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function power point presentation for class 11 and 12 for jee

M
MohanSonawane

The document contains solutions and explanations for various problems related to functions in mathematics, including concepts such as one-to-one, onto, and bijective functions. It provides specific examples and calculation methods for determining the properties and domains of several functions, alongside key results from competitive exams like JEE. Additionally, it includes the range and functional equations to reinforce the understanding of real-valued functions.

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FUNCTIONS PCQs FUNCTIONS
FUNCTIONS PCQs
FUNCTIONS PCQs Solution : 𝟏) 𝒃𝒐𝒕𝒉 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒂𝒏𝒅 𝒐𝒏𝒕𝒐 𝟐)𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒃𝒖𝒕 𝒏𝒐𝒕 𝒐𝒏𝒕𝒐 𝟑) 𝒏𝒆𝒊𝒕𝒉𝒆𝒓 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒏𝒐𝒓 𝒐𝒏𝒕𝒐 𝟒) 𝒐𝒏𝒕𝒐 but not 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝟏. 𝐋𝐞𝐭 𝐟: 𝐑 → 𝑹 𝒃𝒆 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒃𝒚 𝐟 𝐱 = 𝒙 − 𝟏 𝒙 + 𝟏 𝒕𝒉𝒆 𝒇 𝒊𝒔 𝐰𝐞 𝐡𝐚𝐯𝐞 𝐟 𝐱 = 𝒙 − 𝟏 𝒙 + 𝟏 [𝐉𝐞𝐞 − 𝟐𝟎𝟏𝟒] 𝒄𝒍𝒆𝒂𝒓𝒍𝒚 , 𝒇 𝒊𝒔 𝒏𝒐𝒕 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒂𝒔 𝒇 −𝟏 = 𝟎 = 𝒇 𝟏 𝒂𝒏𝒅 𝒓𝒂𝒏𝒈𝒆 ≠ 𝒄𝒐𝒅𝒐𝒎𝒂𝒊𝒏 𝒊. 𝒆 𝒏𝒐𝒕 𝒐𝒏𝒕𝒐
FUNCTIONS PCQs Solution : 𝟏) 𝟑 𝟐 + 𝟐 𝟑 𝟐. 𝐋𝐞𝐭 𝐟 𝒃𝒆 𝒂𝒏 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒐𝒇 𝒓𝒆𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝒇𝒐𝒓 𝒙 ≥ 𝟎, 𝐟 𝐱 =3sinx+4cosx . Then 𝐟 𝐱 at x= −11𝛑 𝟔 𝐟 −𝟏𝟏𝝅 𝟔 = −𝐟 𝟏𝟏𝝅 𝟔 [𝐉𝐞𝐞 − 𝟐𝟎𝟏𝟒] 𝟐) − 𝟑 𝟐 + 𝟐 𝟑 𝟑) 𝟑 𝟐 − 𝟐 𝟑 𝟒) − 𝟑 𝟐 − 𝟐 𝟑 = 𝟑𝒔𝒊𝒏 𝟏𝟏𝝅 𝟔 + 𝟒𝒄𝒐𝒔 𝟏𝟏𝝅 𝟔
FUNCTIONS PCQs = − 𝟑𝒔𝒊𝒏 𝟐𝝅 − 𝝅 𝟔 + 𝟒𝒄𝒐𝒔 𝟐𝝅 − 𝝅 𝟔 = −𝟑 𝟐 + 𝟒 𝟑 𝟐 = 𝟑 𝟐 − 𝟐 𝟑 Key : 3
FUNCTIONS PCQs Solution : 𝟏) (−∞, ∞) 𝟐) (𝟎, ∞) 𝟑) (−∞, 𝟎) 𝟒) −∞, ∞ − {𝟎} 𝟑. 𝐓𝐡𝐞 𝐝𝐨𝐦𝐚𝐢𝐧 𝐨𝐟 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 = 𝟏 𝐱 − 𝐱 [𝐀 − 𝟐𝟎𝟎𝟏] 𝐟 𝐱 = 𝟏 𝐱 − 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐱 − 𝐱 > 𝟎 ⇒ 𝐱 > 𝐱 𝐱 ∈ (−∞ 𝟎)
FUNCTIONS PCQs Solution : 𝟏) 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝐑 𝟐) 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐑 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟑) 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝐑 𝟒) 𝐟 𝐢𝐬 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐨𝐧𝐭𝐨 𝐑 𝟒. 𝐅𝐨𝐫 𝐫𝐞𝐚𝐥 𝐱, 𝐥𝐞𝐭 𝐟 𝐱 = 𝐱𝟑 + 𝟓𝐱 + 𝟏, 𝐭𝐡𝐞𝐧 [𝐀 − 𝟐𝟎𝟎𝟗] 𝐟 𝐱 = 𝐱𝟑 + 𝟓𝐱 + 𝟏 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐋𝐞𝐭 𝐲 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐲 = 𝐱𝟑 + 𝟓𝐱 + 𝟏 ⇒ 𝐱𝟑 + 𝟓𝐱 + 𝟏 − 𝐲 = 𝟎
FUNCTIONS PCQs 𝐀𝐬 𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥 𝐨𝐟 𝐨𝐝𝐝 𝐝𝐞𝐠𝐫𝐞𝐞 𝐡𝐚𝐬 𝐚𝐥𝐰𝐚𝐲𝐬 𝐚𝐭 𝐥𝐞𝐚𝐬𝐭 𝐨𝐧𝐞 𝐫𝐞𝐚𝐥 𝐫𝐨𝐨𝐭 𝐂𝐨𝐫𝐫𝐞𝐬𝐩𝐨𝐧𝐝𝐢𝐧𝐠 𝐭𝐨 𝐚𝐧𝐲 𝐲 ∈ 𝐜𝐨. 𝐝𝐨𝐦𝐚𝐢𝐧 𝐒𝐨 ∃ 𝐬𝐨𝐦𝐞 𝐱 ∈ 𝐝𝐨𝐦𝐚𝐢𝐧 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐟 𝐱 = 𝐲. 𝐡𝐞𝐧𝐜𝐞 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐀𝐥𝐬𝐨 𝐟 𝐢𝐬 𝐜𝐨𝐧𝐭𝐢𝐧𝐨𝐮𝐬 𝐨𝐧 𝐑, 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝐢𝐭 𝐢𝐬 𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐍𝐨𝐰 𝐟′ 𝐱 = 𝟑𝐱𝟐 + 𝟓 > 𝟎 ∴ 𝐟 𝐢𝐬 𝐬𝐭𝐫𝐢𝐜𝐭𝐥𝐲 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 Key : 3
FUNCTIONS PCQs 𝟓. 𝐋𝐞𝐭 𝐟 𝐱 = (𝐱 + 𝟏)𝟐−𝟏, 𝐱 ≥ −𝟏 [𝐀 − 𝟐𝟎𝟎𝟗] 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 − 𝟏: 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐱: 𝐟 𝐱 = 𝐟−𝟏 𝐱 = {𝟎, −𝟏} 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 − 𝟐: 𝐟 𝐢𝐬 𝐚 𝐛𝐢𝐣𝐞𝐜𝐭𝐢𝐨𝐧 𝟏) statement –1 is true , statement – 2 is true ; statement – 2 is the correct explanation of statement - 1 𝟐) statement –1 is true statement – 2 is true ; statement – 2 is not the correct explanation of statement - 1 𝟑) statement –1 is true , statement – 2 is false 𝟒) statement –I is false , statement – II is true
FUNCTIONS PCQs Solution : 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = (𝐱 + 𝟏)𝟐−𝟏 → 𝐬𝟏 ∶ 𝐟 𝐱 = 𝐟′(𝐱) ⇒ (𝐱 + 𝟏)𝟐 −𝟏 = 𝐟′ 𝐱 ⇒ 𝐟 𝐱𝟐 + 𝟐𝐱 = 𝐱 ⇒ (𝐱𝟐 + 𝟐𝐱 + 𝟏)𝟐= 𝐱 + 𝟏 ⇒ (𝐱 + 𝟏)𝟒= 𝐱 + 𝟏 ⇒ 𝐱 + 𝟏 𝐱 + 𝟏 𝟑 − 𝟏 = 𝟎 ⇒ 𝐱 + 𝟏 = 𝟎 𝐨𝐫 (𝐱 + 𝟏)𝟑= 𝟏 ⇒ 𝐱 = −𝟏 𝐨𝐫 𝐱 = 𝟎 ∴ 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 ′ 𝟏′𝐢𝐬 𝐭𝐫𝐮𝐞
FUNCTIONS PCQs → 𝐬𝟐 ∶ 𝐂𝐨𝐧𝐜𝐥𝐮𝐬𝐢𝐨𝐧 𝐨𝐟 ′ 𝐟′𝐢𝐬 𝐧𝐨𝐭 𝐬𝐩𝐞𝐜𝐟𝐢𝐞𝐝 ⇒ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 ⇒ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐛𝐢𝐣𝐞𝐜𝐭𝐢𝐨𝐧 ∴ 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 ′ 𝟐′𝐢𝐬 𝐟𝐚𝐥𝐬𝐞 Key : 3
FUNCTIONS PCQs Solution : 𝟏) [𝟎, 𝛑] 𝟐) −𝛑 𝟐 , 𝛑 𝟐 𝟑) −𝛑 𝟒 , 𝛑 𝟐 𝟒) 𝟎, 𝛑 𝟐 𝟔. 𝐓𝐡𝐞 𝐥𝐚𝐫𝐠𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 𝐥𝐲𝐢𝐧𝐠 𝐢𝐧 −𝛑 𝟐 , 𝛑 𝟐 𝐟𝐨𝐫 𝐰𝐡𝐢𝐜𝐡 𝐲𝐡𝐫 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 = 𝟒−𝐱𝟐 + 𝐜𝐨𝐬−𝟏 x 2 − 𝟏 + 𝐥𝐨𝐠 𝐜𝐨𝐬 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 , 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟕] 𝐟 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐢𝐟 − 𝟏 ≤ 𝐱 𝟐 − 𝟏 ≤ 𝟏
FUNCTIONS PCQs ⇒ 𝟎 ≤ 𝐱 𝟐 ≤ 𝟐 ⇒ 𝟎 ≤ 𝐱 ≤ 𝟒 _____(𝟏) 𝐜𝐨𝐬 𝐱 > 𝟎 ⇒ − 𝛑 𝟐 < 𝐱 < 𝛑 𝟐 ______(𝟐) 𝐟𝐫𝐨𝐦 𝟏 𝐚𝐧𝐝 (𝟐) 𝐟 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝟎 ≤ 𝐱 < 𝛑 𝟐 𝐇𝐞𝐧𝐜𝐞 𝛑 𝟐 ≅ 𝟏. 𝟓𝟕 Key : 4
FUNCTIONS PCQs Solution : 𝟏) − 𝛑 𝟐 , 𝛑 𝟐 𝟐) 𝛑 𝟐 , 𝛑 𝟐 𝟑) 𝟎, 𝛑 𝟐 𝟒) 𝟎, − 𝛑 𝟐 𝟕. 𝐋𝐞𝐭 𝐟: −𝟏, 𝟏 → 𝐁, 𝐛𝐞 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐝𝐞𝐭𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐭𝐚𝐧−𝟏 𝟐𝐱 𝟏 − 𝐱𝟐 , 𝐭𝐡𝐞𝐧 𝐟 𝐢𝐬 𝐛𝐨𝐭𝐡 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝐰𝐡𝐞𝐧 𝐁 𝐢𝐬 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 [𝐀 − 𝟐𝟎𝟎𝟓] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐭𝐚𝐧−𝟏 𝟐𝐱 𝟏 − 𝐱𝟐 = 𝟐 𝐭𝐚𝐧−𝟏𝐱
FUNCTIONS PCQs 𝐆𝐢𝐯𝐞𝐧 𝐱 ∈ −𝟏, 𝟏 ⇒ 𝐭𝐚𝐧 −𝟏 𝐱 ∈ −𝛑 𝟒 , 𝛑 𝟒 ⇒ 𝟐𝐭𝐚𝐧 −𝟏 𝐱 ∈ −𝛑 𝟐 , 𝛑 𝟐 ∴ 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐗 = − 𝛑 𝟐 𝛑 𝟐 Key : 1
FUNCTIONS PCQs Solution : 𝟏) 𝐟 𝐱 + 𝟐 = 𝐟(𝐱 − 𝟐) 𝟐) 𝐟 𝐱 + 𝟐 = 𝐟(𝟐 − 𝐱) 𝟑) 𝐟 𝐱 = 𝐟(−𝐱) 𝟒) 𝐟 𝐱 = −𝐟(−𝐱) 𝟖. 𝐈𝐟 𝐭𝐡𝐞 𝐠𝐫𝐚𝐩𝐡 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐲 = 𝐟 𝐱 𝐢𝐬 𝐬𝐲𝐦𝐦𝐞𝐭𝐫𝐢𝐜𝐚𝐥 𝐚𝐛𝐨𝐮𝐭 𝐭𝐡𝐞 𝐥𝐢𝐧𝐞 𝐱 = 𝟐, 𝐭𝐡𝐞𝐧 [𝐀 − 𝟐𝟎𝟎𝟒] 𝐥𝐞𝐭 𝐱 = 𝐱 − 𝟐; 𝐟 𝐱 = 𝐟(𝐱 + 𝟐) 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 + 𝟐 = 𝐲 𝐢𝐬 𝐬𝐲𝐦𝐦𝐞𝐭𝐫𝐢𝐜 𝐚𝐛𝐨𝐮𝐭 𝐲 − 𝐚𝐱𝐢𝐬 𝐟 𝐱 = 𝐟(−𝐱) ⇒ 𝐟 𝐱 + 𝟐 = 𝐟(𝟐 − 𝐱)
FUNCTIONS PCQs Solution : 𝟏) 𝐟(−𝐱) 𝟐) 𝐟 𝐚 + 𝐟(𝐚 − 𝐱) 𝟑) 𝐟(𝐱) 𝟒) − 𝐟(𝐱) 𝟗. 𝐀 𝐫𝐞𝐚𝐥 𝐯𝐚𝐥𝐮𝐞𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 𝐬𝐚𝐭𝐢𝐬𝐟𝐢𝐞𝐬 𝐭𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐟 𝐱 − 𝐲 = 𝐟 𝐱 𝐟 𝐲 − 𝐟 𝐚 − 𝐱 𝐟(𝐚 + 𝐲) 𝐰𝐡𝐞𝐫𝐞 𝐚 𝐢𝐬 𝐚 𝐠𝐢𝐯𝐞𝐧 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐚𝐧𝐝 𝐟 𝟎 = 𝟏, 𝐟 𝟐𝐚 − 𝐱 𝐢𝐬 𝐞𝐪𝐮𝐚𝐥 𝐭𝐨 [𝐀 − 𝟐𝟎𝟎𝟒] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝟎 = 𝟏 ⇒ 𝐟 𝟎 = 𝐟 𝟎 − 𝟎 = 𝐟 𝟎 . 𝐟 𝟎 − 𝐟 𝐚 − 𝟎 . 𝐟(𝐚 + 𝟎) = 𝟏 − [𝐟(𝐚)]𝟐 ⇒ [𝐟(𝐚)]𝟐= 𝟎 ⇒ 𝐟 𝐚 = 𝟎
FUNCTIONS PCQs 𝐍𝐨𝐰 𝐟 𝟐𝐚 − 𝐱 = 𝐟 𝐚 − 𝐱 − 𝐚 = 𝐟 𝐚 . 𝐟 𝐱 − 𝐚 − 𝐟 𝐚 − 𝐚 . 𝐟(𝐚 + 𝐱 − 𝐚) = −𝐟(𝐱) Key : 4
FUNCTIONS PCQs Solution : 𝟏) [𝟎, 𝟑] 𝟐) [−𝟏, 𝟑] 𝟑) [𝟎, 𝟏] 𝟒) [−𝟏, 𝟏] 𝟏𝟎. 𝐈𝐟 𝐟: 𝐑 → 𝐒, 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 − 𝟑𝐜𝐨𝐬 𝐱 + 𝟏 𝐢𝐬 𝐨𝐧𝐭𝐨, 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 𝐨𝐟 𝐒 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟒] 𝐟: 𝐑 → 𝐒 𝐢𝐬 𝐨𝐧𝐭𝐨 ⇒ 𝐒 = 𝐑𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 = 𝟏 − 𝟏 + 𝟑, 𝟏 + 𝟏 + 𝟑 = 𝟏 − 𝟐, 𝟏 + 𝟐 = −𝟏, 𝟑
FUNCTIONS PCQs Solution : 𝟏) (𝟏, 𝟐) 𝟐) −𝟏, 𝟎 ∪ (𝟏, 𝟐) 𝟑) 𝟏, 𝟐 ∪ (𝟐, ∞) 𝟒) −𝟏, 𝟎 ∪ 𝟏, 𝟐 ∪ (𝟐, ∞) 𝟏𝟏. 𝐓𝐡𝐞 𝐝𝐨𝐦𝐚𝐢𝐧 𝐨𝐟 𝐟 𝐱 = 𝟑 𝟒 − 𝐱𝟐 + 𝐥𝐨𝐠𝟏𝟎 𝐱𝟑 − 𝐱 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟑] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝟑 𝟒 − 𝐱𝟐 + 𝐥𝐨𝐠𝟏𝟎 𝐱𝟑 − 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝟒 − 𝐱𝟐 ≠ 𝟎 ⇒ 𝐱𝟐 ≠ 𝟒 ⇒ 𝐱 ≠ ±𝟐_____(𝟏) 𝐚𝐧𝐝 𝐱𝟑 − 𝐱 > 𝟎 ⇒ 𝐱 𝐱 − 𝟏 > 𝐱 + 𝟏 > 𝟎
FUNCTIONS PCQs ⇒ 𝐱 ∈ −𝟏, 𝟎 ∪ 𝟏, ∞ ____(𝟐) 𝐟𝐫𝐨𝐦 𝟏 𝐚𝐧𝐝 (𝟐) 𝐝𝐨𝐦𝐚𝐢𝐧 𝐨𝐟 𝐟 𝐱 = −𝟏 𝟎 ∪ 𝟏 𝟐 ∪ (𝟐 ∞) Key : 4
FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝟐) 𝐨𝐧𝐭𝐨 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟑) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐨𝐧𝐭𝐨 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝟏𝟐. 𝐀 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐍 → 𝐙 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐧 = 𝐧 − 𝟏 𝟐 , 𝐰𝐡𝐞𝐧 ′ 𝐧′ 𝐢𝐬 𝐨𝐝𝐝 𝐚𝐧𝐝 𝐟 𝐧 = −𝐧 𝟐 𝐰𝐡𝐞𝐧 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧. 𝐭𝐡𝐚𝐧 𝐟 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟑] 𝐟 −𝟏 = 𝟎; 𝐟 𝟐 = −𝟏; 𝐟 𝟑 = 𝟏, 𝐟 𝟒 = −𝟐 … . 𝐚𝐭 𝐧 𝐢𝐬 𝐨𝐝𝐝 → 𝐟 𝐱 𝐢𝐬 + 𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫
FUNCTIONS PCQs 𝐚𝐭 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧 → 𝐟 𝐱 𝐢𝐬 − 𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 ∴ 𝐟: 𝐍 → 𝐙 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 Key : 3
FUNCTIONS PCQs Solution : 𝟏) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝟐) 𝐨𝐧𝐭𝐨 𝟑) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟒) 𝐦𝐚𝐧𝐲 𝐨𝐧𝐞 𝟏𝟑. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 = 𝐥𝐨𝐠 𝐱 + 𝐱𝟐 + 𝟏 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟑] 𝐟 −𝐱 = 𝐥𝐨𝐠[−𝐱 + 𝐱𝟐 − 𝟏 ] = 𝐥𝐨𝐠[ 𝐱𝟐 − 𝟏 − 𝐱 ] = 𝐥𝐨𝐠 𝐱𝟐 + 𝟏 − 𝐱 𝐱𝟐 + 𝟏 + 𝐱 𝐱𝟐 + 𝟏 + 𝐱
FUNCTIONS PCQs = 𝐥𝐨𝐠 𝐱𝟐 + 𝟏 − 𝐱𝟐 𝐱𝟐 + 𝟏 + 𝐱 = 𝐥𝐨𝐠 𝟏 𝐱 + 𝐱𝟐 + 𝟏 = 𝐥𝐨𝐠 𝐱 + 𝐱𝟐 + 𝟏 −𝟏 = −𝐥𝐨𝐠 𝐱 + 𝐱𝟐 + 𝟏 = −𝐟(𝐱) ∴ 𝐟 𝐱 𝐢𝐬 𝐚𝐧 𝐨𝐝𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 Key : 2
