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QUESTION: 1

A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like :

Solution:

QUESTION: 2

The temperature of an open room of volume 30 m^{3} increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 10^{5} Pa. If n_{i} and n_{f} are the number of molecules in the room before and after heating, then n_{f }– n_{i} will be

Solution:

n_{1} = initial number of moles;

n_{2} = final number of moles

Change of number of molecules

n_{f }– n_{i} = (n_{2} – n_{1}) × 6.023 × 10^{23}

≈ – 2.5 × 10^{25}

QUESTION: 3

Which of the following statements is false?

Solution:

In a balanced Wheatstone bridge, the null point remains unchanged even if cell and galvanometer are interchanged.

QUESTION: 4

The following observations were taken for determining surface tension T of water by capillary method:

diameter of capillary, D = 1.25 × 10^{–2} m

rise of water, h = 1.45 × 10^{–2} m.

Using g = 9.80 m/s^{2} and the simplified relation the possible error in surface tension is closest to

Solution:

= 0.8 + 0.689

= 1.489

≈ 1.5%

QUESTION: 5

In amplitude modulation, sinusoidal carrier frequency used is denoted by ω_{c} and the signal frequency is denoted by ω_{m}. The bandwidth (Δω_{m}) of the signal is such that Δω_{m} << ω_{mc}. Which of the following frequencies is not contained in the modulated wave?

Solution:

Modulated wave has frequency range.

ω_{c }± ω_{m}

∴ Since ω_{c} >> ω_{m}

∴ ω_{m} is excluded.

QUESTION: 6

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is

Solution:

For converging lens

u = –40 cm which is equal to 2f

∴ Image will be real and at a distance of 40 cm from convergent lens.

QUESTION: 7

The moment of inertia of a uniform cylinder of length ℓ and radius R about its perpendicular bisector is I. What is the ratio ℓ /R such that the moment of inertia is minimum?

Solution:

QUESTION: 8

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If λ_{min} is the smallest possible wavelength of X-ray in the spectrum, the variation of log λ_{min} with log V is correctly represented in

Solution:

In X-ray tube

Slope is negative

Intercept on y-axis is positive

QUESTION: 9

A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by

Solution:

QUESTION: 10

An electric dipole has a fixed dipole moment which makes angle θ with respect to x-axis. When subjected to an electric field it experiences a torque When subjected to another electric field it experiences a torque The angle θ is

Solution:

From (i) and (ii)

QUESTION: 11

In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be

Solution:

In common emitter configuration for n-p-n transistor, phase difference between output and input voltage is 180°.

QUESTION: 12

C_{p} and C_{v} are specific heats at constant pressure and constant volume respectively. It is observed that

C_{p} – C_{v }= a for hydrogen gas

C_{p} – C_{v } = b for nitrogen gas

The correct relation between a and b is :

Solution:

Let molar heat capacity at constant pressure = X_{p }and molar heat capacity at constant volume = Xv

X_{p} – X_{v} =R

MC_{p} – MC_{v} = R

C_{p} – C_{v} = R/M

For hydrogen; a = R/2

For N_{2}; b = R/28

a/b = 14

a = 14b

QUESTION: 13

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by :

(Given : room temperature = 30°C, specific heat of copper = 0.1 cal/gm°C)

Solution:

100 × 0.1 × (t – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

10t – 750 = 450 + 7650

10t = 1200 + 7650

10t = 8850

t = 885°C

QUESTION: 14

A body of mass m = 10^{–2 }kg is moving in a medium and experiences a frictional force F = –kv^{2}. Its initial speed is v_{0} = 10 ms^{–1}. If, after 10 s, its energy is 1/8 mv_{0}^{2} the value of k will be

Solution:

QUESTION: 15

When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into a voltmeter of range 0-10 V is

Solution:

i_{g }= 5 × 10^{–3} A

G = 15 Ω

Let series resistance be R.

