JEE Advanced 2021 · previous year paper

JEE Advanced 2021 — Paper 2

50 questions with the verified answer key. Open any question for its step-by-step solution.

  1. Question 1Physics· Rotational Dynamics

    One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q , at a height L above the hinge at point O . A block of weight αW\alpha \mathrm{W} is attached at the point P of the beam, as shown in the figure (not to scale). The rope can sustain a maximum tension of (22)W(2 \sqrt{2}) \mathrm{W}. Which of the following statement(s) is(are) correct?

    Question 1 figure
    1. Option A:

      The vertical component of reaction force at O does not depend on α\alpha

    2. Option B:

      The horizontal component of reaction force at O is equal to W for α=0.5\alpha=0.5

    3. Option C:

      The tension in the rope is 2 W for α=0.5\alpha=0.5

    4. Option D:

      The rope breaks if α>1.5\alpha>1.5

  2. Question 2Physics· Sound Waves

    A source, approaching with speed uu towards the open end of a stationary pipe of length LL, is emitting a sound of frequency fsf_{s}.

    The farther end of the pipe is closed. The speed of sound in air is vv and f0f_{0} is the fundamental frequency of the pipe.

    For which of the following combination(s) of uu and fsf_{s}, will the sound reaching the pipe lead to a resonance?

    1. Option A:

      u=0.8vu=0.8 v and fs=f0f_{s}=f_{0}

    2. Option B:

      u=0.8v\mathrm{u}=0.8 \mathrm{v} and fs=2f0\mathrm{f}_{\mathrm{s}}=2 \mathrm{f}_{0}

    3. Option C:

      u=0.8vu=0.8 v and fs=0.5f0f_{s}=0.5 f_{0}

    4. Option D:

      u=0.5vu=0.5 v and fs=1.5f0f_{s}=1.5 f_{0}

  3. Question 3Physics· Geometrical Optics

    For a prism of prism angle θ=60∘\theta=60^{\circ}, the refractive indices of the left half and the right half are, respectively, n1\mathrm{n}_{1} and n2(n2≥n1)\mathrm{n}_{2}\left(\mathrm{n}_{2} \geq \mathrm{n}_{1}\right) as shown in the figure. The angle of incidence ii is chosen such that the incident light rays will have minimum deviation if n1=n2=n=1.5n_{1}=n_{2}=n=1.5. For the case of unequal refractive indices, n1=nn_{1}=n and n2=n_{2}= n+Δn\mathrm{n}+\Delta \mathrm{n} (where Δn≪n\Delta \mathrm{n} \ll \mathrm{n} ), the angle of emergence e=i+Δee=i+\Delta e. Which of the following statement(s) is(are) correct?

    Question 3 figure
    1. Option A:

      The value of Δe\Delta e (in radians) is greater than that of Δn\Delta \mathrm{n}

    2. Option B:

      Δe\Delta e is proportional to Δn\Delta \mathrm{n}

    3. Option C:

      Δe\Delta e lies between 2.0 and 3.0 milliradians, if Δn=2.8×10−3\Delta \mathrm{n}=2.8 \times 10^{-3}

    4. Option D:

      Δe\Delta e lies between 1.0 and 1.6 milliradians, if Δn=2.8×10−3\Delta \mathrm{n}=2.8 \times 10^{-3}

  4. Question 4Physics· Units, Dimensions & Error Analysis

    A physical quantity S⃗\vec{S} is defined as S⃗=(E⃗×B⃗)/μ0\vec{S}=(\vec{E} \times \vec{B}) / \mu_{0}, where E⃗\vec{E} is electric field, B⃗\vec{B} is magnetic field and

    μ0\mu_{0} is the permeability of free space. The dimensions of S⃗\vec{S} are the same as the dimensions of which of the following quantity(ies)?

    1. Option A:

       Energy  Charge × Current \frac{\text { Energy }}{\text { Charge } \times \text { Current }}

    2. Option B:

       Force  Lenght × Time \frac{\text { Force }}{\text { Lenght } \times \text { Time }}

    3. Option C:

       Energy  Volume \frac{\text { Energy }}{\text { Volume }}

    4. Option D:

       Power  Area \frac{\text { Power }}{\text { Area }}

  5. Question 5Physics· Nuclear Physics

    A heavy nucleus NN, at rest, undergoes fission N→P+QN \rightarrow P+Q, where PP and Q are two lighter nuclei.

    Let δ=MN\delta=\mathrm{M}_{\mathrm{N}} −MP−MQ-M_{P}-M_{Q}, where MP,MQM_{P}, M_{Q} and MNM_{N} are the masses of P,QP, Q and NN, respectively.

    EPE_{P} and EQE_{Q} are the kinetic energies of PP and QQ, respectively. The speeds of PP and QQ are vPv_{P} and vQv_{Q}, respectively.

    If cc is the speed of light, which of the following statement(s) is(are) correct ?

    1. Option A:

      EP+EQ=c2δE_{P}+E_{Q}=c^{2} \delta

    2. Option B:

      EP=(MPMP+MQ)c2δ\mathrm{E}_{\mathrm{P}}=\left(\frac{\mathrm{M}_{\mathrm{P}}}{\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}}\right) \mathrm{c}^{2} \delta

    3. Option C:

      vPvQ=MQMP\frac{\mathrm{v}_{\mathrm{P}}}{\mathrm{v}_{\mathrm{Q}}}=\frac{\mathrm{M}_{\mathrm{Q}}}{\mathrm{M}_{\mathrm{P}}}

    4. Option D:

      The magnitude of momentum for P as well as Q is c2μδ\mathrm{c} \sqrt{2 \mu \delta}, where μ=MPMQ(MP+MQ)\mu=\frac{\mathrm{M}_{\mathrm{P}} \mathrm{M}_{\mathrm{Q}}}{\left(\mathrm{M}_{\mathrm{P}}+\mathrm{M}_{\mathrm{Q}}\right)}

  6. Question 6Physics· Moving Charges and Magnetic Field

    Two concentric circular loops, one of radius R and the other of radius 2 R , lie in the xy-plane with the origin as their common centre, as shown in the figure.

