𝑥. 𝑒tan−1𝑦 = ∫ 𝑒t. 𝑡𝑑𝑡 Applying by parts 𝑥. 𝑒tan−1𝑦 = 𝑡(𝑒𝑡) − ∫ 𝑒t. { 𝑑 𝑑𝑡 (𝑡)}𝑑𝑡 𝑥. 𝑒tan−1𝑦 = 𝑡(𝑒𝑡) − ∫ 𝑒t𝑑𝑡 𝑥. 𝑒tan−1𝑦 = 𝑡𝑒𝑡 − 𝑒𝑡 + 𝐶 𝑥. 𝑒tan−1𝑦 = tan−1𝑦(𝑒tan−1𝑦) − 𝑒tan−1𝑦 + 𝐶 When 𝑥 = 0, 𝑦 = 0 0 = tan−10(𝑒tan−10) − 𝑒tan−10 + 𝐶 0 = −𝑒0 + 𝐶 𝐶 = 1 Q82.The differential equation of the family of parabolas having vertex at the origin and axis along positive y-axis is A. xy'−y=0 B. xy'+2y=0 C. xy'−7y=0 D. xy'−2y=0 Answer: D Explanation: Vertex = (0, 0) The equation of the parabola 𝑥2 = 4𝑎𝑦     . . . . (1) On differentiation 2𝑥 = 4𝑎𝑦′ 𝑥 = 2𝑎𝑦′ 𝑎 = 𝑥 2𝑦′ Put value of a in equation (1) 𝑥2 = 4 𝑥 2𝑦′ 𝑦 𝑦′𝑥2 = 2𝑥𝑦 𝑦′𝑥 = 2𝑦 𝑥𝑦′ − 2𝑦 = 0 This is the required differential equation. Q83.Write the differential equation representing the family of curves y = m x, where m is an arbitrary constant. A. xdy−ydx=0 B. xdy−dx=0 C. xdy+ydx=0 D. dy−ydx=0 Answer: A Explanation: We have, Answer: B Explanation: This can be solved as follows: projection of a + b on c = (𝑎 + 𝑏) ⋅ 𝑐 |𝑐| = (5𝑖 − 3𝑗 + 3𝑘) ⋅ (𝑖 + 2𝑗 + 2𝑘) √1 + 4 + 4 = 5 √9 = 5 3 Q88. 𝑰𝒇 𝒂 → , 𝒃 → , 𝒄 → 𝒂𝒓𝒆 𝒕𝒉𝒆 𝒎𝒖𝒕𝒖𝒂𝒍𝒍𝒚 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝒗𝒆𝒄𝒕𝒐𝒓𝒔 𝒆𝒂𝒄𝒉 𝒐𝒇 𝒎𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆 𝒖𝒏𝒊𝒕𝒚, 𝒕𝒉𝒆𝒏 |𝒂 → + 𝒃 → + 𝒄 → | = A. 1 B. 3 C. √3 D. 3√3 Answer: C Explanation: 𝑆𝑖𝑛𝑐𝑒 𝑎 → , 𝑏 → , 𝑐 → 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑚𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑒𝑎𝑐ℎ 𝑜𝑓 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑢𝑛𝑖𝑡𝑦, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, |𝑎| = |𝑏| = |𝑐| = 1 𝑎 ⋅ 𝑏 = 𝑎 ⋅ 𝑐 = 𝑐 ⋅ 𝑏 = 0 𝑁𝑜𝑤, |𝑎 → + 𝑏 → + 𝑐 → |2 = |𝑎|2 + |𝑏|2 + |𝑐|2 + 2𝑎 ⋅ 𝑏 + 2𝑎 ⋅ 𝑐 + 2𝑐 ⋅ 𝑏 ⇒ |𝑎 → + 𝑏 → + 𝑐 → |2 = 1 + 1 + 1 + 0 + 0 + 0 ⇒ |𝑎 → + 𝑏 → + 𝑐 → |2 = 3 ⇒ |𝑎 → + 𝑏 → + 𝑐 → |2 = √3 Q89. 𝑰𝒇 𝒗𝒆𝒄𝒕𝒐𝒓𝒔 𝒊 ^ + 𝟐𝒋 ^ + 𝟑𝒌 ^  𝒂𝒏𝒅 𝟑𝒊 ^ − 𝟐𝒋 ^ + 𝒌 ^ 𝒓𝒆𝒑𝒓𝒆𝒔𝒆𝒏𝒕𝒔 𝒕𝒉𝒆 𝒂𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝒔𝒊𝒅𝒆𝒔 𝒐𝒇 𝒂 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍𝒐𝒈𝒓𝒂𝒎, 𝒕𝒉𝒆 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍𝒐𝒈𝒓𝒂𝒎 𝒊𝒔 A. 4√3 B. 6√3 C. 8√3 D. 16√3 Answer: C Explanation: We have, (𝑎 × 𝑏) = | 𝑖 𝑗 𝑘 1 2 3 3 −2 1 | (𝑎 × 𝑏) = 𝑖(2 + 6) − 𝑗(1 − 9) + 𝑘(−2 − 6) = 8𝑖 + 8𝑗 − 8𝑘 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑎𝑟𝑒𝑎 |(𝑎 × 𝑏)| = |8𝑖 + 8𝑗 − 8𝑘| = √|82 + 82 + 82| = 8√3 Q90.If the position vector of three consecutive vertices of any parallelogram are respectively i + j + k, i + 3j + 5k, 7i + 9j + 11k, then the position vector of fourth vertex is - A. 