JEE Advanced Question Paper 2021 Maths Paper 2

JEE Advanced 2021 Maths question paper for Paper 2 is available on this page. Students who are preparing for the entrance exam can refer to the questions and solutions given here to learn the answers in detail, get a better knowledge of the formulas used in a particular question and understand the proper logic behind solving each question. In essence, students will have a more effective exam preparation by solving the questions themselves. At the same time, they will be getting an idea of the important topics from an examination point of view.
Students can instantly download the solved JEE Advanced 2021 Maths question paper for Paper 2 in PDF format for offline practice. We have also provided analysis videos of both Paper 1 and Paper 2 for the exam that was held on October 3rd. Students will further get helpful insights about the paper pattern, question types, weightage of marks and more. More significantly, engaging in this study exercise will keep every JEE aspirant in good stead as the exam approaches nearer and nearer.

JEE Advanced 2021 Paper 2 Maths Question Paper

Question 1: Let;

S1 = {(i, j, k) : i, j, k ∈ {1,2,...,10}}

S2 = {(i, j) : 1 ≤ i < j + 2 ≤ 10,i, j ∈ {1,2,...,10}}

S3 = {(i, j, k, l) : 1 ≤ i < j < k < l, i, j, k, l ∈ {1,2,...,10}}

S4 = {(i, j,k,l ) : i, j, k and l are distinct elements in {1,2,...,10}}.

If the total number of elements in the set Sr is nr, r = 1, 2,3,4, then which of the following statements is (are) TRUE?

  1. a. n1 = 1000
  2. b. n2 = 44
  3. c. n3 = 220
  4. d. n4/12 = 420

Solution:

  1. Answer: (a, b, d)

    Number of elements in S1 = 10 × 10 × 10 = 1000

    Number of elements in S2 = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 44

    Number of elements in S3 = 10C4 = 210

    Number of elements in S4 = 10P4 = 210 × 4! = 5040


Question 2: Consider a triangle PQR having sides of lengths p, q, and r opposite to the angles P, Q, and R, respectively. Then which of the following statements is (are) TRUE?

  1. a. cos P ≥ 1 - p2/2qr
  2. b. cos R ≥ ((q-r)/(p+q))cos P + ((p-r)/(p+q))cos Q
  3. c. (q+r)/p < 2√(sin Q sin R)/sin P
  4. d. if p<q and p<r, then cos Q > p/r and cos R > p/q

Solution:

  1. Answer: (a, b)

    JEE Advanced Question Paper 2021 Maths Paper 2

    (a) cos P = (q2+r2 - p2)/2qr

    And (q2+r2)/2 ≥ √(q2. r2) (AM ≥ GM)

    ⇒ (q2+r2) ≥ 2qr

    So cos P ≥ (2qr - p2)/2qr

    cos P ≥ 1 - p2/2qr

    (b) ((q-r) cos P + (p-r) cos Q)/(p+q) = ((q cos P + p cos Q) - r(cos P + cos Q))/(p+q)

    = r(1- cos P - cos Q)/(p+q)

    = (r(q-p cos R) - (p-q cos R))/(p+q)

    = ((r-p-q)+(p+q)cos R)/(p+q)

    = cos R + (r-q-p)/(p+q) ≤ cos R (since r < p+q)

    (c) (q+r)/p = (sin Q + sin R)/sin P ≥ 2√(sin Q sin R)/sin P

    (d) If p < q and q < r

    So, p is the smallest side, therefore one of Q or R can be obtuse

    So, one of cos Q or cos R can be negative

    Therefore, cos Q > p/r and cos R > p/q cannot hold always.


Question 3: Let f: [-π/2, π/2] → R be a continuous function such that f(0) = 1 and ∫0π/3 f(t) dt = 0. Then which of the following statements is (are) TRUE?