FUNCTIONS PCQs Solution : 𝟏) 𝐚𝐧 𝐞𝐯𝐞𝐧 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟐) 𝐚𝐧 𝐨𝐝𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟑) 𝐩𝐞𝐫𝐢𝐨𝐝𝐢𝐜 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐞𝐯𝐞𝐧 𝐧𝐨𝐫 𝐨𝐝𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟏𝟒. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟐] 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐬𝐢𝐧𝐱 𝐢𝐬 𝐦𝐚𝐧𝐲 𝐨𝐧𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝐬𝐢𝐧𝐱 𝐢𝐬 𝐚 𝐩𝐞𝐫𝐢𝐨𝐝𝐢𝐜 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐰𝐢𝐭𝐡 𝐩𝐞𝐫𝐢𝐨𝐝 𝟐𝛑
FUNCTIONS PCQs
FUNCTIONS PCQs 1. If f(x) is a real function defined on [-1, 1], then the real function g(x) = f(5x+4) is defined on the interval (TS E – 2015) 1) −𝟒, 𝟗 2) [-1, 9] 3) [-2, 9] 4) [-3, 9 ] Solution : −𝟏 ≤ 𝟓𝒙 + 𝟒 ≤ 𝟏  −𝟏 ≤ 𝐱 ≤ −3 5 −𝟏 ≤ 𝒙 ≤ 𝟏  −𝟓 + 𝟒 ≤ 𝟓𝒙 + 𝟒 ≤ 𝟗 [-1, 9]
FUNCTIONS PCQs 2. If f: N R is defined by f(1) =-1 and f(n+1)=3f(n)+2 for n>1 then f is (TS E – 2015) 1) One – One 2) Onto 3) a constant function 4) f(n)>0 for n>1 Solution : 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
FUNCTIONS PCQs 𝟑) 𝒇: 𝑹 − 𝑹, 𝒈: 𝑹 → 𝑹 defined by f(x) = 5x – 3 , g(x) = x2 + 3 then 𝒈𝒐𝒇−𝟏 𝟑 = (E – 2015) 1) 𝟐𝟓 𝟗 2) 𝟏𝟏𝟏 𝟐𝟓 3) 𝟗 𝟐𝟓 4) 𝟐𝟓 𝟏𝟏𝟏 Solution : 𝒇−𝟏 𝒙 = 𝒙 + 𝟑 𝟓  g (6/5)= 𝟑𝟔 𝟐𝟓 + 𝟑 = 𝟏𝟏𝟏 𝟐𝟓
FUNCTIONS PCQs 4) 𝑨 = 𝒙 ∈ 𝑹/  𝟒 ≤ 𝒙 ≤ 𝝅 𝟑 and f(x) = sin x – x  f(A)= (E – 2015) 1) 𝟑 𝟐 − 𝝅 𝟑 , 𝟏 𝟐 − 𝝅 𝟒 2) −𝟏 𝟐 − 𝝅 𝟒 , 𝟑 𝟐 − 𝝅 𝟑 3) −𝝅 𝟑 , −𝝅 𝟒 4) 𝝅 𝟒 , 𝝅 𝟑 Solution : f(x) = sin x - x f1(x) = cosx – 1<0 . f is decreasing function range is 𝒇 𝝅 𝟑 , 𝒇 𝝅 𝟒 = 𝟑 𝟐 − 𝝅 𝟑 , 𝟏 𝟐 − 𝝅 𝟒
FUNCTIONS PCQs 𝟓) 𝑰𝒇 𝑹 𝒊𝒔 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒐𝒇 𝒂𝒍𝒍 𝒓𝒆𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒂𝒏𝒅 𝒊𝒇 𝒇: 𝑹 − 𝟐 → 𝑹 is defined by f(x) = 𝟐+𝒙 𝟐−𝒙 for x ∈ R – {2} then the range of f is (E – 2014) 1) R 2) R – {2} 3) R – {-1} 4) R – {-2} Solution : 𝒚 𝟏 = 𝟐 + 𝒙 𝟐 − 𝒙 ⇒ 2+x = 2y-xy ⇒ x+xy = 2 𝒚 − 𝟏 ⇒ x = 𝟐 𝒚−𝟏 𝒚+𝟏 Range = R – {-1}
FUNCTIONS PCQs 𝟔) 𝑳𝒆𝒕 𝑸 𝒃𝒆 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒐𝒇 𝒂𝒍𝒍 𝒓𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒊𝒏 𝟎, 𝟏 𝒂𝒏𝒅 𝒇: 𝟎, 𝟏 → 𝟎, 𝟏 𝒃𝒆 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒃𝒚 𝒇(𝒙) = 𝒙 𝒇𝒐𝒓 𝒙 ∈ 𝑸 𝟏 − 𝒙 𝒇𝒐𝒓 𝒙 ∉ 𝑸 then the set S = {𝒙 ∈ [0,1] : (fof)(x) = x} is equal to (E – 2014) Solution : 1) Q 2) [0,1] - Q 3) (0,1) 4) [0,1] f(x) = f-1(x) (i.e) co-domain = Range = [0,1]
FUNCTIONS PCQs Solution : 𝐟 𝐟(𝐱) =f [ 𝑷 − 𝒙𝒏 𝟏 𝒏] 7). 𝐈𝒇 𝒇 𝒙 = 𝑷 − 𝒙𝒏 𝟏 𝒏, 𝐏 > 0 and n is a positive integer then f(f(x)) = (E – 2013) 1) x 2) xn 3) 𝑷 𝟏 𝒏 4) P - xn = 𝑷 − 𝒑 − 𝒙𝒏 𝟏 𝒏 𝒏 𝟏 𝒏 = 𝑷 − 𝑷 + 𝒙𝒏 𝟏 𝒏 = 𝒙𝒏 𝟏 𝒏 = x
FUNCTIONS PCQs Solution : 𝟏. 𝟔 𝟏−𝒙𝟐 − 𝟎. 𝟔𝟐𝟓 𝟔(𝟏+𝒙) > 0 8). 𝒙 ∈ 𝑹/𝒍𝒐𝒈 𝟏. 𝟔 𝟏−𝒙𝟐 − 𝟎. 𝟔𝟐𝟓 𝟔(𝟏+𝒙) ∈ 𝑹 = (E – 2013) 1) −∞, −𝟏 ∪ 𝟕, ∞ 2) (-1,5) 3) (1,7) 4) (-1,7) ⇒ 𝟖 𝟓 𝟏−𝒙𝟐 > 𝟓 𝟖 𝟔(𝟏+𝒙) ⇒ 𝟏 − 𝒙𝟐 > - 6 𝟏 + 𝒙 ⇒ 𝒙𝟐 − 𝟔𝒙 − 𝟕 < 𝟎 ⇒ 𝒙 − 𝟕 𝒙 + 𝟏 <0 ⇒ 𝒙 ∈ −𝟏, 𝟕
FUNCTIONS PCQs Solution : f(x+y)= f(x).f(y) 9). Let f be a non-zero real valued continuous function satisfying f(x+y)= f(x).f(y) for all x,y∈R. If f (2) = 9, then f(6) = (E – 2013) 1) 𝟑𝟐 2) 36 3) 34 4) 33 ⇒ f(x) = ax f(2) = 9 ⇒ a2 = 9 ⇒ a=3 ∴ f(6) = 36
FUNCTIONS PCQs Solution : 𝟏) 𝐟 𝐱 = 𝐱𝟐, 𝐠 𝐱 = 𝐬𝐢𝐧 𝐱 𝟐) 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱, 𝐠 𝐱 = 𝐱 𝟑)𝐟 𝐱 = 𝐬𝐢𝐧𝟐𝐱, 𝐠 𝐱 = 𝐱 𝟒) 𝐟 𝐱 = 𝐱𝟐, 𝐠 𝐱 = 𝐱 𝟏𝟎. 𝐈𝐟 𝐟: 𝐑 → 𝐑+ 𝐚𝐧𝐝 𝐠: 𝐑+ → 𝐑 𝐚𝐫𝐞 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐠 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 𝐚𝐧𝐝 𝐟 𝐠 𝐱 = (𝐬𝐢𝐧 𝐱)𝟐, 𝐭𝐡𝐞𝐧 𝐚 𝐩𝐨𝐬𝐬𝐢𝐛𝐥𝐞 𝐜𝐡𝐨𝐢𝐜𝐞 𝐟 𝐚𝐧𝐝 𝐠 𝐢𝐬 [E-2012] 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 ; 𝐠 𝐱 = 𝐱 ⇒ 𝐠 𝐟 𝐱 = 𝐠 𝐬𝐢𝐧 𝐱 = 𝐬𝐢𝐧 𝐱 = 𝐬𝐢𝐧 𝐱 𝐟 𝐠 𝐱 = 𝐟 𝐱 = 𝐬𝐢𝐧𝟐 𝐱 𝐛𝐲 𝐯𝐞𝐫𝐢𝐟𝐢𝐜𝐚𝐭𝐢𝐨𝐧
FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐭𝐨 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝟐) 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝟑) 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝟏𝟏. 𝐈𝐟 𝐙 → 𝐙 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲𝐟 𝐱 = 𝐱 𝟐 𝐢𝐟 𝐱 𝐢𝐬 𝐞𝐯𝐞𝐧 𝟎 𝐢𝐟 𝐱 𝐨𝐝𝐝 𝐭𝐡𝐞𝐧 𝐟 𝐢𝐬 [𝐄 − 𝟐𝟎𝟏𝟐] 𝐟 𝟎 = 𝟎 ; 𝐟 ±𝟏 = 𝟎 ; 𝐟 ±𝟑 = 𝟎 … … & 𝐟 ±𝟐 = ±𝟏 ; 𝐟 ±𝟒 = ±𝟐 … …
FUNCTIONS PCQs 𝐒𝐨 𝟎, ±𝟏, ±𝟑 … 𝐡𝐚𝐯𝐞 𝐒𝐚𝐦𝐞 𝐢𝐦𝐚𝐠𝐞 ∴ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐁𝐮𝐭 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝐞𝐯𝐞𝐫𝐲 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐡𝐚𝐯𝐞 𝐚 𝐩𝐫𝐞 − 𝐢𝐦𝐚𝐠𝐞 Key : 1
FUNCTIONS PCQs Solution : 𝟏) [𝟎, ∞) 𝟐) [𝟏, ∞) 𝟑) [𝟒, ∞) 𝟒) [𝟓, ∞) 𝟏𝟐. 𝐈𝐟 𝐟: 𝟐, ∞ → 𝐁 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 − 𝟒𝐱 + 𝟓 𝐢𝐬 𝐛𝐢𝐣𝐞𝐜𝐭𝐢𝐨𝐧, 𝐭𝐡𝐞𝐧 𝐁 = [𝐄 − 𝟐𝟎𝟏𝟏] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐱𝟐 − 𝟒𝐱 + 𝟓 = 𝐱𝟐 − 𝟒𝐱 + 𝟓 + 𝟏 = (𝐱 − 𝟐)𝟐+𝟏 ∴ 𝐑𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐱 = [𝟏, ∞)
FUNCTIONS PCQs Solution : 𝟏) {−𝟏𝟒, −𝟏𝟑, … . . , 𝟎, … . . , 𝟏𝟑, 𝟏𝟒} 𝟐) {−𝟏𝟒, −𝟏𝟑, … . . , 𝟎, … . . , 𝟏𝟒, 𝟏𝟓} 𝟑) {−𝟏𝟓, −𝟏𝟒, … . . , 𝟎, … . . , 𝟏𝟒, 𝟏𝟓} 𝟒) {−𝟏𝟓, −𝟏𝟒, … . . , 𝟎, … . . , 𝟏𝟑, 𝟏𝟒} 𝟏𝟑. 𝐈𝐟 𝐟 = 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 𝟓 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐰𝐡𝐞𝐫𝐞 𝐲 , 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐲 , 𝐭𝐡𝐞𝐧 𝐟 𝐱 : 𝐱 < 𝟕𝟏 = [𝐄 − 𝟐𝟎𝟏𝟏] 𝐱 < 𝟕𝟏 ⇒ −𝟕𝟏 < 𝐱 < 𝟕𝟏
FUNCTIONS PCQs ⇒ − 𝟕𝟏 𝟓 < 𝐱 𝟓 < 𝟕𝟏 𝟓 ⇒ −𝟏𝟒. 𝟐 < 𝐱 𝟓 < 𝟏𝟒. 𝟐 ⇒ 𝐟 𝐱 : 𝐱 < 𝟕𝟏 ∵ 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐱 𝟓 = {−𝟏𝟓, −𝟏𝟒, −𝟏𝟑, … . 𝟎, … 𝟏𝟑, 𝟏𝟒} Key : 4
FUNCTIONS PCQs Solution : 𝟏) 𝟏𝟐 𝟐) 𝟏𝟑 𝟑) 𝟏𝟒 𝟒) 𝟏𝟎 𝟏𝟒. 𝐈𝐟 𝐟 𝟎 = 𝟎, 𝐟 𝟏 = 𝟏, 𝐟 𝟐 = 𝟐 𝐚𝐧𝐝 𝐟 𝐱 = 𝐟 𝐱 − 𝟐 + 𝐟 𝐱 − 𝟑 𝐟𝐨𝐫 𝐱 = 𝟑, 𝟒, 𝟓, … … 𝐭𝐡𝐞𝐧 𝐟 𝟗 = [𝐄 − 𝟐𝟎𝟏𝟎] 𝐟 𝟎 = 𝟎, 𝐟 𝟏 = 𝟏, 𝐟 𝟐 = 𝟐 𝐚𝐧𝐝 𝐟 𝐱 = 𝐟 𝐱 − 𝟐 + 𝐟 𝐱 − 𝟑 𝐍𝐨𝐰 𝐟 𝟑 = 𝐟 𝟏 + 𝐟 𝟎 = 𝟏 + 𝟎 = 𝟏 𝐟 𝟒 = 𝐟 𝟐 + 𝐟 𝟏 = 𝟏 + 𝟐 = 𝟑 𝐟 𝟓 = 𝐟 𝟑 + 𝐟 𝟐 = 𝟏 + 𝟐 = 𝟑
FUNCTIONS PCQs 𝐟 𝟓 = 𝐟 𝟑 + 𝐟 𝟐 = 𝟏 + 𝟐 = 𝟑 𝐟 𝟔 = 𝐟 𝟒 + 𝐟 𝟑 = 𝟑 + 𝟏 = 𝟒 𝐟 𝟕 = 𝐟 𝟓 + 𝐟 𝟒 = 𝟑 + 𝟑 = 𝟔 𝐟 𝟖 = 𝐟 𝟔 + 𝐟 𝟓 = 𝟒 + 𝟑 = 𝟕 𝐟 𝟗 = 𝐟 𝟕 + 𝐟 𝟔 = 𝟔 + 𝟒 = 𝟏𝟎 Key : 4
FUNCTIONS PCQs 𝐋𝐢𝐬𝐭 𝐈 𝐋𝐢𝐬𝐭 𝐈𝐈 𝟏𝟓. 𝐋𝐞𝐭 𝐑 𝐝𝐞𝐧𝐨𝐭𝐞 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐚𝐧𝐝 𝐑+ 𝐝𝐞𝐧𝐨𝐭𝐞 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬. 𝐅𝐨𝐫 𝐭𝐡𝐞 𝐬𝐮𝐛𝐬𝐞𝐭𝐬 𝐀 𝐚𝐧𝐝 𝐁 𝐨𝐟 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞 𝐟: 𝐀 → 𝐁 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 𝐟𝐨𝐫 𝐱 ∈ 𝐀 . 𝐨𝐛𝐬𝐞𝐫𝐯𝐞 𝐭𝐡𝐞 𝐭𝐰𝐨 𝐥𝐢𝐬𝐭 given below [𝐄 − 𝟐𝟎𝟏𝟎] (i) f is one-one and onto if (a) A=𝐑+, B=R (ii) f is one-one but not onto if (b) A=B=R (iii) f is onto but not one-one if (c) A=R, B=𝐑+ (iv) f is neither one – one nor onto if (d) A=B=𝐑+
FUNCTIONS PCQs (i) (ii) (iii) (iv) 1) a b c d 2) d b a c 3) d a c b 4) d b c a
FUNCTIONS PCQs 𝐢 𝐟 𝐢𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 − 𝐨𝐧𝐭𝐨 → 𝐀 = 𝐑+ ; 𝐁 = 𝐑+ Solution : 𝐢𝐢 𝐟 𝐢𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 → 𝐀 = 𝐑+ ; 𝐁 = 𝐑 𝐢𝐢𝐢 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 𝐨𝐧𝐞 → 𝐀 = 𝐑 ; 𝐁 = 𝐑+ 𝐢𝐯 𝐟 𝐢𝐬 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 → 𝐀 = 𝐑 ; 𝐁 = 𝐑 Key : 3
FUNCTIONS PCQs Solution : 𝟏) [𝟏, 𝟏𝟐] 𝟐) [𝟏𝟐, 𝟑𝟒] 𝟑) [𝟑𝟓, 𝟓𝟎] 𝟒) [−𝟏𝟐, 𝟏𝟐] 𝟏𝟔. 𝐈𝐟 𝐟: 𝟐, 𝟑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟑 + 𝟑𝐱 − 𝟐, 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐟(𝐱) is contained in the interval: [𝐄 − 𝟐𝟎𝟎𝟗] 𝐈𝐟 𝐟: 𝟐, 𝟑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟑 + 𝟑𝐱 − 𝟐 𝐍𝐨𝐰 𝐟 𝟐 = 𝟖 + 𝟔 − 𝟐 = 𝟏𝟐 𝐟 𝟑 = 𝟐𝟕 + 𝟗 − 𝟐 = 𝟑𝟒 ∴ 𝐑𝐚𝐧𝐠𝐞 [𝟏𝟐, 𝟑𝟒]
FUNCTIONS PCQs Solution : 𝟏) 𝐑 − {𝟎} 𝟐) 𝐑 − {𝟎, 𝟏, 𝟑} 1𝟕. 𝐱 ∈ 𝐑: 𝟐𝐱−𝟏 𝐱𝟑+𝟒𝐱𝟐+𝟑𝐱 = [𝐄 − 𝟐𝟎𝟎𝟗] 𝟑) 𝐑 − {𝟎, −𝟏, −𝟑} 𝟒) 𝐑 − 𝟎, −𝟏, −𝟑, + 𝟏 𝟐 𝐱𝟑 + 𝟒𝐱𝟐 + 𝟑𝐱 ≠ 𝟎 ⇒ 𝐱(𝐱𝟐 + 𝟒𝐱 + 𝐱) ≠ 𝟎 ⇒ 𝐱(𝐱 + 𝟏)(𝐱 + 𝟑) ≠ 𝟎
FUNCTIONS PCQs ⇒ 𝐱 ≠ 𝟎, −𝟏, −𝟑 ∴ 𝐑𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝐬𝐞𝐭 𝐢𝐬 𝐑 − {−𝟏, −𝟑, 𝟎} Key : 3
FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟐) 𝐨𝐧𝐭𝐨 1𝟖. 𝐈𝐟 𝐟: 𝐑 → 𝐂 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐞𝟐𝐢𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧, 𝐟 𝐢𝐬 (𝐰𝐡𝐞𝐫𝐞 𝐂 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐜𝐨𝐦𝐩𝐥𝐞𝐱 𝐧𝐮𝐦𝐛𝐞𝐫𝐬) [𝐄 − 𝟐𝟎𝟎𝟗] 𝟑) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝐟 𝐱 = 𝐞𝟐𝐢𝐱 = 𝐜𝐨𝐬 𝟐𝐱 + 𝐢 𝐬𝐢𝐧 𝟐𝐱 → 𝐟 𝟎 = 𝐟 𝐱 = 𝟎 𝐒𝐨 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 → 𝐓𝐡𝐞𝐫𝐞 𝐞𝐱𝐢𝐭 𝐧𝐨 𝐱 ∈ 𝐑 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐟 𝐱 = 𝟐 𝐒𝐨 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨
FUNCTIONS PCQs Solution : 𝟏) {𝟎, 𝟏} 𝟐) {𝟏, 𝟐} 𝟏𝟗. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱 − 𝟑 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐭𝐡𝐞𝐧 𝐠 𝐟 𝐱 : − 8 5 < 𝐱 < 8 5 = [𝐄 − 𝟐𝟎𝟎𝟖] 𝟑) {−𝟑, −𝟐} 𝟒) {𝟐, 𝟑} − 8 5 < 𝐱 < 8 5 ⇒ 𝟎 < 𝐱 < 𝟖 𝟓 ⇒ −𝟑 ≤ 𝐱 − 𝟑 < −𝟕 𝟓
FUNCTIONS PCQs ⇒ −𝟑 ≤ 𝐱 − 𝟑 < −𝟏. 𝟒 ⇒ 𝐱 − 𝟑 = −𝟑 𝐨𝐫 − 𝟐 ∵ −𝟑 < 𝐱 − 𝟑 < −𝟐 ⇒ 𝐠𝐨𝐟 𝐟 𝐱 = −𝟑 −𝟐 < 𝐱 − 𝟑 < −𝟏. 𝟒 ⇒ 𝐠𝐨𝐟 𝐟 𝐱 = −𝟐 Key : 3
FUNCTIONS PCQs Solution : 𝟏) 𝐟(𝟒 𝟐) 𝟐) 𝐟(𝟑 𝟐) 𝟐𝟎. 𝐈𝐟 𝐟: −𝟔, 𝟔 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 − 𝟑 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐭𝐡𝐚𝐧 𝐟𝟎𝐟𝟎𝐟 −𝟏 + 𝐟𝟎𝐟𝟎𝐟 𝟎 + 𝐟𝟎𝐟𝟎𝐟 𝟏 = [𝐄 − 𝟐𝟎𝟎𝟖] 𝟑) 𝐟(𝟐 𝟐) 𝟒) 𝐟( 𝟐) → 𝐟𝐨𝐟𝐨𝐟 −𝟏 = 𝐟𝐨𝐟 𝐟 −𝟏 = 𝐟𝐨𝐟[𝟏 − 𝟑] = 𝐟𝐨𝐟 −𝟐 = 𝐟 𝟒 − 𝟑 = 𝐟 𝟏 = 𝟏 − 𝟑 = −𝟐 → 𝐟𝐨𝐟𝐨𝐟 𝟎 = 𝐟𝐨𝐟 𝐟 𝟎 = 𝐟𝐨𝐟(−𝟑) = 𝐟 𝐟 −𝟑 = 𝐟 𝟗 − 𝟑 = 𝐟 𝟔 = 𝟑𝟑
FUNCTIONS PCQs → 𝐟𝐨𝐟𝐨𝐟 𝟏 = 𝐟𝐨𝐟 𝐟 𝟏 = 𝐟𝐨𝐟 𝟏 − 𝟑 = 𝐟𝐨𝐟 −𝟐 = 𝐟 𝟒 − 𝟑 = 𝐟(𝟏) = 𝟏 − 𝟑 = −𝟐 𝐒𝐨 𝐟𝐨𝐟𝐨𝐟 −𝟏 + 𝐟𝐨𝐟𝐨𝐟 𝟎 + 𝐟𝟎𝐟𝟎𝐟(𝟏) = −𝟐 + 𝟑𝟑 − 𝟐 = 𝟐𝟗 = 𝟑𝟐 − 𝟑 = (𝟒 𝟐)𝟐−𝟑 = 𝐟(𝟒 𝟐) Key : 1
FUNCTIONS PCQs 𝐈) 𝐟 𝐩 𝐪 𝐢𝐬 𝐫𝐞𝐚𝐥 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 𝐩 𝐪 ∈ 𝐐 𝟐𝟏. 𝐈𝐟 𝐐 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐚𝐧𝐝 𝐟 𝐩 𝐪 = 𝐩𝟐 − 𝐪𝟐 𝐟𝐨𝐫 𝐚𝐧𝐲 p q ∈ 𝐐, 𝐭𝐡𝐞𝐧 𝐨𝐛𝐬𝐞𝐫𝐯𝐞 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 [𝐄 − 𝟐𝟎𝟎𝟖] 𝐈𝐈) 𝐟 𝐩 𝐪 𝐢𝐬 𝐚 𝐜𝐨𝐦𝐩𝐥𝐞𝐱 𝐧𝐮𝐦𝐛𝐞𝐫 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 p q ∈ 𝐐 Which of the following is correct ?
