• Motivational
  • Board Exams
  • Exam Prep Tips
  • Science & Tech
Monday, January 30, 2023
  • Login
Aakash BYJU'S Blog
  • Aakash Home
  • MEDICAL
  • ENGINEERING
  • FOUNDATIONS
  • TOPPERS SPEAK
  • Exam
    • NEET
      • NEET 2023 Eligibility Criteria
      • NEET 2023 Dates
      • NEET 2023 Exam Pattern
      • NEET 2023 Syllabus
      • NEET 2023 Application
      • NEET 2023 Admit Card
      • NEET UG 2023 Result
      • NEET 2023 Cut Off
      • NEET 2023 FAQ
    • JEE Main
      • JEE Main 2023 Eligibility Criteria
      • JEE Main 2023 Dates
      • JEE Main 2023 Exam Pattern
      • JEE Main 2023 Syllabus
      • JEE Main 2023 Application
      • JEE Main 2023 Admit Card
      • JEE Main 2023 Counselling
    • JEE Advanced
      • JEE Advanced 2023 Eligibility Criteria
      • JEE Advanced 2023 Dates
      • JEE Advanced 2023 Application
      • JEE Advanced 2023 Syllabus
      • JEE Advanced 2023 Maths Syllabus
      • JEE Advanced 2023 Physics Syllabus
      • JEE Advanced 2023 Chemistry Syllabus
  • NCERT Solutions
  • NEET PG
    • INI CET
No Result
View All Result
  • Aakash Home
  • MEDICAL
  • ENGINEERING
  • FOUNDATIONS
  • TOPPERS SPEAK
  • Exam
    • NEET
      • NEET 2023 Eligibility Criteria
      • NEET 2023 Dates
      • NEET 2023 Exam Pattern
      • NEET 2023 Syllabus
      • NEET 2023 Application
      • NEET 2023 Admit Card
      • NEET UG 2023 Result
      • NEET 2023 Cut Off
      • NEET 2023 FAQ
    • JEE Main
      • JEE Main 2023 Eligibility Criteria
      • JEE Main 2023 Dates
      • JEE Main 2023 Exam Pattern
      • JEE Main 2023 Syllabus
      • JEE Main 2023 Application
      • JEE Main 2023 Admit Card
      • JEE Main 2023 Counselling
    • JEE Advanced
      • JEE Advanced 2023 Eligibility Criteria
      • JEE Advanced 2023 Dates
      • JEE Advanced 2023 Application
      • JEE Advanced 2023 Syllabus
      • JEE Advanced 2023 Maths Syllabus
      • JEE Advanced 2023 Physics Syllabus
      • JEE Advanced 2023 Chemistry Syllabus
  • NCERT Solutions
  • NEET PG
    • INI CET
No Result
View All Result
Aakash Institute Logo
No Result
View All Result

Home » Engineering Exam Prep » Algebra: Tough questions with concept revision for JEE Main 2022 Maths

    Talk to our expert



    Resend OTP

    By submitting up, I agree to receive all the Whatsapp communication on my registered number and Aakash terms and conditions and privacy policy

    Algebra: Tough questions with concept revision for JEE Main 2022 Maths

    Tough questions with concept revision from the topic algebra while preparing for JEE Main 2022 maths section

    by Team @Aakash
    Apr 27, 2022, 10:30 AM IST
    in Engineering Exam Prep
    0
    Algebra for JEE Mains: Tough questions from Algebra with concept revision for JEE Main 2022 Maths

    Algebra: Tough questions with concept revision for JEE Main 2022 Math

    0
    SHARES
    656
    VIEWS
    Share on FacebookShare on Twitter

    Mathematics is one of the most interesting subjects of all time. For a few students, algebra is one of the trickiest chapters for JEE. However, it’s easy to learn and understand. You need to make sure you practice a lot of problems and keep the standard tricks right up your sleeves.

    We are here to help you with the top tough questions with the solutions to help you brush up on your concepts. Make sure to practise them thoroughly and regularly to score more in the upcoming JEE Mains 2022 exams. 