  1. a. The equation f(x) - 3 cos 3x = 0 has at least one solution in (0, π/3)
  2. b. The equation f(x) - 3 sin 3x = -6/π has at least one solution in (0, π/3)
  3. c. limx0x0xf(t)dt1ex2=1\lim_{x\to 0}\frac{x\int_{0}^{x}f(t)dt}{1-e^{x^{2}}}=-1
  4. d. limx0sinx0xf(t)dtx2=1\lim_{x\to 0}\frac{\sin x\int_{0}^{x}f(t)dt}{{x^{2}}}=-1

Solution:

  1. Answer: (a, b, c)

    f(0) = 1, ∫0π/3 f(t) dt = 0

    (a) Consider a function g(x) = ∫0x f(t)dt - sin 3x. g(x) is continuous and differentiable function

    And g(0) = 0

    g(π/3) = 0

    By Rolle's theorem g’(x) = 0 has at least one solution in (0, π/3)

    f(x) - 3 cos 3x = 0 for some x ∈ (0, π/3)

    (b) Consider a function

    h(x) = ∫0x f(t)dt + cos 3x + 6x/π

    h(x) is continuous and differentiable function and h(0) = 1

    h(π/3) = 1

    By Rolle's theorem h’(x) = 0 for at least one x ∈ (0, π/3)

    f(x) - 3 sin 3x + 6/π = 0 for some x ∈ (0, π/3)

    (c) limx0x0xf(t)dt1ex2\lim_{x\to 0}\frac{x\int_{0}^{x}f(t)dt}{1-e^{x^{2}}}

    (0/0 form)

    By L’Hospital rule

    = limx0xf(x)+0xf(t)dt2xex2\lim_{x\to 0}\frac{xf(x)+\int_{0}^{x}f(t)dt}{-2xe^{x^{2}}} , (0/0 form)

    = limx0xf(x)+f(x)+f(x)4x2ex22ex2\lim_{x\to 0}\frac{xf'(x)+f(x)+f(x)}{-4x^{2}e^{x^{2}}-2e^{x^{2}}}

    = (0+2f(0))/(0-2)

    = -1

    (d) limx0sinx0xf(t)dtx2\lim_{x\to 0}\frac{\sin x\int_{0}^{x}f(t)dt}{{x^{2}}} , (0/0 form)

    = limx0sinx.f(x)+cosx0xf(t)dt2x\lim_{x\to 0}\frac{\sin x\: .f(x)+\cos x\int_{0}^{x}f(t)dt}{2x}

    = limx0cosx.f(x)+sinx.f(x)+cosx.f(x)sinx.0xf(t)dt2\lim_{x\to 0}\frac{\cos x\: .f(x)+\sin x.f'(x)+\cos x.f(x)-\sin x.\int_{0}^{x}f(t)dt}{2}

    = (1+0+1-0)/2

    = 1


Question 4: For any real numbers α and β, let yα,β(x), x ∈ R, be the solution of the differential equation dy/dx + αy = xeβx, y(1) = 1. Let S = {yα,β(x), α, β ∈ R } . Then which of the following functions belong(s) to the set S?

  1. a. f(x) = (x2/2)e-x + (e - ½)e-x
  2. b. f(x) = (-x2/2)e-x + (e + ½)e-x
  3. c. f(x) = (ex/2)(x-½) + (e - e2/4)e-x
  4. d. f(x) = (ex/2)(½ -x) + (e + e2/4)e-x

Solution:

  1. Answer: (a, c)

    dy/dx + αy = xeβx

    Integrating factor (I.F) = e∫αdx = eαx

    So, the solution is y.eαx = ∫xeβx eαx dx

    y.eαx = ∫xe(β+α)x dx

    If α + β ≠ 0

    yeαx = x e(α+β)x/(α+β) - e(α+β)x/(α+β)2 + C

    y = [eβx/(α+β)][x - 1/(α+β)] + Ce-αx …(i)