FUNCTIONS PCQs Solution : 1) 𝐁𝐨𝐭𝐡 𝐈 𝐚𝐧𝐝 𝐈𝐈 𝐚𝐫𝐞 𝐭𝐫𝐮𝐞 2) 𝐈 𝐢𝐬 𝐭𝐫𝐮𝐞 , 𝐈𝐈 𝐢𝐬 𝐟𝐚𝐥𝐬𝐞 3) 𝐈 𝐭𝐫𝐮𝐞 , 𝐈𝐈 𝐢𝐬 𝐟𝐚𝐥𝐬𝐞 4) 𝐁𝐨𝐭𝐡 𝐈 𝐚𝐧𝐝 𝐈𝐈 𝐚𝐫𝐞 𝐟𝐚𝐥𝐬𝐞 𝐁𝐲 𝐯𝐞𝐫𝐢𝐟𝐢𝐜𝐚𝐭𝐢𝐨𝐧 𝐰𝐞 𝐭𝐚𝐤𝐞 𝐚𝐧𝐲 𝐩𝐫𝐨𝐩𝐞𝐫𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐩 𝐪 ∈ 𝐐 𝐟 𝐩 𝐪 = 𝐩𝟐 − 𝐪𝟐 𝐢𝐬 𝐜𝐨𝐦𝐩𝐥𝐞𝐱 𝐧𝐮𝐦𝐛𝐞𝐫
FUNCTIONS PCQs Solution : 𝟏) 𝟏 𝟑 , 𝟏 𝟐) 𝟏 𝟑 , 𝟏 𝟐𝟐. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟏 (𝟐 − 𝐜𝐨𝐬𝟑𝐱) 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 𝐱 ∈ 𝐑, 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐢𝐬 [𝐄 − 𝟐𝟎𝟎𝟕] 𝟑) (𝟏, 𝟐) 𝟒) [𝟏, 𝟐] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝟏 𝟐 − 𝐜𝐨𝐬𝟑𝐱 𝐰. 𝐤. 𝐭 − 𝟏 ≤ −𝐜𝐨𝐬𝟑𝐱 ≤ 𝟏 𝟐 − 𝟏 ≤ 𝟐 − 𝐜𝐨𝐬𝟑𝐱 ≤ 𝟐 + 𝟏
FUNCTIONS PCQs 𝟏 ≤ 𝟐 − 𝐜𝐨𝐬𝟑𝐱 ≤ 𝟑 𝟏 𝟑 ≤ 𝟏 𝟐 − 𝐜𝐨𝐬𝟑𝐱 ≤ 𝟏 ∴ 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐱 = 𝟏 𝟑 , 𝟏 Key : 2
FUNCTIONS PCQs Solution : 𝟏) 𝐱 𝟐) 𝟎 𝟐𝟑. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 − 𝐱 𝐚𝐧𝐝 𝐠 𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐰𝐡𝐞𝐫𝐞 𝐱 𝐢𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐱, 𝐭𝐡𝐞𝐧 𝐟𝐨𝐫 𝐞𝐯𝐞𝐫𝐲 𝐱 ∈ 𝐑, 𝐟 𝐠 𝐱 = [𝐄 − 𝟐𝟎𝟎𝟕] 𝟑) 𝐟(𝐱) 𝟒) 𝐠(𝐱) 𝐟 𝐠 𝐱 = 𝐠 𝐱 − [𝐠 𝐱 ] = 𝐱 − 𝐱 = 𝐱 − 𝐱 = 𝟎 ∵ 𝐟 𝐱 = 𝐱 − 𝐱 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱
FUNCTIONS PCQs Solution : 𝟏) 𝐙, 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐢𝐧𝐭𝐞𝐠𝐞𝐫𝐬 𝟐) 𝐍, 𝐢𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐧𝐚𝐭𝐮𝐫𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 24. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 − 𝐱 − 𝟏 𝟐 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐰𝐡𝐞𝐫𝐞 𝐱 𝐢𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐱, 𝐭𝐡𝐞𝐧 𝐱 ∈ 𝐑: 𝐟 𝐱 − 1 2 = [𝐄 − 𝟐𝟎𝟎𝟕] 𝟑) ∅ , 𝐭𝐡𝐞 𝐞𝐦𝐩𝐭𝐲 𝐬𝐞𝐭 𝟒) 𝐑 𝟎 ≤ 𝐱 − 𝐱 < 𝟏
FUNCTIONS PCQs ⇒ − 𝟏 𝟐 ≤ 𝐱 − 𝐱 − 𝟏 𝟐 < 𝟏 𝟐 ⇒ − 𝟏 𝟐 ≤ 𝐟(𝐱) < 𝟏 𝟐 ⇒ 𝐟 𝐱 ≠ 𝟏 𝟐 ∀𝐱 ⇒ 𝐱 ∈ 𝐑 ∶ 𝐟 𝐱 = 𝟏 𝟐 = ∅ Key : 3
FUNCTIONS PCQs Solution : 𝟏) {𝐱 ∈ 𝐑: 𝟎 ≤ 𝐱 ≤ 𝟏 𝟐) {𝟎, 𝟏} 𝟐𝟓. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟐𝐱 − 𝟐 𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 , 𝐰𝐡𝐞𝐫𝐞 𝐟[𝐱] 𝐢𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐱, 𝐭𝐡𝐚𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐢𝐬 ∶ [𝐄 − 𝟐𝟎𝟎𝟔] 𝟑) {𝐱 ∈ 𝐑: 𝐱 > 𝟎} 𝟒) {𝐱 ∈ 𝐑: 𝐱 ≤ 𝟎} 𝐱 ∈ 𝐑 ⇒ ∃ 𝐧 ∈ 𝐙 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐧 ≤ 𝐱 < 𝐧 + 𝟏 ⇒ 𝐱 = 𝐧 ⇒ 𝟐𝐧 ≤ 𝟐𝐱 < 𝟐𝐧 + 𝟏 → 𝟐𝐱 = 𝟐𝐧 𝐨𝐫 𝟐𝐧 + 𝟏
FUNCTIONS PCQs ⇒ 𝟐𝐱 = 𝟐 𝐱 𝐨𝐫 𝟐 𝐱 + 𝟏 ⇒ 𝟐𝐱 − 𝟐 𝐱 = 𝟎 𝐨𝐫 𝟏 ⇒ 𝐟 𝐱 = 𝟎 𝐨𝐫 𝟏 ⇒ {𝟎, 𝟏} Key : 2
FUNCTIONS PCQs Solution : 𝟏) 𝐑, 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝟐) ∅, 𝐭𝐡𝐞 𝐞𝐦𝐩𝐭𝐲 𝐬𝐞𝐭 𝟐𝟔. {𝐱 ∈ 𝐑: 𝐱 − 𝐱 = 𝟓} [𝐄 − 𝟐𝟎𝟎𝟓] 𝟑) {𝐱 ∈ 𝐑: 𝐱 < 𝟎} 𝟒) {𝐱 ∈ 𝐑: 𝐱 ≥ 𝟎} 𝐈𝐟 𝐱 ≥ 𝟎 𝐭𝐡𝐞𝐧 𝐱 = 𝐱 ⇒ 𝐱 − 𝐱 = 𝟎 ⇒ 𝐱 − 𝐱 = 𝟎 ⇒ 𝐱 − 𝐱 ≠ 𝟓 𝐈𝐟 𝐱 < 𝟎 𝐭𝐡𝐞𝐧 𝐱 = −𝐱 ⇒ 𝐱 − 𝐱 = 𝟐𝐱 < 𝟎 ⇒ 𝐱 − 𝐱 ≠ 𝟓 ∴ 𝐱 ∈ 𝐑 ∶ 𝐱 − 𝐱 = 𝟓 = ∅
FUNCTIONS PCQs Solution : 𝟏) 𝐚 = 𝐜 𝟐) 𝐛 = 𝐝 𝟐𝟕. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐂 → 𝐂 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐚𝐱 + 𝐛 𝐜𝐱 + 𝐝 𝐟𝐨𝐫 𝐱 ∈ 𝐂 𝐰𝐡𝐞𝐫𝐞 𝐛𝐝 ≠ 𝟎 𝐫𝐞𝐝𝐮𝐜𝐞𝐬 𝐭𝐨 𝐚 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐢𝐟 ∶ [𝐄 − 𝟐𝟎𝟎𝟓] 𝟑) 𝐚𝐝 = 𝐛𝐜 𝟒) 𝐚𝐛 = 𝐜𝐝 𝐟 𝐱 = 𝐚𝐱 + 𝐛 𝐜𝐱 + 𝐝 = 𝐚 𝐜 + 𝐛𝐜 − 𝐚𝐝 𝐜[𝐜𝐱 + 𝐝] 𝐢𝐬 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ⇒ 𝐛𝐜 − 𝐚𝐝 = 𝟎 ⇒ 𝐛𝐜 = 𝐚𝐝
FUNCTIONS PCQs Solution : 𝟏) 𝟐𝐤+𝟏 − 𝟏 𝟐) 𝟐(𝟐𝐤+𝟏 − 𝟏) 𝟐𝟖. 𝐈𝐟 𝐍 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫𝐬 𝐚𝐧𝐝 𝐢𝐟 𝐟: 𝐍 → 𝐍 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐧 = 𝐭𝐡𝐞 𝐬𝐮𝐦 𝐨𝐟 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝐧 𝐭𝐡𝐞𝐧 𝐟 𝟐𝐤 , 𝟑 , 𝐰𝐡𝐞𝐫𝐞 𝐤 𝐢𝐬 𝐚 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫𝐬, 𝐢𝐬: [𝐄 − 𝟐𝟎𝟎𝟓] 𝟑) 𝟑(𝟐𝐤+𝟏 − 𝟏) 𝟒) 𝟒(𝟐𝐤+𝟏 − 𝟏) 𝐟𝐫𝐨𝐦 𝐩𝐞𝐫𝐦𝐮𝐭𝐚𝐭𝐢𝐨𝐧𝐬 𝐚𝐧𝐝 𝐜𝐨𝐦𝐛𝐢𝐧𝐚𝐭𝐢𝐨𝐧𝐬 𝐍 = 𝐏𝟏 𝛂𝟏. 𝐏𝟐 𝛂𝟐. 𝐏𝟑 𝛂𝟑 … . . 𝐏𝐤 𝛂𝐤 𝐰𝐡𝐞𝐫𝐞 𝐏𝟏, 𝐏𝟐, 𝐏𝟑 … … 𝐏𝐤 𝐚𝐫𝐞 𝐩𝐫𝐢𝐦𝐞𝐬
FUNCTIONS PCQs & 𝐬𝐮𝐦 𝐨𝐟 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝐍 = 𝐏𝟏 𝛂𝟏+𝟏 𝐏𝟏 − 𝟏 𝐏𝟐 𝛂𝟐+𝟏 𝐏𝟐 − 𝟏 𝐏𝟑 𝛂𝟑+𝟏 𝐏𝟑 − 𝟏 … . . 𝐍𝐨𝐰 𝐟 𝟐𝐤 . 𝟑 = 𝐭𝐡𝐞 𝐬𝐮𝐦 𝐨𝐟 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 = 𝟐𝐤+𝟏 − 𝟏 𝟐 − 𝟏 = 𝟑𝟐 − 𝟏 𝟑 − 𝟏 = 𝟒(𝟐𝐤+𝟏 − 𝟏) 𝐓𝐡𝐞𝐧 𝐧𝐨. 𝐨𝐟 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝐍 = 𝛂𝟏 + 𝟏 𝛂𝟐 + 𝟏 𝛂𝟑 + 𝟏 … … 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝟐𝐤. 𝟑 Key : 4
FUNCTIONS PCQs Solution : 𝟏) {𝟑, 𝟔, 𝟒} 𝟐) {𝟏, 𝟒, 𝟕} 𝟐𝟗. 𝐟: 𝐍 → 𝐙 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐧 = 𝟐 𝐈𝐟 𝐧 = 𝟑𝐤, 𝐤 ∈ 𝐳 𝟏𝟎 − 𝐧 𝐢𝐟 𝐧 = 𝟑𝐤 + 𝟏, 𝐤 ∈ 𝐳 𝟎 𝐧 = 𝟑𝐤 + 𝟐 𝐤 ∈ 𝐳 𝐭𝐡𝐞𝐧 𝐧 ∈ 𝐍: 𝐟 𝐧 > 𝟐 = [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) {𝟒, 𝟕} 𝟒) {𝟕} 𝐧 ∈ 𝐍: 𝐟 𝐧 > 𝟐 = 𝐧 ∈ 𝐍: 𝟑𝐤 + 𝟏 = 𝐧 , 𝐤 ∈ 𝐳, 𝟏𝟎 − 𝐧 > 𝟐 = 𝟏, 𝟒, 𝟕 ∴ 𝐟 𝟏 = 𝟏𝟎 − 𝟏 = 𝟗 > 𝟐 𝐟 𝟒 = 𝟏𝟎 − 𝟒 = 𝟔 > 𝟐 𝐟 𝟕 = 𝟏𝟎 − 𝟕 = 𝟑 > 𝟐
FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐥𝐲 𝐈, 𝐈𝐈 𝟐) 𝐨𝐧𝐥𝐲 𝐈𝐈, 𝐈𝐈𝐈 𝟑𝟎. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟑−𝐱 𝐨𝐛𝐬𝐞𝐫𝐯𝐞 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) 𝐨𝐧𝐥𝐲 𝐈, 𝐈𝐈𝐈 𝟒) 𝐈, 𝐈𝐈, 𝐈𝐈𝐈 𝐈) 𝐟 𝐢𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 𝐈𝐈) 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐈𝐈𝐈) 𝐟 𝐢𝐬 𝐚 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐨𝐮𝐭 𝐨𝐟 𝐭𝐡𝐞𝐬𝐞 𝐭𝐫𝐮𝐞 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 𝐚𝐫𝐞 → 𝐋𝐞𝐭 𝐱𝟏 𝐱𝟐 ∈ 𝐑 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐟 𝐱𝟏 = 𝐟(𝐱𝟐) ⇒ 𝟑−𝐱𝟏 = 𝟑−𝐱𝟐 ⇒ 𝐱𝟏 = 𝐱𝟐 ∴ 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞
FUNCTIONS PCQs → 𝐰𝐞 𝐭𝐚𝐤𝐞 𝐚𝐧𝐲 − 𝐯𝐞 𝐯𝐚𝐥𝐮𝐞 𝐢𝐧 𝐜𝐨. 𝐝𝐨𝐦𝐚𝐢𝐧 𝐭𝐡𝐞𝐲 𝐡𝐚𝐯𝐞 𝐧𝐨. −𝐩𝐫𝐞 𝐢𝐦𝐚𝐠𝐞 𝐢𝐬 𝐝𝐨𝐦𝐚𝐢𝐧 ∴ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 → 𝐟 𝐱 = 𝟑−𝐱 𝐢𝐬 𝐚 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ∴ 𝐈 & 𝐈𝐈 𝐚𝐫𝐞 𝐭𝐫𝐮𝐞 Key : 1
FUNCTIONS PCQs Solution : 𝟏) (−𝟏, 𝟑) 𝟐) [−𝟏, 𝟑) 𝟑𝟏. 𝐈𝐟 𝐟 𝐱 = 𝐱 𝐢𝐟 − 𝟑 < 𝐱 ≤ −𝟏 𝐱 𝐢𝐟 − 𝟏 < 𝐱 < 𝟏 [𝐱] 𝐢𝐟 𝟏 ≤ 𝐱 < 𝟑 𝐭𝐡𝐞𝐧 𝐱: 𝐟 𝐱 ≥ 𝟎 = [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) (−𝟏, 𝟑] 𝟒) [−𝟏, 𝟑] → −𝟑 < 𝐱 ≤ −𝟏 ⇒ 𝐟 𝐱 = 𝐱 < 𝟎 → −𝟏 < 𝐱 ≤ 𝟏 ⇒ 𝐟 𝐱 = 𝐱 ≥ 𝟎 → 𝟏 ≤ 𝐱 < 𝟑 ⇒ −𝟑 < −𝐱 ≤ −𝟏 ⇒ 𝐟 𝐱 = −𝐱 > 𝟎
FUNCTIONS PCQs ∴ 𝐱: 𝐟 𝐱 ≥ 𝟎 = −𝟏, 𝟏 ∪ 𝟏, 𝟑 = (−𝟏, 𝟑) Key : 1
FUNCTIONS PCQs Solution : 𝟏) {−𝟏} 𝟐) {𝟎} 3𝟐. 𝐒𝐮𝐩𝐩𝐨𝐬𝐞 𝐟: −𝟐, 𝟐 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = −𝟏 𝐟𝐨𝐫 − 𝟐 ≤ 𝐱 ≤ 𝟎 = 𝐱 − 𝟏 𝐟𝐨𝐫 𝟎 ≤ 𝐱 ≤ 𝟐 𝐭𝐡𝐞𝐧 𝐱 ∈ −𝟐, 𝟐 : 𝐱 ≤ 𝟎 𝐚𝐧𝐝 𝐟 𝐱 = 𝐱 = [𝐄 − 𝟐𝟎𝟎𝟑] 𝟑) − 𝟏 𝟐 𝟒) ∅ 𝐱 ≤ 𝟎 𝐟 𝐱 = 𝐱 ⇒ 𝐟 −𝐱 = 𝐱 ⇒ −𝐱 − 𝟏 = 𝐱 ⇒ 𝟐𝐱 = −𝟏 ⇒ 𝐱 = − 𝟏 𝟐
FUNCTIONS PCQs Solution : 𝟏) 𝐙 ∪ (−∞, 𝟎) 𝟐) (−∞, 𝟎) 𝟑𝟑. 𝐟: 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐠𝐢𝐯𝐞𝐧 𝐛𝐲 𝐟 𝐱 = 𝐱 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐱 ∈ 𝐑 ∶ 𝐠[𝐟 𝐱 ] ≤ 𝐟[𝐠 𝐱 ] [𝐄 − 𝟐𝟎𝟎𝟑] 𝟑) 𝐙 𝟒) 𝐑 𝐠[𝐟 𝐱 ] ≤ 𝐟[𝐠 𝐱 ] ⇒ 𝐱 ≤ [𝐱] 𝐈𝐟 𝐱 ≥ 𝟎 𝐭𝐡𝐞𝐧 𝐱 = 𝐱 ⇒ 𝐱 = 𝐱 = [𝐱] 𝐈𝐟 𝐱 < 𝟎 𝐭𝐡𝐞𝐧 𝐱 = −𝐱 ⇒ 𝐱 ≤ −[𝐱] ⇒ 𝐱 ≤ − 𝐱 ≤ [𝐱] ∴ 𝐑𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝐬𝐞𝐭 𝐢𝐬 𝐈𝐑
FUNCTIONS PCQs Solution : 𝟏) 𝟎. 𝟓 𝟐) 𝟎. 𝟔 𝟑𝟒. 𝐈𝐟 𝐞𝐟(𝐱) = 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 (𝐗 ∈ −𝟏𝟎, 𝟏𝟎 𝐚𝐧𝐝 𝐟 𝐱 = 𝐤𝐟 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎 + 𝐱𝟐 𝐭𝐡𝐞𝐧 𝐤 [𝐄 − 𝟐𝟎𝟎𝟑] 𝟑) 𝟎. 𝟕 𝟒) 𝟎. 𝟖 𝐆𝐢𝐯𝐞𝐧 𝐞𝐟(𝐱) = 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 ⇒ 𝐟(𝐱) = 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 𝐭𝐚𝐤𝐞 𝐟 𝐱 = 𝐤. 𝐟 𝟐𝟎𝟎 𝟏𝟎𝟎 + 𝐱𝟐
FUNCTIONS PCQs = 𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎 + 𝐱𝟐 𝟏𝟎 − 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎 + 𝐱𝟐 = 𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎𝟎𝟎 + 𝟏𝟎𝐱𝟐 + 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎𝟎 + 𝟏𝟎𝐱𝟐 − 𝟐𝟎𝟎𝐱 = 𝐤. 𝐥𝐨𝐠𝐞 𝐱𝟐 + 𝟐𝟎𝐱 + 𝟏𝟎𝟎 𝐱𝟐 − 𝟐𝟎𝐱 + 𝟏𝟎𝟎 = 𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 𝟐 = 𝟐𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 = 𝟐𝐤. 𝐟(𝐱) ⇒ 𝟏 = 𝟐𝐤 ⇒ 𝐤 = 𝟏 𝟐 Key : 1
FUNCTIONS PCQs Solution : 𝟏) 𝟏 𝟐) 𝟐 𝟑𝟓. 𝐈𝐟 𝐟 𝐱 = 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟒𝐱 𝐬𝐢𝐧𝟐𝐱 + 𝐜𝐨𝐬𝟒𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐟 𝟐𝟎𝟎𝟐 = [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) 𝟑 𝟒) 𝟒 𝐟 𝐱 = 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟒𝐱 𝐬𝐢𝐧𝟐𝐱 + 𝐜𝐨𝐬𝟒𝐱 = 𝐬𝐢𝐧𝟐 𝐱. 𝐬𝐢𝐧𝟐 𝐱 + 𝐜𝐨𝐬𝟐 𝐱 𝐜𝐨𝐬𝟐𝐱. 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱 = (𝟏 − 𝐜𝐨𝐬𝟐 𝐱)𝐬𝐢𝐧𝟐 𝐱 + 𝐜𝐨𝐬𝟐 𝐱 (𝟏 − 𝐬𝐢𝐧𝟐𝐱)𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱
FUNCTIONS PCQs = 𝐬𝐢𝐧𝟐𝐱 − 𝐬𝐢𝐧𝟐𝐱 𝐜𝒐𝒔𝟐𝐱 + 𝐜𝐨𝐬𝟐𝐱 𝐜𝐨𝐬𝟐𝐱 − 𝐬𝐢𝐧𝟐𝐱 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱 = 𝟏 𝐒𝐨 𝐟 𝐱 𝐢𝐬 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ∴ 𝐟 𝟐𝟎𝟎𝟐 = 𝟏 Key : 1
