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Part Test - 8 (JEE Advanced)


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72 Questions MCQ Test JEE Main & Advanced Mock Test Series | Part Test - 8 (JEE Advanced)

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Part Test - 8 (JEE Advanced) - Question 1

When a galvanometer is shunted with a 4 resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 wire, the further reduction in the deflection will be (the main current remains the same).

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 1

Part Test - 8 (JEE Advanced) - Question 2

An inductor (xL = 2) a capacitor (xC = 8) and a resistance (8) is connected in series with an A.C. source. The voltage output of A.C. source is given by v = 10 cos 100πt. The instantaneous potential difference between A and B, when it is half of the voltage output from source at that instant will be : 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 2

Part Test - 8 (JEE Advanced) - Question 3

Figure shows a system of three concentric metal shells A, B and C with radii a, 2a and 3a respectively. Shell B is earthed and shell C is given a charge Q. Now if shell C is connected to shell A, then the final charge on the shell B, is equal to : 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 3

Part Test - 8 (JEE Advanced) - Question 4

A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is : 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 4

Part Test - 8 (JEE Advanced) - Question 5

Four infinite ladder network containing identical resistances of R each, are combined as shown in figure. The equivalent resistance between A and B is RAB and between A and C is RAC. Then the value of  is :

 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 5

Let the equivalent resistance of one infinite ladder be x. Then the complete network reduces to 

Part Test - 8 (JEE Advanced) - Question 6

Two small balls, each having equal positive charge Q are suspended by two insulating strings of equal length L from a hook fixed to a stand. If the whole set-up is transferred to a satellite in orbit around the earth, the tension in equilibrium in each string is equal to 

 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 6

A satellite is in a state of free fall & hence weightlessness. Thus only electric force is responsible for the tension 

Part Test - 8 (JEE Advanced) - Question 7

In a practical wheat stone bridge circuit as shown, when one more resistance of 100 ? is connected in parallel with unknown resistance ' x ', then ratio l1/l2 becomes '2'. l1 is balance length. AB is a uniform wire. Then value of ' x ' must be : 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 7

Part Test - 8 (JEE Advanced) - Question 8

Loop A of radius (r << R) moves towards loop B with a constant velocity V in such a way that their  planes are always parallel. What is the distance between the two loops (x) when the induced emf in loop A is maximum 

  

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 8

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 9

The figure shows, a graph of the current in a discharging circuit of a capacitor through a resistor of resistance 10.    

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 9

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 10

A single circular loop of wire with radius 0.02 m carries a current of  8.0 A. It is placed at the centre of a solenoid that has length 0.65 m, radius 0.080 m and 1300 turns.    

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 10

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 11

In front of an earthed conductor a point charge + q is placed as shown in figure : 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 11

Charge is distributed over the surface of conductor in such a way that net field due to this charge and outside charge q is zero inside the conductor. Field due to only q is non-zero.
 

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 12

In the figure shown the key is switched on at t = 0. Let I1 and I2 be the currents through inductors having self inductances L1 & L2 at any time t respectively. The magnetic energy stored in the inductors 1 and 2 be U1 and U2. Then  at any instant of time is :

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 12

Part Test - 8 (JEE Advanced) - Question 13

Statement – 1
Four point charges q1, q2, q3 and q4 are as shown in figure. The flux over the shown Gaussian surface depends only on charges q1 and q2.   

Statement – 2
Electric field at all points on Gaussian surface depends only on charges q1 and q2.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 13

Statement ? is true directly from Gauss Theorem.
Statement 2 is false. Electric field at any point an Gaussian surface depends on all four charges.
Statement-1 is True, Statement-2 is False

Part Test - 8 (JEE Advanced) - Question 14

Statement 1  : A direct uniformly distributed current flows through a solid long metallic cylinder along its length. It produces magnetic field only outside the cylinder .

Statement 2 : A thin long cylindrical tube carrying uniformly distributed current along its length does not produce a magnetic field inside it. Moreover, a solid cylinder can be supposed to be made up of many thin cylindrical tubes.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 14

The current through solid metallic cylinder also produces magnetic field inside the cylinder. Hence statement-1
is false

Part Test - 8 (JEE Advanced) - Question 15

Statement–1 : Magnitude of potential difference across the terminals of a non-ideal battery in a circuit cannot be greater than its emf.  