V = i_{g} (R + G)

10 = 5 × 10^{–3} (R + 15)

R = 2000 – 15 = 1985 = 1.985 × 10^{3 }Ω

QUESTION: 16

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is

Solution:

Torque at angle θ

QUESTION: 17

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ_{1}/λ_{2}, is given by

Solution:

From energy level diagram

QUESTION: 18

A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of

Solution:

∵ Density remains same

So, mass ∝ Volume

QUESTION: 19

In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is

Solution:

Magnitude of change in flux = R × area under current vs time graph

QUESTION: 20

In a Young's double slit experiment, slit s are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is

Solution:

QUESTION: 21

A magnetic needle of magnetic moment 6.7 × 10^{–2} Am^{2} and moment of inertia 7.5 × 10^{–6} kg m^{2} is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is

Solution:

For 10 oscillations,

t = 10T = 2π × 1.06

= 6.6568 ≈ 6.65 s

QUESTION: 22

The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius) :

Solution:

Variation of g inside earth surface

QUESTION: 23

In the above circuit the current in each resistance is

Solution:

The potential difference in each loop is zero.

∴ No current will flow.

QUESTION: 24

with a particle B of mass m/2 which is at rest. Thecollision is head on, and elastic. The ratio of the de-Broglie wavelengths λ_{A} to λ_{B} after the collision is

Solution:

QUESTION: 25

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and a is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by :

Solution:

QUESTION: 26

A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 second will be

Solution:

QUESTION: 27

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 10^{8} ms^{–1})

Solution:

For relativistic motion

; v = relative speed of approach

QUESTION: 28

In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :

Solution:

In steady state, flow of current through capacitor will be zero.

QUESTION: 29

A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V.

The minimum number of capacitors required to achieve this is

Solution:

Following arrangement will do the needful :

8 capacitors of 1μF in parallel with four such branches in series.

QUESTION: 30

A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?

Solution:

Acceleration is constant and negative

QUESTION: 31

Which of the following compounds will form significant amount of meta product during mono-nitration reaction?

Solution:

(a) Nitration reactions take place in presence of concentrated HNO_{3}^{+} concentrated H_{2}SO_{4}.

(b) Aniline acts as a base. In presence of H_{2}SO_{4}, its protonation takes place and anilinium ion is formed

(c)Anilinium ion is a strongly deactivating group and meta directing in nature so it gives meta nitration product in a significant amount.

QUESTION: 32

ΔU is equal to

Solution:

For adiabatic process, q = 0

∴ As per 1^{st} law of thermodynamics,

ΔU = W

QUESTION: 33

The increasing order of the reactivity of the following halides for the S_{N}1 reaction is

Solution:

Rate of S_{N}1 reaction ∝ stability of carbocation

So, II < I < III

Increase stability of carbocation and hence increase reactivity of halides.

QUESTION: 34

The radius of the second Bohr orbit for hydrogen atom is

(Planck's Const. h = 6.6262 × 10^{–34} Js;

mass of electron = 9.1091 × 10^{–31} kg;

charge of electron e = 1.60210 × 10^{–19} C;

permittivity of vacuum

ε_{0} = 8.854185 × 10^{–12} kg^{–1} m^{–3} A^{2})

Solution:

= 2.12 Å

QUESTION: 35

pK_{a} of a weak acid (HA) and pK_{b} of a weak base(BOH) are 3.2 and 3.4, respectively, The pH of their salt (AB) solution is

Solution:

= 6.9

QUESTION: 36

The formation of which of the following polymers involves hydrolysis reaction?

Solution:

Caprolactam is hydrolysed to produce caproic acid which undergoes condensation to produce Nylon-6.

QUESTION: 37

The most abundant elements by mass in the body of a healthy human adult are :

Oxygen (61.4%); Carbon (22.9%); Hydrogen (10.0%) and Nitrogen (2.6%).

The weight which a 75 kg person would gain if all ^{1}H atoms are replaced by ^{2}H atoms is

Solution:

Mass of hydrogen = 10/100 x 75 = 7.5 kg

Replacing ^{1}H by ^{2}H would replace 7.5 kg with 15 kg

∴ Net gain = 7.5 kg

QUESTION: 38

Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourize the colour of bromine?

Solution:

The above product does not have any C = C or C ≡ C bond, so, it will not give Br_{2}-water test.