FUNCTIONS PCQs Solution : 𝟏) 𝟏 𝟐) 𝟏 𝟐 𝟑𝟔. 𝐈𝐟 𝐭𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬 𝐟 𝐚𝐧𝐝 𝐠 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟑𝐱 − 𝟒, 𝐠 𝐱 = 𝟐 + 𝟑𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐫𝐞𝐬𝐩𝐞𝐜𝐭𝐢𝐯𝐞𝐥𝐲 𝐭𝐡𝐞𝐧 (𝐠−𝟏 𝐨𝐟−𝟏 )(𝟓)) = [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) 𝟏 𝟑 𝟒) 𝟏 𝟒 𝐲 = 𝐟 𝐱 = 𝟑𝐱 − 𝟒 ⇒ 𝐱 = 𝐲 + 𝟒 𝟑 ⇒ 𝐟−𝟏(𝐱) = 𝐱 + 𝟒 𝟑 𝐲 = 𝐠 𝐱 = 𝟑𝐱 + 𝟐 ⇒ 𝐱 = 𝐲 − 𝟐 𝟑 ⇒ 𝐠−𝟏(𝐱) = 𝐱 − 𝟐 𝟑
FUNCTIONS PCQs 𝐍𝐨𝐰 𝐠−𝟏 𝐨𝐟−𝟏 𝟓 = 𝐠−𝟏 𝐟−𝟏 𝟓 = 𝐠−𝟏 𝟓 + 𝟒 𝟑 = 𝐠−𝟏(𝟑) = 𝟑 − 𝟐 𝟑 = 𝟏 𝟑 Key : 3
FUNCTIONS PCQs Solution : 𝟏) {𝟏, −𝟏} 𝟐) {𝐱: 𝟎 ≤ 𝐱 ≤ 𝟒} 3𝟕. 𝐋𝐞𝐭 𝐀 = 𝐱 ∈ 𝐑: 𝐱 ≠ 𝟎, −𝟒 ≤ 𝐱 ≤ 𝟗 𝐚𝐧𝐝 𝐟: 𝐀 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = x x 𝐟𝐨𝐫 𝐱 ∈ 𝐀 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐢𝐬 [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) {𝟏} 𝟒) {𝐱: −𝟒 ≤ 𝐱 ≤ 𝟎} 𝐁𝐲 𝐝𝐞𝐟𝐢𝐧𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐦𝐨𝐝𝐮𝐥𝐞𝐬 𝐱 = 𝐱 𝐢𝐟 𝐱 > 𝟎 = −𝐱 𝐢𝐟 𝐱 < 𝟎 𝐒𝐨 𝐟 𝐱 = 𝐱 𝐱 = ±𝐱 𝐱 = ±𝟏 ∴ 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐱 = −𝟏, 𝟏
FUNCTIONS PCQs Solution : 𝟏) 𝟑 𝟒 , 𝟏 𝟐) 𝟑 𝟒 , 𝟏 𝟑𝟖. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐜𝐨𝐬𝟐 𝐱 + 𝐬𝐢𝐧𝟒 𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐟(𝐑) [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) 𝟑 𝟒 , 𝟏 𝟒) 𝟑 𝟒 , 𝟏 𝐟 𝐱 = 𝐜𝐨𝐬𝟐 𝐱 + 𝐬𝐢𝐧𝟒 𝐱 = 𝐜𝐨𝐬𝟐 𝐱 + 𝐬𝐢𝐧𝟐 𝐱(𝟏 − 𝐜𝐨𝐬𝟐 𝐱) = 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱 − 𝐬𝐢𝐧𝟐𝐱 𝐜𝐨𝐬𝟐𝐱 = 𝟏 − 𝟏 𝟒 𝐬𝐢𝐧𝟐𝟐𝐱
FUNCTIONS PCQs 𝐰. 𝐤. 𝐓 𝟎 ≤ 𝐬𝐢𝐧𝟐𝟐𝐱 ≤ 𝟏 −𝟏 𝟒 ≤ −𝟏 𝟒 𝐬𝐢𝐧𝟐 𝟐𝐱 ≤ 𝟎 𝟏 − 𝟏 𝟒 ≤ 𝟏 − 𝟏 𝟒 𝐬𝐢𝐧𝟐 𝟐𝐱 ≤ 𝟎 + 𝟏 𝟑 𝟒 ≤ 𝐟 𝐱 ≤ 𝟏 Key : 1
FUNCTIONS PCQs Solution : 𝟏) 𝟐−𝟒 𝟐) 𝟐−𝟑 𝟑𝟗. 𝐈𝐟 𝐟 𝐱 = (𝟐𝟓 − 𝐱𝟒) 𝟏 𝟒 𝐟𝐨𝐫 𝟎 < 𝐱 < 𝟓 𝐭𝐡𝐞𝐧 𝐟 𝐟 𝟏 𝟐 = [𝐄 − 𝟐𝟎𝟎𝟏] 𝟑) 𝟐−𝟐 𝟒) 𝟐−𝟏 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = (𝟐𝟓 − 𝐱𝟒 ) 𝟏 𝟒 𝐍𝐨𝐰 𝐟 𝐟 𝟏 𝟐 = 𝐟 𝟐𝟓 − 𝟏 𝟏𝟔 𝟏 𝟒 = 𝐟 𝟑𝟗𝟗 𝟏𝟔 𝟏 𝟒
FUNCTIONS PCQs = 𝟐𝟓 − 𝟑𝟗𝟗 𝟏𝟔 𝟏 𝟒 = 𝟏 𝟏𝟔 𝟏 𝟒 = 𝟏 𝟐𝟒 𝟏 𝟒 = 𝟏 𝟐 = 𝟐−𝟏 Key : 4
FUNCTIONS PCQs Solution : 𝟏) − 𝟏 𝟐) 𝟎 𝟒𝟎. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐟𝐨𝐥𝐥𝐨𝐰𝐬 [𝐄 − 𝟐𝟎𝟎𝟏] 𝟑) 𝟏 𝟒) 𝟐 𝐟 𝐱 = 𝟎(𝐱 𝐢𝐬 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) = 𝟏(𝐱 𝐢𝐬 𝐢𝐫𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) 𝐠 𝐱 = −𝟏(𝐱 𝐢𝐬 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) = 𝟎(𝐱 𝐢𝐬 𝐢𝐫𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) 𝐭𝐡𝐞𝐧 𝐟𝐨𝐠 𝛑 + 𝐠𝐨𝐟(𝐞) 𝐟𝐨𝐠 𝛑 + 𝐠𝐨𝐟(𝐞) = 𝐟 𝐠 𝛑 + 𝐠 𝐟 𝐞 ∴ 𝛑 𝐚𝐧𝐝 𝐞 𝐚𝐫𝐞 𝐢𝐫𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞𝐬
FUNCTIONS PCQs = 𝐟 𝟎 + 𝐠[𝟏] = 𝟎 − 𝟏 = −𝟏 ∴ 𝟎&𝟏 𝐚𝐫𝐞 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞𝐬) Key : 1
FUNCTIONS PCQs Solution : 𝟏) 𝟎 𝟐) 𝟐 𝟒𝟏. 𝐋𝐞𝐭 𝐟: 𝐑 → 𝐑 𝐛𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 + 𝟐, 𝐱 ≤ −𝟏 = 𝐱𝟐(−𝟏 ≤ 𝐱 ≤ 𝟏) = 𝟐 − 𝐱 𝐱 ≥ 𝟏 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐟 −𝟏. 𝟕𝟓 + 𝐟 𝟎. 𝟓 + 𝐟 𝟏. 𝟓 = [𝐄 − 𝟐𝟎𝟎𝟏] 𝟑) 𝟏 𝟒) − 𝟏 𝐟 −𝟏. 𝟕𝟓 + 𝐟 𝟎. 𝟓 + 𝐟 𝟏. 𝟓 = −𝟏. 𝟕𝟓 + 𝟐 + 𝟎. 𝟓 𝟐 + 𝟐 − 𝟏. 𝟓 = 𝟎. 𝟐𝟓 + 𝟎. 𝟐𝟓 + 𝟎. 𝟓 = 𝟏
FUNCTIONS PCQs Solution : 𝟏) 𝐗 = 𝐘 = 𝐑+ 𝟐) 𝐗 = 𝐑, 𝐘 = 𝐑+ 𝟒𝟐. 𝐋𝐞𝐭 𝐗 𝐚𝐧𝐝 𝐘 𝐛𝐞 𝐬𝐮𝐛𝐬𝐞𝐭𝐬 𝐨𝐟 𝐑, 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐭𝐡𝐞 𝐟: 𝐗 → 𝐘 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 𝐟𝐨𝐫 𝐱 ∈ 𝐗 𝐨𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 onto if [𝐄 − 𝟐𝟎𝟎𝟎] 𝟑) 𝐗 = 𝐑+, 𝐲 = 𝐑 𝟒) 𝐗 = 𝐘 = 𝐑 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐱𝟐 𝐚𝐧𝐝 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝐁𝐲 𝐯𝐞𝐫𝐢𝐟𝐢𝐜𝐚𝐭𝐢𝐨𝐧 𝐗 ∈ 𝐑+; 𝐲 ∈ 𝐑 𝐟 𝐱 = 𝐱𝟐 is satisfies one-one but not onto
FUNCTIONS PCQs Solution : 𝟏) ± 𝟏 𝟐) ± 𝟐 𝟒𝟑. 𝐟 = 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟐𝐱 + 𝟑 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱𝟐 + 𝟕 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞𝐬 𝐨𝐟 𝐱 𝐟𝐨𝐫 𝐰𝐡𝐢𝐜𝐡 𝐟 𝐠 𝐱 = 𝟐𝟓 𝐚𝐫𝐞 [𝐄 − 𝟐𝟎𝟎𝟎] 𝟑) ± 𝟑 𝟒) ± 𝟒 𝐆𝐢𝐯𝐞𝐧 𝐟: 𝐑 → 𝐑; 𝐠: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟐𝐱 + 𝟑 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱𝟐 + 𝟕 𝐍𝐨𝐰 𝐰𝐞 𝐭𝐚𝐤𝐞 𝐟 𝐠 𝐱 = 𝟐𝟓 ⇒ 𝐟 𝐱𝟐 + 𝟕 = 𝟐𝟓
FUNCTIONS PCQs ⇒ 𝟐 𝐱𝟐 + 𝟕 + 𝟑 = 𝟐𝟓 ⇒ 𝟐𝐱𝟐 + 𝟏𝟕 = 𝟐𝟓 ⇒ 𝟐𝐱𝟐 = 𝟖 ⇒ 𝐱𝟐 = 𝟒 ⇒ 𝐱 = ±𝟐 Key : 2
FUNCTIONS PCQs Solution : 𝟏)𝟏 𝟐)𝟐 44. For any integer n1, the number of positive divisors of n is denoted by d(n). Then for a prime p, d(d(d(𝒑𝟕))) = [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) 𝟑 𝟒) 𝟒 𝒅 𝒑𝟕 = 𝟖 = 𝟐𝟑 ⇒ 𝒅 𝒅 𝒑𝟕 = 𝟒 = 𝟐𝟐 ⇒ 𝒅 𝒅 𝒅 𝒑𝟕 = 𝟑 Key : 3
FUNCTIONS PCQs Solution : 𝟏)𝟐𝟒 𝟐)𝟐𝟔 45. A function f satisfies the relation 𝒇 𝒏𝟐 = 𝒇 𝒏 + 𝟔 for n 2 and f (2) = 8 then the value of f(256) is [𝐊𝐞𝐫𝐚𝐥𝐚 − 𝟐𝟎𝟏𝟓] 𝟑) 𝟐𝟐 𝟒) 𝟒𝟖 𝒇 𝟐𝟓𝟔 = 𝒇 𝟏𝟔𝟐 = 𝒇(𝟏𝟔) + 𝟔 𝒇 𝟏𝟔 = 𝒇 𝟒𝟐 = 𝒇 𝟒 + 𝟔 𝒇 𝟒 = 𝒇 𝟐𝟐 = 𝒇 𝟐 + 𝟔 𝑩𝒖𝒕 𝒇 𝟐 = 𝟖 f(4)= 14, f(16)=20, f(256)= 26 Key : 2
FUNCTIONS PCQs Solution : 𝟏)𝟏 𝟐) − 𝟏𝟓 46. If f(1) = 1, f(2n) = f(n) and f(2n+1)= 𝒇 𝒏 𝟐 − 𝟐 for n = 1,2,3…. Then the value of f(1)+f(2)+ …… +f(25) is equal to [𝐊𝐞𝐫𝐚𝐥𝐚 − 𝟐𝟎𝟏𝟓] 𝟑) − 𝟏𝟕 𝟒) − 𝟏 𝒇 𝟐𝒏 + 𝟏 = −𝟏 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒏 𝒇 𝟏 = 𝒇 𝟐 = 𝟏 𝒂𝒏𝒅 𝒇 𝟐𝒏 = 𝟏  f(4)= f(8)= f(16)=1 𝒇 𝟐𝒏 = −𝟏 when ‘n’ is odd 𝒔𝒖𝒎 = 𝟓 − 𝟐𝟎 = −𝟏𝟓 Key : 2
FUNCTIONS PCQs Solution : 𝟏) f is periodic 𝟐) f is an odd function 47. Let f(x) = 𝒙 − 𝟐 , 𝐰𝐡𝐞𝐫𝐞 ′ 𝐱′𝐢𝐬 𝐚 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬, which one of the following is true [𝐊𝐞𝐫𝐚𝐥𝐚 − 𝟐𝟎𝟏𝟒] 𝟑) f is not a 1-1 function 𝟒) f is an equal function 𝒇 𝟏 = 𝟏 − 𝟐 = 𝟏 𝒇 𝟑 = 𝟑 − 𝟐 = 𝟏  𝒇 𝒊𝒔 𝒏𝒐𝒕 𝟏 − 𝟏 Key : 3
FUNCTIONS PCQs 𝐋𝐢𝐬𝐭 − 𝐈 𝟐𝟕. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 + 𝟒 𝐟𝐨𝐫 𝐱 < −𝟒 𝟑𝐱 + 𝟐 𝐟𝐨𝐫 − 𝟒 ≤ 𝐱 < 𝟒 𝐱 − 𝟒 𝐟𝐨𝐫 𝐱 ≥ 𝟒 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐜𝐨𝐫𝐫𝐞𝐜𝐭 𝐦𝐚𝐭𝐜𝐡𝐢𝐧𝐠 𝐨𝐟 𝐋𝐢𝐬𝐭 − 𝐈 𝐟𝐫𝐨𝐦 𝐋𝐢𝐬𝐭 – 𝐈𝐈 𝐢𝐬 [𝐄 − 𝟐𝟎𝟎𝟔] 𝐋𝐢𝐬𝐭 − 𝐈I 𝐀) 𝐟 −𝟓 + 𝐟(−𝟒) 𝐢) 𝟏𝟒 𝐁) 𝐟( 𝐟 −𝟖 ) 𝐢𝐢) 𝟒 𝐂) 𝐟(𝐟 −𝟕 + 𝐟 𝟑 ) 𝐢𝐢𝐢) − 𝟏𝟏 𝐃) 𝐟 𝐟 𝐟 𝐟 𝟎 + 𝟏 𝐢𝐯) − 𝟏 𝐯) 𝟏 𝐯𝐢) 𝟎
FUNCTIONS PCQs Solution : 𝐀 𝐁 𝐂 𝐃 1) iii vi ii v 2) iii iv ii v 3) iv iii ii i 4) iii vi v ii 𝐀 𝐟 −𝟓 + 𝐟 −𝟒 = −𝟓 + 𝟒 + 𝟑 −𝟒 + 𝟐 = −𝟏 + (−𝟏𝟎) = −𝟏𝟏
FUNCTIONS PCQs 𝐁 𝐟 𝐟 −𝟖 = 𝐟 −𝟖 + 𝟒 = 𝐟 𝟒 = 𝟒 − 𝟒 = 𝟎 𝐂 𝐟 𝐟 −𝟕 + 𝐟 𝟑 = 𝐟[ −𝟕 + 𝟒 + (𝟑 𝟑 + 𝟐] = 𝐟 −𝟑 + 𝟏𝟏 = 𝐟 𝟖 = 𝟖 − 𝟒 = 𝟒 𝐃 𝐟(𝐟 𝐟 𝟑 𝟎 + 𝟐 ]) + 𝟏 = 𝐟(𝐟[𝐟 𝟐 ]) + 𝟏 = 𝐟 𝐟 𝟑 𝟐 + 𝟐 + 𝟏 = 𝐟 𝐟 𝟖 + 𝟏 = 𝐟 𝟒 + 𝟏 = 𝟒 − 𝟒 + 𝟏 = 𝟏 Key : 1
FUNCTIONS PCQs Thank you…

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function power point presentation for class 11 and 12 for jee

  • 3. FUNCTIONS PCQs Solution : 𝟏) 𝒃𝒐𝒕𝒉 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒂𝒏𝒅 𝒐𝒏𝒕𝒐 𝟐)𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒃𝒖𝒕 𝒏𝒐𝒕 𝒐𝒏𝒕𝒐 𝟑) 𝒏𝒆𝒊𝒕𝒉𝒆𝒓 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒏𝒐𝒓 𝒐𝒏𝒕𝒐 𝟒) 𝒐𝒏𝒕𝒐 but not 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝟏. 𝐋𝐞𝐭 𝐟: 𝐑 → 𝑹 𝒃𝒆 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒃𝒚 𝐟 𝐱 = 𝒙 − 𝟏 𝒙 + 𝟏 𝒕𝒉𝒆 𝒇 𝒊𝒔 𝐰𝐞 𝐡𝐚𝐯𝐞 𝐟 𝐱 = 𝒙 − 𝟏 𝒙 + 𝟏 [𝐉𝐞𝐞 − 𝟐𝟎𝟏𝟒] 𝒄𝒍𝒆𝒂𝒓𝒍𝒚 , 𝒇 𝒊𝒔 𝒏𝒐𝒕 𝒐𝒏𝒆 − 𝒐𝒏𝒆 𝒂𝒔 𝒇 −𝟏 = 𝟎 = 𝒇 𝟏 𝒂𝒏𝒅 𝒓𝒂𝒏𝒈𝒆 ≠ 𝒄𝒐𝒅𝒐𝒎𝒂𝒊𝒏 𝒊. 𝒆 𝒏𝒐𝒕 𝒐𝒏𝒕𝒐
  • 4. FUNCTIONS PCQs Solution : 𝟏) 𝟑 𝟐 + 𝟐 𝟑 𝟐. 𝐋𝐞𝐭 𝐟 𝒃𝒆 𝒂𝒏 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒐𝒇 𝒓𝒆𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝒇𝒐𝒓 𝒙 ≥ 𝟎, 𝐟 𝐱 =3sinx+4cosx . Then 𝐟 𝐱 at x= −11𝛑 𝟔 𝐟 −𝟏𝟏𝝅 𝟔 = −𝐟 𝟏𝟏𝝅 𝟔 [𝐉𝐞𝐞 − 𝟐𝟎𝟏𝟒] 𝟐) − 𝟑 𝟐 + 𝟐 𝟑 𝟑) 𝟑 𝟐 − 𝟐 𝟑 𝟒) − 𝟑 𝟐 − 𝟐 𝟑 = 𝟑𝒔𝒊𝒏 𝟏𝟏𝝅 𝟔 + 𝟒𝒄𝒐𝒔 𝟏𝟏𝝅 𝟔
  • 5. FUNCTIONS PCQs = − 𝟑𝒔𝒊𝒏 𝟐𝝅 − 𝝅 𝟔 + 𝟒𝒄𝒐𝒔 𝟐𝝅 − 𝝅 𝟔 = −𝟑 𝟐 + 𝟒 𝟑 𝟐 = 𝟑 𝟐 − 𝟐 𝟑 Key : 3
  • 6. FUNCTIONS PCQs Solution : 𝟏) (−∞, ∞) 𝟐) (𝟎, ∞) 𝟑) (−∞, 𝟎) 𝟒) −∞, ∞ − {𝟎} 𝟑. 𝐓𝐡𝐞 𝐝𝐨𝐦𝐚𝐢𝐧 𝐨𝐟 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 = 𝟏 𝐱 − 𝐱 [𝐀 − 𝟐𝟎𝟎𝟏] 𝐟 𝐱 = 𝟏 𝐱 − 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐱 − 𝐱 > 𝟎 ⇒ 𝐱 > 𝐱 𝐱 ∈ (−∞ 𝟎)
  • 7. FUNCTIONS PCQs Solution : 𝟏) 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝐑 𝟐) 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐑 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟑) 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝐑 𝟒) 𝐟 𝐢𝐬 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐨𝐧𝐭𝐨 𝐑 𝟒. 𝐅𝐨𝐫 𝐫𝐞𝐚𝐥 𝐱, 𝐥𝐞𝐭 𝐟 𝐱 = 𝐱𝟑 + 𝟓𝐱 + 𝟏, 𝐭𝐡𝐞𝐧 [𝐀 − 𝟐𝟎𝟎𝟗] 𝐟 𝐱 = 𝐱𝟑 + 𝟓𝐱 + 𝟏 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐋𝐞𝐭 𝐲 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐲 = 𝐱𝟑 + 𝟓𝐱 + 𝟏 ⇒ 𝐱𝟑 + 𝟓𝐱 + 𝟏 − 𝐲 = 𝟎
