Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

JEE Main 2021 August 26 Shift 1 Maths Question Paper with Solutions

JEE Main Question Paper solutions for August 26 Shift 1 Maths are available here. Candidates can find the JEE Main 2021 August 26 Shift 1 Maths Question Paper with Solutions to evaluate their performance. Students can directly download the PDF version of the question paper to check solutions offline. This set of solved questions given in JEE Main 2021 session 4 Maths question paper for August 26 will help you to formulate the correct answers. The JEE Main 2021 Shift 1 Maths question paper for August 26 given here will aid students to eventually be well prepared to face the upcoming exam.

JEE Main 2021 August 26th Shift 1 Maths Question Paper

Question 1. f(x) =

\(cos((2 tan^{-1}((sin((cot^{-1}(\sqrt{((1-x)/x) )} ) )\)
then;

  1. a. (1-x)2 f'(x) - 2(f(x)) 2 = 0
  2. b. (1-x) 2 f'(x) + 2(f(x)) 2= 0
  3. c. (1+x) 2 f'(x) - 2(f(x)) 2= 0
  4. d. (1+x) 2 f'(x)+2(f(x)) 2= 0

Solution:

  1. Answer: (b)

    f(x) =

    \(cos((2 tan^{-1}((sin((cot^{-1}(\sqrt{((1-x)/x) )} ) )\)
    then

    f(x) =

    \(cos((2 tan^{-1}((sin((sin^{-1}(\sqrt{x} ) ) ) \)

    = cos((2 tan-1(√x)

    = cos((cos-1 (((1 - x)/(1 + x)) ) = (1 - x)/(1 + x)

    f'(x) = -2/(1+x)2

    2(f(x))2 = 2((1-x)/(1+x))2

    (1 - x) 2 f'(x) + 2(f(x)) 2= 0


Question 2. How many 3 digit numbers by using the digits 0,1,3,4,6,7 can be possible if repetition are allowed?

    Solution:

    1. Answer: 180

      digits 0,1,3,4,6

      So, 5 x 6x 6 = 180


    Question 3. The sum of the square of the distance of a point from (0, 0), (0, 1), (1, 1), (1, 0) is 18 and its locus is a circle. If d is the diameter of the circle, then find d2.

      Solution:

      1. Answer: 16

        JEE Main 2021 August 26 Shift 1 Maths Paper Question 3 Solution

        (OP) 2 + (AP) 2 + (PC) 2 + (PB) 2=18

        => x2 + y2 + x2 + (y - 1) 2 + (x - 1) 2 + y2 + (x - 1) 2 + (y - 1) 2 =18

        => 4x2 + 4y2 - 4x - 4y=14

        => x2 + y2 - x - y=7/2

        => r=2

        => d = 4 => d2 = 16


      Question 4: If loge (x+y) = 4xy. Find (d2y)/(dx2) at x = 0.

        Solution:

        1. Answer: 40

          loge (x+y) = 4xy

          or 1/((x + y)) [1 + (dy/dx)] = 4[x (dy/dx) + y]

          or 1 + (dy/dx) = 4(x + y) [x (dy/dx) + y]⋯(i)

          If x = 0, then y = 1

          From (i)

          1 + dy/dx = 4(0 + 1)[0 + 1] = 4

          Or dy/dx = 3

          From (i), again differentiate w.r.t. x

          0 + (d2 y)/(dx2 ) = 4(x + y)[x (d2y)/(dx2 ) + 2 (dy/dx)] + 4[x (dy/dx) + y](1 + (dy/dx))

          At x = 0, y = 1, dy/dx = 3

          (d2 y)/(dx2) = 4(0 + 1)[0 + 2x3]+4[0 + 1](1 + 3) = 40

          Therefore, (d2 y)/(dx2 ) = 40


        Question 5: Find value of (5 - e2 ) for ellipse x2/8 + y2/4 = 1

          Solution:

          1. Answer: 9/2

            x2/8 + y2/4 = 1

            a2 = 8, b2 = 4

            e = c/a = √((a2 - b2)/a2 ) = √(1-4/8) = √(1/2)

            (5 - e2 ) = 5 - 1/2 = 9/2


          Question 6: In a hospital, 89% has disease A and 98% has disease B, K% is the number of patients who has both, then K can’t be the subset of ____.