    Question 1

    The value of a for which one root of the quadratic equation (a2 – 5a + 3) x2 + (3a – 1) x + 2 = 0 is twice as large as other, is 

    (A) –2/3 

    (B) ⅓

    (C) –1/3 

    (D) 2/3 

    Answer:

    (D) Let  α and 2 be the roots of the given equation , 

    then (a^2 – 5a + 3)α^2  + (3a – 1)α   + 2 = 0 … (i) 

    and (a^2 – 5a + 3) (4 α ^2 ) + (3a – 1) (2(A) + 2 = 0 … (ii) 

    Solving (i) and (ii), we get α  = -3/(3a-1) 

    put this value in (ii), we get a = 2/3 

    Question 2

    Let b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively are given by _________.

    Answer:

    Let r = λb + μc and c = ± (xi + yj).

    Since c and b are perpendicular, we have 4x + 3y = 0

    ⇒ c = ±x (i − 43j), {Because, y = [−4 / 3]x}

    Now, projection of r on b = [r . b] / [|b|] = 1

    ⇒ [(λb + μc) . b] / [|b|]

    = [λb . B] / [|b|] = 1

    ⇒ λ = 1 / 5

    Again, projection of r on c = [r . c] / [|c|] = 2

    This gives μx = [6 / 5]

    ⇒ r = [1 / 5] (4i + 3j) + [6 / 5] (i − [4 / 3]j)

    = 2i−j or

    r = [1 / 5] (4i + 3j) − [6 / 5] (i − [4 / 3]j)

    = [−2 / 5] i + [11 / 5] j

    Question 3

    A unit vector makes an angle π / 4 with a z-axis. If a + i + j is a unit vector, then a is equal to _________.

    Answer:

    Let a = li + mj + nk, where l2 + m2 + n2 = 1. a makes an angle π / 4 with z−axis.

    Hence, n = 1 / √2, l2 + m2 = 1 / 2 …..(i)

    Therefore, a = li + mj + k / √2

    a + i + j = (l + 1) i + (m + 1) j + k / √2

    Its magnitude is 1, hence (l + 1)2 + (m + 1)2 = 1 / 2 …..(ii)

    From (i) and (ii),

    2lm = 1 / 2

    ⇒ l = m = −1 / 2

    Hence, a = [−i / 2] − [j / 2] + [k / √2].

    Question 4

    Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation p × {(x − q) × p} + q × {(x − r) × q} + r × {(x − p) × r} = 0, then x is given by ____________.

    Answer:

    |p| = |q| = |r| = c, (say) and

    p . q = 0 = p . r = q . r

    p × |( x − q) × p |+ q × |(x − r) × q| + r × |( x − p) × r| = 0

    ⇒ (p . p) (x − q) − {p . (x − q)} p + . . . . . . . . . = 0

    ⇒ c2 (x − q + x − r + x − p) − (p . x) p − (q . x) q − (r . x) r = 0

    ⇒ c2 {3x − (p + q + r)} − [(p . x) p + (q . x) q + (r . x) r] = 0

    which is satisfied by x = [1 / 2] (p + q+ r).

    Question 5

    If a,b,c are real and x3 − 3b2x + 2c3 is divisible by x − a and x − b, then what is a?

    Answer:

    As f(x) = x3 − 3b2x + 2c3 is divisible by x − a and x − b,

    Therefore, f(a) = 0 ⇒ a3 − 3b2a + 2c3 = 0 …..(i) and

    f(b)=0

    b3 − 3b3 + 2c3 = 0 …..(ii)

    From (ii), b = c

    From (i), a3 − 3ab2 +2b3 = 0 (Putting b=c) ⇒ (a−b) (a2 + ab − 2b2) = 0 ⇒ a = b or a2 + ab = 2b

    Thus a = b = c or a2 + ab = 2b2 and b = ca2 + ab = 2b2 is satisfied by a = −2b.

    But b = c , a2 + ab − 2b2 and b = c is equivalent to a = −2b = −2c

    Question 6

    Find the number of real values of x for which the equality ∣3x2 + 12x + 6∣ = 5x + 16 holds good.