Statement–2 : When a current of magnitude I is passing through a battery of emf E and internal resistance r as shown, the magnitude of potential difference (V) across the battery is given by V = E– I r

   

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 15

Statement-1 is obviously false if the current is sent in opposite direction given in figure of statement-2

Part Test - 8 (JEE Advanced) - Question 16

Statement–1 : No electric current will be present within a region having uniform and constant magnetic field.

Statement–2 : Within a region of uniform and constant magnetic field , the path integral of magnetic field  along any closed path is zero. Hence from Ampere circuital law 

(where the given terms have usual meaning), no current can be present within a region having uniform and constant magnetic field. 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 16

 along any closed path within a uniform magnetic field is always zero. Hence the closed path can be chosen of any size, even very small size enclosing a very small area. Hence we can prove that net current through each area of infinitesimally small size within region of uniform magnetic field is zero. Hence we can say no current (rather than no net current) flows through region of uniform magnetic field. Hence statement -2 is correct explanation of statement-1.

Part Test - 8 (JEE Advanced) - Question 17

In the circuit given below, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

Q. The value of emf E1 is : 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 17

Part Test - 8 (JEE Advanced) - Question 18

In the circuit given below, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

Q. The resistance R1 has value : 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 18

Part Test - 8 (JEE Advanced) - Question 19

In the circuit given below, both batteries are ideal. EMF E1 of battery 1 has a fixed value, but emf E2 of battery 2 can be varied between 1.0 V and 10.0 V. The graph gives the currents through the two batteries as a function of E2, but are not marked as which plot corresponds to which battery. But for both plots, current is assumed to be negative when the direction of the current through the battery is opposite the direction of that battery's emf. (Direction of emf is from negative to positive)

Q. The resistance R2 is equal to :

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 19

Part Test - 8 (JEE Advanced) - Question 20

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Q. Which wire has the greatest radius ?

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 20

Part Test - 8 (JEE Advanced) - Question 21

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Q. Which wire has the greatest magnitude of the magnetic field on the surface ?  

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 21

Part Test - 8 (JEE Advanced) - Question 22

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of the magnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. All curves start from the origin.

Q. The current density in wire a is     

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 22

Part Test - 8 (JEE Advanced) - Question 23

Match the following :

The following table gives the lengths of four copper rods at the same temperature, their diameters, and the potential differences between their ends.

Correctly match the physical quantities mentioned in the left column with the rods as marked. 
    (A) Rod 1            (p) Greatest Drift speed of the electrons.        
    (B) Rod 2            (q) Greatest Current                    
    (C) Rod 3            (r) Second greatest current
    (D) Rod 4            (s) Greatest Electric field                
                              (t) greatest resistance

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 23

Part Test - 8 (JEE Advanced) - Question 24

A  square loop of uniform conducting wire is as shown in figure. A current-I ( in amperes) enters the loop from one end and exits the loop from opposite end as shown in figure. The length of one side of square loop is l metre. The wire has uniform cross section area and uniform linear mass density. In four situations of column-I, the loop is subjected to four different uniform and constant magnetic field. Under the conditions of column-I, match the column-I with corresponding results of column-II ( Bo in column I is a positive nonzero constant)

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 24

Part Test - 8 (JEE Advanced) - Question 25

In an f.c.c. crystal, which of the following shaded planes contains the following type of arrangement of atoms?

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 25

Shown arrangement is hexagonally closed pack plane and these type of planes are arranged perpendicular to
body diagonal of fcc unit cell as shown.

Part Test - 8 (JEE Advanced) - Question 26

In an f.c.c. unit cell, atoms are numbered as shown below. The atoms not touching each other are :

(Atom numbered 3 is face centre of front face).

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 26

Atoms along one edge or at corners do not touch each other in fcc cell.

Part Test - 8 (JEE Advanced) - Question 27

Electrolysis of a solution of HSO4- ions produces S2O8– –. Assuming 75% current efficiency, what  current  should  be  employed  to  achieve  a  production  rate  of 1 mole of S2O8- - per hour ? 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 27

Part Test - 8 (JEE Advanced) - Question 28

You are given the following cell at 298 K,  = – 0.76 V. Which of the following amounts of NaOH (equivalent weight = 40) will just make the pH of cathodic compartment to be equal to 7.0 :

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 28

Part Test - 8 (JEE Advanced) - Question 29

For the following parallel chain reaction

what will be that value of overall half-life of A in minutes?     