6(i+j+k) B. 7(i+j+k) C. 2j−4k D. 6i+8j+10k Answer: B Explanation: Let the position vector of 4 vertices are 𝐴(𝑖 + 𝑗 + 𝑘), 𝐵(𝑖 + 3𝑗 + 5𝑘), 𝐶(7𝑖 + 9𝑗 + 11𝑘) 𝐷(𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘) Since the diagonals of parallelogram bisects, so the midpoint of AC and BD coincides, therefore, 7 + 1 2 = 1 + 𝑥 2 ⇒ 𝑥 = 7 9 + 1 2 = 3 + 𝑦 2 ⇒ 𝑦 = 7 11 + 1 2 = 5 + 𝑧 2 ⇒ 𝑧 = 7 Therefore, fourth position vector is 7i + 7j + 7k Q91.If vector 3j + 2j + 8k and 2i + xj + k are perpendicular then x is equal to A. 7 B. -7 C. 5 D. -4 Answer: B Explanation: For perpendicular vectors, 3 ⋅ 2 + 2 ⋅ 𝑥 + 8 ⋅ 1 = 0 ⇒ 6 + 2𝑥 + 8 = 0 ⇒ 2𝑥 = −14 ⇒ 𝑥 = −7 Q92.If i + 2j + 3k is parallel to sum of the vector 3i + λj + 2k and -2i + 3j + k, then equals to: A. 1 B. -1 C. 2 D. -2 Answer: B Explanation: According to given condition 𝑖 + 2𝑗 + 3𝑘 = 3𝑖 + 𝜆𝑗 + 2𝑘  + −2𝑖 + 3𝑗 + 𝑘 ⇒ 𝑖 + 2𝑗 + 3𝑘 = 𝑖 + (3 + 𝜆)𝑗 + 3𝑘 ⇒ 𝑖 + 2𝑗 + 3𝑘 = 𝑖 + (3 + 𝜆)𝑗 + 3𝑘 For vectors to be parallel 𝑥 − 1 2 = 𝑦 − 2 3 = 𝑧 − 3 4 = 𝜆 𝑥 = 2𝜆 + 1 𝑦 = 3𝜆 + 2 𝑧 = 4𝜆 + 3 The coordinates of any point on second line are given by 𝑥 − 4 5 = 𝑦 − 1 2 = 𝑧 = 𝜇 𝑥 = 5𝜇 + 4 𝑦 = 2𝜇 + 1 𝑧 = 𝜇 The point of intersection can be calculated by considering corresponding coordinates equal as, 2𝜆 + 1 = 5𝜇 + 4 ⇒ 2𝜆 − 5𝜇 = 3 3𝜆 + 2 = 2𝜇 + 1 ⇒ 3𝜆 − 2𝜇 = −1 4𝜆 + 3 = 𝜇 ⇒ 4𝜆 − 𝜇 = −3 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦𝑖𝑛𝑔 2𝜆 − 5𝜇 = 3 𝑏𝑦 2 𝑎𝑠, 2𝜆 − 5𝜇 = 3 ⇒ 2(2𝜆 − 5𝜇 = 3) ⇒ 4𝜆 − 10𝜇 = 6 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑛𝑔 4𝜆 − 𝜇 = −3 𝑓𝑟𝑜𝑚 4𝜆 − 10𝜇 = 6 𝑎𝑠, 4𝜆 − 10𝜇 = 6 −(4𝜆 − 𝜇 = −3) _ −9𝜇 = 9 ⇒ 𝜇 = −1 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝜇 = −1 𝑖𝑛 4𝜆 − 𝜇 = −3 𝑤𝑒 𝑔𝑒𝑡 4𝜆 − (−1) = −3 ⇒ 4𝜆 + 1 = −3 ⇒ 4𝜆 = −4 ⇒ 𝜆 = −1 𝑇ℎ𝑒 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒 𝑥 = 2(−1) + 1 = −1 𝑦 = 3(−1) + 2 = −1 𝑧 = 4(−1) + 3 = −1 𝐼. 𝑒 (−1, −1, −1) Q95.The equation of the plane through the line of intersection of the planes x + y + z and 2x + 3y + 4z = 5 which is ⊥ of the plane x - y + z = 0 is: A. x−z+2=0 B. x+z+2=0 C. y+z+2=0 D. y+z−2=0 Answer: A Explanation: The equation of a plane passing through the line of intersection of the planes 𝑥 + 𝑦 + 𝑧 = 1 𝑎𝑛𝑑 2𝑥 + 3𝑦 + 4𝑧 = 5 𝑖𝑠: (𝒙 + 𝒚 + 𝒛 − 𝟏) + 𝝀(𝟐𝒙 + 𝟑𝒚 + 𝟒𝒛 − 𝟓) = 𝟎 ⇒ (𝟏 + 𝟐𝝀)𝒙 + (𝟏 + 𝟑𝝀)𝒚 + (𝟏 + 𝟒𝝀)𝒛 − 𝟏 − 𝟓𝝀 = 𝟎. . . . . . . . . . (𝟏) 𝑺𝒊𝒏𝒄𝒆 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝒕𝒐 𝒕𝒉𝒆 𝒑𝒍𝒂𝒏𝒆 𝒙 − 𝒚 + 𝒛 = 𝟎 … … … … … … (𝟐) 𝑺𝒊𝒏𝒄𝒆 𝒇𝒐𝒓 𝒕𝒘𝒐 𝒑𝒍𝒂𝒏𝒆𝒔, 𝒂𝟏𝒙 + 𝒃𝟏𝒚 + 𝒄𝟏𝒛 + 𝒅𝟏 = 𝟎 𝒂𝟐𝒙 + 