NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.2) Exercise 6.2

The subject specialists at the Extramarks’ website have created the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2  and the solutions for all the other chapters as well to help equip students with a clear and accurate understanding of all the topics taught in Class 12. These NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 include thorough, step-by-step explanations for all the problems found in the textbooks. These can be of great help for board examination preparations.

First Derivative Test, Maximum and Lowest Values of a Function in a Closed Interval, Approximations, Maxima and Minima, Growing and Decreasing Functions, Tangents and Normals, and many such topics are all covered in Chapter 6 of the NCERT textbook. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 contain solutions to all the questions related to these topics that are there in the second exercise of this chapter.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 can work as a crucial tool for the students to study Mathematics for exams and also help with the homework.

For children to succeed in the Class 12 board exams and other competitive exams that they appear for after Class 12, Mathematics is essential. Students can use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 to aid in their preparation for undergraduate engineering admission exams like JEE Mains, BITSAT, VITEEE, and other such competitive exams.

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Founded in 1961 as a literary, scientific, and philanthropic society under the Societies Registration Act, the National Council of Educational Research and Training (NCERT) is a stand-alone entity of the Government of India. It is responsible for establishing the curriculum for all nationwide schools that adhere to the Central Board of Secondary Education (CBSE). One of the ideal resources to aid in a student’s preparation for the CBSE exam, as well as engineering entrance exams like the JEE etc., is the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2.

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.2) Exercise 6.2

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 are the solutions to the second exercise of Chapter 6 of Class 12 Mathematics. In this chapter, students get to examine the derivative’s uses in a variety of academic domains, including engineering, science, social science, and many more.

For instance, students will discover how the derivative can be used to 1) calculate the rate of change of quantities, 2, find the equations of tangent and normal to a curve at a point, and 3, find turning points on the graph of a function. All of this will assist the students in identifying the points at which the largest or smallest value (locally) of a function occurs.

Additionally, students will get to use a function’s derivative to determine the intervals at which it is increasing or decreasing.

The derivative is then used to determine the approximate value of various quantities.

The chapter is divided into different exercises that contain questions that the students can practice understanding the concepts that are taught in the chapter thoroughly. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 are solutions to these questions. These are specifically for the Class 12 maths Chapter 6 Ex 6.2, but there are solutions available on the Extramarks’ website to all the exercises in the NCERT books.

What Is The Distance Between Two Points

In each chapter there will be some topics that the students depending on their calibre might find more difficult than other topics. Finding the distance between two points can be one of those difficult topics but with the help of the right tools, the students can achieve good marks in it as well.

Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 and all the other solutions prepared by the experts at the Extramarks’ website are given so that the students can get a clear understanding of how to solve the questions that the students might find difficult. These exercises are generally done in the class when the chapter is being taught. Going through the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 can also help the students complete their homework for the questions which are not covered in the class. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 can also help students prepare for class tests as those are also prepared based on the questions that are there in these textbook exercises.

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NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2

The students can create a strong base for their preparation of individual chapters depending on the weightage that is given to that particular chapter by the NCERT. Abiding by the guidance of NCERT, the students can find the list of chapters along with the weightage assigned to them on the Extramarks’ website.

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Q.1 Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Ans

Function is given by f(x)=3x + 17Differentiating w.r.t. x, we get    f‘(x)=3>0 in every interval of R.Thus, the function is strictly increasing on R.

Q.2 Show that the function given by f(x) = e2x is strictly increasing on R

Ans

Let x1 and x2 be any two numbers in R.Then, we have:x1 < x2⇒2x1 < 2x2⇒   e2x1 < e2x2⇒f(x1)<f(x2)Hence, f is strictly increasing on R.

