{"id":17493,"date":"2022-05-24T20:30:19","date_gmt":"2022-05-24T15:00:19","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17493"},"modified":"2022-05-16T15:30:45","modified_gmt":"2022-05-16T10:00:45","slug":"mcq-questions-for-class-11-maths-chapter-13","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-13\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Limits and Derivatives Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 13 Limits and Derivatives Objective Questions.<\/p>\n

Limits and Derivatives Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Limits and Derivatives Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Limits and Derivatives Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Limits and Derivatives Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nThe expansion of log(1 – x) is
\n(a) x – x\u00b2 \/2 + x\u00b3 \/3 – ……..
\n(b) x + x\u00b2 \/2 + x\u00b3 \/3 + ……..
\n(c) -x + x\u00b2 \/2 – x\u00b3 \/3 + ……..
\n(d) -x – x\u00b2 \/2 – x\u00b3 \/3 – ……..<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) -x – x\u00b2 \/2 – x\u00b3 \/3 – ……..
\nlog(1 – x) = -x – x\u00b2 \/2 – x\u00b3 \/3 – ……..<\/p>\n<\/details>\n

\n

Question 2.
\nThe value of Limx\u2192a<\/sub> (a \u00d7 sin x – x \u00d7 sin a)\/(ax\u00b2 – xa\u00b2) is
\n(a) = (a \u00d7 cos a + sin a)\/a\u00b2
\n(b) = (a \u00d7 cos a – sin a)\/a\u00b2
\n(c) = (a \u00d7 cos a + sin a)\/a
\n(d) = (a \u00d7 cos a – sin a)\/a<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) = (a \u00d7 cos a – sin a)\/a\u00b2
\nGiven,
\nLimx\u2192a<\/sub> (a \u00d7 sin x – x \u00d7 sin a)\/(ax\u00b2 – xa\u00b2)
\nWhen we put x = a in the expression, we get 0\/0 form.
\nNow apply L. Hospital rule, we get
\nLimx\u2192a<\/sub> (a \u00d7 cos x – sin a)\/(2ax – a\u00b2)
\n= (a \u00d7 cos a – sin a)\/(2a \u00d7 a – a\u00b2)
\n= (a \u00d7 cos a – sin a)\/(2a\u00b2 – a\u00b2)
\n= (a \u00d7 cos a – sin a)\/a\u00b2
\nSo, Limx\u2192a<\/sub> (a \u00d7 sin x – x \u00d7 sin a)\/(ax\u00b2 – xa\u00b2) = (a \u00d7 cos a – sin a)\/a\u00b2<\/p>\n<\/details>\n

\n

Question 3.
\nLimx\u2192-1<\/sub> [1 + x + x\u00b2 + ……….+ x10<\/sup>] is
\n(a) 0
\n(b) 1
\n(c) -1
\n(d) 2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 1
\nGiven, Limx\u2192-1<\/sub> [1 + x + x\u00b2 + ……….+ x10<\/sup>]
\n= 1 + (-1) + (-1)\u00b2 + ……….+ (-1)10<\/sup>
\n= 1 – 1 + 1 – ……. + 1
\n= 1<\/p>\n<\/details>\n

\n

Question 4.
\nThe value of the limit Limx\u21920<\/sub> {log(1 + ax)}\/x is
\n(a) 0
\n(b) 1
\n(c) a
\n(d) 1\/a<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) a
\nGiven, Limx\u21920<\/sub> {log(1 + ax)}\/x
\n= Limx\u21920<\/sub> {ax – (ax)\u00b2 \/2 + (ax)\u00b3 \/3 – (ax)4<\/sup> \/4 + …….}\/x
\n= Limx\u21920<\/sub> {ax – a\u00b2 x\u00b2 \/2 + a\u00b3 x\u00b3 \/3 – a4<\/sup> x4<\/sup> \/4 + …….}\/x
\n= Limx\u21920<\/sub> {a – a\u00b2 x \/2 + a\u00b3 x\u00b2 \/3 – a4<\/sup> x\u00b3 \/4 + …….}
\n= a – 0
\n= a<\/p>\n<\/details>\n

\n

Question 5.
\nThe value of the limit Limx\u21920<\/sub> (cos x)cot\u00b2 x<\/sup> is
\n(a) 1
\n(b) e
\n(c) e1\/2<\/sup>
\n(d) e-1\/2<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) e-1\/2<\/sup>
\nGiven, Limx\u21920<\/sub> (cos x)cot\u00b2 x<\/sup>
\n= Limx\u21920<\/sub> (1 + cos x – 1)cot\u00b2 x<\/sup>
\n= eLimx\u21920<\/sub><\/sup> (cos x – 1) \u00d7 cot\u00b2 x
\n= eLimx\u21920<\/sub><\/sup> (cos x – 1)\/tan\u00b2 x
\n= e-1\/2<\/sup><\/p>\n<\/details>\n