𝒃𝟐𝒚 + 𝒄𝟐𝒛 + 𝒅𝟐 = 𝟎 𝑪𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒐𝒇 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓𝒊𝒕𝒚 𝒊𝒔 𝒂𝟏𝒂𝟐 + 𝒃𝟏𝒃𝟐 + 𝒄𝟏𝒄𝟐 = 𝟎 𝑼𝒔𝒊𝒏𝒈 𝒕𝒉𝒊𝒔 𝒊𝒏 𝒂𝒃𝒐𝒗𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 (𝟏) 𝒂𝒏𝒅 (𝟐) 𝒂𝒔, (𝟏 + 𝟐𝝀)𝟏 − (𝟏 + 𝟑𝝀)𝟏 + (𝟏 + 𝟒𝝀)𝟏 = 𝟎 𝟏 + 𝟐𝝀 − 𝟏 − 𝟑𝝀 + 𝟏 + 𝟒𝝀 = 𝟎 𝟑𝝀 = −𝟏 𝝀 = − 𝟏 𝟑 𝑺𝒖𝒃𝒔𝒕𝒊𝒕𝒖𝒕𝒊𝒏𝒈 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝝀 𝒊𝒏 𝒂𝒃𝒐𝒗𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒑𝒍𝒂𝒏𝒆, 𝑻𝒉𝒆 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒑𝒍𝒂𝒏𝒆 𝒊𝒔 : (𝒙 + 𝒚 + 𝒛 − 𝟏) + 𝝀(𝟐𝒙 + 𝟑𝒚 + 𝟒𝒛 − 𝟓) = 𝟎 ⇒ (𝒙 + 𝒚 + 𝒛 − 𝟏) − 𝟏 𝟑 (𝟐𝒙 + 𝟑𝒚 + 𝟒𝒛 − 𝟓) = 𝟎 ⇒ 𝟑𝒙 + 𝟑𝒚 + 𝟑𝒛 − 𝟑 − 𝟐𝒙 − 𝟑𝒚 − 𝟒𝒛 + 𝟓 𝟑 = 𝟎 ⇒ 𝟑𝒙 + 𝟑𝒚 + 𝟑𝒛 − 𝟑 − 𝟐𝒙 − 𝟑𝒚 − 𝟒𝒛 + 𝟓 = 𝟎 ⇒ 𝒙 − 𝒛 + 𝟐 = 𝟎 Q96.If the points (1,1, p) and (-3,0,1)be equidistant from the plane A. 5/3 or -5/3 B. 1/3 or -7/3 C. 2/3 or -7/3 D. 1/3 or -2/3 Answer: B Q97.The position vector of the foot of perpendicular drawn from the point P(1,8,4) to the line joining A(0,-1,3) and B(5,4,4) is: A. (5, 5, 5) B. (5, 4, 4) C. (4, 4, 4) D. (5,-4, 4) Answer: B Q98.The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by A. (2, 0, 0) B. (0, 5, 0) C. (0, 0, 7) D. (0, 5, 7) Answer: A Explanation: Since on x-axis y and z coordinates are zero, therefore, coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) would be (2, 0, 0) Q99.A line makes equal angles with co-ordinate axis. Direction cosines of this line are: A. 1 √2 , 1 √2 , 1 √2 B. 1 √3 , 1 √3 , 1 √3 C. 1 2 , 1 2 , 1 2 D. 1 5 , 1 5 , 1 5 Answer: B Explanation: Let the line makes angle α with each of the axis. Then, its direction cosines are cos α, cos α, cos α 𝑆𝑖𝑛𝑐𝑒 𝑐𝑜𝑠2𝛼 + 𝑐𝑜𝑠2𝛼 + 𝑐𝑜𝑠2𝛼 = 1. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑐𝑜𝑠2𝛼 + 𝑐𝑜𝑠2𝛼 + 𝑐𝑜𝑠2𝛼 = 1 ⇒ 3𝑐𝑜𝑠2𝛼 = 1 ⇒ 𝑐𝑜𝑠2𝛼 = 1 3 ⇒ 𝑐𝑜𝑠𝛼 = 1 √3 Q101.The feasible solution for a LPP is shown in Fig.. Let Z = 3x - 4y be the objective function. Minimum of Z occurs at A. (0, 0) B. (0, 8) C. (5, 0) D. (4, 10) Answer: B Explanation: Corner points Z value for (0,0) 0 (5,0) 15 (6,5) -2 (6,8) -14 (4,10) -28 (0,8) -32 (minimum) Q102.The common region determined by all the linear constraints of a LPP is/are called: A. corner points B. Feasible region C. unbounded region D. Bounded region Answer: B Explanation: Feasible region Q103.Any point in the feasible region that gives the maximum or minimum value of the objective function is also known as A. optimal solution B. infeasible solution C. constraints D. linear values