Q.3 Show that the function given by f(x) = sin x is (a) strictly increasing in ( 0 , π 2 ) , (b) strictly decreasing in ( π 2 , π ) , (c) neither increasing nor decreasing in ( 0 , π )

Ans

The given function is f(x) = sin x.∴f‘(x)=cosx(a)Since for each  x∈(0,π2), cosx>0       we have f‘(x)>0.Hence, f is strictly increasing in (0,π2).(b)Since for each  x∈(π2,π), cosx<0       we have f‘(x)<0.Hence, f is strictly decreasing in (π2,π).(c)From the results obtained in (a) and (b), it is clear that f is      neither increasing nor decreasing in (0, π).

Q.4 Find the intervals in which the function f given by f(x) = 2x2 − 3x is

(a) strictly increasing (b) strictly decreasing

Ans

The given function is f(x)=2x2− 3x.      f‘(x)=4x−3Therefore, f‘(x)=0⇒4x−3=0⇒   x=34Now the point x=34 divides the real line into two disjoint intervals namely, (−∞, 34) and (34,∞).

In the interval ( −∞,  3 4 ), f’( x )=4x−3<0 Therefore, f is strictly deacreasing in this interval. Also, in the interval   ( 3 4 ,∞ ), f’( x )>0​ and so the function f is strictly increasing in this interval. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@EC9D@

Q.5 Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing

Ans

∴f‘x=0⇒6 (x−3) (x+2)=0⇒x=−2,3The points x =−2 and x = 3 divide the real line into three disjoint intervals i.e., (−∞,−2), (−2,3) and (3,∞).

In intervals  (−∞,−2) and (3,∞), f‘(x) is positive while in interval (−2,3), f‘(x) is negative. Hence, the given function (f) is strictly increasing in intervals (−∞,−2) and (3,∞), while function (f) is strictly decreasing in interval  −2,3 .

Q.6 Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x2 + 2x − 5 (b)10 − 6x − 2x2

(c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9x − x2

(e) (x + 1)3 (x − 3)3

Ans

(a) We have,  f(x)=x2+ 2x−5Differentiating both sides with respect to x, we getf‘(x)=2x+2∴f‘(x)=0⇒2x+2=0⇒      x=−1Point x =−1 divides the real line into two disjoint intervals i.e., (−∞,−1) and (−1,∞).In interval(−∞,−1), f‘(x)=2x+2<0∴f is strictly decreasing in interval (−∞,−1).Thus, f is strictly decreasing for x < −1.In interval  (−1,∞), f‘(x)=2x+2>0∴f is strictly increasing in interval (−1,∞).Thus, f is strictly increasing for x > −1.(b)We have,  f(x)=10−6x−2x2Differentiating both sides with respect to x, we getf‘(x)=− 6−4x∴f‘(x)=0⇒− 6−4x=0⇒      x=−32Point x =−32 divides the real line into two disjoint intervalsi.e., (−∞,−32) and (−32,∞).In interval(−∞,−32), f‘(x)=− 6−4x>0∴f is strictly increasing in interval (−∞,−32).Thus, f is strictly increasing for x<−32.In interval  (−32,∞), f‘(x)=−6−4x<0∴f is strictly decreasing in interval (−32,∞).Thus, f is strictly decreasing for x >−32.(c)We have,  f(x)=−2x3−9x2−12x+1Differentiating both sides with respect to x, we getf‘(x)=− 6x2−18x−12=−6(x2+3x+2)∴f‘(x)=0⇒−6(x2+3x+2)=0⇒(x+1)(x+2)=0⇒      x=−1,−2The points x =−1 and x =−2 divide the real line into three disjoint intervals i.e., (−∞,−2),(−2,−1) and (−1,∞).In intervals  (−∞,−2) and (−1,∞), f‘(x) is negative. Hence, the given function (f) is strictly decreasing in intervals(−∞,−2) and (−1,∞),i.e., function (f) is strictly decreasing when x<−2 and x>−1.