    Answer:

    Equation is |3x2 + 12x + 6|= 5x + 16 …….(i)

    when 3x2 + 12x + 6 ≥ 0

    ⇔ x2+ 4x ≥ −2

    ⇔ |x + 2|2 ≥ 4 − 2

    ⇔ |x + 2| ≥ (√2)2

    ⇔ x + 2 ≤ √−2 or x + 2 ≥ √2 …….(ii)

    Then (i) becomes 3x2 + 12x + 6 = 5x + 16

    ⇔3x2 + 7x −10 = 0

    ⇒x = 1, −10 / 3

    But x = −10 / 3 does not satisfy (ii).

    When 3x2 + 12x + 6 < 0 ⇒ x2+ 4x < −2

    |x + 2| ≤ √2 ⇒ [−√2 −2] ≤ x ≤ [−2 + √2] ? …….(iii)

    Then (i) becomes ⇒ 3×2 + 12x + 6 = −(5 x + 16)

    ⇒3x2 + 17x + 22 = 0

    ⇒x =−2,−11 / 3

    But x = −11 / 3 does not satisfy (iii).

    So, 1 and – 2 are the only solutions.

    Question 7

    What is the set of all real numbers x for which x2−|x + 2|+ x > 0?

    Answer:

    Case I:

    When x + 2 ≥ 0 i.e. x ≥ −2, then given inequality becomes x2−|x + 2|+ x > 0

    x2−2 > 0 ⇒|x| > √2

    x<√−2 or x > √2

    As x ≥ −2, therefore, in this case the part of the solution set is [−2 ,√−2) ∪ (√2 , ∞).

    Case II: When x + 2 ≤ 0 i.e. x ≤ −2, then given inequality becomes x2 + (x + 2) + x > 0

    ⇒ x2 + 2x + 2 > 0 ⇒ (x + 1)2 + 1 > 0, which is true for all real x.

    Hence, the part of the solution set in this case is (−∞, −2].

    Combining the two cases, the solution set is (−∞, −2) ∪ ([−2, √−2] ∪ (√2, ∞) = (−∞, √−2) ∪ (√2, ∞).

    Question 8

    If a, b are the roots of x2 + px + 1 = 0 and c, d are the roots of x2 + qx + 1 = 0, then the value of (a – (C) (b – (C) (a + (D) (b + (D) is 

    (A) p 2 – q 2 

    (B) q 2 – p 2 

    (C) q 2 + p2 

    (D) none of these 

    Answer:

    (B) 

    Given, x^2+ px + 1 = (x – (A) (x – (B) => a + b = -p 

    and ab = 1 and x^2 + qx + 1 = (x – (C) (x – (D) => c + d = -q and cd = 1  

    Therefore, (a – (C) (b – (C) (a + (D) (b + (D) = (c – (A) (c – (B) (-d – (A) (-d – (B) 

    = (c^2 + pc + 1) (d2 – pd + 1) 

    But c^2 + qc + 1 = 0 and d^2 + qd + 1 = 0. 

    Hence,

    (a – (C) (b – (C) (a + (D) (b + (D) = (-qc + p(C) (-qd – p(D) = cd (q – p) (q + p) = q^2 – p^2

    Question 9

    If the coefficient of x7 in (ax2 + [1/bx])11 is equal to the coefficient of x−7 in (ax − 1/bx2)11, then ab = __________.

    Answer:

    In the expansion of (ax2 + [1/bx])11, the general term is

    Tr + 1 = 11Cr (ax2)11 − r (1/bx)r

    = 11Cr * a11 − r * [ 1/br ] * x22 − 3r

    For x7, we must have 22 – 3r = 7

    r = 5, and the coefficient of x7 = 11C5 * a11−5 * [1/b5] = 11C5 * a6 * b5

    Similarly, in the expansion of (ax−1bx2)11, the general term is

    Tr + 1 = 11Cr * (−1)r * [a11 − r/br ] * x11 − 3r

    For x-7 we must have, 11 – 3r = -7 , r = 6, and the coefficient of x−7 is 11C6 * [a5/b6] = 11C5 * a5 * b6.

