Wiley s Problems in Mathematics for IIT JEE Main and Advanced Vol II 2 Maestro Series with Summarized Concepts

662 Pages·2019·136.84 MB·English

by  Wiley Expert Teachers

WILEY
PROBLEMS
MATHEMATICS
IN
JEE
FOR
with Summarized Concepts
VOLUME –II
WILEY
PROBLEMS
INMATHEMATICS
FO RJ EE
VOLUME – II with Summarized Concepts
Copyright © 2018 by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002.
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First Edition: 2018
ISBN: 978-81-265-7630-2
ISBN: 978-81-265-8690-5 (ebk)
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Printed at:
Note to the Student
Wiley Mathematics Problem Book is specifically designed to meet the needs of engineering (JEE) aspirants and give an edge to
their preparation. The book offers complete coverage of the mathematics curriculum (of Class 12 syllabus) for JEE. It is enriched with
unique elements and features that help students recapitulate the concepts, build problem-solving skills and apply them to solve all
question-types asked in the engineering entrance examinations. The book is a valuable resource for both JEE (Main) and JEE (Advanced)
aspirants. The chapter flow of the book is aligned with JEE Main syllabus and its coverage in the classroom. However, topics specific to
JEE (Advanced) and advanced level questions are also covered both as solved examples and practice exercises.
We will now walk you through the target examinations and some key features of the book that enhance the learning experience.
TARGET EXAMINATION
Admission to Undergraduate Engineering Programs at IITs, NITs and other Center and State (participating) funded Technical
Institutions use the Joint Entrance Examination Main (JEE Main) score as eligibility/merit criteria. The JEE (Main) is also an eligibility test
for the Joint Entrance Examination Advanced [JEE (Advanced)], which is mandatory for the candidate if he/she is aspiring for admission
to the undergraduate program offered by the IITs. The JEE (Advanced) scores are used as an eligibility criteria for admission into IITs.
An effective exam strategy for success in these examinations can be based on the detailed analysis of previous years question papers
and planning your preparation accordingly. The Mathematics Question Paper of these examinations is a judicious mix of easy, moderate
and tough questions. The analysis of question distribution over the units of mathematics syllabus for these examinations is given below.
EXAM ANALYSIS OF PAPERS
Mathematics question paper comes as an amalgamation of easy, moderate and tough questions. This section shows the unit-wise as
well as chapter-wise analysis of previous 9 years (2010-2018) JEE Main and JEE Advanced papers.
JEE Main
Year
Unit
2010 2011 2012 2013 2014 2015 2016 2017 2018
Algebra 14 13 13 12 12 11 12 13 12
Calculus 8 10 9 8 9 8 7 10 8
Trigonometry 2 1 1 3 2 3 3 2 3
Analytical Geometry 6 6 7 7 7 8 8 5 7
JEE Advanced
Year
Unit
2010 2011 2012 2013 2014 2015 2016 2017 2018
Algebra 16 17 12 14 12 6 12 10 8
Trigonometry 5 1 2 4 3 1 2 1 1
Analytical Geometry 13 8 9 10 7 3 9 7 9
Differential Calculus 2 7 6 2 11 5 7 8 12
Integral Calculus 8 7 10 7 5 4 5 7 4
Vector 3 3 2 3 2 1 1 3 2
Prelims_Volume II.indd 3 27-Jul-18 6:21:54 PM
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MATHE Chapter Complex Numbers and Quadratic Equati Permutations and Combinations Binomial Theorem Sequences and Series Statistics Mathematical Reasoning Matrices and Determinants Vector Algebra Probability Sets, Relations and Functions Limits, Continuity and Differentiability Application of Derivatives Integrals Application of Integrals Differential Equations Trignometric Functions Inverse Trignometric Functions Conic Sections Three-Dimensional Geometry
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Chapter Complex Numbers Quadratic Equations Permutations and Combinations Sequence and Series Binomial Theorem Logarithms Matrices and Determinants Probability Properties and Solution of Triangles (Heights and Distances) Trigonometric Equations Trigonometric Ratios and Identities Inverse Trigonometric Function (Principal Values Only) Rectangular Coordinate System Straight Lines and Pair of Lines Conics Circle Three-Dimensional Geometry Sets and Relations Limits Functions Continuity and Differentiability Differentiation
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Prelims_Volume II.indd 6 27-Jul-18 6:21:55 PM
FEATURES OF THE BOOK
A. Understand the Concepts 21.2 Tangent and Normal
1. All the concepts as per the JEE curriculum A tangent to a point is a line which touches the curve at that point.
A normal to a point is the line which is perpendicular to the tan-
are explained in simple steps to develop
gent at that point.