    Put α = β = 1 in (i)

    y = (ex/2)(x - ½) + Ce-x

    y(1) = 1

    1 = (e/2)(½) +C/e

    ⇒ C = e - e2/4

    So, y = (ex/2)(x-½) + (e - e2/4)e-x

    If α + β = 0 and α = 1

    dy/dx + y = xe-x

    I.F = ex

    yex = ∫x dx

    yex = x2/2 + C

    y = e-xx2/2 + Ce-x

    y(1) = 1

    1 = 1/2e + C/e

    ⇒ C = e - ½

    y = e-xx2/2 + (e - ½)e-x


Question 5: Let O be the origin and OA=2i^+2j^+k^\overrightarrow{OA}=2\hat{i}+2\hat{j}+\hat{k}, OB=i^2j^+2k^\overrightarrow{OB}=\hat{i}-2\hat{j}+2\hat{k} and OC=12(OBλOA)\overrightarrow{OC}=\frac{1}{2}(\overrightarrow{OB}-\lambda \overrightarrow{OA}) for some λ> 0. If OB×OC=92\left | \overrightarrow{OB}\times \overrightarrow{OC}\right |=\frac{9}{2}, then which of the following statements is (are) TRUE ?

  1. a. Projection of OC\overrightarrow{OC} on OA\overrightarrow{OA} is -3/2
  2. b. Area of the triangle OAB is 9/2
  3. c. Area of the triangle ABC is 9/2
  4. d. The acute angle between the diagonals of the parallelogram with adjacent sides OA\overrightarrow{OA} and OC\overrightarrow{OC} is π/3

Solution:

  1. Answer: (a, b, c)

    OA=2i^+2j^+k^\overrightarrow{OA}=2\hat{i}+2\hat{j}+\hat{k}

    OB=i^2j^+2k^\overrightarrow{OB}=\hat{i}-2\hat{j}+2\hat{k}

    OC=12(OBλOA)\overrightarrow{OC}=\frac{1}{2}(\overrightarrow{OB}-\lambda \overrightarrow{OA})

    OB×OC=OB×12(OBλOA)\overrightarrow{OB}\times \overrightarrow{OC}=\overrightarrow{OB}\times \frac{1}{2}(\overrightarrow{OB}-\lambda \overrightarrow{OA})

    = λ2OB×OA=λ2(OA×OB)\frac{-\lambda }{2}\overrightarrow{OB}\times \overrightarrow{OA}=\frac{\lambda }{2}(\overrightarrow{OA}\times \overrightarrow{OB})

    Now, OA×OB=i^j^k^221122=6i^3j^6k^\overrightarrow{OA}\times \overrightarrow{OB}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2& 2 &1 \\ 1& -2&2 \end{vmatrix}=6\hat{i}-3\hat{j}-6\hat{k}

    So, OB×OC=3λ2(2i^j^2k^)\overrightarrow{OB}\times \overrightarrow{OC}=\frac{3\lambda }{2}(2\hat{i}-\hat{j}-2\hat{k})\\

    OB×OC=9λ2=92\left | \overrightarrow{OB}\times \overrightarrow{OC} \right |=\left | \frac{9\lambda }{2} \right |=\frac{9}{2}

    So, λ = 1 (since λ >0)

    OC=12(OBOA)\overrightarrow{OC}=\frac{1}{2}(\overrightarrow{OB}-\overrightarrow{OA})\\

    OC=12(i^4j^+k^)\overrightarrow{OC}=\frac{1}{2}(-\hat{i}-4\hat{j}+\hat{k})\\

    (a) Projection of vector OC on vector OA = OC.OAOA\frac{\overrightarrow{OC}.\overrightarrow{OA}}{\left | \overrightarrow{OA} \right |}

    = ½ (-2-8+1)/3

    = -3/2

    (b) Area of triangle OAB = 12OA×OB\frac{1}{2}\left | \overrightarrow{OA}\times \overrightarrow{OB} \right | = 9/2

    (c) Area of the triangle ABC is = 12AB×AC=12i^j^k^14152412\frac{1}{2}\left | \overrightarrow{AB}\times \overrightarrow{AC} \right |=\frac{1}{2}\left |\begin{vmatrix} \hat{i} & \hat{j}&\hat{k} \\ -1& -4 & 1\\ -\frac{5}{2}& -4 &- \frac{1}{2} \end{vmatrix} \right |