Answer: A Explanation: Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution. Q104.The inequations or equations in the variables of linear programming problems which describes the condition under which the optimization (maximization or minimization) is to be accomplished are called A. objective functions B. objective variables C. constraints D. decision variables Answer: C Explanation: The in equations or equations in the variables of linear programming problems which describes the condition under which the optimization (maximization or minimization) is to be accomplished are called constraints. Q11.Linear function Z = ax + by, where a, b are constants, which has to be maximized or minimized is called a linear objective function. Here, x and y are known as A. Constraints B. Decision variables C. Objective variables D. None of these Answer: B Explanation: Linear function Z = ax + by, where a, b are constants, which has to be maximised or minimized is called a linear objective function. Variables x and y are known as decision variables. Q105.Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z A. only has a minimum value on R B. only has a minimum value on R C. may have a maximum and a minimum value on R D. must has both a maximum and a minimum value on R Answer: D Explanation: Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R Q106.How many times must a fair coin be tossed so that the probability of getting at least one head is more than 80%. A. 5 B. 4 C. 3 D. 2 Answer: C Explanation: We have given Probability of getting at least one head is 80%. This implies the value of head can be 1 or 2 but not 0. This implies Q109.A cricket player has an average score of 40 runs for 52 innings played by him. In an innings his highest score exceeds his lowest score by 100 runs. If these two innings are excluded, his average of the remaining 50 innings is 38 runs. Find his highest score in an innings. A. 80 B. 40 C. 140 D. 60 Answer: C Explanation: Let the lowest score of the cricketer be X. Cricketer’s Highest score = X + 100 X + X + 100 = 40 × 52 -50 × 38 2X+100 = 2080 -1900 2X = 80 X = 40 Highest score is 140 runs. Q110.Of the four numbers, whose average is 60, the first is one-fourth of the sum of the last three. The second number is one-third of the sum of other three, and the third is half of the other three. Find the fourth number. A. 52 B. 48 C. 80 D. 60 Answer: A Explanation: 𝑳𝒆𝒕 𝒕𝒉𝒆 𝒏𝒐𝒔. 