In intervals  ( −2,−1 ), f‘( x ) is positive. So, function (f) is strictly increasing in interval  ( −2,−1 ), i.e.,  function (f) is strictly increasing when −2<x<−1.  ( d )We have,   f( x )=6−9x− x 2 Differentiating both sides with respect to x, we get f’( x )=−9−2x ∴f’( x )=0⇒−9−2x=0 ⇒      x=− 9 2 Point x =− 9 2 divides the real line into two disjoint intervals i.e., ( −∞,− 9 2 ) and ( − 9 2 ,∞ ). In interval( −∞,− 9 2 ), f‘( x )=−9−2x>0 ∴f is strictly increasing in interval ( −∞,− 9 2 ). Thus, f is strictly increasing for x < − 9 2 . In interval  ( − 9 2 ,∞ ), f‘( x )=−9−2x<0 ∴f is strictly decreasing in interval ( − 9 2 ,∞ ). Thus, f is strictly decreasing for x > − 9 2 . 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( e )We have, f(x)= (x + 1) 3 (x− 3) 3 Differentiating both sides with respect to x, we get        f’( x )= ( x + 1 ) 3 d dx ( x−3 ) 3 + ( x−3 ) 3 d dx ( x + 1 ) 3     [ By product rule ] ⇒f’( x )= ( x + 1 ) 3 3 ( x−3 ) 2 d dx ( x−3 )+ ( x−3 ) 3 3 ( x + 1 ) 2 d dx ( x + 1 ) [ By chain rule ] ⇒f’( x )=3 ( x + 1 ) 3 ( x−3 ) 2 ( 1−0 )+3 ( x−3 ) 3 ( x + 1 ) 2 ( 1+0 ) ⇒f’( x )=3 ( x + 1 ) 2 ( x−3 ) 2 ( x+1+x−3 ) ⇒f’( x )=3 ( x + 1 ) 2 ( x−3 ) 2 ( 2x−2 )       =6 ( x + 1 ) 2 ( x−3 ) 2 ( x−1 ) Now, f’( x )=0⇒x=−1, 1,  3 The points x=−1, x=1, and x=3 divide the real line into four disjoint intervals i.e., ( −∞,−1 ),( −1, 1 ),( 1,3 ) and ( 3,∞ ). In intervals  ( −∞,−1 )  and (−1, 1), 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 f’( x )=6 ( x + 1 ) 2 ( x−3 ) 2 ( x−1 )<0 ∴f is strictly decreasing in intervals ( −∞,−1 )  and ( −1, 1 ). In intervals (1, 3) and ( 3,∞ ), f’( x )=6 ( x + 1 ) 2 ( x−3 ) 2 ( x−1 )>0 ∴f is strictly increasing in intervals ( 1, 3 ) and ( 3,∞ ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@EF8C@

Q.7

Showthat y=log(1+x)−2x2+x,x> −1, is an increasing function of x throughout its domain.

Ans

We have,  y=log(1+x)−2x2+xDifferentiating both sides with respect to x, we get   dydx=ddxlog(1+x)−ddx(2x2+x)=11+xddx(1+x)−(2+x)ddx2x−2xddx(2+x)(2+x)2=11+x×(0+1)−2(2+x)−2x(0+1)(2+x)2=11+x−4(2+x)2=4+4x+x2−4−4x(1+x)(2+x)2=x2(1+x)(2+x)2Now,                   dydx=0⇒ x2(1+x)(2+x)2=0⇒   x2=0[(1+x)≠0 and (2+x)2≠0]⇒     x=0Since x >−1, point x = 0 divides the domain (−1, ∞) in two disjoint intervals i.e.,−1 < x < 0 and x > 0.When −1< x < 0, we have:x< 0⇒x2>0x>−1⇒x+2>0⇒(x+2)2>0∴dydx=x2(x+2)2>0Also, when x > 0:x>0⇒x2>0,  (2+x2)>0∴dydx=x2(x+2)2>0Hence, function f is increasing throughout this domain.

Q.8

Find the values of x for whichy=[x(x−2)]2is an increasingfunction.