    As given, 11C5 [a6/b5] = 11C5 * [a5/b6]

    ⇒ ab = 1

    Question 10

    Let R = (5√5 + 11)2n + 1 and f = R − [R], where [.] denotes the greatest integer function. The value of R * f is

    Answer:

    Since (5√5 − 11) (5√5 + 11) = 4

    5√5 − 11 = 4 / (5√5 + 11), Because 0 < 5√5 − 11 < 1 ⇒ 0 < (5√5 − 11)2n + 1 < 1, for positive integer n.

    Again, (5√5 + 11)2n + 1 − (5√5 − 11)2n + 1 = 2 {2n+1C1 (5√5)2n * 11 + 2n + 1C3(5√5)2n − 2 × 113 + . . .. +

    2n + 1C2n+1 112n+1}

    = 2 {2n+1C1(125)n * 11 + 2n+1C3(125)n−1 113 +. . . . . . + 2n+1C2n+1 112n+1}

    = 2k, (for some positive integer k)

    Let f′= (5√5 − 11)2n + 1 , then [R] + f − f′=2k

    f − f′ = 2k − [R]

    ⇒ f − f′ is an integer.

    But, 0 ≤ f < 1; 0 < f′<1

    ⇒ −1 < f − f′ < 1

    f − f′ = 0 (integer)

    f = f′

    Therefore, Rf = Rf′ = (5√5 + 11)2n + 1 * (5√5 − 11)2n + 1

    = ([5√5]2 + 112)2n + 1

    = 42n+1

    Question 11

    If the coefficients of pth, (p + 1)th and (p + 2)th terms in the expansion of (1 + x )n are in A.P., then find the equation in terms of n.

    Answer:

    Coefficient of pth, (p + 1)th and (p + 2)th terms in expansion of (1 + x )n are nCp−1, nCp ,nCp+1. Then 2nCp = nCp−1 + nCp+1

    ⇒n2 − n (4p + 1) + 4p2 − 2 = 0

    Trick: Let p = 1, hence nC0, nC1 and nC2 are in A.P.

    ⇒ 2 * nC1 = nC0 + nC2

    ⇒ 2n = 1 + [n (n − 1)] / [2]

    ⇒ 4n = 2 + n2 − n

    ⇒ n2 − 5n + 2 = 0

    Question 12

    If f (x) = cos (log x), then find the value of f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)].

    Answer:

    f (x) = cos (log x)

    Now let y = f (x) * f (4) − [1 / 2] * [f (x / 4) + f (4x)]

    y = cos (log x) * cos (log 4) − [1 / 2] * [cos log (x / 4) + cos (log 4x)]

    y = cos (log x) cos (log 4) − [1 / 2] * [cos (log x −log 4) + cos (log x + log 4)]

    y = cos (log x) cos (log 4) − [1 / 2] * [2 cos (log x) cos (log 4)]

    y = 0

    Question 13

     If log10 5 = a and log10 3 = b then

    (a) log30 8 = 3(1-a)/(b+1)

    (b) log 40 15 = (a+b)/(3-2a)

    (c) log 243 32 = (1-a)/b

    (d) none of these

    Solution:

    Given log10 5 = a and log10 3 = b

    We check all the given options.

    log30 8 = log 23/log (3×10)

    = 3 log 2/( log 3 + log 10)

    = 3 log (10/5)/(1 + log 3)

    = 3 (1 – log 5)/(1 + log 3)

    = 3(1 – a)/(1+b)

    Hence option a is correct.

    log 40 15 = log 15/log 40

    = log(3×5)/log(10×4)

    = (log 3 + log 5)/(1 + log 22)

    = (log 3 + log 5)/(1 +2 log 2)

    = (log 3 + log 5)/(1 +2 log (10/5))

    = (log 3 + log 5)/(1 +2 (1 – log 5))

    = (log 3 + log 5)/(1 +2 – 2 log 5)

    = (b+a)/(3 – 2a)

    Hence option b is correct.

    log 32-243 = log 32/log 243

    = log 25/log 35

    = 5 log 2/5 log 3

    = log 2/log 3

    = (1 – log 5)/log 3 (since log 2 = log (10/5) = log 10 – log 5 = 1 – log 5)

    = (1-a)/b

    So option c is correct.