  • 8. FUNCTIONS PCQs 𝐀𝐬 𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥 𝐨𝐟 𝐨𝐝𝐝 𝐝𝐞𝐠𝐫𝐞𝐞 𝐡𝐚𝐬 𝐚𝐥𝐰𝐚𝐲𝐬 𝐚𝐭 𝐥𝐞𝐚𝐬𝐭 𝐨𝐧𝐞 𝐫𝐞𝐚𝐥 𝐫𝐨𝐨𝐭 𝐂𝐨𝐫𝐫𝐞𝐬𝐩𝐨𝐧𝐝𝐢𝐧𝐠 𝐭𝐨 𝐚𝐧𝐲 𝐲 ∈ 𝐜𝐨. 𝐝𝐨𝐦𝐚𝐢𝐧 𝐒𝐨 ∃ 𝐬𝐨𝐦𝐞 𝐱 ∈ 𝐝𝐨𝐦𝐚𝐢𝐧 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐟 𝐱 = 𝐲. 𝐡𝐞𝐧𝐜𝐞 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐀𝐥𝐬𝐨 𝐟 𝐢𝐬 𝐜𝐨𝐧𝐭𝐢𝐧𝐨𝐮𝐬 𝐨𝐧 𝐑, 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝐢𝐭 𝐢𝐬 𝐚 𝐩𝐨𝐥𝐲𝐧𝐨𝐦𝐢𝐚𝐥 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐍𝐨𝐰 𝐟′ 𝐱 = 𝟑𝐱𝟐 + 𝟓 > 𝟎 ∴ 𝐟 𝐢𝐬 𝐬𝐭𝐫𝐢𝐜𝐭𝐥𝐲 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 Key : 3
  • 9. FUNCTIONS PCQs 𝟓. 𝐋𝐞𝐭 𝐟 𝐱 = (𝐱 + 𝟏)𝟐−𝟏, 𝐱 ≥ −𝟏 [𝐀 − 𝟐𝟎𝟎𝟗] 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 − 𝟏: 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐱: 𝐟 𝐱 = 𝐟−𝟏 𝐱 = {𝟎, −𝟏} 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 − 𝟐: 𝐟 𝐢𝐬 𝐚 𝐛𝐢𝐣𝐞𝐜𝐭𝐢𝐨𝐧 𝟏) statement –1 is true , statement – 2 is true ; statement – 2 is the correct explanation of statement - 1 𝟐) statement –1 is true statement – 2 is true ; statement – 2 is not the correct explanation of statement - 1 𝟑) statement –1 is true , statement – 2 is false 𝟒) statement –I is false , statement – II is true
  • 10. FUNCTIONS PCQs Solution : 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = (𝐱 + 𝟏)𝟐−𝟏 → 𝐬𝟏 ∶ 𝐟 𝐱 = 𝐟′(𝐱) ⇒ (𝐱 + 𝟏)𝟐 −𝟏 = 𝐟′ 𝐱 ⇒ 𝐟 𝐱𝟐 + 𝟐𝐱 = 𝐱 ⇒ (𝐱𝟐 + 𝟐𝐱 + 𝟏)𝟐= 𝐱 + 𝟏 ⇒ (𝐱 + 𝟏)𝟒= 𝐱 + 𝟏 ⇒ 𝐱 + 𝟏 𝐱 + 𝟏 𝟑 − 𝟏 = 𝟎 ⇒ 𝐱 + 𝟏 = 𝟎 𝐨𝐫 (𝐱 + 𝟏)𝟑= 𝟏 ⇒ 𝐱 = −𝟏 𝐨𝐫 𝐱 = 𝟎 ∴ 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 ′ 𝟏′𝐢𝐬 𝐭𝐫𝐮𝐞
  • 11. FUNCTIONS PCQs → 𝐬𝟐 ∶ 𝐂𝐨𝐧𝐜𝐥𝐮𝐬𝐢𝐨𝐧 𝐨𝐟 ′ 𝐟′𝐢𝐬 𝐧𝐨𝐭 𝐬𝐩𝐞𝐜𝐟𝐢𝐞𝐝 ⇒ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 ⇒ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐛𝐢𝐣𝐞𝐜𝐭𝐢𝐨𝐧 ∴ 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭 ′ 𝟐′𝐢𝐬 𝐟𝐚𝐥𝐬𝐞 Key : 3
  • 12. FUNCTIONS PCQs Solution : 𝟏) [𝟎, 𝛑] 𝟐) −𝛑 𝟐 , 𝛑 𝟐 𝟑) −𝛑 𝟒 , 𝛑 𝟐 𝟒) 𝟎, 𝛑 𝟐 𝟔. 𝐓𝐡𝐞 𝐥𝐚𝐫𝐠𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 𝐥𝐲𝐢𝐧𝐠 𝐢𝐧 −𝛑 𝟐 , 𝛑 𝟐 𝐟𝐨𝐫 𝐰𝐡𝐢𝐜𝐡 𝐲𝐡𝐫 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 = 𝟒−𝐱𝟐 + 𝐜𝐨𝐬−𝟏 x 2 − 𝟏 + 𝐥𝐨𝐠 𝐜𝐨𝐬 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 , 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟕] 𝐟 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐢𝐟 − 𝟏 ≤ 𝐱 𝟐 − 𝟏 ≤ 𝟏
  • 13. FUNCTIONS PCQs ⇒ 𝟎 ≤ 𝐱 𝟐 ≤ 𝟐 ⇒ 𝟎 ≤ 𝐱 ≤ 𝟒 _____(𝟏) 𝐜𝐨𝐬 𝐱 > 𝟎 ⇒ − 𝛑 𝟐 < 𝐱 < 𝛑 𝟐 ______(𝟐) 𝐟𝐫𝐨𝐦 𝟏 𝐚𝐧𝐝 (𝟐) 𝐟 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝟎 ≤ 𝐱 < 𝛑 𝟐 𝐇𝐞𝐧𝐜𝐞 𝛑 𝟐 ≅ 𝟏. 𝟓𝟕 Key : 4
  • 14. FUNCTIONS PCQs Solution : 𝟏) − 𝛑 𝟐 , 𝛑 𝟐 𝟐) 𝛑 𝟐 , 𝛑 𝟐 𝟑) 𝟎, 𝛑 𝟐 𝟒) 𝟎, − 𝛑 𝟐 𝟕. 𝐋𝐞𝐭 𝐟: −𝟏, 𝟏 → 𝐁, 𝐛𝐞 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐝𝐞𝐭𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐭𝐚𝐧−𝟏 𝟐𝐱 𝟏 − 𝐱𝟐 , 𝐭𝐡𝐞𝐧 𝐟 𝐢𝐬 𝐛𝐨𝐭𝐡 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝐰𝐡𝐞𝐧 𝐁 𝐢𝐬 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 [𝐀 − 𝟐𝟎𝟎𝟓] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐭𝐚𝐧−𝟏 𝟐𝐱 𝟏 − 𝐱𝟐 = 𝟐 𝐭𝐚𝐧−𝟏𝐱
  • 15. FUNCTIONS PCQs 𝐆𝐢𝐯𝐞𝐧 𝐱 ∈ −𝟏, 𝟏 ⇒ 𝐭𝐚𝐧 −𝟏 𝐱 ∈ −𝛑 𝟒 , 𝛑 𝟒 ⇒ 𝟐𝐭𝐚𝐧 −𝟏 𝐱 ∈ −𝛑 𝟐 , 𝛑 𝟐 ∴ 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐗 = − 𝛑 𝟐 𝛑 𝟐 Key : 1
  • 16. FUNCTIONS PCQs Solution : 𝟏) 𝐟 𝐱 + 𝟐 = 𝐟(𝐱 − 𝟐) 𝟐) 𝐟 𝐱 + 𝟐 = 𝐟(𝟐 − 𝐱) 𝟑) 𝐟 𝐱 = 𝐟(−𝐱) 𝟒) 𝐟 𝐱 = −𝐟(−𝐱) 𝟖. 𝐈𝐟 𝐭𝐡𝐞 𝐠𝐫𝐚𝐩𝐡 𝐨𝐟 𝐭𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐲 = 𝐟 𝐱 𝐢𝐬 𝐬𝐲𝐦𝐦𝐞𝐭𝐫𝐢𝐜𝐚𝐥 𝐚𝐛𝐨𝐮𝐭 𝐭𝐡𝐞 𝐥𝐢𝐧𝐞 𝐱 = 𝟐, 𝐭𝐡𝐞𝐧 [𝐀 − 𝟐𝟎𝟎𝟒] 𝐥𝐞𝐭 𝐱 = 𝐱 − 𝟐; 𝐟 𝐱 = 𝐟(𝐱 + 𝟐) 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 + 𝟐 = 𝐲 𝐢𝐬 𝐬𝐲𝐦𝐦𝐞𝐭𝐫𝐢𝐜 𝐚𝐛𝐨𝐮𝐭 𝐲 − 𝐚𝐱𝐢𝐬 𝐟 𝐱 = 𝐟(−𝐱) ⇒ 𝐟 𝐱 + 𝟐 = 𝐟(𝟐 − 𝐱)
  • 17. FUNCTIONS PCQs Solution : 𝟏) 𝐟(−𝐱) 𝟐) 𝐟 𝐚 + 𝐟(𝐚 − 𝐱) 𝟑) 𝐟(𝐱) 𝟒) − 𝐟(𝐱) 𝟗. 𝐀 𝐫𝐞𝐚𝐥 𝐯𝐚𝐥𝐮𝐞𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 𝐬𝐚𝐭𝐢𝐬𝐟𝐢𝐞𝐬 𝐭𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐚𝐥 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝐟 𝐱 − 𝐲 = 𝐟 𝐱 𝐟 𝐲 − 𝐟 𝐚 − 𝐱 𝐟(𝐚 + 𝐲) 𝐰𝐡𝐞𝐫𝐞 𝐚 𝐢𝐬 𝐚 𝐠𝐢𝐯𝐞𝐧 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐚𝐧𝐝 𝐟 𝟎 = 𝟏, 𝐟 𝟐𝐚 − 𝐱 𝐢𝐬 𝐞𝐪𝐮𝐚𝐥 𝐭𝐨 [𝐀 − 𝟐𝟎𝟎𝟒] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝟎 = 𝟏 ⇒ 𝐟 𝟎 = 𝐟 𝟎 − 𝟎 = 𝐟 𝟎 . 𝐟 𝟎 − 𝐟 𝐚 − 𝟎 . 𝐟(𝐚 + 𝟎) = 𝟏 − [𝐟(𝐚)]𝟐 ⇒ [𝐟(𝐚)]𝟐= 𝟎 ⇒ 𝐟 𝐚 = 𝟎
  • 18. FUNCTIONS PCQs 𝐍𝐨𝐰 𝐟 𝟐𝐚 − 𝐱 = 𝐟 𝐚 − 𝐱 − 𝐚 = 𝐟 𝐚 . 𝐟 𝐱 − 𝐚 − 𝐟 𝐚 − 𝐚 . 𝐟(𝐚 + 𝐱 − 𝐚) = −𝐟(𝐱) Key : 4
  • 19. FUNCTIONS PCQs Solution : 𝟏) [𝟎, 𝟑] 𝟐) [−𝟏, 𝟑] 𝟑) [𝟎, 𝟏] 𝟒) [−𝟏, 𝟏] 𝟏𝟎. 𝐈𝐟 𝐟: 𝐑 → 𝐒, 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 − 𝟑𝐜𝐨𝐬 𝐱 + 𝟏 𝐢𝐬 𝐨𝐧𝐭𝐨, 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐚𝐥 𝐨𝐟 𝐒 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟒] 𝐟: 𝐑 → 𝐒 𝐢𝐬 𝐨𝐧𝐭𝐨 ⇒ 𝐒 = 𝐑𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 = 𝟏 − 𝟏 + 𝟑, 𝟏 + 𝟏 + 𝟑 = 𝟏 − 𝟐, 𝟏 + 𝟐 = −𝟏, 𝟑
  • 20. FUNCTIONS PCQs Solution : 𝟏) (𝟏, 𝟐) 𝟐) −𝟏, 𝟎 ∪ (𝟏, 𝟐) 𝟑) 𝟏, 𝟐 ∪ (𝟐, ∞) 𝟒) −𝟏, 𝟎 ∪ 𝟏, 𝟐 ∪ (𝟐, ∞) 𝟏𝟏. 𝐓𝐡𝐞 𝐝𝐨𝐦𝐚𝐢𝐧 𝐨𝐟 𝐟 𝐱 = 𝟑 𝟒 − 𝐱𝟐 + 𝐥𝐨𝐠𝟏𝟎 𝐱𝟑 − 𝐱 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟑] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝟑 𝟒 − 𝐱𝟐 + 𝐥𝐨𝐠𝟏𝟎 𝐱𝟑 − 𝐱 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝟒 − 𝐱𝟐 ≠ 𝟎 ⇒ 𝐱𝟐 ≠ 𝟒 ⇒ 𝐱 ≠ ±𝟐_____(𝟏) 𝐚𝐧𝐝 𝐱𝟑 − 𝐱 > 𝟎 ⇒ 𝐱 𝐱 − 𝟏 > 𝐱 + 𝟏 > 𝟎
  • 21. FUNCTIONS PCQs ⇒ 𝐱 ∈ −𝟏, 𝟎 ∪ 𝟏, ∞ ____(𝟐) 𝐟𝐫𝐨𝐦 𝟏 𝐚𝐧𝐝 (𝟐) 𝐝𝐨𝐦𝐚𝐢𝐧 𝐨𝐟 𝐟 𝐱 = −𝟏 𝟎 ∪ 𝟏 𝟐 ∪ (𝟐 ∞) Key : 4
  • 22. FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝟐) 𝐨𝐧𝐭𝐨 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟑) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐨𝐧𝐭𝐨 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝟏𝟐. 𝐀 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐍 → 𝐙 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐧 = 𝐧 − 𝟏 𝟐 , 𝐰𝐡𝐞𝐧 ′ 𝐧′ 𝐢𝐬 𝐨𝐝𝐝 𝐚𝐧𝐝 𝐟 𝐧 = −𝐧 𝟐 𝐰𝐡𝐞𝐧 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧. 𝐭𝐡𝐚𝐧 𝐟 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟑] 𝐟 −𝟏 = 𝟎; 𝐟 𝟐 = −𝟏; 𝐟 𝟑 = 𝟏, 𝐟 𝟒 = −𝟐 … . 𝐚𝐭 𝐧 𝐢𝐬 𝐨𝐝𝐝 → 𝐟 𝐱 𝐢𝐬 + 𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫
  • 23. FUNCTIONS PCQs 𝐚𝐭 𝐧 𝐢𝐬 𝐞𝐯𝐞𝐧 → 𝐟 𝐱 𝐢𝐬 − 𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 ∴ 𝐟: 𝐍 → 𝐙 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 Key : 3
  • 24. FUNCTIONS PCQs Solution : 𝟏) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝟐) 𝐨𝐧𝐭𝐨 𝟑) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟒) 𝐦𝐚𝐧𝐲 𝐨𝐧𝐞 𝟏𝟑. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟 𝐱 = 𝐥𝐨𝐠 𝐱 + 𝐱𝟐 + 𝟏 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟑] 𝐟 −𝐱 = 𝐥𝐨𝐠[−𝐱 + 𝐱𝟐 − 𝟏 ] = 𝐥𝐨𝐠[ 𝐱𝟐 − 𝟏 − 𝐱 ] = 𝐥𝐨𝐠 𝐱𝟐 + 𝟏 − 𝐱 𝐱𝟐 + 𝟏 + 𝐱 𝐱𝟐 + 𝟏 + 𝐱
  • 25. FUNCTIONS PCQs = 𝐥𝐨𝐠 𝐱𝟐 + 𝟏 − 𝐱𝟐 𝐱𝟐 + 𝟏 + 𝐱 = 𝐥𝐨𝐠 𝟏 𝐱 + 𝐱𝟐 + 𝟏 = 𝐥𝐨𝐠 𝐱 + 𝐱𝟐 + 𝟏 −𝟏 = −𝐥𝐨𝐠 𝐱 + 𝐱𝟐 + 𝟏 = −𝐟(𝐱) ∴ 𝐟 𝐱 𝐢𝐬 𝐚𝐧 𝐨𝐝𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 Key : 2
  • 26. FUNCTIONS PCQs Solution : 𝟏) 𝐚𝐧 𝐞𝐯𝐞𝐧 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟐) 𝐚𝐧 𝐨𝐝𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟑) 𝐩𝐞𝐫𝐢𝐨𝐝𝐢𝐜 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐞𝐯𝐞𝐧 𝐧𝐨𝐫 𝐨𝐝𝐝 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝟏𝟒. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 𝐢𝐬 [𝐀 − 𝟐𝟎𝟎𝟐] 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐬𝐢𝐧𝐱 𝐢𝐬 𝐦𝐚𝐧𝐲 𝐨𝐧𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝐬𝐢𝐧𝐱 𝐢𝐬 𝐚 𝐩𝐞𝐫𝐢𝐨𝐝𝐢𝐜 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐰𝐢𝐭𝐡 𝐩𝐞𝐫𝐢𝐨𝐝 𝟐𝛑
  • 28. FUNCTIONS PCQs 1. If f(x) is a real function defined on [-1, 1], then the real function g(x) = f(5x+4) is defined on the interval (TS E – 2015) 1) −𝟒, 𝟗 2) [-1, 9] 3) [-2, 9] 4) [-3, 9 ] Solution : −𝟏 ≤ 𝟓𝒙 + 𝟒 ≤ 𝟏  −𝟏 ≤ 𝐱 ≤ −3 5 −𝟏 ≤ 𝒙 ≤ 𝟏  −𝟓 + 𝟒 ≤ 𝟓𝒙 + 𝟒 ≤ 𝟗 [-1, 9]
  • 29. FUNCTIONS PCQs 2. If f: N R is defined by f(1) =-1 and f(n+1)=3f(n)+2 for n>1 then f is (TS E – 2015) 1) One – One 2) Onto 3) a constant function 4) f(n)>0 for n>1 Solution : 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
  • 30. FUNCTIONS PCQs 𝟑) 𝒇: 𝑹 − 𝑹, 𝒈: 𝑹 → 𝑹 defined by f(x) = 5x – 3 , g(x) = x2 + 3 then 𝒈𝒐𝒇−𝟏 𝟑 = (E – 2015) 1) 𝟐𝟓 𝟗 2) 𝟏𝟏𝟏 𝟐𝟓 3) 𝟗 𝟐𝟓 4) 𝟐𝟓 𝟏𝟏𝟏 Solution : 𝒇−𝟏 𝒙 = 𝒙 + 𝟑 𝟓  g (6/5)= 𝟑𝟔 𝟐𝟓 + 𝟑 = 𝟏𝟏𝟏 𝟐𝟓
  • 31. FUNCTIONS PCQs 4) 𝑨 = 𝒙 ∈ 𝑹/  𝟒 ≤ 𝒙 ≤ 𝝅 𝟑 and f(x) = sin x – x  f(A)= (E – 2015) 1) 𝟑 𝟐 − 𝝅 𝟑 , 𝟏 𝟐 − 𝝅 𝟒 2) −𝟏 𝟐 − 𝝅 𝟒 , 𝟑 𝟐 − 𝝅 𝟑 3) −𝝅 𝟑 , −𝝅 𝟒 4) 𝝅 𝟒 , 𝝅 𝟑 Solution : f(x) = sin x - x f1(x) = cosx – 1<0 . f is decreasing function range is 𝒇 𝝅 𝟑 , 𝒇 𝝅 𝟒 = 𝟑 𝟐 − 𝝅 𝟑 , 𝟏 𝟐 − 𝝅 𝟒