    The smaller loop carries current I1\mathrm{I}_{1} in the anti-clockwise direction and the larger loop carries current I2I_{2} in the clock wise direction, with

    I2>2I1I_{2}>2 I_{1}. B⃗(x,y)\vec{B}(x, y) denotes the magnetic field at a point ( x,y\mathrm{x}, \mathrm{y} ) in the xy-plane.

    Which of the following statement(s) is(are) correct?

    Question 6 figure
    1. Option A:

      B⃗(x,y)\vec{B}(x, y) is perpendicular to the xy-plane at any point in the plane

    2. Option B:

      ∣B⃗(x,y)∣|\vec{B}(x, y)| depends on xx and yy only through the radial distance r=x2+y2r=\sqrt{x^{2}+y^{2}}

    3. Option C:

      ∣B⃗(x,y)∣|\vec{B}(x, y)| is non-zero at all points for r<Rr<R

    4. Option D:

      B→(x,y)\overrightarrow{\mathrm{B}}(\mathrm{x}, \mathrm{y}) points normally outward from the xy-plane for all the points between the two loops

  7. Question 7Physics· Current Electricity

    In order to measure the internal resistance r1r_{1} of a cell of emf EE, a meter bridge of wire resistance R0=50ΩR_{0}=50 \Omega, a resistance

    R0/2R_{0} / 2, another cell of emf E/2E / 2 (internal resistance rr ) and a galvanometer GG are used in a circuit, as shown in the figure.

    If the null point is found at ℓ=72 cm\ell=72 \mathrm{~cm}, then the value of r1=\mathrm{r}_{1}= \qquad Ω\Omega.

    Question 7 figure
  8. Question 8Physics· Gravitation

    The distance between two stars of masses 3MS3 \mathrm{M}_{\mathrm{S}} and 6MS6 \mathrm{M}_{\mathrm{S}} is 9R9 R. Here RR is the mean distance between the centers of the Earth and the Sun, and MS\mathrm{M}_{\mathrm{S}} is the mass of the Sun. The two stars orbit around their common centre of mass in circular orbits with period nT , where T is the period of Earth's revolution around the Sun. The value of nn is \qquad

  9. Question 9Physics· Atomic Physics

    In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P,Q\mathrm{P}, \mathrm{Q} and R are

    EP,EQE_{P}, E_{Q} and ERE_{R}, respectively, and they are related by EP=2EQ=2ERE_{P}=2 E_{Q}=2 E_{R}. In this experiment, the same source of monochromatic light is used for metal P and Q while a different source of monochromatic light is used for the metal R.The work functions for metals P,Q\mathrm{P}, \mathrm{Q} and R are 4.0eV,4.5eV4.0 \mathrm{eV}, 4.5 \mathrm{eV} and 5.5 eV ,

    respectively. The energy of the incident photon used for metal RR, in eV , is \qquad _.

  10. Question 10Chemistry· Alkyl and Aryl Halides

    Correct option(s) for the following sequence of reactions is (are)

    Question 10 figure
    1. Option A:

      Q=KNO2, W=LiA1H4\mathrm{Q}=\mathrm{KNO}_{2}, \mathrm{~W}=\mathrm{LiA1H}_{4}

    2. Option B:

      R=\mathrm{R}= benzenamine, V=KCN\mathrm{V}=\mathrm{KCN}

    3. Option C:

      Q=AgNO2,R=\mathrm{Q}=\mathrm{AgNO}_{2}, \mathrm{R}= phenylmethanamine

    4. Option D:

      W=LiAlH4, V=AgCN\mathrm{W}=\mathrm{LiAlH}_{4}, \mathrm{~V}=\mathrm{AgCN}

  11. Question 11Chemistry· Chemical Kinetics

    For the following reaction 2X+Y→kP2 \mathrm{X}+\mathrm{Y} \xrightarrow{\mathrm{k}} \mathrm{P} the rate of reaction is d[P]dt=k[X]\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{X}]. Two moles of X\mathbf{X} are mixed with one mole of Y\mathbf{Y} to make 1.0 L of solution. At 50 s,0.550 \mathrm{~s}, 0.5 mole of Y\mathbf{Y} is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: ln⁡2=0.693\ln 2=0.693 )

    1. Option A:

      The rate constant, k , of the reaction is 13.86×10−4 s−113.86 \times 10^{-4} \mathrm{~s}^{-1}

    2. Option B:

      Half-life of X is 50 s

    3. Option C:

      At 50 s,−d[X]dt=13.86×10−3 mol L−1 s−150 \mathrm{~s},-\frac{\mathrm{d}[\mathrm{X}]}{\mathrm{dt}}=13.86 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}.

    4. Option D:

      At 100 s,−d[Y]dt=3.46×10−3 mol L−1 s−1100 \mathrm{~s},-\frac{\mathrm{d}[\mathrm{Y}]}{\mathrm{dt}}=3.46 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}.