Ans

We have,      y=[x(x−2)]2=x2(x−2)2Differentiating w.r.t. x, we get   dydx=ddxx2(x−2)2=x2ddx(x−2)2+(x−2)2ddxx2=x2.2(x−2)ddx(x−2)+(x−2)2.2x=2x2(x−2)(1−0)+2x(x−2)2=2x(x−2)(x+x−2)=2x(x−2)(2x−2)=4x(x−2)(x−1)∴dydx=0⇒x=0,1,2The points x = 0, x = 1, and x = 2 divide the real line into fourdisjoint intervals i.e.,(−∞,0),(0,1),(1,2),(2,∞).In intervals (−∞,0)and  (1,2), dydx<0∴y is strictly decreasing in intervals (−∞,0)and  (1,2).However, in intervals (0, 1) and (2,∞), dydx>0∴y is strictly increasing in intervals (0, 1) and (2,∞).∴y is strictly increasing for 0 < x < 1 and x > 2.

Q.9

Prove that y=4sinθ(2+cosθ)−θ is an increasing function of θ in [0,π2].

Ans

We have,  y=4sinθ(2+cosθ)−θDifferentiating w.r.t. θ, we getdydθ=ddθ{4sinθ(2+cosθ)−θ}        =(2+cosθ)ddθ(4sinθ)−4sinθddθ(2+cosθ)(2+cosθ)2−ddθθ        =4(2+cosθ)cosθ−4sinθ(0−sinθ)(2+cosθ)2−1        =4(2+cosθ)cosθ+4sin2θ(2+cosθ)2−1        =8cosθ+4(cos2θ+sin2θ)(2+cosθ)2−1        =8cosθ+4(2+cosθ)2−1Now, dydθ=0⇒8cosθ+4(2+cosθ)2−1=0⇒8cosθ+4(2+cosθ)2=1⇒8cosθ+4=(2+cosθ)2⇒8cosθ+4=4+4cosθ+cos2θ⇒cos2θ−4cosθ=0⇒cosθ(cosθ−4)=0⇒cosθ=0  or cosθ=4 Since cos θ≠4∴cosθ=0⇒θ=π2Now,   dydθ=8cosθ+4(2+cosθ)2−1=8cosθ+4−(2+cosθ)2(2+cosθ)2=8cosθ+4−4−4cosθ−cos2θ(2+cosθ)2=cosθ(4−cosθ)(2+cosθ)2In interval  (0,π2),  we have cosθ > 0. Also, 4 > cosθ⇒4−cosθ > 0.∴cosθ(4−cosθ)>0 and also (2+cosθ)2>0⇒cosθ(4−cosθ)(2+cosθ)2>0⇒dydθ>0Therefore, y is strictly increasing in interval  (0,π2).Also, the given function is continuous at  x=0 and x=π2.Hence, y is increasing in interval (0,π2).

Q.10

Provethatthelogarithmicfunctionisstrictly increasingon (0, ∞).

Ans

The given function if f(x)=logx∴      f‘(x)=1xIt is clear that for x > 0,f‘(x)=1x>0Hence, f(x) = log x is strictly increasing in interval (0,∞).

Q.11 Prove that the function f given by f(x)= x2 − x + 1 is neither strictly increasing
nor strictly decreasing on (−1, 1).

Ans

The given function is f(x) = x2−x + 1.Differentiating with respect to x, we getf’(x)=2x−1Now, f‘(x)=0⇒2x−1=0⇒x=12The point 12  divides the interval (−1, 1) into two disjoint intervals i.e., (−1, 12)  and (12 , 1).Now, for interval (−1, 12) , f‘(x)=2x−1<0Therefore, f is strictly decreasing in interval (−1, 12).Now, for interval (12 , 1) , f‘(x)=2x−1>0So, f is strictly increasing in interval (12 , 1).Hence, f is neither strictly increasing nor decreasing in interval (−1, 1).

Q.12

Which of the following functions are strictly decreasing on(0,π2)?