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 29

Part Test - 8 (JEE Advanced) - Question 30

Decomposition of A follows first order kinetics by the following equation.
        4A(g) ——> B(g)  +  2C(g)

If initially, total pressure was 800 mm of Hg and after 10 minutes it is found to be 650 mm of Hg. What is half-life of A ? (Assume only A is present initially)

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 30

Part Test - 8 (JEE Advanced) - Question 31

Assuming the formation of an ideal solution, determine the boiling point of a mixture containing 1560 g benzene (molar mass = 78) and 1125 g chlorobenzene (molar mass = 112.5) using the following against an external pressure of 1000 Torr:

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 31

Part Test - 8 (JEE Advanced) - Question 32

A complex containing K+, Pt(IV) and Cl is 100% ionised. Given that i = 3. Thus, complex is :

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 32

and in [PtCl6]2-ion oxidation state of Pt is +4.

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 33

A current of 2.68 A is passed for one hour through an aqueous solution of CuSO4 using copper electrodes . Select the correct statement(s) from the following :

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 33

Increase in mass of cathode = decrease in mass of Anode =

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 34

Decomposition of 3 A(g) ——> 2 B(g) + 2C(g) follows first order kinetics. Initially only A is present in the container. Pressure developed after 20 min. and infinite time are 3.5 and 4 atm respectively. Which of the following is true.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 34

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 35

Which of the following is/are correct for an ideal binary solution of two volatile liquids (eg. benzene and toluene)? 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 35

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 36

Which of the following is a mismatch :

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 36

Part Test - 8 (JEE Advanced) - Question 37

Statement-1 : Time taken for the completion of 75% of a Ist order reaction is double than its t½.

Statement-2 : Time taken for completion of any fraction of a Ist order reaction is proportional to the extent of reaction completed.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 37

Statement-1 is correct but statement-2 is incorrect. The time taken for completion of any fraction of a Ist order
reaction is a fixed maximum value for a particular reaction.

Part Test - 8 (JEE Advanced) - Question 38

Statement-1 : The difference in the boiling points of equimolar solution of HCl and HF decreases as their molarity is decreased.

Statement-2 : The extent of dissociation decreases steadily with increasing dilution.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 38

Statement-1 is correct because As molarity decreases degree of dissociation of both HCI and HF will increase and approach unity.
Statement-2 is false because degree of dissociation increases with dilution.

Part Test - 8 (JEE Advanced) - Question 39

Statement-1 : A gas with higher critical temperature is absorbed more than a gas with lower critical temperature on the same absorbent.

Statement-2 :    Higher critical temperature implies that the gas is more easily liquifiable.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 39

Both the statements are individually correct but they are not related.

Part Test - 8 (JEE Advanced) - Question 40

Statement-1 : Specific conductance decreases with dilution whereas equivalent conductance increases.

Statement-2 : On dilution number of ions per millilitre decreases but total number of ions increases considerably.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 40

Total number of ions will increase slightly on dilution (Not considerably).

Part Test - 8 (JEE Advanced) - Question 41

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation 

 

When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.

Q. What is the molar conductance of NaCl at infinite dilution ?

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 41

Part Test - 8 (JEE Advanced) - Question 42

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation 

 

When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.

Q. What is the cell constant of the conductivity cell (C) ? 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 42

Part Test - 8 (JEE Advanced) - Question 43

The molar conductance of NaCl varies with the concentration as shown in the following table and all values follows the equation 

 

When a certain conductivity cell (C) was filled with 25 x 10–4 (M) NaCl solution. The resistance of the cell was found to be 1000 ohm. At Infinite dilution, conductance of Cl and SO4–2 are 80 ohm–1 cm2 mole–1 and 160 ohm–1 cm2 mole–1 respectively.

Q. If the cell (C) is filled with 5 x 10–3 (N) Na2SO4 the obserbed resistance was 400 ohm. What is the molar conductance of Na2SO4 ?

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 43

Part Test - 8 (JEE Advanced) - Question 44

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where  n is the number of products (ions or molecules) obtained per mole of the reactant & i =  for association.