  12. Question 12Chemistry· Electrochemistry

    Some standard electrode potentials at 298 K are given below:

    Pb2+/Pb−0.13 VNi2+/Ni−0.24 VCd2+/Cd−0.40 VFe2+/Fe−0.44 V\begin{aligned} & \mathrm{Pb}^{2+} / \mathrm{Pb}-0.13 \mathrm{~V} \\& \mathrm{Ni}^{2+/} \mathrm{Ni}-0.24 \mathrm{~V} \\& \mathrm{Cd}^{2+} / \mathrm{Cd}-0.40 \mathrm{~V} \\& \mathrm{Fe}^{2+} / \mathrm{Fe}-0.44 \mathrm{~V} \end{aligned}

    To a solution containing 0.001 M of X2+\mathbf{X}^{2+} and 0.1 M of Y2+\mathbf{Y}^{2+}, the metal rods X\mathbf{X} and Y\mathbf{Y} are inserted (at 298 K ) and connected by a conducting wire. This resulted in dissolution of X\mathbf{X}. The correct combination(s) of X\mathbf{X} and Y\mathbf{Y}, respectively, is(are) (Given: Gas constant, R=8.314 J K−1 mol−1\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, Faraday constant, F=96500Cmol−1\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1} )

    1. Option A:

      Cd and Ni

    2. Option B:

      Cd and Fe

    3. Option C:

      Ni and Pb

    4. Option D:

      Ni and Fe

  13. Question 13Chemistry· Coordination Compounds

    The pair(s) of complexes wherein both exhibit tetrahedral geometry is(are) (Note: py = pyridine Given: Atomic numbers of Fe,Co,Ni\mathrm{Fe}, \mathrm{Co}, \mathrm{Ni} and Cu are 26, 27, 28 and 29, respectively)

    1. Option A:

      [FeCl4]−\left[\mathrm{FeCl}_{4}\right]^{-}and [Fe(CO)4]2−\left[\mathrm{Fe}(\mathrm{CO})_{4}\right]^{2-}

    2. Option B:

      [Co(CO)4]−\left[\mathrm{Co}(\mathrm{CO})_{4}\right]^{-}and [CoCl4]2−\left[\mathrm{CoCl}_{4}\right]^{2-}

    3. Option C:

      [Ni(CO)4]\left[\mathrm{Ni}(\mathrm{CO})_{4}\right] and [Ni(CN)4]2−\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}

    4. Option D:

      [Cu(py)4]+\left[\mathrm{Cu}(\mathrm{py})_{4}\right]^{+}and [Cu(CN)4]3−\left[\mathrm{Cu}(\mathrm{CN})_{4}\right]^{3-}

  14. Question 14Chemistry· p-Block Elements (Group 15-18)

    The correct statement(s) related to oxoacids of phosphorous is(are)

    1. Option A:

      Upon heating, H3PO3\mathrm{H}_{3} \mathrm{PO}_{3} undergoes disproportionation reaction to produce H3PO4\mathrm{H}_{3} \mathrm{PO}_{4} and PH3\mathrm{PH}_{3}.

    2. Option B:

      While H3PO3\mathrm{H}_{3} \mathrm{PO}_{3} can act as reducing agent, H3PO4\mathrm{H}_{3} \mathrm{PO}_{4} cannot.

    3. Option C:

      H3PO3\mathrm{H}_{3} \mathrm{PO}_{3} is a monobasic acid.

    4. Option D:

      The H atom of P−H\mathrm{P}-\mathrm{H} bond in H3PO3\mathrm{H}_{3} \mathrm{PO}_{3} is not ionizable in water.

  15. Question 15Chemistry· Thermodynamics & Thermochemistry

    One mole of an ideal gas at 900 K , undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two process are same, the value of ln⁡V3 V2\ln \frac{\mathrm{V}_{3}}{\mathrm{~V}_{2}}

  16. Question 16Chemistry· Structure of Atom

    Consider a helium (He) atom that absorbs a photon of wavelength 330 nm . The change in the velocity (in cms−1\mathrm{cm} \mathrm{s}^{-1} ) of He atom after the photon absorption is \qquad . (Assume: Momentum is conserved when photon is absorbed. Use: Planck constant =6.6×10−34 J s=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}, Avogadro number =6×1023 mol−1=6 \times 10^{23} \mathrm{~mol}^{-1}, Molar mass of He=4 g mol−1\mathrm{He}=4 \mathrm{~g} \mathrm{~mol}^{-1} )

  17. Question 17Chemistry· p-Block Elements (Group 15-18)

    Ozonolysis of ClO2\mathrm{ClO}_{2} produces an oxide of chlorine. The average oxidation state of chlorine in this oxide is

  18. Question 18Mathematics· Permutations and Combinations

    Let

    S1={(i,j,k):i,j,k∈{1,2,…,10}},S_{1} = \{(i,j,k) : i,j,k \in \{1,2,\ldots,10\}\}, S2={(i,j):1≤i<j+2≤10,  i,j∈{1,2,…,10}},S_{2} = \{(i,j) : 1 \leq i < j+2 \leq 10, \; i,j \in \{1,2,\ldots,10\}\}, S3={(i,j,k,l):1≤i<j<k<l,  i,j,k,l∈{1,2,…,10}},S_{3} = \{(i,j,k,l) : 1 \leq i < j < k < l, \; i,j,k,l \in \{1,2,\ldots,10\}\},

    and

    S4={(i,j,k,l):i,j,k,l  are   distinct   elements   in   {1,2,…,10}}.S_{4} = \{(i,j,k,l) : i,j,k,l \; \text{are\; distinct\; elements\; in\; } \{1,2,\ldots,10\}\}.

    If the total number of elements in the set SrS_{r} is nrn_{r}, r=1,2,3,4r = 1,2,3,4, then which of the following statements is (are) \textbf{TRUE}?