    Hence option a, b and c are correct.

    These above-mentioned questions are from important chapters in algebra which includes Vector Algebra, Quadratic Equations, Logarithm, Binomial Theorem and Complex Numbers.

    This subject requires you to be quick with solving the problems and this can be achieved if you practice more and more problems. Because there is a limited number of formulas in algebra, a pro tip will be to make a note of all the important formulas in one place. Revise this sheet on a daily basis. Keep updating your list with new formulas you come across. Algebra tests how fluent you are with the application of the concepts. All the best.

    FAQs

    1. Whether NTA has changed the syllabus of JEE Main 2022?

    No, JEE Main 2022 has the same syllabus as last year. However, NTA has decided that there will be 90 Questions in the Question Paper, and candidates will be required to attempt 75 questions only. There will be no negative marking for 15 optional questions.

    .JEE Mains 2022 Syllabus has been released by National Testing Agency (NTA) on the official website. To know the details, visit jeemain.nta.nic.in.

    2. What is a good score in JEE Mains 2022?

    A good JEE Main score is something that will make you eligible to appear for the JEE Advanced exam. A good score in JEE Main can be 250+. This also depends on the college where you are seeking admission. With a good JEE Main 2022 score, you can be admitted among India’s highly ranked engineering universities, such as IITs, NITs, IIITs, GFTIs, etc.

    With approximately 22 Lakhs applicants and many competitors, you need to buckle up. Tips from experts will help you reach your target score. Also, take a look at the best way on how to attempt JEE Main 2022

    3. How to attempt Maths questions In JEE Main 2022?

    Maths is considered the most challenging and important section of the JEE 2022 exam. Candidates need to put in extra effort in the Maths section. This will help them enhance their overall scorecard. You can also check for the Maths syllabus at JEE Main 2022 Maths Syllabus.

    Tips:

    • Plan your schedule on how to manage time between sections in your exam.
    • Practice solving questions quickly and within your decided timeframe.
    • The mathematics paper will be quite lengthy, but there will be around eight questions that you can attempt easily. So, your prime focus must be on those questions. Try to take about 15-20 minutes for these eight questions. Then, you can move to the moderate questions.

    4. What will be the advantages of Multiple Sessions in JEE Main 2022?

    Multiple Sessions allow the candidates to improve their scores in the examination, and this also gives the candidates a chance to make multiple attempts in a year. In their first attempt, the candidates will get a firsthand experience of taking the JEE examination and know where they are lacking and how much more practice they need.

    This will also reduce the chances of wasting a year, and droppers would not have to wait a full year to reappear again.

    Tags: jee mainJEE Main 2022JEE Main 2022 Math
    Previous Post

    NEET Exam 2022 Updates: Aspirants To Get Extra 20 Minutes Time To Attempt Questions

    Next Post

    CBSE Class 10 Term 2 English Paper 2022: Exam Paper Analysis, Marking Scheme & Question Paper PDF

    Next Post
    Class 10 English exam analysis

    CBSE Class 10 Term 2 English Paper 2022: Exam Paper Analysis, Marking Scheme & Question Paper PDF

    Leave a Reply Cancel reply

    Your email address will not be published. Required fields are marked *

    Contact form

      Talk to our expert



      Resend OTP

      By submitting up, I agree to receive all the Whatsapp communication on my registered number and Aakash terms and conditions and privacy policy

      Recommended

      JEE Main 2023 Chemistry January 29 Shift 2 Question Paper and Solutions

      JEE Main 2023 Chemistry January 29 Shift 2 Question Paper and Solutions

      Jan 30, 2023, 2:39 AM IST
      JEE Main 2023 Maths January 29 Shift 2 Question Paper and Solutions

      JEE Main 2023 Maths January 29 Shift 2 Question Paper and Solutions

      Jan 30, 2023, 2:38 AM IST

      Trending

      JEE Main 2022 Marks vs Rank vs Percentile

      JEE Main Marks vs Percentile vs Ranks

      Jan 27, 2023, 10:30 AM IST
      JEE Main 2023 Question Paper 29 January Shift-1

      JEE Main 2023 January 29 – Shift 1 Question Paper with Solutions

      Jan 30, 2023, 2:09 AM IST

      Popular

      ANTHE'19 DPT

      I registered for ANTHE 2019. How to access the Daily Practice Tests (DPT)?