    = 126i^3j^6k^\frac{1}{2}\left | 6\hat{i}-3 \hat{j}-6\hat{k}\right |

    = 9/2

    (d) Acute angle between the diagonals of the parallelogram with adjacent sides

    OA\overrightarrow{OA} and OC\overrightarrow{OC} = θ

    (OA+OC).(OAOC)OA+OCOAOC=cosθ\frac{(\overrightarrow{OA}+\overrightarrow{OC}).(\overrightarrow{OA}-\overrightarrow{OC})}{\left | \overrightarrow{OA}+\overrightarrow{OC}\right |\left | \overrightarrow{OA}-\overrightarrow{OC}\right |}=\cos \theta

    cos θ = (32i^+32k^).(52i^+4j^+12k^)322×904\frac{(\frac{3}{2}\hat{i}+\frac{3}{2}\hat{k}).(\frac{5}{2}\hat{i}+4\hat{j}+\frac{1}{2}\hat{k})}{\frac{3}{2}\sqrt{2}\times \sqrt{\frac{90}{4}}}

    = 18/3√2 ×√90

    θ ≠ π/3


Question 6: Let E denote the parabola y2 = 8x. Let P = (-2, 4), and let Q and Q’ be two distinct points on E such that the lines PQ and PQ’ are tangents to E. Let F be the focus of E. Then which of the following statements is (are) TRUE?

  1. a. The triangle PFQ is a right-angled triangle
  2. b. The triangle QPQ’ is a right-angled triangle
  3. c. The distance between P and F is 5√2
  4. d. F lies on the line joining Q and Q’

Solution:

  1. Answer: (a, b, d)

    E : y2 = 8x

    P : (-2, 4)

    JEE Advanced 2021 Question Paper Maths Paper 2

    Point P (-2, 4) lies on directrix (x = -2) of parabola y2 = 8x

    So, ∠QPQ’ = π/2 and chord QQ’ is a focal chord and segment PQ subtends a right angle at the focus.

    Slope of QQ’ = 2/(t1+t2) = 1

    Slope of PF = -1

    PF = 4√2


Question Stem for Question Nos. 7 and 8


Consider the region R = {(x,y)∈ R×R : x ≥ 0 and y2 ≤ 4 - x. Let F be the family of all circles that are contained in R and have centres on the x-axis. Let C be the circle that has the largest radius among the circles in F. Let (α, β) be a point where circle C meets the curve y2 = 4 - x.

Question 7: The radius of the circle C is

    Solution:

    1. Answer: (1.50)


    Question 8: The value of α is

      Solution:

      1. Answer: (2.00)

        Sol: For comprehension Question 7 and Question 8

        JEE Advanced Question Paper 2021 Maths Paper 2 Solution

        Let the circle be,

        (x - a)2 + y2 = r2

        Solving it with parabola

        y2 = 4 - x we get

        (x - a)2 + 4 – x = r2

        x2 – x(2a + 1) + (a2 + 4 - r2) = 0 ...(1)

        D = 0

        ⇒ 4r2 + 4a - 15 = 0

        Clearly a ≥ r

        So 4r2 + 4r - 15 ≤ 0

        ⇒ rmax = 3/2 = a

        Radius of circle C is 3/2

        From (1) x2 - 4x + 4 = 0

        ⇒ x = 2 = α


      Question Stem for Question Nos. 9 and 10


      Let f1 : (0, ∞) → R and f2 : (0, ∞) → R be defined by f1(x) = 0xj=121(tj)jdt\int_{0}^{x}\prod_{j=1}^{21}(t-j)^{j}dt, x>0 and f2(x) = 98(x - 1)50 - 600(x - 1)49 + 2450, x > 0, where, for any positive integer n and real number a1, a2,...an. i=1nai\prod_{i=1}^{n}a_{i} denotes the product of a1, a2,..an. Let mi and ni, respectively, denote the number of points of local minima and the number of points of local maxima of function fi, i = 1, 2, in the interval (0, ∞).