Where n is number of reactant particles associated to give 1 mole product. 

A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.

Q. The vant Hoff factor will be [ if we start with one mole of X ]

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 44

Part Test - 8 (JEE Advanced) - Question 45

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where  n is the number of products (ions or molecules) obtained per mole of the reactant & i =  for association.

Where n is number of reactant particles associated to give 1 mole product. 

A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.

Q. The colligative properties observed will be     

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 45

(A) During association 

Part Test - 8 (JEE Advanced) - Question 46

Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociation where  n is the number of products (ions or molecules) obtained per mole of the reactant & i =  for association.

Where n is number of reactant particles associated to give 1 mole product. 

A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.

Q. The equilibrium constant for the process can be expressed as

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 46

Part Test - 8 (JEE Advanced) - Question 47

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 47

Part Test - 8 (JEE Advanced) - Question 48

For A + B → C in column - II the graphs given can be from any of these four types.
(p) –   Vs time (x axis) (q) t1/2 Vs initial concentration (x axis)         (s) Conc. Vs time (x axis) (t) log k Vs 1/Temperature (x axis)  

  Match the graphs in Column–II for the given order of reactions in Column – I 

 

                                                                                   

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 48

(A) First order :– t1/2 is constant. (p)
concentration and rate decrease exponentially.(q)
Arhennius equation is applicable for every order reaction (t)
(B) Zero order :- Rate is constant. (p)
Concentration decreases linearly with t. (r)

Arhennius equation is applicable for every order reaction (t)

(C) Second order :-

Arhennius equation is applicable for every order reaction (t)

(D) Same as (A).

Part Test - 8 (JEE Advanced) - Question 49

Two roads OA and OB intersect at an angle of 60º.  A car driver approaches O from A where OA = 800m at a uniform speed of 20m/s. Simultaneously another car moves from O towards B at a uniform speed of 25m/s. If t is time when two cars are closest, find t.

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 49

Let the distance between two cars after t seconds be s. Distance covered by first car in t second is 20t meter, so its distance from O will be (800 – 20t) meter and distance covered by second car will be 25t meter.

Part Test - 8 (JEE Advanced) - Question 50

If f(x) satisfies x + |f (x)| = 2f (x) then f–1 (x) satisfies

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 50

Part Test - 8 (JEE Advanced) - Question 51

Range of the function y = sin  + cos  is 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 51

Part Test - 8 (JEE Advanced) - Question 52

Let f : R → R and g : R → R be two one - one and onto functions such that they are mirror images of each other about the line y = 0, then h (x) = f(x) + g(x) is

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 52

f : R → R & g : R →R be two one-one onto functions such that f & g are mirror images of each other about line                                            y = 0. It means one is –ve of the other
i.e. f(x) = – g(x)
⇒ f(x) + g(x) = 0
⇒ h(x) = 0

h(x) is not onto as well as not one-one

Part Test - 8 (JEE Advanced) - Question 53

The function f (x) = (x2 – 1) |x2 – 3x + 2| + cos |x| is not differentiable at x = 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 53

Part Test - 8 (JEE Advanced) - Question 54

The function f (x) = (x – [x]) sin p x is, (where [.] denotes greatest integer function)

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 54

f(x) is continuous at every integer & also f(x) is continuous for any other real number. So f(x) is continuous for all real numbers.To check differentiability

Part Test - 8 (JEE Advanced) - Question 55

If f(x) = x3 + ax2 + ax + x(tanθ + cotθ) is increasing for all real x and if    then

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 55

Part Test - 8 (JEE Advanced) - Question 56

For what values of a, m and b, Lagrange's mean value theorem is applicable to the function f(x) for  x [0, 2]

     

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 56

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 57

 does not have critical points if 'a' equal to

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 57

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 58

Let f(x) = Then f(x) is strictly increasing in R for

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 58

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 59

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 59

*Multiple options can be correct
Part Test - 8 (JEE Advanced) - Question 60

Which of the following is/are true 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 60

If f is continuous at a point. Then |f| will also continuous at that point.
If f is differentiable then f|f| will also be differentiable

Part Test - 8 (JEE Advanced) - Question 61

Statement-1 : If a 4th degree polynomial function has four distinct real roots, then its graph have exactly  two inflection points.