  • 32. FUNCTIONS PCQs 𝟓) 𝑰𝒇 𝑹 𝒊𝒔 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒐𝒇 𝒂𝒍𝒍 𝒓𝒆𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒂𝒏𝒅 𝒊𝒇 𝒇: 𝑹 − 𝟐 → 𝑹 is defined by f(x) = 𝟐+𝒙 𝟐−𝒙 for x ∈ R – {2} then the range of f is (E – 2014) 1) R 2) R – {2} 3) R – {-1} 4) R – {-2} Solution : 𝒚 𝟏 = 𝟐 + 𝒙 𝟐 − 𝒙 ⇒ 2+x = 2y-xy ⇒ x+xy = 2 𝒚 − 𝟏 ⇒ x = 𝟐 𝒚−𝟏 𝒚+𝟏 Range = R – {-1}
  • 33. FUNCTIONS PCQs 𝟔) 𝑳𝒆𝒕 𝑸 𝒃𝒆 𝒕𝒉𝒆 𝒔𝒆𝒕 𝒐𝒇 𝒂𝒍𝒍 𝒓𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒊𝒏 𝟎, 𝟏 𝒂𝒏𝒅 𝒇: 𝟎, 𝟏 → 𝟎, 𝟏 𝒃𝒆 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒃𝒚 𝒇(𝒙) = 𝒙 𝒇𝒐𝒓 𝒙 ∈ 𝑸 𝟏 − 𝒙 𝒇𝒐𝒓 𝒙 ∉ 𝑸 then the set S = {𝒙 ∈ [0,1] : (fof)(x) = x} is equal to (E – 2014) Solution : 1) Q 2) [0,1] - Q 3) (0,1) 4) [0,1] f(x) = f-1(x) (i.e) co-domain = Range = [0,1]
  • 34. FUNCTIONS PCQs Solution : 𝐟 𝐟(𝐱) =f [ 𝑷 − 𝒙𝒏 𝟏 𝒏] 7). 𝐈𝒇 𝒇 𝒙 = 𝑷 − 𝒙𝒏 𝟏 𝒏, 𝐏 > 0 and n is a positive integer then f(f(x)) = (E – 2013) 1) x 2) xn 3) 𝑷 𝟏 𝒏 4) P - xn = 𝑷 − 𝒑 − 𝒙𝒏 𝟏 𝒏 𝒏 𝟏 𝒏 = 𝑷 − 𝑷 + 𝒙𝒏 𝟏 𝒏 = 𝒙𝒏 𝟏 𝒏 = x
  • 35. FUNCTIONS PCQs Solution : 𝟏. 𝟔 𝟏−𝒙𝟐 − 𝟎. 𝟔𝟐𝟓 𝟔(𝟏+𝒙) > 0 8). 𝒙 ∈ 𝑹/𝒍𝒐𝒈 𝟏. 𝟔 𝟏−𝒙𝟐 − 𝟎. 𝟔𝟐𝟓 𝟔(𝟏+𝒙) ∈ 𝑹 = (E – 2013) 1) −∞, −𝟏 ∪ 𝟕, ∞ 2) (-1,5) 3) (1,7) 4) (-1,7) ⇒ 𝟖 𝟓 𝟏−𝒙𝟐 > 𝟓 𝟖 𝟔(𝟏+𝒙) ⇒ 𝟏 − 𝒙𝟐 > - 6 𝟏 + 𝒙 ⇒ 𝒙𝟐 − 𝟔𝒙 − 𝟕 < 𝟎 ⇒ 𝒙 − 𝟕 𝒙 + 𝟏 <0 ⇒ 𝒙 ∈ −𝟏, 𝟕
  • 36. FUNCTIONS PCQs Solution : f(x+y)= f(x).f(y) 9). Let f be a non-zero real valued continuous function satisfying f(x+y)= f(x).f(y) for all x,y∈R. If f (2) = 9, then f(6) = (E – 2013) 1) 𝟑𝟐 2) 36 3) 34 4) 33 ⇒ f(x) = ax f(2) = 9 ⇒ a2 = 9 ⇒ a=3 ∴ f(6) = 36
  • 37. FUNCTIONS PCQs Solution : 𝟏) 𝐟 𝐱 = 𝐱𝟐, 𝐠 𝐱 = 𝐬𝐢𝐧 𝐱 𝟐) 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱, 𝐠 𝐱 = 𝐱 𝟑)𝐟 𝐱 = 𝐬𝐢𝐧𝟐𝐱, 𝐠 𝐱 = 𝐱 𝟒) 𝐟 𝐱 = 𝐱𝟐, 𝐠 𝐱 = 𝐱 𝟏𝟎. 𝐈𝐟 𝐟: 𝐑 → 𝐑+ 𝐚𝐧𝐝 𝐠: 𝐑+ → 𝐑 𝐚𝐫𝐞 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐠 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 𝐚𝐧𝐝 𝐟 𝐠 𝐱 = (𝐬𝐢𝐧 𝐱)𝟐, 𝐭𝐡𝐞𝐧 𝐚 𝐩𝐨𝐬𝐬𝐢𝐛𝐥𝐞 𝐜𝐡𝐨𝐢𝐜𝐞 𝐟 𝐚𝐧𝐝 𝐠 𝐢𝐬 [E-2012] 𝐟 𝐱 = 𝐬𝐢𝐧 𝐱 ; 𝐠 𝐱 = 𝐱 ⇒ 𝐠 𝐟 𝐱 = 𝐠 𝐬𝐢𝐧 𝐱 = 𝐬𝐢𝐧 𝐱 = 𝐬𝐢𝐧 𝐱 𝐟 𝐠 𝐱 = 𝐟 𝐱 = 𝐬𝐢𝐧𝟐 𝐱 𝐛𝐲 𝐯𝐞𝐫𝐢𝐟𝐢𝐜𝐚𝐭𝐢𝐨𝐧
  • 38. FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐭𝐨 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝟐) 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝟑) 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 𝐭𝐨 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝟏𝟏. 𝐈𝐟 𝐙 → 𝐙 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲𝐟 𝐱 = 𝐱 𝟐 𝐢𝐟 𝐱 𝐢𝐬 𝐞𝐯𝐞𝐧 𝟎 𝐢𝐟 𝐱 𝐨𝐝𝐝 𝐭𝐡𝐞𝐧 𝐟 𝐢𝐬 [𝐄 − 𝟐𝟎𝟏𝟐] 𝐟 𝟎 = 𝟎 ; 𝐟 ±𝟏 = 𝟎 ; 𝐟 ±𝟑 = 𝟎 … … & 𝐟 ±𝟐 = ±𝟏 ; 𝐟 ±𝟒 = ±𝟐 … …
  • 39. FUNCTIONS PCQs 𝐒𝐨 𝟎, ±𝟏, ±𝟑 … 𝐡𝐚𝐯𝐞 𝐒𝐚𝐦𝐞 𝐢𝐦𝐚𝐠𝐞 ∴ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐁𝐮𝐭 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝐞𝐯𝐞𝐫𝐲 𝐞𝐥𝐞𝐦𝐞𝐧𝐭 𝐡𝐚𝐯𝐞 𝐚 𝐩𝐫𝐞 − 𝐢𝐦𝐚𝐠𝐞 Key : 1
  • 40. FUNCTIONS PCQs Solution : 𝟏) [𝟎, ∞) 𝟐) [𝟏, ∞) 𝟑) [𝟒, ∞) 𝟒) [𝟓, ∞) 𝟏𝟐. 𝐈𝐟 𝐟: 𝟐, ∞ → 𝐁 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 − 𝟒𝐱 + 𝟓 𝐢𝐬 𝐛𝐢𝐣𝐞𝐜𝐭𝐢𝐨𝐧, 𝐭𝐡𝐞𝐧 𝐁 = [𝐄 − 𝟐𝟎𝟏𝟏] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐱𝟐 − 𝟒𝐱 + 𝟓 = 𝐱𝟐 − 𝟒𝐱 + 𝟓 + 𝟏 = (𝐱 − 𝟐)𝟐+𝟏 ∴ 𝐑𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐱 = [𝟏, ∞)
  • 41. FUNCTIONS PCQs Solution : 𝟏) {−𝟏𝟒, −𝟏𝟑, … . . , 𝟎, … . . , 𝟏𝟑, 𝟏𝟒} 𝟐) {−𝟏𝟒, −𝟏𝟑, … . . , 𝟎, … . . , 𝟏𝟒, 𝟏𝟓} 𝟑) {−𝟏𝟓, −𝟏𝟒, … . . , 𝟎, … . . , 𝟏𝟒, 𝟏𝟓} 𝟒) {−𝟏𝟓, −𝟏𝟒, … . . , 𝟎, … . . , 𝟏𝟑, 𝟏𝟒} 𝟏𝟑. 𝐈𝐟 𝐟 = 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 𝟓 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐰𝐡𝐞𝐫𝐞 𝐲 , 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐲 , 𝐭𝐡𝐞𝐧 𝐟 𝐱 : 𝐱 < 𝟕𝟏 = [𝐄 − 𝟐𝟎𝟏𝟏] 𝐱 < 𝟕𝟏 ⇒ −𝟕𝟏 < 𝐱 < 𝟕𝟏
  • 42. FUNCTIONS PCQs ⇒ − 𝟕𝟏 𝟓 < 𝐱 𝟓 < 𝟕𝟏 𝟓 ⇒ −𝟏𝟒. 𝟐 < 𝐱 𝟓 < 𝟏𝟒. 𝟐 ⇒ 𝐟 𝐱 : 𝐱 < 𝟕𝟏 ∵ 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐱 𝟓 = {−𝟏𝟓, −𝟏𝟒, −𝟏𝟑, … . 𝟎, … 𝟏𝟑, 𝟏𝟒} Key : 4
  • 43. FUNCTIONS PCQs Solution : 𝟏) 𝟏𝟐 𝟐) 𝟏𝟑 𝟑) 𝟏𝟒 𝟒) 𝟏𝟎 𝟏𝟒. 𝐈𝐟 𝐟 𝟎 = 𝟎, 𝐟 𝟏 = 𝟏, 𝐟 𝟐 = 𝟐 𝐚𝐧𝐝 𝐟 𝐱 = 𝐟 𝐱 − 𝟐 + 𝐟 𝐱 − 𝟑 𝐟𝐨𝐫 𝐱 = 𝟑, 𝟒, 𝟓, … … 𝐭𝐡𝐞𝐧 𝐟 𝟗 = [𝐄 − 𝟐𝟎𝟏𝟎] 𝐟 𝟎 = 𝟎, 𝐟 𝟏 = 𝟏, 𝐟 𝟐 = 𝟐 𝐚𝐧𝐝 𝐟 𝐱 = 𝐟 𝐱 − 𝟐 + 𝐟 𝐱 − 𝟑 𝐍𝐨𝐰 𝐟 𝟑 = 𝐟 𝟏 + 𝐟 𝟎 = 𝟏 + 𝟎 = 𝟏 𝐟 𝟒 = 𝐟 𝟐 + 𝐟 𝟏 = 𝟏 + 𝟐 = 𝟑 𝐟 𝟓 = 𝐟 𝟑 + 𝐟 𝟐 = 𝟏 + 𝟐 = 𝟑
  • 44. FUNCTIONS PCQs 𝐟 𝟓 = 𝐟 𝟑 + 𝐟 𝟐 = 𝟏 + 𝟐 = 𝟑 𝐟 𝟔 = 𝐟 𝟒 + 𝐟 𝟑 = 𝟑 + 𝟏 = 𝟒 𝐟 𝟕 = 𝐟 𝟓 + 𝐟 𝟒 = 𝟑 + 𝟑 = 𝟔 𝐟 𝟖 = 𝐟 𝟔 + 𝐟 𝟓 = 𝟒 + 𝟑 = 𝟕 𝐟 𝟗 = 𝐟 𝟕 + 𝐟 𝟔 = 𝟔 + 𝟒 = 𝟏𝟎 Key : 4
  • 45. FUNCTIONS PCQs 𝐋𝐢𝐬𝐭 𝐈 𝐋𝐢𝐬𝐭 𝐈𝐈 𝟏𝟓. 𝐋𝐞𝐭 𝐑 𝐝𝐞𝐧𝐨𝐭𝐞 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐚𝐧𝐝 𝐑+ 𝐝𝐞𝐧𝐨𝐭𝐞 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬. 𝐅𝐨𝐫 𝐭𝐡𝐞 𝐬𝐮𝐛𝐬𝐞𝐭𝐬 𝐀 𝐚𝐧𝐝 𝐁 𝐨𝐟 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞 𝐟: 𝐀 → 𝐁 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 𝐟𝐨𝐫 𝐱 ∈ 𝐀 . 𝐨𝐛𝐬𝐞𝐫𝐯𝐞 𝐭𝐡𝐞 𝐭𝐰𝐨 𝐥𝐢𝐬𝐭 given below [𝐄 − 𝟐𝟎𝟏𝟎] (i) f is one-one and onto if (a) A=𝐑+, B=R (ii) f is one-one but not onto if (b) A=B=R (iii) f is onto but not one-one if (c) A=R, B=𝐑+ (iv) f is neither one – one nor onto if (d) A=B=𝐑+
  • 46. FUNCTIONS PCQs (i) (ii) (iii) (iv) 1) a b c d 2) d b a c 3) d a c b 4) d b c a
  • 47. FUNCTIONS PCQs 𝐢 𝐟 𝐢𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 − 𝐨𝐧𝐭𝐨 → 𝐀 = 𝐑+ ; 𝐁 = 𝐑+ Solution : 𝐢𝐢 𝐟 𝐢𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 → 𝐀 = 𝐑+ ; 𝐁 = 𝐑 𝐢𝐢𝐢 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐞 𝐨𝐧𝐞 → 𝐀 = 𝐑 ; 𝐁 = 𝐑+ 𝐢𝐯 𝐟 𝐢𝐬 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 → 𝐀 = 𝐑 ; 𝐁 = 𝐑 Key : 3
  • 48. FUNCTIONS PCQs Solution : 𝟏) [𝟏, 𝟏𝟐] 𝟐) [𝟏𝟐, 𝟑𝟒] 𝟑) [𝟑𝟓, 𝟓𝟎] 𝟒) [−𝟏𝟐, 𝟏𝟐] 𝟏𝟔. 𝐈𝐟 𝐟: 𝟐, 𝟑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟑 + 𝟑𝐱 − 𝟐, 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐟(𝐱) is contained in the interval: [𝐄 − 𝟐𝟎𝟎𝟗] 𝐈𝐟 𝐟: 𝟐, 𝟑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟑 + 𝟑𝐱 − 𝟐 𝐍𝐨𝐰 𝐟 𝟐 = 𝟖 + 𝟔 − 𝟐 = 𝟏𝟐 𝐟 𝟑 = 𝟐𝟕 + 𝟗 − 𝟐 = 𝟑𝟒 ∴ 𝐑𝐚𝐧𝐠𝐞 [𝟏𝟐, 𝟑𝟒]
  • 49. FUNCTIONS PCQs Solution : 𝟏) 𝐑 − {𝟎} 𝟐) 𝐑 − {𝟎, 𝟏, 𝟑} 1𝟕. 𝐱 ∈ 𝐑: 𝟐𝐱−𝟏 𝐱𝟑+𝟒𝐱𝟐+𝟑𝐱 = [𝐄 − 𝟐𝟎𝟎𝟗] 𝟑) 𝐑 − {𝟎, −𝟏, −𝟑} 𝟒) 𝐑 − 𝟎, −𝟏, −𝟑, + 𝟏 𝟐 𝐱𝟑 + 𝟒𝐱𝟐 + 𝟑𝐱 ≠ 𝟎 ⇒ 𝐱(𝐱𝟐 + 𝟒𝐱 + 𝐱) ≠ 𝟎 ⇒ 𝐱(𝐱 + 𝟏)(𝐱 + 𝟑) ≠ 𝟎
  • 50. FUNCTIONS PCQs ⇒ 𝐱 ≠ 𝟎, −𝟏, −𝟑 ∴ 𝐑𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝐬𝐞𝐭 𝐢𝐬 𝐑 − {−𝟏, −𝟑, 𝟎} Key : 3
  • 51. FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝟐) 𝐨𝐧𝐭𝐨 1𝟖. 𝐈𝐟 𝐟: 𝐑 → 𝐂 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐞𝟐𝐢𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧, 𝐟 𝐢𝐬 (𝐰𝐡𝐞𝐫𝐞 𝐂 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐜𝐨𝐦𝐩𝐥𝐞𝐱 𝐧𝐮𝐦𝐛𝐞𝐫𝐬) [𝐄 − 𝟐𝟎𝟎𝟗] 𝟑) 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐚𝐧𝐝 𝐨𝐧𝐭𝐨 𝟒) 𝐧𝐞𝐢𝐭𝐡𝐞𝐫 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐧𝐨𝐫 𝐨𝐧𝐭𝐨 𝐟 𝐱 = 𝐞𝟐𝐢𝐱 = 𝐜𝐨𝐬 𝟐𝐱 + 𝐢 𝐬𝐢𝐧 𝟐𝐱 → 𝐟 𝟎 = 𝐟 𝐱 = 𝟎 𝐒𝐨 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐞 − 𝐨𝐧𝐞 → 𝐓𝐡𝐞𝐫𝐞 𝐞𝐱𝐢𝐭 𝐧𝐨 𝐱 ∈ 𝐑 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐟 𝐱 = 𝟐 𝐒𝐨 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨
  • 52. FUNCTIONS PCQs Solution : 𝟏) {𝟎, 𝟏} 𝟐) {𝟏, 𝟐} 𝟏𝟗. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱 − 𝟑 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐭𝐡𝐞𝐧 𝐠 𝐟 𝐱 : − 8 5 < 𝐱 < 8 5 = [𝐄 − 𝟐𝟎𝟎𝟖] 𝟑) {−𝟑, −𝟐} 𝟒) {𝟐, 𝟑} − 8 5 < 𝐱 < 8 5 ⇒ 𝟎 < 𝐱 < 𝟖 𝟓 ⇒ −𝟑 ≤ 𝐱 − 𝟑 < −𝟕 𝟓
  • 53. FUNCTIONS PCQs ⇒ −𝟑 ≤ 𝐱 − 𝟑 < −𝟏. 𝟒 ⇒ 𝐱 − 𝟑 = −𝟑 𝐨𝐫 − 𝟐 ∵ −𝟑 < 𝐱 − 𝟑 < −𝟐 ⇒ 𝐠𝐨𝐟 𝐟 𝐱 = −𝟑 −𝟐 < 𝐱 − 𝟑 < −𝟏. 𝟒 ⇒ 𝐠𝐨𝐟 𝐟 𝐱 = −𝟐 Key : 3
  • 54. FUNCTIONS PCQs Solution : 𝟏) 𝐟(𝟒 𝟐) 𝟐) 𝐟(𝟑 𝟐) 𝟐𝟎. 𝐈𝐟 𝐟: −𝟔, 𝟔 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 − 𝟑 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐭𝐡𝐚𝐧 𝐟𝟎𝐟𝟎𝐟 −𝟏 + 𝐟𝟎𝐟𝟎𝐟 𝟎 + 𝐟𝟎𝐟𝟎𝐟 𝟏 = [𝐄 − 𝟐𝟎𝟎𝟖] 𝟑) 𝐟(𝟐 𝟐) 𝟒) 𝐟( 𝟐) → 𝐟𝐨𝐟𝐨𝐟 −𝟏 = 𝐟𝐨𝐟 𝐟 −𝟏 = 𝐟𝐨𝐟[𝟏 − 𝟑] = 𝐟𝐨𝐟 −𝟐 = 𝐟 𝟒 − 𝟑 = 𝐟 𝟏 = 𝟏 − 𝟑 = −𝟐 → 𝐟𝐨𝐟𝐨𝐟 𝟎 = 𝐟𝐨𝐟 𝐟 𝟎 = 𝐟𝐨𝐟(−𝟑) = 𝐟 𝐟 −𝟑 = 𝐟 𝟗 − 𝟑 = 𝐟 𝟔 = 𝟑𝟑
  • 55. FUNCTIONS PCQs → 𝐟𝐨𝐟𝐨𝐟 𝟏 = 𝐟𝐨𝐟 𝐟 𝟏 = 𝐟𝐨𝐟 𝟏 − 𝟑 = 𝐟𝐨𝐟 −𝟐 = 𝐟 𝟒 − 𝟑 = 𝐟(𝟏) = 𝟏 − 𝟑 = −𝟐 𝐒𝐨 𝐟𝐨𝐟𝐨𝐟 −𝟏 + 𝐟𝐨𝐟𝐨𝐟 𝟎 + 𝐟𝟎𝐟𝟎𝐟(𝟏) = −𝟐 + 𝟑𝟑 − 𝟐 = 𝟐𝟗 = 𝟑𝟐 − 𝟑 = (𝟒 𝟐)𝟐−𝟑 = 𝐟(𝟒 𝟐) Key : 1
  • 56. FUNCTIONS PCQs 𝐈) 𝐟 𝐩 𝐪 𝐢𝐬 𝐫𝐞𝐚𝐥 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 𝐩 𝐪 ∈ 𝐐 𝟐𝟏. 𝐈𝐟 𝐐 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐚𝐧𝐝 𝐟 𝐩 𝐪 = 𝐩𝟐 − 𝐪𝟐 𝐟𝐨𝐫 𝐚𝐧𝐲 p q ∈ 𝐐, 𝐭𝐡𝐞𝐧 𝐨𝐛𝐬𝐞𝐫𝐯𝐞 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 [𝐄 − 𝟐𝟎𝟎𝟖] 𝐈𝐈) 𝐟 𝐩 𝐪 𝐢𝐬 𝐚 𝐜𝐨𝐦𝐩𝐥𝐞𝐱 𝐧𝐮𝐦𝐛𝐞𝐫 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 p q ∈ 𝐐 Which of the following is correct ?