\n

Question 6.
\nThen value of Limx\u21921<\/sub> (1 + log x – x)}\/(1 – 2x + x\u00b2) is
\n(a) 0
\n(b) 1
\n(c) 1\/2
\n(d) -1\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) -1\/2
\nGiven, Limx\u21921<\/sub> (1 + log x – x)}\/(1 – 2x + x\u00b2)
\n= Limx\u21921<\/sub> (1\/x – 1)}\/(-2 + 2x) {Using L. Hospital Rule}
\n= Limx\u21921<\/sub> (1 – x)\/{2x(x – 1)}
\n= Limx\u21921<\/sub> (-1\/2x)
\n= -1\/2<\/p>\n<\/details>\n

\n

Question 7.
\nThe value of limy\u21920<\/sub> {(x + y) \u00d7 sec (x + y) – x \u00d7 sec x}\/y is
\n(a) x \u00d7 tan x \u00d7 sec x
\n(b) x \u00d7 tan x \u00d7 sec x + x \u00d7 sec x
\n(c) tan x \u00d7 sec x + sec x
\n(d) x \u00d7 tan x \u00d7 sec x + sec x<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) x \u00d7 tan x \u00d7 sec x + sec x
\nGiven, limy\u21920<\/sub> {(x + y) \u00d7 sec (x + y) – x \u00d7 sec x}\/y
\n= limy\u21920<\/sub> {x sec (x + y) + y sec (x + y) – x \u00d7 sec x}\/y
\n= limy\u21920<\/sub> [x{ sec (x + y) – sec x} + y sec (x + y)]\/y
\n= limy\u21920<\/sub> x{ sec (x + y) – sec x}\/y + limy\u21920<\/sub> {y sec (x + y)}\/y
\n= limy\u21920<\/sub> x{1\/cos (x + y) – 1\/cos x}\/y + limy\u21920<\/sub> {y sec (x + y)}\/y
\n= limy\u21920<\/sub> [{cos x – cos (x + y)} \u00d7 x\/{y \u00d7 cos (x + y) \u00d7 cos x}] + limy\u21920<\/sub> {y sec (x + y)}\/y
\n= limy\u21920<\/sub> [{2sin (x + y\/2) \u00d7 sin(y\/2)} \u00d7 2x\/{2y \u00d7 cos (x + y) \u00d7 cos x}] + limy\u21920<\/sub> {y sec (x + y)}\/y
\n= limy\u21920<\/sub> {sin (x + y\/2) \u00d7 limy\u21920<\/sub> {sin(y\/2)\/(2y\/2)} \u00d7 limy\u21920<\/sub> { x\/{y \u00d7 cos (x + y) \u00d7 cos x}] + sec x
\n= sin x \u00d7 1 \u00d7 x\/cos\u00b2 x + sec x
\n= x \u00d7 tan x \u00d7 sec x + sec x
\nSo, limy\u21920<\/sub> {(x + y) \u00d7 sec (x + y) – x \u00d7 sec x}\/y = x \u00d7 tan x \u00d7 sec x + sec x<\/p>\n<\/details>\n

\n

Question 8.
\nLimx\u21920<\/sub> (ex\u00b2<\/sup> – cos x)\/x\u00b2 is equals to
\n(a) 0
\n(b) 1
\n(c) 2\/3
\n(d) 3\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 3\/2
\nGiven, Limx\u21920<\/sub> (ex\u00b2<\/sup> – cos x)\/x\u00b2
\n= Limx\u21920<\/sub> (ex\u00b2<\/sup> – cos x – 1 + 1)\/x\u00b2
\n= Limx\u21920<\/sub> {(ex\u00b2<\/sup> – 1)\/x\u00b2 + (1 – cos x)}\/x\u00b2
\n= Limx\u21920<\/sub> {(ex\u00b2<\/sup> – 1)\/x\u00b2 + Limx\u21920<\/sub> (1 – cos x)}\/x\u00b2
\n= 1 + 1\/2
\n= (2 + 1)\/2
\n= 3\/2<\/p>\n<\/details>\n

\n

Question 9.
\nThe expansion of ax<\/sup> is
\n(a) ax<\/sup> = 1 + x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 + x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..
\n(b) ax<\/sup> = 1 – x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 – x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..
\n(c) ax<\/sup> = 1 + x\/1 \u00d7 (log a) + x\u00b2 \/2 \u00d7 (log a)\u00b2 + x\u00b3 \/3 \u00d7 (log a)\u00b3 + ………..
\n(d) ax<\/sup> = 1 – x\/1 \u00d7 (log a) + x\u00b2 \/2 \u00d7 (log a)\u00b2 – x\u00b3 \/3 \u00d7 (log a)\u00b3 + ………..<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) ax<\/sup> = 1 + x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 + x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..
\nax<\/sup> = 1 + x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 + x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..<\/p>\n<\/details>\n