    1. Option A:

      n1=1000n_{1}=1000

    2. Option B:

      n2=44\mathrm{n}_{2}=44

    3. Option C:

      n3=220n_{3}=220

    4. Option D:

      n412=420\frac{\mathrm{n}_{4}}{12}=420

  19. Question 19Mathematics· Limits, Continuity and Differentiability

    Let f:[−π2,π2]→Rf:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow R be a continuous function such that f(0)=1f(0)=1 and ∫0π3f(t)dt=0\int_{0}^{\frac{\pi}{3}} f(t) d t=0. Then which of the following statements is(are) TRUE?

    1. Option A:

      The equation f(x)−3cos⁡3x=0f(x)-3 \cos 3 x=0 has at least one solution in (0,π3)\left(0, \frac{\pi}{3}\right)

    2. Option B:

      The equation f(x)−3sin⁡3x=−6πf(x)-3 \sin 3 x=-\frac{6}{\pi} has at least one solution in (0,π3)\left(0, \frac{\pi}{3}\right)

    3. Option C:

      lim⁡x→0x∫0xf(t)dt1−ex2=−1\lim _{x \rightarrow 0} \frac{x \int_{0}^{x} f(t) d t}{1-e^{x^{2}}}=-1

    4. Option D:

      lim⁡x→0sin⁡x∫0xf(t)dtx2=−1\lim _{x \rightarrow 0} \frac{\sin x \int_{0}^{x} f(t) d t}{x^{2}}=-1

  20. Question 20Mathematics· Differential Equations

    For any real numbers α\alpha and β\beta, let yα,β(x),x∈Ry_{\alpha, \beta}(x), x \in R, be the solution of the differential equation

    dydx+αy=xeβx,y(1)=1\frac{d y}{d x}+\alpha y=x e^{\beta x}, y(1)=1

    Let S={yα,β(x):α,β∈R}S=\left\{y_{\alpha, \beta}(x): \alpha, \beta \in R\right\}. Then which of the following functions belong(s) the set SS ?

    1. Option A:

      f(x)=x22e−x+(e−12)e−xf(x)=\frac{x^{2}}{2} e^{-x}+\left(e-\frac{1}{2}\right) e^{-x}

    2. Option B:

      f(x)=−x22e−x+(e+12)e−xf(x)=-\frac{x^{2}}{2} e^{-x}+\left(e+\frac{1}{2}\right) e^{-x}

    3. Option C:

      f(x)=ex2(x−12)+(e−e24)e−xf(x)=\frac{e^{x}}{2}\left(x-\frac{1}{2}\right)+\left(e-\frac{e^{2}}{4}\right) e^{-x}

    4. Option D:

      f(x)=ex2(12−x)+(e+e24)e−xf(x)=\frac{e^{x}}{2}\left(\frac{1}{2}-x\right)+\left(e+\frac{e^{2}}{4}\right) e^{-x}

  21. Question 21Mathematics· Vector Algebra

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

    1. Option A:

      Projection of OC→\overrightarrow{\mathrm{OC}} on OA→\overrightarrow{\mathrm{OA}} is −32-\frac{3}{2}

    2. Option B:

      Area of the triangle OAB is 92\frac{9}{2}

    3. Option C:

      Area of the triangle ABC is 92\frac{9}{2}

    4. Option D:

      The acute angle between the diagonals of the parallelogram with adjacent sides OA→\overrightarrow{\mathrm{OA}} and OC→\overrightarrow{\mathrm{OC}} is π3\frac{\pi}{3}

  22. Question 22Mathematics· Parabola

    Let E denote the parabola y2=8x\mathrm{y}^{2}=8 \mathrm{x}. Let P=(−2,4)\mathrm{P}=(-2,4) and let Q and Q′\mathrm{Q}^{\prime} be two distinct points on E such that the lines PQ and PQ′\mathrm{PQ}^{\prime} are tangents to E . Let F be the focus of E . Then which of the following statements is(are) TRUE?

    1. Option A:

      The triangle PFQ is a right-angled triangle

    2. Option B:

      The triangle QPQ' ′^{\prime} is a right-angle triangle

    3. Option C:

      The distance between PP and FF is 525 \sqrt{2}

    4. Option D:

      F lies on the line joining Q and Q′\mathrm{Q}^{\prime}

  23. Question 23Mathematics· Probability

    A number is chosen at random from the set {1,2,3,…..,2000}\{1,2,3, \ldots . ., 2000\}. Let pp be the probability that the chosen number is a multiple of 3 or a multiple of 7 . Then the value of 500 p is \qquad

  24. Question 24Mathematics· Ellipse

    Let EE be the ellipse x216+y29=1\frac{x^{2}}{16}+\frac{y^{2}}{9}=1. For any three distinct points P,QP, Q and Q′Q^{\prime} on EE, let M(P,Q)M(P, Q) be the midpoint of the line segment joining PP and QQ, and M(P,Q′)M\left(P, Q^{\prime}\right) be the mid-point of the line segment joining PP and Q′\mathrm{Q}^{\prime}. Then the maximum possible value of the distance between M(P,Q)\mathrm{M}(\mathrm{P}, \mathrm{Q}) and M(P,Q′)\mathrm{M}\left(\mathrm{P}, \mathrm{Q}^{\prime}\right), as P,Q\mathrm{P}, \mathrm{Q} and Q′\mathrm{Q}^{\prime} vary on E , is \qquad

  25. Question 25Mathematics· Definite Integration

    For any real number x , let [x[\mathrm{x} ] denote the largest integer less than or equal to x . If

    I=∫010[10xx+1]dxI=\int_{0}^{10}\left[\sqrt{\frac{10 x}{x+1}}\right] d x

    then the value of 9I is \qquad

  26. Question 26Physics· Kinetic Theory of Gases

    A Soft plastic bottle, filled with water of density 1gm/cc1 \mathrm{gm} / \mathrm{cc}, carries an inverted glass test-tube with some air (ideal gas)

    trapped as shown in the figure. The test-tube has a mass of 5 gm , and it is made of a thick glass of density 2.5gm/cc2.5 \mathrm{gm} / \mathrm{cc}.