  • 57. FUNCTIONS PCQs Solution : 1) 𝐁𝐨𝐭𝐡 𝐈 𝐚𝐧𝐝 𝐈𝐈 𝐚𝐫𝐞 𝐭𝐫𝐮𝐞 2) 𝐈 𝐢𝐬 𝐭𝐫𝐮𝐞 , 𝐈𝐈 𝐢𝐬 𝐟𝐚𝐥𝐬𝐞 3) 𝐈 𝐭𝐫𝐮𝐞 , 𝐈𝐈 𝐢𝐬 𝐟𝐚𝐥𝐬𝐞 4) 𝐁𝐨𝐭𝐡 𝐈 𝐚𝐧𝐝 𝐈𝐈 𝐚𝐫𝐞 𝐟𝐚𝐥𝐬𝐞 𝐁𝐲 𝐯𝐞𝐫𝐢𝐟𝐢𝐜𝐚𝐭𝐢𝐨𝐧 𝐰𝐞 𝐭𝐚𝐤𝐞 𝐚𝐧𝐲 𝐩𝐫𝐨𝐩𝐞𝐫𝐟𝐫𝐚𝐜𝐭𝐢𝐨𝐧 𝐩 𝐪 ∈ 𝐐 𝐟 𝐩 𝐪 = 𝐩𝟐 − 𝐪𝟐 𝐢𝐬 𝐜𝐨𝐦𝐩𝐥𝐞𝐱 𝐧𝐮𝐦𝐛𝐞𝐫
  • 58. FUNCTIONS PCQs Solution : 𝟏) 𝟏 𝟑 , 𝟏 𝟐) 𝟏 𝟑 , 𝟏 𝟐𝟐. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟏 (𝟐 − 𝐜𝐨𝐬𝟑𝐱) 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 𝐱 ∈ 𝐑, 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐢𝐬 [𝐄 − 𝟐𝟎𝟎𝟕] 𝟑) (𝟏, 𝟐) 𝟒) [𝟏, 𝟐] 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝟏 𝟐 − 𝐜𝐨𝐬𝟑𝐱 𝐰. 𝐤. 𝐭 − 𝟏 ≤ −𝐜𝐨𝐬𝟑𝐱 ≤ 𝟏 𝟐 − 𝟏 ≤ 𝟐 − 𝐜𝐨𝐬𝟑𝐱 ≤ 𝟐 + 𝟏
  • 59. FUNCTIONS PCQs 𝟏 ≤ 𝟐 − 𝐜𝐨𝐬𝟑𝐱 ≤ 𝟑 𝟏 𝟑 ≤ 𝟏 𝟐 − 𝐜𝐨𝐬𝟑𝐱 ≤ 𝟏 ∴ 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐱 = 𝟏 𝟑 , 𝟏 Key : 2
  • 60. FUNCTIONS PCQs Solution : 𝟏) 𝐱 𝟐) 𝟎 𝟐𝟑. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 − 𝐱 𝐚𝐧𝐝 𝐠 𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐰𝐡𝐞𝐫𝐞 𝐱 𝐢𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐱, 𝐭𝐡𝐞𝐧 𝐟𝐨𝐫 𝐞𝐯𝐞𝐫𝐲 𝐱 ∈ 𝐑, 𝐟 𝐠 𝐱 = [𝐄 − 𝟐𝟎𝟎𝟕] 𝟑) 𝐟(𝐱) 𝟒) 𝐠(𝐱) 𝐟 𝐠 𝐱 = 𝐠 𝐱 − [𝐠 𝐱 ] = 𝐱 − 𝐱 = 𝐱 − 𝐱 = 𝟎 ∵ 𝐟 𝐱 = 𝐱 − 𝐱 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱
  • 61. FUNCTIONS PCQs Solution : 𝟏) 𝐙, 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐢𝐧𝐭𝐞𝐠𝐞𝐫𝐬 𝟐) 𝐍, 𝐢𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐧𝐚𝐭𝐮𝐫𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 24. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 − 𝐱 − 𝟏 𝟐 𝐟𝐨𝐫 𝐱 ∈ 𝐑, 𝐰𝐡𝐞𝐫𝐞 𝐱 𝐢𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐱, 𝐭𝐡𝐞𝐧 𝐱 ∈ 𝐑: 𝐟 𝐱 − 1 2 = [𝐄 − 𝟐𝟎𝟎𝟕] 𝟑) ∅ , 𝐭𝐡𝐞 𝐞𝐦𝐩𝐭𝐲 𝐬𝐞𝐭 𝟒) 𝐑 𝟎 ≤ 𝐱 − 𝐱 < 𝟏
  • 62. FUNCTIONS PCQs ⇒ − 𝟏 𝟐 ≤ 𝐱 − 𝐱 − 𝟏 𝟐 < 𝟏 𝟐 ⇒ − 𝟏 𝟐 ≤ 𝐟(𝐱) < 𝟏 𝟐 ⇒ 𝐟 𝐱 ≠ 𝟏 𝟐 ∀𝐱 ⇒ 𝐱 ∈ 𝐑 ∶ 𝐟 𝐱 = 𝟏 𝟐 = ∅ Key : 3
  • 63. FUNCTIONS PCQs Solution : 𝟏) {𝐱 ∈ 𝐑: 𝟎 ≤ 𝐱 ≤ 𝟏 𝟐) {𝟎, 𝟏} 𝟐𝟓. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟐𝐱 − 𝟐 𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 , 𝐰𝐡𝐞𝐫𝐞 𝐟[𝐱] 𝐢𝐬 𝐭𝐡𝐞 𝐠𝐫𝐞𝐚𝐭𝐞𝐬𝐭 𝐢𝐧𝐭𝐞𝐠𝐞𝐫 𝐧𝐨𝐭 𝐞𝐱𝐜𝐞𝐞𝐝𝐢𝐧𝐠 𝐱, 𝐭𝐡𝐚𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐢𝐬 ∶ [𝐄 − 𝟐𝟎𝟎𝟔] 𝟑) {𝐱 ∈ 𝐑: 𝐱 > 𝟎} 𝟒) {𝐱 ∈ 𝐑: 𝐱 ≤ 𝟎} 𝐱 ∈ 𝐑 ⇒ ∃ 𝐧 ∈ 𝐙 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐧 ≤ 𝐱 < 𝐧 + 𝟏 ⇒ 𝐱 = 𝐧 ⇒ 𝟐𝐧 ≤ 𝟐𝐱 < 𝟐𝐧 + 𝟏 → 𝟐𝐱 = 𝟐𝐧 𝐨𝐫 𝟐𝐧 + 𝟏
  • 64. FUNCTIONS PCQs ⇒ 𝟐𝐱 = 𝟐 𝐱 𝐨𝐫 𝟐 𝐱 + 𝟏 ⇒ 𝟐𝐱 − 𝟐 𝐱 = 𝟎 𝐨𝐫 𝟏 ⇒ 𝐟 𝐱 = 𝟎 𝐨𝐫 𝟏 ⇒ {𝟎, 𝟏} Key : 2
  • 65. FUNCTIONS PCQs Solution : 𝟏) 𝐑, 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝟐) ∅, 𝐭𝐡𝐞 𝐞𝐦𝐩𝐭𝐲 𝐬𝐞𝐭 𝟐𝟔. {𝐱 ∈ 𝐑: 𝐱 − 𝐱 = 𝟓} [𝐄 − 𝟐𝟎𝟎𝟓] 𝟑) {𝐱 ∈ 𝐑: 𝐱 < 𝟎} 𝟒) {𝐱 ∈ 𝐑: 𝐱 ≥ 𝟎} 𝐈𝐟 𝐱 ≥ 𝟎 𝐭𝐡𝐞𝐧 𝐱 = 𝐱 ⇒ 𝐱 − 𝐱 = 𝟎 ⇒ 𝐱 − 𝐱 = 𝟎 ⇒ 𝐱 − 𝐱 ≠ 𝟓 𝐈𝐟 𝐱 < 𝟎 𝐭𝐡𝐞𝐧 𝐱 = −𝐱 ⇒ 𝐱 − 𝐱 = 𝟐𝐱 < 𝟎 ⇒ 𝐱 − 𝐱 ≠ 𝟓 ∴ 𝐱 ∈ 𝐑 ∶ 𝐱 − 𝐱 = 𝟓 = ∅
  • 66. FUNCTIONS PCQs Solution : 𝟏) 𝐚 = 𝐜 𝟐) 𝐛 = 𝐝 𝟐𝟕. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐂 → 𝐂 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐚𝐱 + 𝐛 𝐜𝐱 + 𝐝 𝐟𝐨𝐫 𝐱 ∈ 𝐂 𝐰𝐡𝐞𝐫𝐞 𝐛𝐝 ≠ 𝟎 𝐫𝐞𝐝𝐮𝐜𝐞𝐬 𝐭𝐨 𝐚 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐢𝐟 ∶ [𝐄 − 𝟐𝟎𝟎𝟓] 𝟑) 𝐚𝐝 = 𝐛𝐜 𝟒) 𝐚𝐛 = 𝐜𝐝 𝐟 𝐱 = 𝐚𝐱 + 𝐛 𝐜𝐱 + 𝐝 = 𝐚 𝐜 + 𝐛𝐜 − 𝐚𝐝 𝐜[𝐜𝐱 + 𝐝] 𝐢𝐬 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ⇒ 𝐛𝐜 − 𝐚𝐝 = 𝟎 ⇒ 𝐛𝐜 = 𝐚𝐝
  • 67. FUNCTIONS PCQs Solution : 𝟏) 𝟐𝐤+𝟏 − 𝟏 𝟐) 𝟐(𝟐𝐤+𝟏 − 𝟏) 𝟐𝟖. 𝐈𝐟 𝐍 𝐝𝐞𝐧𝐨𝐭𝐞𝐬 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫𝐬 𝐚𝐧𝐝 𝐢𝐟 𝐟: 𝐍 → 𝐍 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐧 = 𝐭𝐡𝐞 𝐬𝐮𝐦 𝐨𝐟 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝐧 𝐭𝐡𝐞𝐧 𝐟 𝟐𝐤 , 𝟑 , 𝐰𝐡𝐞𝐫𝐞 𝐤 𝐢𝐬 𝐚 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 𝐢𝐧𝐭𝐞𝐠𝐞𝐫𝐬, 𝐢𝐬: [𝐄 − 𝟐𝟎𝟎𝟓] 𝟑) 𝟑(𝟐𝐤+𝟏 − 𝟏) 𝟒) 𝟒(𝟐𝐤+𝟏 − 𝟏) 𝐟𝐫𝐨𝐦 𝐩𝐞𝐫𝐦𝐮𝐭𝐚𝐭𝐢𝐨𝐧𝐬 𝐚𝐧𝐝 𝐜𝐨𝐦𝐛𝐢𝐧𝐚𝐭𝐢𝐨𝐧𝐬 𝐍 = 𝐏𝟏 𝛂𝟏. 𝐏𝟐 𝛂𝟐. 𝐏𝟑 𝛂𝟑 … . . 𝐏𝐤 𝛂𝐤 𝐰𝐡𝐞𝐫𝐞 𝐏𝟏, 𝐏𝟐, 𝐏𝟑 … … 𝐏𝐤 𝐚𝐫𝐞 𝐩𝐫𝐢𝐦𝐞𝐬
  • 68. FUNCTIONS PCQs & 𝐬𝐮𝐦 𝐨𝐟 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝐍 = 𝐏𝟏 𝛂𝟏+𝟏 𝐏𝟏 − 𝟏 𝐏𝟐 𝛂𝟐+𝟏 𝐏𝟐 − 𝟏 𝐏𝟑 𝛂𝟑+𝟏 𝐏𝟑 − 𝟏 … . . 𝐍𝐨𝐰 𝐟 𝟐𝐤 . 𝟑 = 𝐭𝐡𝐞 𝐬𝐮𝐦 𝐨𝐟 𝐩𝐨𝐬𝐢𝐭𝐢𝐯𝐞 = 𝟐𝐤+𝟏 − 𝟏 𝟐 − 𝟏 = 𝟑𝟐 − 𝟏 𝟑 − 𝟏 = 𝟒(𝟐𝐤+𝟏 − 𝟏) 𝐓𝐡𝐞𝐧 𝐧𝐨. 𝐨𝐟 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝐍 = 𝛂𝟏 + 𝟏 𝛂𝟐 + 𝟏 𝛂𝟑 + 𝟏 … … 𝐝𝐢𝐯𝐢𝐬𝐨𝐫𝐬 𝐨𝐟 𝟐𝐤. 𝟑 Key : 4
  • 69. FUNCTIONS PCQs Solution : 𝟏) {𝟑, 𝟔, 𝟒} 𝟐) {𝟏, 𝟒, 𝟕} 𝟐𝟗. 𝐟: 𝐍 → 𝐙 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐧 = 𝟐 𝐈𝐟 𝐧 = 𝟑𝐤, 𝐤 ∈ 𝐳 𝟏𝟎 − 𝐧 𝐢𝐟 𝐧 = 𝟑𝐤 + 𝟏, 𝐤 ∈ 𝐳 𝟎 𝐧 = 𝟑𝐤 + 𝟐 𝐤 ∈ 𝐳 𝐭𝐡𝐞𝐧 𝐧 ∈ 𝐍: 𝐟 𝐧 > 𝟐 = [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) {𝟒, 𝟕} 𝟒) {𝟕} 𝐧 ∈ 𝐍: 𝐟 𝐧 > 𝟐 = 𝐧 ∈ 𝐍: 𝟑𝐤 + 𝟏 = 𝐧 , 𝐤 ∈ 𝐳, 𝟏𝟎 − 𝐧 > 𝟐 = 𝟏, 𝟒, 𝟕 ∴ 𝐟 𝟏 = 𝟏𝟎 − 𝟏 = 𝟗 > 𝟐 𝐟 𝟒 = 𝟏𝟎 − 𝟒 = 𝟔 > 𝟐 𝐟 𝟕 = 𝟏𝟎 − 𝟕 = 𝟑 > 𝟐
  • 70. FUNCTIONS PCQs Solution : 𝟏) 𝐨𝐧𝐥𝐲 𝐈, 𝐈𝐈 𝟐) 𝐨𝐧𝐥𝐲 𝐈𝐈, 𝐈𝐈𝐈 𝟑𝟎. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟑−𝐱 𝐨𝐛𝐬𝐞𝐫𝐯𝐞 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) 𝐨𝐧𝐥𝐲 𝐈, 𝐈𝐈𝐈 𝟒) 𝐈, 𝐈𝐈, 𝐈𝐈𝐈 𝐈) 𝐟 𝐢𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 𝐈𝐈) 𝐟 𝐢𝐬 𝐨𝐧𝐭𝐨 𝐈𝐈𝐈) 𝐟 𝐢𝐬 𝐚 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐨𝐮𝐭 𝐨𝐟 𝐭𝐡𝐞𝐬𝐞 𝐭𝐫𝐮𝐞 𝐬𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭𝐬 𝐚𝐫𝐞 → 𝐋𝐞𝐭 𝐱𝟏 𝐱𝟐 ∈ 𝐑 𝐬𝐮𝐜𝐡 𝐭𝐡𝐚𝐭 𝐟 𝐱𝟏 = 𝐟(𝐱𝟐) ⇒ 𝟑−𝐱𝟏 = 𝟑−𝐱𝟐 ⇒ 𝐱𝟏 = 𝐱𝟐 ∴ 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞
  • 71. FUNCTIONS PCQs → 𝐰𝐞 𝐭𝐚𝐤𝐞 𝐚𝐧𝐲 − 𝐯𝐞 𝐯𝐚𝐥𝐮𝐞 𝐢𝐧 𝐜𝐨. 𝐝𝐨𝐦𝐚𝐢𝐧 𝐭𝐡𝐞𝐲 𝐡𝐚𝐯𝐞 𝐧𝐨. −𝐩𝐫𝐞 𝐢𝐦𝐚𝐠𝐞 𝐢𝐬 𝐝𝐨𝐦𝐚𝐢𝐧 ∴ 𝐟 𝐢𝐬 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 → 𝐟 𝐱 = 𝟑−𝐱 𝐢𝐬 𝐚 𝐝𝐞𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ∴ 𝐈 & 𝐈𝐈 𝐚𝐫𝐞 𝐭𝐫𝐮𝐞 Key : 1
  • 72. FUNCTIONS PCQs Solution : 𝟏) (−𝟏, 𝟑) 𝟐) [−𝟏, 𝟑) 𝟑𝟏. 𝐈𝐟 𝐟 𝐱 = 𝐱 𝐢𝐟 − 𝟑 < 𝐱 ≤ −𝟏 𝐱 𝐢𝐟 − 𝟏 < 𝐱 < 𝟏 [𝐱] 𝐢𝐟 𝟏 ≤ 𝐱 < 𝟑 𝐭𝐡𝐞𝐧 𝐱: 𝐟 𝐱 ≥ 𝟎 = [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) (−𝟏, 𝟑] 𝟒) [−𝟏, 𝟑] → −𝟑 < 𝐱 ≤ −𝟏 ⇒ 𝐟 𝐱 = 𝐱 < 𝟎 → −𝟏 < 𝐱 ≤ 𝟏 ⇒ 𝐟 𝐱 = 𝐱 ≥ 𝟎 → 𝟏 ≤ 𝐱 < 𝟑 ⇒ −𝟑 < −𝐱 ≤ −𝟏 ⇒ 𝐟 𝐱 = −𝐱 > 𝟎
  • 73. FUNCTIONS PCQs ∴ 𝐱: 𝐟 𝐱 ≥ 𝟎 = −𝟏, 𝟏 ∪ 𝟏, 𝟑 = (−𝟏, 𝟑) Key : 1
  • 74. FUNCTIONS PCQs Solution : 𝟏) {−𝟏} 𝟐) {𝟎} 3𝟐. 𝐒𝐮𝐩𝐩𝐨𝐬𝐞 𝐟: −𝟐, 𝟐 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = −𝟏 𝐟𝐨𝐫 − 𝟐 ≤ 𝐱 ≤ 𝟎 = 𝐱 − 𝟏 𝐟𝐨𝐫 𝟎 ≤ 𝐱 ≤ 𝟐 𝐭𝐡𝐞𝐧 𝐱 ∈ −𝟐, 𝟐 : 𝐱 ≤ 𝟎 𝐚𝐧𝐝 𝐟 𝐱 = 𝐱 = [𝐄 − 𝟐𝟎𝟎𝟑] 𝟑) − 𝟏 𝟐 𝟒) ∅ 𝐱 ≤ 𝟎 𝐟 𝐱 = 𝐱 ⇒ 𝐟 −𝐱 = 𝐱 ⇒ −𝐱 − 𝟏 = 𝐱 ⇒ 𝟐𝐱 = −𝟏 ⇒ 𝐱 = − 𝟏 𝟐
  • 75. FUNCTIONS PCQs Solution : 𝟏) 𝐙 ∪ (−∞, 𝟎) 𝟐) (−∞, 𝟎) 𝟑𝟑. 𝐟: 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐠𝐢𝐯𝐞𝐧 𝐛𝐲 𝐟 𝐱 = 𝐱 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱 𝐟𝐨𝐫 𝐞𝐚𝐜𝐡 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐱 ∈ 𝐑 ∶ 𝐠[𝐟 𝐱 ] ≤ 𝐟[𝐠 𝐱 ] [𝐄 − 𝟐𝟎𝟎𝟑] 𝟑) 𝐙 𝟒) 𝐑 𝐠[𝐟 𝐱 ] ≤ 𝐟[𝐠 𝐱 ] ⇒ 𝐱 ≤ [𝐱] 𝐈𝐟 𝐱 ≥ 𝟎 𝐭𝐡𝐞𝐧 𝐱 = 𝐱 ⇒ 𝐱 = 𝐱 = [𝐱] 𝐈𝐟 𝐱 < 𝟎 𝐭𝐡𝐞𝐧 𝐱 = −𝐱 ⇒ 𝐱 ≤ −[𝐱] ⇒ 𝐱 ≤ − 𝐱 ≤ [𝐱] ∴ 𝐑𝐞𝐪𝐮𝐢𝐫𝐞𝐝 𝐬𝐞𝐭 𝐢𝐬 𝐈𝐑
  • 76. FUNCTIONS PCQs Solution : 𝟏) 𝟎. 𝟓 𝟐) 𝟎. 𝟔 𝟑𝟒. 𝐈𝐟 𝐞𝐟(𝐱) = 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 (𝐗 ∈ −𝟏𝟎, 𝟏𝟎 𝐚𝐧𝐝 𝐟 𝐱 = 𝐤𝐟 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎 + 𝐱𝟐 𝐭𝐡𝐞𝐧 𝐤 [𝐄 − 𝟐𝟎𝟎𝟑] 𝟑) 𝟎. 𝟕 𝟒) 𝟎. 𝟖 𝐆𝐢𝐯𝐞𝐧 𝐞𝐟(𝐱) = 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 ⇒ 𝐟(𝐱) = 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 𝐭𝐚𝐤𝐞 𝐟 𝐱 = 𝐤. 𝐟 𝟐𝟎𝟎 𝟏𝟎𝟎 + 𝐱𝟐