𝒃𝒆 𝒂, 𝒃, 𝒄, 𝒅 𝒂 = 𝟏 𝟒 (𝒃 + 𝒄 + 𝒅) ⇒ 𝟒𝒂 = 𝒃 + 𝒄 + 𝒅 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 = 𝟔𝟎 𝒂+𝒃+𝒄+𝒅 𝟒 = 𝟔𝟎 𝟓𝒂 𝟒 = 𝟔𝟎 ⇒ 𝒂 = 𝟒𝟖 𝑺𝒊𝒎𝒊𝒍𝒂𝒓𝒍𝒚, 𝒃 = 𝟔𝟎 & 𝒄 = 𝟖𝟎 𝑨𝒍𝒔𝒐, 𝒊𝒕 𝒊𝒔 𝒌𝒏𝒐𝒘𝒏 𝒕𝒉𝒂𝒕 𝒂𝒗𝒆𝒓𝒂𝒈𝒆 = 𝟔𝟎, => 𝒂 + 𝒃 + 𝒄 + 𝒅 = 𝟐𝟒𝟎 => 𝒅 = 𝟓𝟐 Q111.The average annual income (in Rs.) of certain group of illiterate workers is A and that of other workers is W. The number of illiterate workers is 11 times that of other workers. Then the average monthly income (in Rs.) of all the workers is : A. 𝐴+𝑊 2 B. 𝐴+11𝑊 2 C. 1 11𝐴 + 𝑊 D. 11𝐴+𝑊 12 Answer: D Explanation: 𝐿𝑒𝑡 𝑡ℎ𝑒 𝑛𝑜. 𝑜𝑓 𝑜𝑡ℎ𝑒𝑟 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 𝑥 𝑁𝑜. 𝑜𝑓 𝑖𝑙𝑙𝑖𝑡𝑒𝑟𝑎𝑡𝑒 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 11𝑥 𝑇𝑜𝑡𝑎𝑙 𝑛𝑜 𝑜𝑓 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 11𝑥 + 𝑥 = 12𝑥 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 𝑜𝑓 𝑖𝑙𝑙𝑖𝑡𝑒𝑟𝑎𝑡𝑒 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 11𝑊𝑥 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 𝑜𝑓 𝑜𝑡ℎ𝑒𝑟 𝑤𝑜𝑟𝑘𝑒𝑟𝑠 = 𝑥 × 𝑊 = 𝑥𝑊 𝐶𝑜𝑚𝑏𝑖𝑛𝑒𝑑 𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑐𝑜𝑚𝑒 = 11𝐴𝑥 + 𝑊𝑥 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 = 11𝐴𝑥 + 𝑊𝑥 12𝑥 = 11𝐴 + 𝑊 12 Q112.The average age of a family of 5 members 4 year ago was 24 years. Mean while a child was born in this family and still the average age of the whole family is same today. The present age of the child is: A. 2 years B. 1 1 2 years C. 4 years D. data insufficient Answer: C Explanation: Total age of family of 5 members = 24 × 5 + 4 × 5 = 140 𝑇𝑜𝑡𝑎𝑙 𝑎𝑔𝑒 𝑜𝑓 𝑓𝑎𝑚𝑖𝑙𝑦 𝑜𝑓 6 𝑚𝑒𝑚𝑏𝑒𝑟𝑠 = 24 𝑥 6 = 144 𝑆𝑜, 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑎𝑔𝑒 𝑜𝑓 𝑐ℎ𝑖𝑙𝑑 = 144— 140 = 4 𝑦𝑒𝑎𝑟𝑠 Q113.The average of five positive numbers is 99. The averages of the first two and the last two numbers are 117 and 92 respectively. What is the third number? A. 46 B. 54 C. 56 D. 77 Answer: D Explanation: Average of 5 numbers = 99 So, the sum of all 5 numbers = 5× 99 = 495 The average of first two numbers = 117 So, sum of first two numbers = 117 × 2 = 234 The average of second two numbers = 92 So, sum of second two numbers = 92×2 = 184 The sum of 1st, 2nd, 4th and 5th numbers = 234+184 = 418 The fifth number = 495-418 = 77 Q114.The average weight of 16 boys in a class is 50.25 kgs of which 8 playing boys is 45.15 kgs. Find the average weight of the non-playing boys in the class. A. 47.55 kgs B. 48 kgs C. 49.25 kgs D. 55.35 kgs Answer: D Explanation: Total Weight of 16 boys = 16 × 50.25 Weight of 8 playing boys = 8 × 45.15