(A)cosx (B)cos2x (C)cos3x (D)tanx

Ans

( A ) Let  f 1 ( x )=cosx Differentiating w.r.t. x, we get     f’ 1 ( x )= d dx cosx     =−sinx In interval ( 0, π 2 ), sinx is positive in first quadrant. So,   f’ 1 ( x )<0 Therefore,  f 1 ( x )=cosx is strictly decreasing in interval ( 0, π 2 ). ( B )Let  f 2 ( x )=cos2x Differentiating w.r.t. x, we get     f’ 2 ( x )= d dx cos2x     =−2sin2x In interval ( 0, π 2 ), sin2x is positive in first quadrant. So,   f’ 2 ( x )<0 Therefore,  f 2 ( x )=cos2x is strictly decreasing in interval ( 0, π 2 ). ( C )Let  f 3 ( x )=cos3x Differentiating w.r.t. x, we get     f’ 3 ( x )= d dx cos3x     =−3sin3x Now,  f’ 3 ( x )=0⇒−3sin3x=0 ⇒        3x=π⇒x= π 3 The point x= π 3   divides the interval ( 0, π 2 )  into two disjoint intervals i.e., ( 0, π 3 ) and ( π 3 , π 2 ). Now,​ in interval  ( 0, π 3 ), f’ 3 ( x )=−3sin3x<0 [ ∵0<x< π 3 ⇒0<3x<π ] ∴ f 3 is strictly decreasing in interval ( 0, π 3 ). Now, in interval  ( π 3 , π 2 ), π 3 <x< π 2 ⇒π<3x< 3π 2 So, sin3x is in third quadrant, where sin3x is negative. ∴ f 3 ( x )=−3sin3x>0 Therefore,   f 3 is strictly increasing in interval ( π 3 , π 2 ). Hence, f 3 is neither increasing nor decreasing in interval ( 0, π 2 ). ( D )Let  f 4 ( x )=tanx Differentiating w.r.t. x, we get f’ 4 ( x )= d dx tanx   = sec 2 x In  interval ( 0, π 2 ),​  secx>0⇒ sec 2 x>0 ∴ f 4 ( x )>0 ∴f4 is strictly increasing in interval ( 0, π 2 ). Therefore, functions cos x and cos 2x are strictly decreasing in ( 0, π 2 ). Hence, the correct answers are A and B. 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Q.13

On which of the following intervals is the function f given byf(x)=x100 +sinx– 1strictly decreasing?(A)(0, 1)(B)(π2, π)(C)(0, π2)(D)None of these

Ans

We  have, f(x)=x100+sinx−1Differentiating w.r.t. x, we getf‘(x)=100x99+cosxf‘(x)>0[∵100x99>0 and cosx>0 in interval(0,1)]Thus, function f is strictly increasing in interval (0, 1).In interval (π2,π),cosx<0 and 100x99>0So,   f‘(x)>0Thus, function f is strictly increasing in interval (π2,π).In interval (0,π2),cosx>0 and 100x99>0⇒100x99+cosx>0So,   f‘(x)>0Thus, function f is strictly increasing in interval (0,π2).Hence, function f is strictly decreasing in none of the intervals.The correct answer is D.

Q.14 Find the least value of a such that the function f given f(x) =x2+ax+1 is strictly increasing on (1, 2).

Ans

We have,f(x)=x2+ax+1Differentiating w.r.t. x, we getf‘(x)=2x+aNow, function f will be increasing in (1,2),if f‘(x)>0⇒2x+a>0     x>−a2Therefore, we have to find the least value of a such that     x>−a2,   when x∈(1,2)⇒ x>−a2,   (when 1<x<2)Thus, the least value of a for f to be increasing on(1, 2) is given by,–a2=1⇒a=−2Hence, the required value of a is −2.

Q.15

Let I be any interval disjoint from (−1,1).Prove that the function fgiven byf(x)=x+1x is strictly increasing onI.