\n

Question 10.
\nThe value of the limit Limn\u21920<\/sub> (1 + an)b\/n<\/sup> is
\n(a) ea<\/sup>
\n(b) eb<\/sup>
\n(c) eab<\/sup>
\n(d) ea\/b<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) eab<\/sup>
\nGiven, Limn\u21920<\/sub> (1 + an)b\/n<\/sup>
\n= eLimn\u21920<\/sub>(an \u00d7 b\/n)<\/sup>
\n= eLimn\u21920<\/sub>(ab)<\/sup>
\n= eab<\/sup><\/p>\n<\/details>\n

\n

Question 11.
\nThe value of Limx\u21920<\/sub> cos x\/(1 + sin x) is
\n(a) 0
\n(b) -1
\n(c) 1
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 1
\nGiven, Limx\u21920<\/sub> cos x\/(1 + sin x)
\n= cos 0\/(1 + sin 0)
\n= 1\/(1 + 0)
\n= 1\/1
\n= 1<\/p>\n<\/details>\n

\n

Question 12.
\nLim tanx\u2192\u03c0\/4<\/sub> tan 2x \u00d7 tan(\u03c0\/4 – x) is
\n(a) 0
\n(b) 1
\n(c) 1\/2
\n(d) 3\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 1\/2
\nGiven, Lim tanx\u2192\u03c0\/4<\/sub> tan 2x \u00d7 tan(\u03c0\/4 – x)
\n= Lim tanh\u21920<\/sub> tan 2(\u03c0\/4 – x) \u00d7 tan(-h)
\n= Lim tanh\u21920<\/sub> -cot 2h\/(-cot h)
\n= Lim tanh\u21920<\/sub> tan h\/tan 2h
\n= (1\/2) \u00d7 Lim tanh\u21920<\/sub> (tan h\/h)\/(2h\/tan 2h)
\n= (1\/2) \u00d7 {Lim tanh\u21920<\/sub> (tan h\/h)}\/{Lim tanh\u21920<\/sub> (2h\/tan 2h)}
\n= (1\/2) \u00d7 1
\n= 1\/2<\/p>\n<\/details>\n

\n

Question 13.
\nLimx\u21922<\/sub> (x\u00b3 – 6x\u00b2 + 11x – 6)\/(x\u00b2 – 6x + 8) =
\n(a) 0
\n(b) 1
\n(c) 1\/2
\n(d) Limit does not exist<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 1\/2
\nWhen x = 2, the expression
\n(x\u00b3 – 6x\u00b2 + 11x – 6)\/(x\u00b2 – 6x + 8) assumes the form 0\/0
\nNow,
\nLimx\u21922<\/sub> (x\u00b3 – 6x\u00b2 + 11x – 6)\/(x\u00b2 – 6x + 8) = Limx\u21922<\/sub> {(x – 1) \u00d7 (x – 2) \u00d7 (x – 3)}\/{(x – 2) \u00d7 (x – 4)}
\n= Limx\u21922<\/sub> {(x – 1) \u00d7 (x – 3)}\/(x – 4)
\n= {(2 – 1) \u00d7 (2 – 3)}\/(2 – 4)
\n= 1\/2<\/p>\n<\/details>\n

\n

Question 14.
\nThe value of the limit Limx\u21922<\/sub> (x – 2)\/\u221a(2 – x) is
\n(a) 0
\n(b) 1
\n(c) -1
\n(d) 2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 0
\nGiven, Limx\u21922<\/sub> (x – 2)\/\u221a(2 – x)
\n= Limx\u21922<\/sub> -(2 – x)\/\u221a(2 – x)
\n= Limx\u21922<\/sub> -{\u221a(2 – x) \u00d7 \u221a(2 – x)}\/\u221a(2 – x)
\n= Limx\u21922<\/sub> -\u221a(2 – x)
\n= -\u221a(2 – 2)
\n= 0<\/p>\n<\/details>\n