fundamental understanding of the subject. If the equation of a curve is y = f(x) and a point A(x, y) lies on it,
1 1
then the equation of the tangent at point A is
dy
y–y1=dx (x−x1)
A
and the equation of the normal at point A is
1
y - y =- (x - x)
1 (dy/dx) 1
A
Key Point:
When the curve is given in parametric form, that is, when x =
g(t) and y = h(t), the equation of tangent at the point t = t is
1
y−h(t)=h′(t1)[x−g(t)] 2. Important points to remember about concepts
1 g′(t) 1
1 highlighted as Key Points.
and the equation of normal is
g¢(t)
y-h(t)=- 1 [x-g(t)]
1 h¢(t) 1
1
B. Every Aspect of the Subject Covered
In form of formulas, figures, graphs and tables to enhance problem-solving skills.
 p p
(i) sin−1(sin θ) = θ, ∀q ∈− , 
 2 2
(ii) cos−1(cos θ) = θ, ∀ θ ∈ [0, π]
 p p
(iii) tan−1(tan θ) = θ, ∀q ∈− , 
 2 2
C
A B(iv) cot−1(cos θ) = θ, ∀ θ ∈ (0, π) Table 17.1 Domain and principal ranges of all the six inverse trig-
p onometric functions
(v) sec−1(sec θ) = θ, ∀q ∈[0,p]− 
2 Function Domain Principal Range
(vi) cosec−1(cosec θ) = θ, ∀q∈−p ,p− {0} (values of x) (values of y)
L N r M y  2 2 é p pù
y = sin−1 x [−1, 1] ê- , ú
ë 2 2û
y y = cos−1 x [−1, 1] [0, π]
æ p pö
y = tan−1 x (−∞, ∞) ç- , ÷
è 2 2ø
P(x, y)
O 1 1
Figure 21.4 ψ
ψ
x
O T M N
Figure 21.2
Features of the book.indd 3 26-Jul-18 12:06:35 PM
C. Reinforce Concepts
Illustration 21.1 Find the slope of tangent at the point that has
1. Illustrations pose a specific problem using the ordinate −3 on the curve x3 = 3y2.
concepts already presented and then work Solution: x,
through the solution. we get
3x2=3´æç2ydyö÷
è dxø
dy x2
⇒ =
dx 2y
Now, to obtain this value, we require abscissa as well. Substituting
y = −3 in the equation of curve, we have
901
Your Turn 1
1. Find the slopes of the curve y = (x + 2)(x − 3) at the points
where it meets x−axis. Ans. −5, 5 2. Your Turn within each chapter is present to
2. Find the points on the curve y = x3 − 2x2 + x − 2 when the gra- reinforce and check the understanding of the
dient is zero.
students.
1 50
Ans. (1, −2) and  ,− 
3 27
3. Find the equation of tangent and normal to the curve
x3 = y2
normal, subtangent and subnormal.
13 13 2 3
Ans. 3x − 2y − 1 = 0, 2x + 3y − 5 = 0, , , ,
3 2 3 2
Additional Solved Examples
3. Additional Solved Examples suitable for JEE 1. The number of real solutions of cos−1 x + cos−1 2x = −π is
exams are provided with in-depth solutions for (A) 0 (B) 1
the students to understand the logic behind (C) 2 (D)
Solution:
and formula used.
cos−1 x = −(π + cos−1 2x)
Range of cos−1 x ∈ [0, π]
Since cos−1 x has a range from [0, π], thus the sum of two cos−1
cannot be equal to −π a negative quantity.
Hence, the correct answer is option (A).
Features of the book.indd 4 26-Jul-18 12:06:36 PM
D. Understanding the Exam Pattern
Through Previous Years’ Solved JEE Main/AIEEE Questions and Previous Years’ Solved JEE Advanced/IIT-JEE Questions.