            Solution:

            1. Answer: Least k =87, Max k = 89

              JEE Main 2021 August 26 Shift 1 Maths Paper Question 6

              Let total =100

              n(A)=89, n(B)=98

              max (n(A),n(B)) ≤ n(AUB) ≤ 100

              98 ≤ n(A)+n(B)-n(A n B) ≤ 100

              => 98 ≤ 98 + 89- k ≤ 100

              Least k =87

              Max k = 89


            Question 7: P(A) = p, P(B) = 2p, P(Exactly one out of A and B occurs) = 5/9. What will be the maximum value of p?

            JEE Main 2021 August 26 Shift 1 Maths Paper Question 7

              Solution:

              1. Answer: 5/9

                Given: P(A) = p and P(B) = 2p

                \(P(\bar{A} \cap B) + P(\bar{B} \cap A) = \frac{5}{9}\)

                P(A)-P(A∩B)+P(B)-P(A∩B) = 5/9

                3p-2P(A∩B) = 5/9

                0 ≤ P(A∩B) ≤ p

                For maximum value: 3p - 2p = 5/9

                So, p = 5/9


              Question 8: Evaluate

              \(\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}[(\frac{1-x}{1+x})^2 + (\frac{1+x}{1-x})^2 - 2]dx\)

                Solution:

                1. Answer: 4 ln ½

                  JEE Main 2021 August 26 Shift 1 Maths Paper Question 8


                Question 9: If cosx/(1+sinx) = |tanx|, then find the number of solutions in [0, 2π].

                  Solution:

                  1. Answer: x = π/6

                    Case 1: tan x ≥ 0

                    cosx/(1+sinx ) = sinx/cosx

                    ⇒cos2x = sin2x+sinx

                    ⇒1-sin2x = sin2x + sinx

                    ⇒2 sin2x + sin x – 1 = 0

                    ⇒sin x = -1 ,1/2

                    sin x = 1/2 ⇒ x = π/6

                    sin x = -1 (not possible)

                    Case 2: tan x < 0

                    (Cos x)/(1+sinx) = -sin x/cos x

                    ⇒cos2x = -sin2x - sinx

                    ⇒1-sin2x = -sin2x - sinx

                    Sin x = -1 (Not possible)

                    Only one solution exists x = π/6


                  Question 10: Sum of infinite terms A, AR, AR2, AR3,.... is 15. If the sum of the square of these terms is 150, then find the sum of AR2, AR4, AR6, ⋯.

                    Solution:

                    1. Answer: ½

                      A + AR + AR2 + ⋯ = 15

                      ⇒A/(1-R) = 15 ⋯(i)

                      Now, A2 + A2 R2 + A2 R4 + ⋯ = 150

                      ⇒A2/(1-R2 ) = 150

                      ⇒(A⋅A)/((1-R)(1 + R)) = 150

                      ⇒15A/(1 + R) = 150

                      ⇒A = 10(1 + R)⋯(ii)

                      Solving (i) & (ii):

                      10(1 + R)/(1-R) = 15

                      ⇒R = 1/5

                      So, A = 12

                      AR2 + AR4 + AR6 + ⋯ = (AR^2)/(1-R2 ) = (12⋅1/25)/(1-1/25) = 1/2


                    Question 11: If

                    \(arg(\frac{z-1}{z+1}) = \frac{\pi}{4}\)
                    , then find equation of circle.

                      Solution:

                      1. Answer: √2

                        JEE Main 2021 August 26 Shift 1 Maths Paper Question 11


                      Question 12: A wire of length 36 units is cut into two parts which are bent respectively to form a square of side x units and a circle of radius of r units. If the sum of areas of the square and the circle so formed is minimum, then find the circumference of the circle.