      Question 9: The value of 2m1 + 3n1 + m1n1 is

        Solution:

        1. Answer: (57.00)


        Question 10: The value of 6m2 + 4n2 + 8m2n2 is

          Solution:

          1. Answer: (06.00)

            Solution for Question 9 and 10

            f1(x)=j=121(xj)jf_{1}'(x)=\prod_{j=1}^{21}(x-j)^{j}\\

            f1’(x) = (x - 1)(x - 2)2 (x - 3)3 ,..., (x - 20)20 (x - 21)21

            Checking the sign scheme of f1’(x) at x = 1, 2, 3, ..., 21, we get

            f1(x) has local minima at x = 1, 5, 9, 13, 17, 21 and local maxima at x = 3, 7, 11, 15, 19

            ⇒ m1 = 6, n1 = 5

            f2(x) = 98(x - 1)50 - 600(x - 1)49 + 2450

            f2’(x) = 98 × 50(x - 1)49- 600 × 49 × (x - 1)48

            = 98 × 50 × (x - 1)48 (x - 7)

            f2(x) has local minimum at x = 7 and no local maximum.

            ⇒ m2 = 1, n2 = 0

            2m1 + 3n1 + m1n1

            = 2 × 6 + 3 × 5 + 6 × 5

            = 57

            6m2 + 4n2 + 8m2n2

            = 6 × 1 + 4 × 0 + 8 × 1 × 0

            = 6


          Question Stem for Question Nos. 11 and 12


          Let gi = [π/8, 3π/8] → R, i = 1, 2 and f: [π/8, 3π/8] → R be the functions such that g1(x) = 1, g2(x) = |4x - π| and f(x) = sin2x, for all x∈[π/8, 3π/8]. Define Si=π83π8f(x).gi(x)dxS_{i}=\int_{\frac{\pi }{8}}^{\frac{3\pi }{8}}f(x).g_{i}(x)dx, i = 1,2.

          Question 11: The value of 16S1/π is

            Solution:

            1. Answer: (2.00)

              S1 = ∫π/83π/8 sin2x . 1 dx

              = ½ ∫π/83π/8(1 - cos 2x)dx

              = ½ (x - sin 2x/x)π/83π/8

              = ½ (π/4 - 0)

              = π/8

              => 16S1/π = 2


            Question 12:The value of 48S22 is

              Solution:

              1. Answer: (1.50)

                S2= ∫π/83π/8 sin2x |4x - π| dx

                = ∫π/83π/8 4 sin2x |x - π/4| dx

                Let x - π/4 = t

                => dx = dt

                S2 = ∫-π/8π/8 4 sin2 (π/4 + t)|t| dt

                = ∫-π/8π/8 2(1 - cos 2(π/4 + t) |t| dt

                = ∫-π/8π/8 (2 + 2 sin 2t) |t| dt

                = 2∫-π/8π/8|t|dt + 2∫-π/8π/8|t| sin 2t dt

                = 4∫0π/8t dt + 0

                S2 = [2t2]0π/8

                = π2/32

                48S22 = 3/2


              Question Paragraph: Let M = {(x, y) ∈ R × R : x2 + y2≤ r2}, where r > 0. Consider the geometric progression an = 1/2n-1, n = 1,2, 3… Let S0 = 0 and, for n ≥ 1, let Sn denote the sum of the first n terms of this progression. For n ≥ 1 , let Cn denote the circle with center (Sn–1, 0) and radius an, and Dn denote the circle with center (Sn–1, Sn–1) and radius an.


              Question 13: Consider M with r = 1025/513. Let k be the number of all those circles Cn that are inside M. Let l be the maximum possible number of circles among these k circles such that no two circles intersect. Then,

              1. a. k + 2l = 22
              2. b. 2k + l = 26
              3. c. 2k + 3l = 34
              4. d. 3k + 2l = 40

              Solution:

              1. Answer: (d)

                an = 1/2n-1

                And Sn = 2(1 - 1/2n)

                For circles Cn to be inside M.