Ans

we​ have,f(x)=x+1xDifferentiating w.r.t. x, we getf‘(x)=1−1x2Now, f‘(x)=0⇒1−1x2=0⇒1=1x2⇒x=±1The points x=1 and x=−1 divide the real line in three disjointintervals i.e., (−∞,−1),(−1,1)  and  (1,∞).In interval (−1, 1), it is noticed that:−1<x<1⇒x2<1⇒1<1x2,  x≠0⇒1−1x2<0,  x≠0f‘(x)=1−1x2<0  on (−1,1)~{0}∴f is strictly decreasing on  (−1,1)~{0}In intervals  (−∞,−1)  and  (1,∞), it is seen thatx<−1 or x>1⇒x2>1⇒1>1x2⇒1−1x2>0∴f‘(x)=1−1x2>0 on  (−∞,−1)  and  (1,∞).∴f is strictly increasing on (−∞,−1)  and  (1,∞).Hence, function f is strictly increasing in interval I disjoint from (−1, 1).

Q.16

Provethatthefunctionfgivenbyf(x) =logsinxisstrictly increasingon (0,π2)andstrictlydecreasingon(π2,π).

Ans

We have,f(x)=logsinxDifferentiating both sides w.r.t. x, we getf‘(x)=1sinxddxsinx[By chain rule]         =1sinxcosx=cotxf‘(x)=cotx>0​  in interval (0,π2)∴f(x) is strictly increasing function in interval (0,π2).f‘(x)=cotx<0​  in interval (π2,π)∴f(x) is strictly decreasing function in interval (π2,π).

Q.17 Prove that the function f given by f(x) = logcos x is strictly decreasing on ( 0 , π 2 ) and strictly increasing on ( π 2 , π ) .

Ans

We have,f(x)=logcosxDifferentiating both sides w.r.t. x, we getf‘(x)=1cosxddxcosx[By chain rule]         =1cosx×−sinx=−tanxf‘(x)=−tanx<0​  in interval (0,π2)∴f(x) is strictly decreasing function in interval (0,π2).f‘(x)=−tanx>0​  in interval (π2,π)∴f(x) is strictly increasing function in interval (π2,π).

Q.18

Prove that the function f given byf(x) = x2−3x2+3x−100is increasinginR.

Ans

We have,f(x)=x3−3x2+3x−100Differentiating w.r.t. x, we getf‘(x)=3x2−6x+3=3(x2−2x+1)=3(x−1)2For any x∈R, (x−1)2>0.Thus, f‘(x)  is always positive in R.Hence, the given function f(x) is increasing in R.

Q.19

The interval in which y=x2e−xis increasing is(A) (−∞, ∞) (B )(−2, 0) (C) (2, ∞) (D) (0, 2)

Ans

We have,y=x2e−xDifferentiating w.r.t. x, we getdydx=x2ddxe−x+e−xddxx2       =−x2e−x+e−x(2x)       =e−x(−x2+2x)Now, dydx=0⇒e−x(−x2+2x)=0⇒x=0,2[∵e−x≠0]The points x=0 and x=2 divide the real line into three disjoint intervals i.e., (−∞,0),(0,2) and  (2,∞).f‘(x)<0 in intervals (−∞,0) and (2,∞).So, f(x)​ is decreasing in intervals (−∞,0) and (2,∞).Now, f‘(x)>0 in interval (0,2)So, f(x)​ is increasing in intervals (0,2) Hence, f is strictly increasing in interval (0, 2).The correct answer is D.

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1. Should the students try to remember the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 word to word?

In subjects like Mathematics, memorising the answers rarely helps. The students are advised to try to understand what the question expects them to answer. The next step should be to understand how to answer that question efficiently. Another way could be to try to solve the question first and then cross-check it with the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 to see if the students are on the right track and following the right way of solving the question. 

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 should be used as a tool to understand the solutions, what formulas are used in it and what is the correct way to solve the question. The questions that are asked in the final board examinations have a high chance of resembling the questions that are there in the NCERT book. Therefore, the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 and other such solutions to the NCERT questions are very important for revision. 

Still, the students should not forget that the questions can be framed differently. 

The students are therefore advised to use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 as a way to conceptually understand the question and the solution and not try to practice rote learning.