                        Solution:

                        1. Answer: 2π x 18/(4 + π) = 36π/(4 + π)

                          Sum of perimeter of square and circle = 36

                          4x + 2πr = 36⇒2x + πr = 18

                          x = (18 - πr)/2 = 9 - πr/2 .....(i)

                          Sum of areas: S = x2 + πr2 ....(ii)

                          S = πr2 + (9 - πr/2) 2 , diff. w.r.to r

                          dS/dr = 2πr + 2(9 - πr/2)( - π/2)

                          dS/dr = 0

                          ⇒r - 9/2 + πr/4

                          ⇒r(π + 4)/4 = 9/2

                          ⇒r = 18/(4 + π)

                          Circumference of circle = 2π x 18/(4 + π) = 36π/(4 + π)


                        Question 13: Solve

                        \(\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}\)

                          Solution:

                          1. Answer: [(1/2)] tan-14 - 0

                            \(\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{r=0}^{2n-1}\frac{1}{1+\frac{4r^2}{n^2}}=\int_0^2 \frac{1}{1+(2x)^2}dx\)

                            = [(1/2)] tan-14 - 0


                          Question 14: If

                          \(\vec{a} . \vec{c} =3 , \vec{a} = \hat{i}+ \hat{j}+ \hat{k}, \vec{b} =\hat{j} - \hat{k}, \vec{a} \times \vec{c} =\vec{b}\)
                          , Find
                          \([\vec{a} \ \; \vec{b} \;\; \vec{c}]\)

                            Solution:

                            1. Answer:

                              \([\vec{a} \ \; \vec{b} \;\; \vec{c}]\)
                              = -2

                              Given

                              \(\vec{a} . \vec{c} =3 , \vec{a} = \hat{i}+ \hat{j}+ \hat{k}, \vec{b} =\hat{j} - \hat{k}, \vec{a} \times \vec{c} =\vec{b}\)

                              This implies,

                              \((\vec{a} \times \vec{c})⋅.\vec{b} =\vec{b}.\vec{b}\)

                              or

                              \([\vec{a} \ \; \vec{b} \;\; \vec{c}]\)
                              = -2


                            Question 15: Evaluate

                            \( \sum_{r=0}^{20} r^2(^{20}C_r)\)

                              Solution:

                              1. Answer: 20 ⋅ 21 ⋅ 218

                                JEE Main Aug 26 Maths Solutions


                              Question 16: Solve (1 + y) tan2x + tan x (dy/dx) + y = 0

                                Solution:

                                1. Answer: y tan x = -tan x + x + c

                                  (1 + y) tan2x + tan x (dy/dx) + y = 0

                                  tan x(dy/dx) + (tan2x + 1)y = -tan2x

                                  dy/dx + (tan x + cot x )y = -tan x

                                  I. F. =

                                  \(e^{\int{tan x + cotx)dx }}\)

                                  = e ln sec x + ln sin x

                                  = e ln sec x . sin x

                                  = tan x

                                  • y tan x = -∫tan2 x dx
                                  • y tan x = -∫(sec2 x-1) dx
                                  • y tan x = -tan x + x + c

                                Question 17: If

                                \(y = \frac{1}{1+x}+\frac{2}{1+x^2}+\frac{2^2}{1+x^4 }+....+\frac{2^100}{1+x^200}\)
                                , then find y at x = 2.

                                  Solution:

                                  1. Answer:

                                    \(y = 1 + \frac{2^{101}}{1-2^{400}}\)

                                    JEE Main Aug 26 Maths Question Paper Solutions


                                  Video Solutions - JEE Main 2021 Question Papers

                                  JEE Main 2021 Maths Paper August 26th Shift 1

                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1
                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1
                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1
                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1
                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1
                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1
                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1
                                  JEE Main 2021 Maths Paper With Solutions July 20 Shift 1