\n

Question 15.
\nThe derivative of the function f(x) = 3x\u00b3 – 2x\u00b3 + 5x – 1 at x = -1 is
\n(a) 0
\n(b) 1
\n(c) -18
\n(d) 18<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 18
\nGiven, function f(x) = 3x\u00b3 – 2x\u00b2 + 5x – 1
\nDifferentiate w.r.t. x, we get
\ndf(x)\/dx = 3 \u00d7 3 \u00d7 x\u00b2 – 2 \u00d7 2 \u00d7 x + 5
\n\u21d2 df(x)\/dx = 9x\u00b2 – 4x + 5
\n\u21d2 {df(x)\/dx}x =-1<\/sub> = 9 \u00d7 (-1)\u00b2 – 4 \u00d7 (-1) + 5
\n\u21d2 {df(x)\/dx}x =-1<\/sub> = 9 + 4 + 5
\n\u21d2 {df(x)\/dx}x =-1<\/sub> = 18<\/p>\n<\/details>\n

\n

Question 16.
\nLimx\u21920<\/sub> sin\u00b2(x\/3)\/x\u00b2 is equals to
\n(a) 1\/2
\n(b) 1\/3
\n(c) 1\/4
\n(d) 1\/9<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 1\/9
\nGiven, Limx\u21920<\/sub> sin\u00b2 (x\/3)\/ x\u00b2
\n= Limx\u21920<\/sub> [sin\u00b2 (x\/3)\/ (x\/3)\u00b2 \u00d7 {(x\/3)\u00b2 \/x\u00b2}]
\n= Limx\u21920<\/sub> [{sin (x\/3)\/ (x\/3)}\u00b2 \u00d7 {(x\u00b2 \/9)\/x\u00b2}]
\n= 1 \u00d7 1\/9
\n= 1\/9<\/p>\n<\/details>\n

\n

Question 17.
\nThe expansion of ax is
\n(a) ax<\/sup> = 1 + x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 + x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..
\n(b) ax<\/sup> = 1 – x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 – x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..
\n(c) ax<\/sup> = 1 + x\/1 \u00d7 (log a) + x\u00b2 \/2 \u00d7 (log a)\u00b2 + x\u00b3 \/3 \u00d7 (log a)\u00b3 + ………..
\n(d) ax<\/sup> = 1 – x\/1 \u00d7 (log a) + x\u00b2 \/2 \u00d7 (log a)\u00b2 – x\u00b3 \/3 \u00d7 (log a)\u00b3 + ………..<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) ax<\/sup> = 1 + x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 + x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..
\nax = 1 + x\/1! \u00d7 (log a) + x\u00b2 \/2! \u00d7 (log a)\u00b2 + x\u00b3 \/3! \u00d7 (log a)\u00b3 + ………..<\/p>\n<\/details>\n

\n

Question 18.
\nDifferentiation of cos \u221ax with respect to x is
\n(a) sin x \/2\u221ax
\n(b) -sin x \/2\u221ax
\n(c) sin \u221ax \/2\u221ax
\n(d) -sin \u221ax \/2\u221ax<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) -sin \u221ax \/2\u221ax
\nLet y = cos \u221ax
\nPut u = \u221ax
\ndu\/dx = 1\/2\u221ax
\nNow, y = cos u
\ndy\/du = -sin u
\ndy\/dx = (dy\/du) \u00d7 (du\/dx)
\n= -sin u \u00d7 (1\/2\u221ax)
\n= -sin \u221ax \/2\u221ax<\/p>\n<\/details>\n

\n

Question 19.
\nDifferentiation of log(sin x) is
\n(a) cosec x
\n(b) cot x
\n(c) sin x
\n(d) cos x<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) cot x
\nLet y = log(sin x)
\nAgain let u = sin x
\ndu\/dx = cos x
\nNow, y = log u
\ndy\/du = 1\/u = 1\/sin x
\nNow, dy\/dx = (dy\/du) \u00d7 (du\/dx)
\n\u21d2 dy\/dx = (1\/sin x) \u00d7 cos x
\n\u21d2 dy\/dx = cos x\/sin x
\n\u21d2 dy\/dx = cot x<\/p>\n<\/details>\n

\n

Question 20.
\nLimx\u2192\u221e<\/sub> {(x + 5)\/(x + 1)}x<\/sup> equals
\n(a) e\u00b2
\n(b) e4<\/sup>
\n(c) e6<\/sup>
\n(d) e8<\/sup><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) e6<\/sup>
\nGiven, Limx\u2192\u221e<\/sub> {(x + 5)\/(x + 1)}x
\n= Limx\u2192\u221e<\/sub> {1 + 6\/(x + 1)}x
\n= eLimx\u2192\u221e<\/sub>6x\/(x + 1)<\/sup>
\n= eLimx\u2192\u221e<\/sub> 6\/(1 + 1\/x)<\/sup>
\n= e6\/(1 + 1\/\u221e)<\/sup>
\n= e6\/(1 + 0)<\/sup>
\n= e6<\/sup><\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can …<\/p>\n

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