    Initially the bottle is sealed at atmospheric pressure p0=105 Pap_{0}=10^{5} \mathrm{~Pa} so that the volume of the trapped air is

    v0=3.3cc\mathrm{v}_{0}=3.3 \mathrm{cc}. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p0+Δpp_{0}+\Delta p without changing its orientation. At this pressure,

    the volume of the trapped air is v0−Δv\mathrm{v}_{0}-\Delta \mathrm{v}. Let Δv=X\Delta \mathrm{v}=\mathrm{X} cc and

    Δp=Y×103 Pa\Delta \mathrm{p}=\mathrm{Y} \times 10^{3} \mathrm{~Pa}.

    The value of X is   ⁣ ⁣  ⁣ ⁣ \text{ }\!\!~\!\!\text{ } .

  27. Question 27Physics· Kinetic Theory of Gases

    A Soft plastic bottle, filled with water of density 1gm/cc1 \mathrm{gm} / \mathrm{cc}, carries an inverted glass test-tube with some air (ideal gas)

    trapped as shown in the figure. The test-tube has a mass of 5 gm , and it is made of a thick glass of density 2.5gm/cc2.5 \mathrm{gm} / \mathrm{cc}.

    Initially the bottle is sealed at atmospheric pressure p0=105 Pap_{0}=10^{5} \mathrm{~Pa} so that the volume of the trapped air is

    v0=3.3cc\mathrm{v}_{0}=3.3 \mathrm{cc}. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p0+Δpp_{0}+\Delta p without changing its orientation. At this pressure,

    the volume of the trapped air is v0−Δv\mathrm{v}_{0}-\Delta \mathrm{v}. Let Δv=X\Delta \mathrm{v}=\mathrm{X} cc and

    Δp=Y×103 Pa\Delta \mathrm{p}=\mathrm{Y} \times 10^{3} \mathrm{~Pa}.

    The value of Y is   ⁣ ⁣  ⁣ ⁣ \text{ }\!\!~\!\!\text{ } .

  28. Question 28Physics· Rotational Dynamics

    A pendulum consists of a bob of mass m=0.1 kgm=0.1 \mathrm{~kg} and a massless inextensible string of length L=1.0 m\mathrm{L}=1.0 \mathrm{~m}.

    It is suspended from a fixed point at height H=0.9 m\mathrm{H}=0.9 \mathrm{~m} above a frictionless horizontal floor. Initially,

    the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P=0.2 kg−m/s\mathrm{P}=0.2 \mathrm{~kg}-\mathrm{m} / \mathrm{s} is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor.

    The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is

    Jkg−m2/s\mathrm{J} \mathrm{kg}-\mathrm{m}^{2} / \mathrm{s}. The kinetic energy of the pendulum just after the lift-off is K Joules.

    The value of J is   ⁣ ⁣  ⁣ ⁣ \text{ }\!\!~\!\!\text{ } .

  29. Question 29Physics· Alternating Current

    In the circuit, a metal filament lamp is connected in series with a capacitor of capacitance CμF\mathrm{C} \mu \mathrm{F} across a 200 V,50200 \mathrm{~V}, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V . Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and supply voltage is ϕ\phi. Assume, π3≈5\pi \sqrt{3} \approx 5.

    The value of C is   ⁣ ⁣  ⁣ ⁣ \text{ }\!\!~\!\!\text{ } .

  30. Question 30Physics· Alternating Current

    In the circuit, a metal filament lamp is connected in series with a capacitor of capacitance CμF\mathrm{C} \mu \mathrm{F} across a 200 V,50200 \mathrm{~V}, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100 V . Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude of the phase-angle (in degrees) between the current and supply voltage is ϕ\phi. Assume, π3≈5\pi \sqrt{3} \approx 5.

    The value of ϕ\phi is   ⁣ ⁣  ⁣ ⁣ \text{ }\!\!~\!\!\text{ } .

  31. Question 31Physics· Electromagnetic Induction

    A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius a, with its centre at the origin. A magnetic dipole of moment mm is brought along the axis of this loop from infinity to a point at distance r(≫r(\gg a) from the centre of the loop with its north pole always facing the loop, as shown in the figure below.

    The magnitude of magnetic field of a dipole mm, at a point on its axis at distance rr, is μ02πmr3\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}},

    where μ0\mu_{0} is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1m_{1} and m2m_{2}, separated by a distance rr on the common axis, with their north poles facing each other, is km1m2r4\frac{k m_{1} m_{2}}{r^{4}}, where kk is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

    When the dipole

    figure

    is placed at a distance

    figure

    from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to

    1. Option A:

      m/r3m/{{r}^{3}}

    2. Option B:

      m2/r2{{\text{m}}^{2}}/{{\text{r}}^{2}}

    3. Option C:

      m/r2\text{m}/{{\text{r}}^{2}}

    4. Option D:

      m2/r{{\text{m}}^{2}}/\text{r}

  32. Question 32Physics· Electromagnetic Induction

    A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius a, with its centre at the origin. A magnetic dipole of moment mm is brought along the axis of this loop from infinity to a point at distance r(≫r(\gg a) from the centre of the loop with its north pole always facing the loop, as shown in the figure below.