Statement-2 : If the equation f"(x) = 0 has distinct real roots, then the equation f(x) = 0  has atleast four real roots. 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 61

If a 4th degree polynomial function has four distinct real roots, then its graph will be like following 
            
    From graph we can say curve has 2 point of inflection.
    Statement 1 is true
    Statement 2 is not true. 

Part Test - 8 (JEE Advanced) - Question 62

Statement-1 : Let 0 < a < b. Then Rolle's theorem is applicable to the function f(x) =  in [a,b] and Rolle's constant is the GM of a and b.

Statement-2 : All the three conditions of Rolle's theorem are satisfied by the above function f(x).

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 62

Part Test - 8 (JEE Advanced) - Question 63

Statement-1    All points of intersection of y = f (x) & y = f – 1 (x) lies on y = x only.

Statement-2    If point P (α, β)  lies on y = f (x) then Q (β, α) lies on y = f – 1 (x).

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 63

Pt. of intersection of y = f(x) & y = f-1(x) lies on y = x. Also they may not lie on y = x. Infact points which does not
lies on y = x are even in numbers.
and statement (2) is true.

Part Test - 8 (JEE Advanced) - Question 64

Statement-1  where [.] represent greatest integer function.
Statement-2  

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 64

Part Test - 8 (JEE Advanced) - Question 65

The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation. 
Since LHS > ⇒  f(x) – g(x) > 0. Now squaring both sides, we get  f2 + g2 + 2|f.g| = f2 + g2 – 2fg
    ⇒ |fg| = – fg. The equation can hold if  f.g < 0 and f > g. 
    This can be simplified to f > 0, g < 0. 

    Answer the following questions on the basis of this method

Q. The complete solution of the equation |x3 – x| + |2 – x| = x3 – 2 is

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 65

Part Test - 8 (JEE Advanced) - Question 66

The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation. 
Since LHS > ⇒  f(x) – g(x) > 0. Now squaring both sides, we get  f2 + g2 + 2|f.g| = f2 + g2 – 2fg
    ⇒ |fg| = – fg. The equation can hold if  f.g < 0 and f > g. 
    This can be simplified to f > 0, g < 0. 

    Answer the following questions on the basis of this method

Q. The complete solution set of the equation |x2 – x| + |x + 3| = |x2 – 2x – 3| is 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 66

Part Test - 8 (JEE Advanced) - Question 67

The general procedure for solving equation containing modulus function is to split the domain into subintervals and solve the various cases. But there are certain structures of equations which can be solved by a different approach. For example, for solving the equation |f(x)| + |g(x)| = f(x) – g(x) one can follow this method. First find the permissible set of values of x for the equation. 
Since LHS > ⇒  f(x) – g(x) > 0. Now squaring both sides, we get  f2 + g2 + 2|f.g| = f2 + g2 – 2fg
    ⇒ |fg| = – fg. The equation can hold if  f.g < 0 and f > g. 
    This can be simplified to f > 0, g < 0. 

    Answer the following questions on the basis of this method

Q. The solution set belonging to (0, 2π) of the equation |sin x – cos x| = |sin x| + |cosx| is

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 67

Part Test - 8 (JEE Advanced) - Question 68

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |ex – 1| is a “two-one” function. y = x3 – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

Q. In the following functions which one is a “two-one” function :-

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 68

Part Test - 8 (JEE Advanced) - Question 69

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |ex – 1| is a “two-one” function. y = x3 – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

Q. Let f(x) = {x} be the fractional part function. For what domain is the function “two-one”?

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 69

From graph it is clear that any horizontal line cut the graph at either one or two points between 

Part Test - 8 (JEE Advanced) - Question 70

Let f be a function defined so that every element of the codomain has at most two pre-images and there is at least one element in the co-domain which has exactly two pre-images we shall call this function as “two-one” function. A two-one function is definitely a many one function but vice-versa is not true. For example, y = |ex – 1| is a “two-one” function. y = x3 – x is a many one function but not a “two-one” function. In the light of above definition answer the following questions:

Q. The values of ‘a’ for which the function f(x) = x4 – ax2 is a “two-one” function are 

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 70

Part Test - 8 (JEE Advanced) - Question 71

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 71

 

Part Test - 8 (JEE Advanced) - Question 72

Detailed Solution for Part Test - 8 (JEE Advanced) - Question 72

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