      Jun 2, 2020, 12:11 PM IST
      This National Level Scholarship Exam offering upto 90% Scholarship for Students in VIII, IX & X Grades

      This National Level Scholarship Exam offering upto 90% Scholarship for Students in VIII, IX & X Grades

      Sep 2, 2022, 6:14 PM IST
      Aakash National Talent Hunt Exam 2018 – A Perfect Start

      Aakash National Talent Hunt Exam 2018 – A Perfect Start

      Jun 2, 2020, 1:10 PM IST
      COVID-19

      The Unsung Heroes of the COVID-19 Pandemic

      Jun 25, 2020, 1:02 PM IST

      Popular Web Stories

    • Documents Required for Medical Admission

    • Dilute vs Concentrated Solution

    • ANTHE Scholarship 2022

    • Destruction of Colloids

    • ANTHE Books

    • Top Medical Colleges in UP

    • Non Biodegradable Items

    • Sum of Circumference of Circles

    • Dithecous Anther

    • Isomers with Molecular Formula c5h12
    • Recent Posts

      • JEE Main 2023 Chemistry January 29 Shift 2 Question Paper and Solutions
      • JEE Main 2023 Maths January 29 Shift 2 Question Paper and Solutions
      • JEE Main 2023 January 29 Shift 2 Physics Question Paper and Solutions
      • JEE Main 2023 January 29 Shift 2 Question Paper and Solutions
      • JEE Main 2023 January 29 Chemistry Shift 1 Question Paper with Solutions

      Follow Us

      • NCERT Solutions for Class 6
      • NCERT Solutions for Class 7
      • NCERT Solutions for Class 8
      • NCERT Solutions for Class 9
      • NCERT Solutions for Class 10
      • NCERT Solutions for Class 11
      • NCERT Solutions for Class 12
      • NCERT Solutions
      • Other Text Book Solutions
      • Important Concepts
      • Ask And Answer
      • Aakash Answers

      Copyright © Aakash Institute

      No Result
      View All Result
      • Aakash Home
      • MEDICAL
      • ENGINEERING
      • FOUNDATIONS
      • TOPPERS SPEAK
      • Exam
        • NEET
          • NEET 2023 Eligibility Criteria
          • NEET 2023 Dates
          • NEET 2023 Exam Pattern
          • NEET 2023 Syllabus
          • NEET 2023 Application
          • NEET 2023 Admit Card
          • NEET UG 2023 Result
          • NEET 2023 Cut Off
          • NEET 2023 FAQ
        • JEE Main
          • JEE Main 2023 Eligibility Criteria
          • JEE Main 2023 Dates
          • JEE Main 2023 Exam Pattern
          • JEE Main 2023 Syllabus
          • JEE Main 2023 Application
          • JEE Main 2023 Admit Card
          • JEE Main 2023 Counselling
        • JEE Advanced
          • JEE Advanced 2023 Eligibility Criteria
          • JEE Advanced 2023 Dates
          • JEE Advanced 2023 Application
          • JEE Advanced 2023 Syllabus
          • JEE Advanced 2023 Maths Syllabus
          • JEE Advanced 2023 Physics Syllabus
          • JEE Advanced 2023 Chemistry Syllabus
      • NCERT Solutions
      • NEET PG
        • INI CET

      Copyright © Aakash Institute

      Welcome Back!

      Login to your account below

      Forgotten Password?

      Retrieve your password

      Please enter your username or email address to reset your password.

      Log In