  • 77. FUNCTIONS PCQs = 𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎 + 𝐱𝟐 𝟏𝟎 − 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎 + 𝐱𝟐 = 𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎𝟎𝟎 + 𝟏𝟎𝐱𝟐 + 𝟐𝟎𝟎𝐱 𝟏𝟎𝟎𝟎 + 𝟏𝟎𝐱𝟐 − 𝟐𝟎𝟎𝐱 = 𝐤. 𝐥𝐨𝐠𝐞 𝐱𝟐 + 𝟐𝟎𝐱 + 𝟏𝟎𝟎 𝐱𝟐 − 𝟐𝟎𝐱 + 𝟏𝟎𝟎 = 𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 𝟐 = 𝟐𝐤. 𝐥𝐨𝐠𝐞 𝟏𝟎 + 𝐱 𝟏𝟎 − 𝐱 = 𝟐𝐤. 𝐟(𝐱) ⇒ 𝟏 = 𝟐𝐤 ⇒ 𝐤 = 𝟏 𝟐 Key : 1
  • 78. FUNCTIONS PCQs Solution : 𝟏) 𝟏 𝟐) 𝟐 𝟑𝟓. 𝐈𝐟 𝐟 𝐱 = 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟒𝐱 𝐬𝐢𝐧𝟐𝐱 + 𝐜𝐨𝐬𝟒𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐟 𝟐𝟎𝟎𝟐 = [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) 𝟑 𝟒) 𝟒 𝐟 𝐱 = 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟒𝐱 𝐬𝐢𝐧𝟐𝐱 + 𝐜𝐨𝐬𝟒𝐱 = 𝐬𝐢𝐧𝟐 𝐱. 𝐬𝐢𝐧𝟐 𝐱 + 𝐜𝐨𝐬𝟐 𝐱 𝐜𝐨𝐬𝟐𝐱. 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱 = (𝟏 − 𝐜𝐨𝐬𝟐 𝐱)𝐬𝐢𝐧𝟐 𝐱 + 𝐜𝐨𝐬𝟐 𝐱 (𝟏 − 𝐬𝐢𝐧𝟐𝐱)𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱
  • 79. FUNCTIONS PCQs = 𝐬𝐢𝐧𝟐𝐱 − 𝐬𝐢𝐧𝟐𝐱 𝐜𝒐𝒔𝟐𝐱 + 𝐜𝐨𝐬𝟐𝐱 𝐜𝐨𝐬𝟐𝐱 − 𝐬𝐢𝐧𝟐𝐱 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱 = 𝟏 𝐒𝐨 𝐟 𝐱 𝐢𝐬 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 ∴ 𝐟 𝟐𝟎𝟎𝟐 = 𝟏 Key : 1
  • 80. FUNCTIONS PCQs Solution : 𝟏) 𝟏 𝟐) 𝟏 𝟐 𝟑𝟔. 𝐈𝐟 𝐭𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬 𝐟 𝐚𝐧𝐝 𝐠 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟑𝐱 − 𝟒, 𝐠 𝐱 = 𝟐 + 𝟑𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐫𝐞𝐬𝐩𝐞𝐜𝐭𝐢𝐯𝐞𝐥𝐲 𝐭𝐡𝐞𝐧 (𝐠−𝟏 𝐨𝐟−𝟏 )(𝟓)) = [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) 𝟏 𝟑 𝟒) 𝟏 𝟒 𝐲 = 𝐟 𝐱 = 𝟑𝐱 − 𝟒 ⇒ 𝐱 = 𝐲 + 𝟒 𝟑 ⇒ 𝐟−𝟏(𝐱) = 𝐱 + 𝟒 𝟑 𝐲 = 𝐠 𝐱 = 𝟑𝐱 + 𝟐 ⇒ 𝐱 = 𝐲 − 𝟐 𝟑 ⇒ 𝐠−𝟏(𝐱) = 𝐱 − 𝟐 𝟑
  • 81. FUNCTIONS PCQs 𝐍𝐨𝐰 𝐠−𝟏 𝐨𝐟−𝟏 𝟓 = 𝐠−𝟏 𝐟−𝟏 𝟓 = 𝐠−𝟏 𝟓 + 𝟒 𝟑 = 𝐠−𝟏(𝟑) = 𝟑 − 𝟐 𝟑 = 𝟏 𝟑 Key : 3
  • 82. FUNCTIONS PCQs Solution : 𝟏) {𝟏, −𝟏} 𝟐) {𝐱: 𝟎 ≤ 𝐱 ≤ 𝟒} 3𝟕. 𝐋𝐞𝐭 𝐀 = 𝐱 ∈ 𝐑: 𝐱 ≠ 𝟎, −𝟒 ≤ 𝐱 ≤ 𝟗 𝐚𝐧𝐝 𝐟: 𝐀 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = x x 𝐟𝐨𝐫 𝐱 ∈ 𝐀 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐢𝐬 [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) {𝟏} 𝟒) {𝐱: −𝟒 ≤ 𝐱 ≤ 𝟎} 𝐁𝐲 𝐝𝐞𝐟𝐢𝐧𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐦𝐨𝐝𝐮𝐥𝐞𝐬 𝐱 = 𝐱 𝐢𝐟 𝐱 > 𝟎 = −𝐱 𝐢𝐟 𝐱 < 𝟎 𝐒𝐨 𝐟 𝐱 = 𝐱 𝐱 = ±𝐱 𝐱 = ±𝟏 ∴ 𝐫𝐚𝐧𝐠𝐞 𝐨𝐟 𝐟 𝐱 = −𝟏, 𝟏
  • 83. FUNCTIONS PCQs Solution : 𝟏) 𝟑 𝟒 , 𝟏 𝟐) 𝟑 𝟒 , 𝟏 𝟑𝟖. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐜𝐨𝐬𝟐 𝐱 + 𝐬𝐢𝐧𝟒 𝐱 𝐟𝐨𝐫 𝐱 ∈ 𝐑 𝐭𝐡𝐞𝐧 𝐟(𝐑) [𝐄 − 𝟐𝟎𝟎𝟐] 𝟑) 𝟑 𝟒 , 𝟏 𝟒) 𝟑 𝟒 , 𝟏 𝐟 𝐱 = 𝐜𝐨𝐬𝟐 𝐱 + 𝐬𝐢𝐧𝟒 𝐱 = 𝐜𝐨𝐬𝟐 𝐱 + 𝐬𝐢𝐧𝟐 𝐱(𝟏 − 𝐜𝐨𝐬𝟐 𝐱) = 𝐜𝐨𝐬𝟐𝐱 + 𝐬𝐢𝐧𝟐𝐱 − 𝐬𝐢𝐧𝟐𝐱 𝐜𝐨𝐬𝟐𝐱 = 𝟏 − 𝟏 𝟒 𝐬𝐢𝐧𝟐𝟐𝐱
  • 84. FUNCTIONS PCQs 𝐰. 𝐤. 𝐓 𝟎 ≤ 𝐬𝐢𝐧𝟐𝟐𝐱 ≤ 𝟏 −𝟏 𝟒 ≤ −𝟏 𝟒 𝐬𝐢𝐧𝟐 𝟐𝐱 ≤ 𝟎 𝟏 − 𝟏 𝟒 ≤ 𝟏 − 𝟏 𝟒 𝐬𝐢𝐧𝟐 𝟐𝐱 ≤ 𝟎 + 𝟏 𝟑 𝟒 ≤ 𝐟 𝐱 ≤ 𝟏 Key : 1
  • 85. FUNCTIONS PCQs Solution : 𝟏) 𝟐−𝟒 𝟐) 𝟐−𝟑 𝟑𝟗. 𝐈𝐟 𝐟 𝐱 = (𝟐𝟓 − 𝐱𝟒) 𝟏 𝟒 𝐟𝐨𝐫 𝟎 < 𝐱 < 𝟓 𝐭𝐡𝐞𝐧 𝐟 𝐟 𝟏 𝟐 = [𝐄 − 𝟐𝟎𝟎𝟏] 𝟑) 𝟐−𝟐 𝟒) 𝟐−𝟏 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = (𝟐𝟓 − 𝐱𝟒 ) 𝟏 𝟒 𝐍𝐨𝐰 𝐟 𝐟 𝟏 𝟐 = 𝐟 𝟐𝟓 − 𝟏 𝟏𝟔 𝟏 𝟒 = 𝐟 𝟑𝟗𝟗 𝟏𝟔 𝟏 𝟒
  • 86. FUNCTIONS PCQs = 𝟐𝟓 − 𝟑𝟗𝟗 𝟏𝟔 𝟏 𝟒 = 𝟏 𝟏𝟔 𝟏 𝟒 = 𝟏 𝟐𝟒 𝟏 𝟒 = 𝟏 𝟐 = 𝟐−𝟏 Key : 4
  • 87. FUNCTIONS PCQs Solution : 𝟏) − 𝟏 𝟐) 𝟎 𝟒𝟎. 𝐓𝐡𝐞 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐟: 𝐑 → 𝐑 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐚𝐬 𝐟𝐨𝐥𝐥𝐨𝐰𝐬 [𝐄 − 𝟐𝟎𝟎𝟏] 𝟑) 𝟏 𝟒) 𝟐 𝐟 𝐱 = 𝟎(𝐱 𝐢𝐬 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) = 𝟏(𝐱 𝐢𝐬 𝐢𝐫𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) 𝐠 𝐱 = −𝟏(𝐱 𝐢𝐬 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) = 𝟎(𝐱 𝐢𝐬 𝐢𝐫𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥) 𝐭𝐡𝐞𝐧 𝐟𝐨𝐠 𝛑 + 𝐠𝐨𝐟(𝐞) 𝐟𝐨𝐠 𝛑 + 𝐠𝐨𝐟(𝐞) = 𝐟 𝐠 𝛑 + 𝐠 𝐟 𝐞 ∴ 𝛑 𝐚𝐧𝐝 𝐞 𝐚𝐫𝐞 𝐢𝐫𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞𝐬
  • 88. FUNCTIONS PCQs = 𝐟 𝟎 + 𝐠[𝟏] = 𝟎 − 𝟏 = −𝟏 ∴ 𝟎&𝟏 𝐚𝐫𝐞 𝐫𝐚𝐭𝐢𝐨𝐧𝐚𝐥 𝐯𝐚𝐥𝐮𝐞𝐬) Key : 1
  • 89. FUNCTIONS PCQs Solution : 𝟏) 𝟎 𝟐) 𝟐 𝟒𝟏. 𝐋𝐞𝐭 𝐟: 𝐑 → 𝐑 𝐛𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 + 𝟐, 𝐱 ≤ −𝟏 = 𝐱𝟐(−𝟏 ≤ 𝐱 ≤ 𝟏) = 𝟐 − 𝐱 𝐱 ≥ 𝟏 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐟 −𝟏. 𝟕𝟓 + 𝐟 𝟎. 𝟓 + 𝐟 𝟏. 𝟓 = [𝐄 − 𝟐𝟎𝟎𝟏] 𝟑) 𝟏 𝟒) − 𝟏 𝐟 −𝟏. 𝟕𝟓 + 𝐟 𝟎. 𝟓 + 𝐟 𝟏. 𝟓 = −𝟏. 𝟕𝟓 + 𝟐 + 𝟎. 𝟓 𝟐 + 𝟐 − 𝟏. 𝟓 = 𝟎. 𝟐𝟓 + 𝟎. 𝟐𝟓 + 𝟎. 𝟓 = 𝟏
  • 90. FUNCTIONS PCQs Solution : 𝟏) 𝐗 = 𝐘 = 𝐑+ 𝟐) 𝐗 = 𝐑, 𝐘 = 𝐑+ 𝟒𝟐. 𝐋𝐞𝐭 𝐗 𝐚𝐧𝐝 𝐘 𝐛𝐞 𝐬𝐮𝐛𝐬𝐞𝐭𝐬 𝐨𝐟 𝐑, 𝐭𝐡𝐞 𝐬𝐞𝐭 𝐨𝐟 𝐚𝐥𝐥 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬 𝐭𝐡𝐞 𝐟: 𝐗 → 𝐘 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱𝟐 𝐟𝐨𝐫 𝐱 ∈ 𝐗 𝐨𝐬 𝐨𝐧𝐞 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 onto if [𝐄 − 𝟐𝟎𝟎𝟎] 𝟑) 𝐗 = 𝐑+, 𝐲 = 𝐑 𝟒) 𝐗 = 𝐘 = 𝐑 𝐆𝐢𝐯𝐞𝐧 𝐟 𝐱 = 𝐱𝟐 𝐚𝐧𝐝 𝐟 𝐢𝐬 𝐨𝐧𝐞 − 𝐨𝐧𝐞 𝐛𝐮𝐭 𝐧𝐨𝐭 𝐨𝐧𝐭𝐨 𝐁𝐲 𝐯𝐞𝐫𝐢𝐟𝐢𝐜𝐚𝐭𝐢𝐨𝐧 𝐗 ∈ 𝐑+; 𝐲 ∈ 𝐑 𝐟 𝐱 = 𝐱𝟐 is satisfies one-one but not onto
  • 91. FUNCTIONS PCQs Solution : 𝟏) ± 𝟏 𝟐) ± 𝟐 𝟒𝟑. 𝐟 = 𝐑 → 𝐑 𝐚𝐧𝐝 𝐠: 𝐑 → 𝐑 𝐚𝐫𝐞 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟐𝐱 + 𝟑 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱𝟐 + 𝟕 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐯𝐚𝐥𝐮𝐞𝐬 𝐨𝐟 𝐱 𝐟𝐨𝐫 𝐰𝐡𝐢𝐜𝐡 𝐟 𝐠 𝐱 = 𝟐𝟓 𝐚𝐫𝐞 [𝐄 − 𝟐𝟎𝟎𝟎] 𝟑) ± 𝟑 𝟒) ± 𝟒 𝐆𝐢𝐯𝐞𝐧 𝐟: 𝐑 → 𝐑; 𝐠: 𝐑 → 𝐑 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝟐𝐱 + 𝟑 𝐚𝐧𝐝 𝐠 𝐱 = 𝐱𝟐 + 𝟕 𝐍𝐨𝐰 𝐰𝐞 𝐭𝐚𝐤𝐞 𝐟 𝐠 𝐱 = 𝟐𝟓 ⇒ 𝐟 𝐱𝟐 + 𝟕 = 𝟐𝟓
  • 92. FUNCTIONS PCQs ⇒ 𝟐 𝐱𝟐 + 𝟕 + 𝟑 = 𝟐𝟓 ⇒ 𝟐𝐱𝟐 + 𝟏𝟕 = 𝟐𝟓 ⇒ 𝟐𝐱𝟐 = 𝟖 ⇒ 𝐱𝟐 = 𝟒 ⇒ 𝐱 = ±𝟐 Key : 2
  • 93. FUNCTIONS PCQs Solution : 𝟏)𝟏 𝟐)𝟐 44. For any integer n1, the number of positive divisors of n is denoted by d(n). Then for a prime p, d(d(d(𝒑𝟕))) = [𝐄 − 𝟐𝟎𝟎𝟒] 𝟑) 𝟑 𝟒) 𝟒 𝒅 𝒑𝟕 = 𝟖 = 𝟐𝟑 ⇒ 𝒅 𝒅 𝒑𝟕 = 𝟒 = 𝟐𝟐 ⇒ 𝒅 𝒅 𝒅 𝒑𝟕 = 𝟑 Key : 3
  • 94. FUNCTIONS PCQs Solution : 𝟏)𝟐𝟒 𝟐)𝟐𝟔 45. A function f satisfies the relation 𝒇 𝒏𝟐 = 𝒇 𝒏 + 𝟔 for n 2 and f (2) = 8 then the value of f(256) is [𝐊𝐞𝐫𝐚𝐥𝐚 − 𝟐𝟎𝟏𝟓] 𝟑) 𝟐𝟐 𝟒) 𝟒𝟖 𝒇 𝟐𝟓𝟔 = 𝒇 𝟏𝟔𝟐 = 𝒇(𝟏𝟔) + 𝟔 𝒇 𝟏𝟔 = 𝒇 𝟒𝟐 = 𝒇 𝟒 + 𝟔 𝒇 𝟒 = 𝒇 𝟐𝟐 = 𝒇 𝟐 + 𝟔 𝑩𝒖𝒕 𝒇 𝟐 = 𝟖 f(4)= 14, f(16)=20, f(256)= 26 Key : 2
  • 95. FUNCTIONS PCQs Solution : 𝟏)𝟏 𝟐) − 𝟏𝟓 46. If f(1) = 1, f(2n) = f(n) and f(2n+1)= 𝒇 𝒏 𝟐 − 𝟐 for n = 1,2,3…. Then the value of f(1)+f(2)+ …… +f(25) is equal to [𝐊𝐞𝐫𝐚𝐥𝐚 − 𝟐𝟎𝟏𝟓] 𝟑) − 𝟏𝟕 𝟒) − 𝟏 𝒇 𝟐𝒏 + 𝟏 = −𝟏 𝒇𝒐𝒓 𝒂𝒍𝒍 𝒏 𝒇 𝟏 = 𝒇 𝟐 = 𝟏 𝒂𝒏𝒅 𝒇 𝟐𝒏 = 𝟏  f(4)= f(8)= f(16)=1 𝒇 𝟐𝒏 = −𝟏 when ‘n’ is odd 𝒔𝒖𝒎 = 𝟓 − 𝟐𝟎 = −𝟏𝟓 Key : 2
  • 96. FUNCTIONS PCQs Solution : 𝟏) f is periodic 𝟐) f is an odd function 47. Let f(x) = 𝒙 − 𝟐 , 𝐰𝐡𝐞𝐫𝐞 ′ 𝐱′𝐢𝐬 𝐚 𝐫𝐞𝐚𝐥 𝐧𝐮𝐦𝐛𝐞𝐫𝐬, which one of the following is true [𝐊𝐞𝐫𝐚𝐥𝐚 − 𝟐𝟎𝟏𝟒] 𝟑) f is not a 1-1 function 𝟒) f is an equal function 𝒇 𝟏 = 𝟏 − 𝟐 = 𝟏 𝒇 𝟑 = 𝟑 − 𝟐 = 𝟏  𝒇 𝒊𝒔 𝒏𝒐𝒕 𝟏 − 𝟏 Key : 3
  • 97. FUNCTIONS PCQs 𝐋𝐢𝐬𝐭 − 𝐈 𝟐𝟕. 𝐈𝐟 𝐟: 𝐑 → 𝐑 𝐢𝐬 𝐝𝐞𝐟𝐢𝐧𝐞𝐝 𝐛𝐲 𝐟 𝐱 = 𝐱 + 𝟒 𝐟𝐨𝐫 𝐱 < −𝟒 𝟑𝐱 + 𝟐 𝐟𝐨𝐫 − 𝟒 ≤ 𝐱 < 𝟒 𝐱 − 𝟒 𝐟𝐨𝐫 𝐱 ≥ 𝟒 𝐭𝐡𝐞𝐧 𝐭𝐡𝐞 𝐜𝐨𝐫𝐫𝐞𝐜𝐭 𝐦𝐚𝐭𝐜𝐡𝐢𝐧𝐠 𝐨𝐟 𝐋𝐢𝐬𝐭 − 𝐈 𝐟𝐫𝐨𝐦 𝐋𝐢𝐬𝐭 – 𝐈𝐈 𝐢𝐬 [𝐄 − 𝟐𝟎𝟎𝟔] 𝐋𝐢𝐬𝐭 − 𝐈I 𝐀) 𝐟 −𝟓 + 𝐟(−𝟒) 𝐢) 𝟏𝟒 𝐁) 𝐟( 𝐟 −𝟖 ) 𝐢𝐢) 𝟒 𝐂) 𝐟(𝐟 −𝟕 + 𝐟 𝟑 ) 𝐢𝐢𝐢) − 𝟏𝟏 𝐃) 𝐟 𝐟 𝐟 𝐟 𝟎 + 𝟏 𝐢𝐯) − 𝟏 𝐯) 𝟏 𝐯𝐢) 𝟎
  • 98. FUNCTIONS PCQs Solution : 𝐀 𝐁 𝐂 𝐃 1) iii vi ii v 2) iii iv ii v 3) iv iii ii i 4) iii vi v ii 𝐀 𝐟 −𝟓 + 𝐟 −𝟒 = −𝟓 + 𝟒 + 𝟑 −𝟒 + 𝟐 = −𝟏 + (−𝟏𝟎) = −𝟏𝟏
  • 99. FUNCTIONS PCQs 𝐁 𝐟 𝐟 −𝟖 = 𝐟 −𝟖 + 𝟒 = 𝐟 𝟒 = 𝟒 − 𝟒 = 𝟎 𝐂 𝐟 𝐟 −𝟕 + 𝐟 𝟑 = 𝐟[ −𝟕 + 𝟒 + (𝟑 𝟑 + 𝟐] = 𝐟 −𝟑 + 𝟏𝟏 = 𝐟 𝟖 = 𝟖 − 𝟒 = 𝟒 𝐃 𝐟(𝐟 𝐟 𝟑 𝟎 + 𝟐 ]) + 𝟏 = 𝐟(𝐟[𝐟 𝟐 ]) + 𝟏 = 𝐟 𝐟 𝟑 𝟐 + 𝟐 + 𝟏 = 𝐟 𝐟 𝟖 + 𝟏 = 𝐟 𝟒 + 𝟏 = 𝟒 − 𝟒 + 𝟏 = 𝟏 Key : 1