                Sn-1 + an < 1025/513

                ⇒ Sn < 1025/513

                ⇒ 1 - 1/2n < 1025/1026

                ⇒ 1 - 1/1026

                ⇒ 2n < 1026

                ⇒ n ≤ 10

                ∴ Number of circles inside be 10 = K

                Clearly, alternate circles do not intersect each other i.e., C1, C3, C5, C7, C9 do not intersect each other as well as C2, C4, C6, C8 and C10 do not intersect each other hence maximum of 5 set of circles do not intersect each other.

                ∴ l = 5

                ∴ 3K + 2l = 40

                ∴ Option (D) is correct


              Question 14: Consider M with r = (2199-1)√2/2198. The number of all those circles Dn that are inside M is;

              1. a. 198
              2. b. 199
              3. c. 200
              4. d. 201

              Solution:

              1. Answer: (B)

                Since r = (2199-1)√2/2198

                Now, √2Sn-1 + an < (2199-1)√2/2198

                2√2(1 - 1/2n-1) + 1/2n-1 < (2199-1)/2198

                ∴ 2√2 - √2/2n-2 + 1/2n-1 < 2√2 - √2/2198

                (1/2n-2)(½ - √2) < -√2/2198

                (2√2-1)/2. 2n-2 > √2/2198

                2n-2 < (2 - 1/√2) 2197

                n ≤ 199

                ∴ Number of circles = 199

                Option (B) is correct.


              Question Paragraph: Let ψ1 = [0, ∞) → R, ψ2 = [0, ∞) → R, f:[0, ∞) → R and g:[0, ∞) → R be functions such that f(0) = g(0) = 0,


              ψ1(x) = e-x + x, x≥0,

              ψ2(x) = x2 - 2x - 2e-x + 2, x≥0

              f(x)=xx(tt2)et2dt,x>0f(x)=\int_{-x}^{x}(\left | t \right |-t^{2})e^{-t^{2}}dt, x> 0

              And g(x)=0x2tetdt,x>0g(x)=\int_{0}^{x^{2}}\sqrt{t}e^{-t}dt,x> 0

              Question 15: Which of the following statements is TRUE?

              1. a. f(√ln 3) + g(√ln 3) = ⅓
              2. b. For every x > 1, there exists an α ∈ (1, x) such that ψ1(x) = 1 + α x
              3. c. For every x > 0, there exists a β ∈ (0, x) such that ψ2(x) = 2x (ψ1(β) -1)
              4. d. f is an increasing function on the interval [0, 3/2]

              Solution:

              1. Answer: (c)

                Since, g(x)=0x2tetdt,x>0g(x)=\int_{0}^{x^{2}}\sqrt{t}e^{-t}dt,x> 0

                Let t = u2

                ⇒ dt = 2u du

                So g(x) = 0xueu2.2udu\int_{0}^{x}ue^{-u^{2}}.2u\: du\\

                = 20xt2et2dt2\int_{0}^{x}t^{2}e^{-t^{2}}dt\\ …(i)

                And f(x)=xx(tt2)et2dt,x>0f(x)=\int_{-x}^{x}(\left | t \right |-t^{2})e^{-t^{2}}dt, x> 0

                Therefore f(x)=20x(tt2)et2dtf(x)=2\int_{0}^{x}(t-t^{2})e^{-t^{2}}dt …(ii)

                From equation (i) + (ii) : f(x) + g(x) = 0x2tet2dt\int_{0}^{x}2te^{-t^{2}}dt

                Let t2 = P

                ⇒ 2t dt = dP

                f(x) + g(x) = 0x2ePdP=[eP]0x2\int_{0}^{x^{2}}e^{-P}dP= \left [ -e^{-P} \right ]_{0}^{x^{2}}

                f(x) + g(x) = 1ex21-e^{-x^{2}} …(iii)

                ∴ f(√ln 3) + g(√ln 3) = 1-e-ln 3

                = 1-⅓

                = ⅔

                ∴ Option (a) is incorrect.