    The magnitude of magnetic field of a dipole mm, at a point on its axis at distance rr, is μ02πmr3\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}},

    where μ0\mu_{0} is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1m_{1} and m2m_{2}, separated by a distance rr on the common axis, with their north poles facing each other, is km1m2r4\frac{k m_{1} m_{2}}{r^{4}}, where kk is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

    The work done in bringing the dipole from infinity to a distance rr from the center of the loop by the given process is proportional to

    1. Option A:

      m/r5\text{m}/{{\text{r}}^{5}}

    2. Option B:

      m2/r5{{m}^{2}}/{{r}^{5}}

    3. Option C:

      m2/r6{{\text{m}}^{2}}/{{\text{r}}^{6}}

    4. Option D:

      m2/r7{{\text{m}}^{2}}/{{\text{r}}^{7}}

  33. Question 33Physics· Thermodynamics

    A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific heat at constant volume, CV=2R\mathrm{C}_{\mathrm{V}}=2 \mathrm{R}. Here, R is the gas constant. Initially, each side has a volume V0\mathrm{V}_{0} and temperature T0T_{0}. The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to V0/2V_{0} / 2. Consequently, the gas temperatures on the left and the right sides become TLT_{L} and TRT_{R}, respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

    The value of QRT0\frac{\text{Q}}{\text{R}{{\text{T}}_{0}}} is

    1. Option A:

      4(22+1)4\left( 2\sqrt{2}+1 \right)

    2. Option B:

      4(22−1)4\left( 2\sqrt{2}-1 \right)

    3. Option C:

      (52+1)\left( 5\sqrt{2}+1 \right)

    4. Option D:

      (52−1)\left( 5\sqrt{2}-1 \right)

  34. Question 34Chemistry· Electrochemistry

    At 298 K, the limiting molar conductivity of a weak monobasic acid is

    4×102  S cm2mol−1.4 \times 10^{2} \; \text{S cm}^{2} \text{mol}^{-1}.

    At 298 K, for an aqueous solution of the acid the degree of dissociation is α\alpha and the molar conductivity is

    y×102  S cm2mol−1.y \times 10^{2} \; \text{S cm}^{2} \text{mol}^{-1}.

    At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes

    3y×102  S cm2mol−1.3y \times 10^{2} \; \text{S cm}^{2} \text{mol}^{-1}.

    The value of α\boldsymbol{\alpha} is \qquad .

  35. Question 35Chemistry· Electrochemistry

    At 298 K, the limiting molar conductivity of a weak monobasic acid is

    4×102  S cm2mol−1.4 \times 10^{2} \; \text{S cm}^{2} \text{mol}^{-1}.

    At 298 K, for an aqueous solution of the acid the degree of dissociation is α\alpha and the molar conductivity is

    y×102  S cm2mol−1.y \times 10^{2} \; \text{S cm}^{2} \text{mol}^{-1}.

    At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes

    3y×102  S cm2mol−1.3y \times 10^{2} \; \text{S cm}^{2} \text{mol}^{-1}.

    The value of y\mathbf{y} is \qquad .

  36. Question 36Chemistry· Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

    Reaction of x gx\,\mathrm{g} of Sn\mathrm{Sn} with HCl\mathrm{HCl} quantitatively produced a salt. Entire amount of the salt reacted with y gy\,\mathrm{g} of nitrobenzene in the presence of required amount of HCl\mathrm{HCl} to produce 1.29 g1.29\,\mathrm{g} of an organic salt (quantitatively).

    (Use molar masses in g mol−1\mathrm{g\,mol^{-1}} of H,C,N,O,Cl,Sn\mathrm{H,C,N,O,Cl,Sn} as 1,12,14,16,35,1191,12,14,16,35,119 respectively.)

    The value of xx is ___\_\_\_ and the value of yy is ___\_\_\_.

    The value of x\mathbf{x} is ______\_\_\_\_\_\_.

  37. Question 37Chemistry· Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

    Reaction of x gx\,\mathrm{g} of Sn\mathrm{Sn} with HCl\mathrm{HCl} quantitatively produced a salt. Entire amount of the salt reacted with y gy\,\mathrm{g} of nitrobenzene in the presence of required amount of HCl\mathrm{HCl} to produce 1.29 g1.29\,\mathrm{g} of an organic salt (quantitatively).

    (Use molar masses in g mol−1\mathrm{g\,mol^{-1}} of H,C,N,O,Cl,Sn\mathrm{H,C,N,O,Cl,Sn} as 1,12,14,16,35,1191,12,14,16,35,119 respectively.)

    The value of xx is ___\_\_\_ and the value of yy is ___\_\_\_.

    The value of y\mathbf{y} is _____\_\_\_\_\_.

  38. Question 38Chemistry· Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

    A sample ( 5.6 g ) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03MKMnO40.03 \mathrm{M} \mathrm{KMnO}_{4} solution to reach the end point. Number of moles of Fe2+\mathrm{Fe}^{2+} present in 250 mL solution is x×10−2\mathbf{x} \times 10^{-2} (consider complete dissolution of FeCl2\mathrm{FeCl}_{2} ). The amount of iron present in the sample is y%\mathbf{y} \% by weight. (Assume: KMnO4\mathrm{KMnO}_{4} reacts only with Fe2+\mathrm{Fe}^{2+} in the solution Use: Molar mass of iron as 56 g mol−156 \mathrm{~g} \mathrm{~mol}^{-1} )

    The value of x\mathbf{x} is \qquad .