Previous Years' Solved JEE Advanced/
IIT-JEE Questions
1. Let (x, y) be such that Previous Years' Solved JEE Main/AIEEE
sin-1(ax)+cos-1(y)+cos-1(bxy)=p Questions
2
Match the statements in Column I with statements in Column I1I.. If sin-1æçxö÷+cosec-1æç5ö÷=p, then a value of x is
è5ø è4ø 2

Column I Column II
(A) 1 (B) 3
(A) If a = 1 and b = 0, then (x, y) (P) lies on the circle x2 + y2 = 1
(C) 4 (D) 5 [AIEEE 2007]
(B) If a = 1 and b = 1, then (x, y) (Q) lies on (x2 − 1)(y2 − 1) = 0
Solution: We have
(C) If a = 1 and b = x, y) (R) lies on y = x x 4 p x 4 x 3
sin-1 +sin-1 = Þsin-1 =cos-1 Þsin-1 =sin-1
(D) If a = 2 and b = 2, then (x, y) (S) lies on (4x2 − 1)(y2 − 1) = 0 5 5 2 5 5 5 5
Therefore, x = 3.
[IIT-JEE 2007]
Hence, the correct answer is option (B).
E. Practice to Complete Your Learning
Through Practice Exercise 1 (JEE Main) and Practice Exercise 2 (JEE Advanced). All questions types as per JEE Main and
Advanced covered.
Matrix Match Type Questions
Practice Exercise 1
24. Match the following:
1. The points on the curve y = 12x − x3 at which the gradient is List I List II
zero are
(A) Circular plate is expanded by the heat from (p) 2
(A) (0, 2), (2, 16) (B) (0, −2), (2, −16)
the radius 5 cm to 5.06 cm. Approximate
(C) (2, −16), (−2, 16) (D) (2, 16), (−2, −16) increase in the area is
2. The area of the triangle formed by the coordinate axes and a (B) If an edge of a cube increases by 1%, then the (q) 0.6 p
tangent to the curve xy=a2at the point (x,y)on it is percentage increase in the volume is
1 1
(A) a2x1 (B) a2y1 (C) 2a2 (D) 4a2 x2 (r) 3
y x (C) If the rate of decrease of y = − 2x + 5 is
1 1 2
3. The slope of tangent to the curve x=t2+3t−8, y=2t2 Ctwoicme pthree rhatee nofs dioecnre Taysep oef Qx, tuheens txi ois neqsual
-2t-5at the point (2, −1) is to (given that the rate of decrease is non−zero)
Paragraph for Questions 9−11: Let a(t) is a function of t such
(D) Rate of increase in the area of the equilateral
(A) 22/7 (B) 6/7 (C) − 6 (D) None of these da 3 3
tthriaatn gle o=f 2s fiodre a1ll5 t hcem v,a lwuehse no f eta acnhd said =e 0i sw h(esn) t = 0. Further
dt 4
increasing at the rate of 0.1 cm/sec; is
y = m(t) x + c(t) is the tangent to the curve y = x2 − 2ax + a2 + a at the
Practice Exercise 2 point whose abscissa is 0. Then (t) 4
9. If the ratIen otfe cghaenrg Tey opf tehe Q duisteasntcieo onf tshe vertex of y = x2 − 2ax
+ a2 + a from the origin with respect to t is k, then k =
Single/Multiple Correct Choice Type Questions x2 y2
(A) 2 29. Let( Bα) b 2e th2e angle in r(aCd)i ans2 b etween (D) +4 2 = 1 and the
1. For the curve represented parametically by the equations, 36 4
x = 2 ln cot t + 1 and y = tan t + cot t 10. If the rate ofc cirhcalen gx2e +o yf 2c =(t )1 w2 iatht trheesipr epcot intot st ,o wf thheen i nt t=e rks,e icst io(cid:31)n, . If a = tan−1
(A) tangent at t = p/4 is parallel to x-axis then k k2.
(B) normal at t = p/4 is parallel to y-axis (A) 16 − 2 22 3 (B) 8 2 + 2 2
(C) tangent at t = p/4 is parallel to the line y = x (C) 10320. + F 2in d the mi nimum v(aDlu) e1 6of (2x1 + − 2 x2)2 + 2−x12 − x92,
(D) tangent and normal intersect at the point (2, 1) where x ∈ (0, 2) and x ∈ R+.
1 2
Features of the book.indd 5 26-Jul-18 12:06:39 PM