                From equation (ii) : f’(x) = f(x)=2(xx2)ex2=2x(1x)ex2f'(x)=2(x-x^{2})e^{-x^{2}}=2x(1-x)e^{-x^{2}}

                Since f(x) is increasing in (0, 1)

                ∴ Option (d) is incorrect

                ψ1(x) = e-x + x

                ⇒ ψ1’(x) = 1- e-x <1 for x>1

                Then for α ∈ (1, x), ψ1(x) = 1+ αx does not true for α > 1.

                ∴ Option (b) is incorrect

                Now ψ2(x) = x2 - 2x - 2e-x + 2

                ψ2’(x) = 2x - 2 + 2e-x

                ∴ ψ2’(x) = 2ψ1(x)-2

                From LMVT

                2(x) - ψ2(0)]/(x-0) = ψ2’(β) for β ∈(∞, x)

                => ψ2(x) = 2x(ψ1(β) - 1)

                Option (c) is correct.


              Question 16: Which of the following statements is TRUE?

              1. a. ψ1(x) ≤ 1, for all x>0
              2. b. ψ2(x) ≤ 0, for all x>0
              3. c. f(x)1ex223x3+25x5f(x)\geq 1-e^{-x^{2}}-\frac{2}{3}x^{3}+\frac{2}{5}x^{5}, for all x∈(0, ½)
              4. d. g(x)≤ (2/3) x3 - (⅖)x5 + (1/7)x7, for all x∈(0, ½)

              Solution:

              1. Answer: (d)

                Since ψ1(x) = e-x+x

                And for all x>0, ψ1(x) > 1

                ∴ (a) is not correct

                ψ2(x) = x2 + 2 - 2(e-x + x) >0 for x > 0

                ∴ (b) is not correct.

                Now, √te-t = √t(1 - t/1! + t2/2! - t3/3! + ….∞)

                And √te-t ≤ t1/2 - t3/2 + ½ t5/2

                0x2tetdt0x2(t12t32+12t52)dt\int_{0}^{x^{2}}\sqrt{t}e^{-t}dt\leq \int_{0}^{x^{2}}(t^{\frac{1}{2}}-t^{\frac{3}{2}}+\frac{1}{2}t^{\frac{5}{2}})dt

                = (⅔) x3 - (⅔) x5 + (1/7) + (1/7) x7

                ∴ Option (d) is correct

                And f(x)=xx(tt2)et2dtf(x)=\int_{-x}^{x}(\left | t \right |-t^{2})e^{-t^{2}}dt

                = 20x(tt2)et2dt2\int_{0}^{x}(t-t^{2})e^{-t^{2}}dt

                = 20x2tet2dt20xt2et2dt2\int_{0}^{x}2te^{-t^{2}}dt-2\int_{0}^{x}t^{2}e^{-t^{2}}dt\\

                = 1ex220xt2et2dt1-e^{-x^{2}}-2\int_{0}^{x}t^{2}e^{-t^{2}}dt

                Therefore f(x)1ex220xt2(1t2)dtf(x)\leq 1-e^{-x^{2}}-2\int_{0}^{x}t^{2}(1-t^{2})dt

                = 1ex22x33+25x51-e^{-x^{2}}-2\frac{x^{3}}{3}+\frac{2}{5}x^{5} for all x∈(0, ½)

                ∴ Option (c) is incorrect.


              Question 17: A number is chosen at random from the set {1, 2, 3......, 2000}. Let p be the probability that the number is a multiple of 3 or a multiple of 7. Then the value of 500p is;

                Solution:

                1. Answer: (214)

                  E = a number which is multiple of 3 or multiple of 7

                  n(E) = (3, 6, 9, ........, 1998) + (7, 14, 21, ..........., 1995) - (21, 42, 63, ......... 1995)

                  n(E) = 666+ 285 - 95

                  n(E) = 856

                  n(E) = 2000

                  P(E) = 856/2000

                  P(E) × 500 = 856/4 = 214


                Question 18: Let E be the ellipse x2/16 + y2/9 = 1. For any three distinct points P, Q and Q’ on E, let M (P, Q) be the mid-point of the line segment joining P and Q, and M(P, Q’) be the mid-point of the line segment joining P and Q’. Then the maximum possible value of the distance between M (P, Q) and M(P, Q’), as P, Q and Q’ vary on E, is