  39. Question 39Chemistry· Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

    A sample ( 5.6 g ) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03MKMnO40.03 \mathrm{M} \mathrm{KMnO}_{4} solution to reach the end point. Number of moles of Fe2+\mathrm{Fe}^{2+} present in 250 mL solution is x×10−2\mathbf{x} \times 10^{-2} (consider complete dissolution of FeCl2\mathrm{FeCl}_{2} ). The amount of iron present in the sample is y%\mathbf{y} \% by weight. (Assume: KMnO4\mathrm{KMnO}_{4} reacts only with Fe2+\mathrm{Fe}^{2+} in the solution Use: Molar mass of iron as 56 g mol−156 \mathrm{~g} \mathrm{~mol}^{-1} )

    The value of y\mathbf{y} is \qquad .

  40. Question 40Chemistry· Thermodynamics & Thermochemistry

    The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:

    HX3C−H(g)→HX3CX∙(g)+HX∙(g)ΔH∘=105 kcal   mol−1\ce{H3C-H(g) -> H3C^{\bullet}(g) + H^{\bullet}(g)} \qquad \Delta H^\circ = 105\ \text{kcal\; mol}^{-1} Cl−Cl(g)→ClX∙(g)+ClX∙(g)ΔH∘=58 kcal   mol−1\ce{Cl-Cl(g) -> Cl^{\bullet}(g) + Cl^{\bullet}(g)} \qquad \Delta H^\circ = 58\ \text{kcal\; mol}^{-1} HX3C−Cl(g)→HX3CX∙(g)+ClX∙(g)ΔH∘=85 kcal   mol−1\ce{H3C-Cl(g) -> H3C^{\bullet}(g) + Cl^{\bullet}(g)} \qquad \Delta H^\circ = 85\ \text{kcal\; mol}^{-1} H−Cl(g)→HX∙(g)+ClX∙(g)ΔH∘=103 kcal   mol−1\ce{H-Cl(g) -> H^{\bullet}(g) + Cl^{\bullet}(g)} \qquad \Delta H^\circ = 103\ \text{kcal\; mol}^{-1}

    For the following reaction CH4( g)+Cl2( g)→ light CH3Cl(g)+HCl(g)\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \xrightarrow{\text { light }} \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g}) The correct statement is

    1. Option A:

      Initiation step is exothermic with ΔH∘=−58kcalmol−1\Delta \mathrm{H}^{\circ}=-58 \mathrm{kcal} \mathrm{mol}^{-1}.

    2. Option B:

      Propagation step involving ∙CH3{ }^{\bullet} \mathrm{CH}_{3} formation is exothermic with ΔH∘=−2kcalmol−1\Delta \mathrm{H}^{\circ}=-2 \mathrm{kcal} \mathrm{mol}^{-1}.

    3. Option C:

      Propagation step involving CH3Cl\mathrm{CH}_{3} \mathrm{Cl} formation is endothermic with ΔH∘=+27kcalmol−1\Delta \mathrm{H}^{\circ}=+27 \mathrm{kcal} \mathrm{mol}^{-1}.

    4. Option D:

      The reaction is exothermic with ΔH∘=−25kcalmol−1\Delta \mathrm{H}^{\circ}=-25 \mathrm{kcal} \mathrm{mol}^{-1}.

  41. Question 41Chemistry· Practical Inorganic chemistry (Qualitative Analysis)

    The reaction of K3[Fe(CN)6]\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] with freshly prepared FeSO4\mathrm{FeSO}_{4} solution produces a dark blue precipitate called Turnbull's blue. Reaction of K4[Fe(CN)6]\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] with the FeSO4\mathrm{FeSO}_{4} solution in complete absence of air produces a white precipitate X\mathbf{X}, which turns blue in air. Mixing the FeSO4\mathrm{FeSO}_{4} solution with NaNO3\mathrm{NaNO}_{3}, followed by a slow addition of concentrated H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} through the side of the test tube produces a brown ring.

    Precipitate X\mathbf{X} is

    1. Option A:

      Fe4[Fe(CN)6]3\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}

    2. Option B:

      Fe[Fe(CN)6]\mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]

    3. Option C:

      K2Fe[Fe(CN)6]\mathrm{K}_{2} \mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]

    4. Option D:

      KFe⁡[Fe⁡(CN)6]\operatorname{KFe}\left[\operatorname{Fe}(\mathrm{CN})_{6}\right]

  42. Question 42Chemistry· Practical Inorganic chemistry (Qualitative Analysis)

    The reaction of K3[Fe(CN)6]\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] with freshly prepared FeSO4\mathrm{FeSO}_{4} solution produces a dark blue precipitate called Turnbull's blue. Reaction of K4[Fe(CN)6]\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] with the FeSO4\mathrm{FeSO}_{4} solution in complete absence of air produces a white precipitate X\mathbf{X}, which turns blue in air. Mixing the FeSO4\mathrm{FeSO}_{4} solution with NaNO3\mathrm{NaNO}_{3}, followed by a slow addition of concentrated H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} through the side of the test tube produces a brown ring.

    Among the following, the brown ring is due to the formation of

    1. Option A:

      [Fe(NO)2(SO4)2]2−\left[\mathrm{Fe}(\mathrm{NO})_{2}\left(\mathrm{SO}_{4}\right)_{2}\right]^{2-}

    2. Option B:

      [Fe(NO)2(H2O)4]3+\left[\mathrm{Fe}(\mathrm{NO})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{3+}

    3. Option C:

      [Fe(NO)4(SO4)2]\left[\mathrm{Fe}(\mathrm{NO})_{4}\left(\mathrm{SO}_{4}\right)_{2}\right]

    4. Option D:

      [Fe(NO)(H2O)5]2+\left[\mathrm{Fe}(\mathrm{NO})\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\right]^{2+}

  43. Question 43Mathematics· Circles

    The radius of the circle C is \qquad

  44. Question 44Mathematics· Circles

    The value of α\alpha is \qquad

  45. Question 45Mathematics· Application of Derivatives

    Let f1:(0,∞)→Rf_{1}:(0, \infty) \rightarrow R and f2:(0,∞)→Rf_{2}:(0, \infty) \rightarrow R be defined by f1(x)=∫0x∏j=121(t−j)jdt,x>0f_{1}(x)=\int_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t, x>0 and f2(x)=98(x−1)50−600(x−1)49+2450,x>0f_{2}(x)=98(x-1)^{50}-600(x-1)^{49}+2450, x>0, where, for any positive integer nn and real numbers a1,a2a_{1}, a_{2}, ……,an,∏i=1nai\ldots \ldots, a_{n}, \prod_{i=1}^{n} a_{i} denotes the product of a1,a2,…..,ana_{1}, a_{2}, \ldots . ., a_{n}. Let mim_{i} and nin_{i}, respectively, denote the number of points of local minima and the number of points of local maxima of function fi,i=1,2f_{i}, i=1,2, in the interval ( 0 , ∞)\infty).

    The value of 2m1+3n1+m1n12 m_{1}+3 n_{1}+m_{1} n_{1} is \qquad

  46. Question 46Mathematics· Application of Derivatives

    Let f1:(0,∞)→Rf_{1}:(0, \infty) \rightarrow R and f2:(0,∞)→Rf_{2}:(0, \infty) \rightarrow R be defined by f1(x)=∫0x∏j=121(t−j)jdt,x>0f_{1}(x)=\int_{0}^{x} \prod_{j=1}^{21}(t-j)^{j} d t, x>0 and f2(x)=98(x−1)50−600(x−1)49+2450,x>0f_{2}(x)=98(x-1)^{50}-600(x-1)^{49}+2450, x>0, where, for any positive integer nn and real numbers a1,a2a_{1}, a_{2}, ……,an,∏i=1nai\ldots \ldots, a_{n}, \prod_{i=1}^{n} a_{i} denotes the product of a1,a2,…..,ana_{1}, a_{2}, \ldots . ., a_{n}. Let mim_{i} and nin_{i}, respectively, denote the number of points of local minima and the number of points of local maxima of function fi,i=1,2f_{i}, i=1,2, in the interval ( 0 , ∞)\infty).

    The value of 6m2+4n2+8m2n26 m_{2}+4 n_{2}+8 m_{2} n_{2} is \qquad

  47. Question 47Mathematics· Definite Integration

    Let gi:[π8,3π8]→R,i=1,2g_{i}:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R, i=1,2 and f:[π8,3π8]→Rf:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R be functions such that g1(x)=1,g2(x)=∣4x−π∣g_{1}(x)=1, g_{2}(x)=|4 x-\pi| and f(x)=sin⁡2xf(x)=\sin ^{2} x, for all x∈[π8,3π8]x \in\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right]. Define Si=∫π83π8f(x)⋅gi(x)dx,i=1,2S_{i}=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f(x) \cdot g_{i}(x) d x, i=1,2

    The value of 16S1π\frac{16 S_{1}}{\pi} is \qquad

  48. Question 48Mathematics· Definite Integration

    Let gi:[π8,3π8]→R,i=1,2g_{i}:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R, i=1,2 and f:[π8,3π8]→Rf:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R be functions such that g1(x)=1,g2(x)=∣4x−π∣g_{1}(x)=1, g_{2}(x)=|4 x-\pi| and f(x)=sin⁡2xf(x)=\sin ^{2} x, for all x∈[π8,3π8]x \in\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right]. Define Si=∫π83π8f(x)⋅gi(x)dx,i=1,2S_{i}=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f(x) \cdot g_{i}(x) d x, i=1,2

    The value of 48 S2π2\frac{48 \mathrm{~S}_{2}}{\pi^{2}} is \qquad

  49. Question 49Mathematics· Application of Derivatives

    Which of the following statements is TRUE?

    1. Option A:

      f(ln⁡3)+g(ln⁡3)=13f(\sqrt{\ln 3})+g(\sqrt{\ln 3})=\frac{1}{3}

    2. Option B:

      For every x>1x > 1, there exists an α∈(1,x)\alpha \in (1, x) such that ψ1(x)=1+αx\psi_1(x) = 1 + \alpha x

    3. Option C:

      For every x>0\mathrm{x}>0, there exists a β∈(0,x)\beta \in(0, \mathrm{x}) such that ψ2(x)=2x(ψ1(β)−1)\psi_{2}(\mathrm{x})=2 \mathrm{x}\left(\psi_{1}(\beta)-1\right)

    4. Option D:

      ff is an increasing function on the interval [0,32]\left[0, \frac{3}{2}\right]

  50. Question 50Mathematics· Definite Integration

    Which of the following statements is TRUE?

    1. Option A:

      ψ1(x)≤1\psi_{1}(\mathrm{x}) \leq 1, for all x>0\mathrm{x}>0

    2. Option B:

      ψ2(x)≤0\psi_{2}(x) \leq 0, for all x>0x>0

    3. Option C:

      f(x)≥1−e−x2−23x3+25x5f(x) \geq 1-e^{-x^{2}}-\frac{2}{3} x^{3}+\frac{2}{5} x^{5}, for all x∈(0,12)x \in\left(0, \frac{1}{2}\right)

    4. Option D:

      g(x)≤23x3−25x5+17x7g(x) \leq \frac{2}{3} x^{3}-\frac{2}{5} x^{5}+\frac{1}{7} x^{7}, for all x∈(0,12)x \in\left(0, \frac{1}{2}\right)

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