                  Solution:

                  1. Answer: (4)

                    Let P(α), Q(θ), Q’(θ’)

                    M = ½ (4 cos α + 4 cos θ), ½ (3 sin α + 3 sin θ)

                    M’ = ½ (4 cos α + 4 cos θ’), ½ (3 sin α + 3 sin θ’)

                    MM’ = ½ √((4 cos θ - 4 cos θ’)2 + (3 sin θ - 3 sin θ’)2)

                    MM’ = ½ distance between Q and Q’

                    MM’ is not depending on P

                    Maximum of QQ’ is possible when QQ’ = major axis

                    QQ’ = 2(4) = 8

                    MM’ = ½ (QQ’)

                    MM’ = 4


                  Question 19: For any real number x, let [x] denote the largest integer less than or equal to x. If l=010[10xx+1]dxl=\int_{0}^{10}\left [ \sqrt{\frac{10x}{x+1}} \right ]dx , then the value of 9I is;

                    Solution:

                    1. Answer: (182.00)

                      l=010[10xx+1]dxl=\int_{0}^{10}\left [ \sqrt{\frac{10x}{x+1}} \right ]dx

                      y = 10x/(x+1), 0 ≤ x≤10

                      xy+y = 10x

                      x = y/(10-y)

                      0≤ y/(10-y) ≤ 10

                      y/(10-y) ≥ 0 and (y/(10-y)) - 10 ≤ 0

                      y/(y-10) ≤ 0 and (11y-100)/(y-10) ≥ 0

                      JEE Advanced Question Paper 2021 Maths Solution Paper 2

                      y ∈ [0, 10) and y ∈ (-∞, 100/11] ⋃ (10, ∞)

                      y ∈ [0, 100/11]

                      √y ∈ [0, 10/√11]

                      ⇒ [√y] = {0, 1, 2, 3}

                      Case 1:

                      0≤10x/(x+1) < 1

                      10x/(x+1) ≥ 0 and 10x/(x+1) - 1 < 0

                      JEE 2021 Advanced Question Paper Maths Solution 2

                      x∈ (-∞, -1) ⋃ [0, ∞) and x∈ (-1, 1/9)

                      x∈ [0, 1/9) then [√(10x/(x+1))] = 0

                      Case 2:

                      1 ≤ 10x/(x+1) < 4

                      10x/(x+1) - 1 ≥ 0 and 10x/(x+1) - 4 < 0

                      (9x-1)/(x+1) ≥ 0 and (6x-4)/(x+1) < 0

                      Maths JEE Advanced 2021 Solution Paper 2

                      x∈ (-∞, -1)⋃ [1/9, ∞) and x∈ (-1, ⅔)

                      x∈[1/9, ⅔), [√(10x/(x+1))] = 1

                      Case 3:

                      4 ≤ (10x/x+1) < 9

                      10x/(x+1) - 4 ≥ 0 and 10x/(x+1) < 9

                      (6x-4)/(x+1) ≥ 0 and (x-9)/(x+1) < 0

                      Maths JEE Advanced 2021 Paper 2 Solutions

                      x∈ (-∞, -1)⋃ [2/3, ∞) and x∈ (-1, 9)

                      x∈ [⅔, 9); [√(10x/(x+1))] = 2

                      Case 4:

                      x∈ [9, 10]

                      ⇒ [√(10x/(x+1))] = 3

                      l = 0190.dx+19231.dx+2392.dx+9103.dx\int_{0}^{\frac{1}{9}}0.dx +\int_{\frac{1}{9}}^{\frac{2}{3}}1.dx+\int_{\frac{2}{3}}^{9}2.dx+\int_{9}^{10}3.dx

                      l = (⅔ - 1/9) +2(9-⅔) + 3(10-9)

                      l = 5/9 + 50/3 + 3

                      9l = 182


                    JEE Advanced 2021 Maths Paper 2 Solutions

                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions
                    JEE Advanced 2021 Maths Paper 2 Question and Solutions