QUESTION: 39

In the following reactions, ZnO is respectively acting as a/an

(a) ZnO + Na_{2}O → Na_{2}ZnO_{2}

(b) ZnO + CO_{2} → ZnCO_{3}

Solution:

In (a), ZnO acts as acidic oxide as Na_{2}

O is basic oxide.

In (b), ZnO acts as basic oxide as CO_{2}

is acidic oxide.

QUESTION: 40

Both lithium and magnesium display several similar properties due to the diagonal relationship, however, the one which is incorrect, is

Solution:

Mg forms basic carbonate

3MgCO_{3} · Mg(OH)_{2} ·3H_{2}O but no such basic carbonate is formed by Li.

QUESTION: 41

3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is

Solution:

Since product (X) contains two chiral centres and it is unsymmetrical.

So, its total stereoisomers = 2^{2} = 4.

QUESTION: 42

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be

Solution:

In FCC, one of the face is like

By ΔABC,

2a^{2} = 16r^{2}

Distance of closest approach = 2r = a/√2

QUESTION: 43

Two reactions R_{1} and R_{2} have identical pre - exponential factors. Activation energy of R_{1 }exceeds that of R_{2 }by 10 kJ mol^{–1}. If k_{1} and k_{2} are rate constants for reactions R_{1} and R_{2} respectively at 300 K, then ln (k_{1}/ k_{2}) is equal to

(R = 8.314 J mole^{–1} K^{–1})

Solution:

QUESTION: 44

The correct sequence of reagents for the following conversion will be

Solution:

QUESTION: 45

The Tyndall effect is observed only when following conditions are satisfied

(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.

(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used

(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude

(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude

Solution:

For Tyndall effect refractive index of dispersion phase and dispersion medium must differ significantly. Secondly, size of dispersed phase should not differ much from wavelength used.

QUESTION: 46

Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution?

Solution:

Sugars in which there is free anomeric –OH group are reducing sugars

QUESTION: 47

Given

C_{(graphite)} + O_{2}(g) →CO_{2}(g);

Δ_{r}Hº = –393.5 kJ mol^{–1}

H_{2}(g) + 1/2 O_{2} (g) → H_{2}O(l);

Δ_{r}H^{0} = -285.8 kJ mol^{-1}

CO_{2}(g) + 2H_{2}O(l) → CH_{4}(g) + 2O_{2}(g);

Δ_{r}H^{0} = +890.3 kJ mol^{-1}

Based on the above thermochemical equations, the value of Δ_{r}Hº at 298 K for the reaction

C_{(graphite)} + 2H_{2}(g) → CH_{4}(g) wil1 be

Solution:

By applying the operation

(i) + 2 × (ii) + (iii), we get

QUESTION: 48

Which of the following reactions is an example of a redox reaction?

Solution:

Xe is oxidised from +4(in XeF_{4}) to +6(in XeF_{6})

Oxygen is reduced from +1 (in O_{2}F_{2}) to zero (in O_{2})

QUESTION: 49

The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are

Solution:

QUESTION: 50

The major product obtained in the following reaction is

Solution:

QUESTION: 51

Sodium salt of an organic acid 'X' produces effervescence with conc. H_{2}SO_{4}. 'X' reacts with the acidified aqueous CaCl_{2} solution to give a white precipitate which decolourises acidic solution of KMnO_{4}. 'X' is

Solution:

QUESTION: 52

Which of the following species is not paramagnetic?

Solution:

CO has 14 electrons (even) ∴ it is diamagnetic

NO has 15e^{-}(odd) ∴ it is paramagnetic and has 1 unpaired electron in π*2p molecular orbital.

B_{2} has 10e^{-} (even) but still paramagnetic and has two unpaired electrons in π2p_{x} and π2p_{y} (s-p mixing).

O_{2} has 16 e^{-} (even) but still paramagnetic and has two unpaired electrons in π*2p_{x} and π*2p_{y} molecular

QUESTION: 53

The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be

(K_{f} for benzene = 5.12 K kg mol^{–1})

Solution:

⇒ i = 0.527

α = 0.946

∴ % association = 94.6%

QUESTION: 54

Which of the following molecules is least resonance stabilized?

Solution:

However, all molecules given in options are stabilised by resonance but compound given in option (4) is least resonance stabilised (other three are aromatic)

QUESTION: 55

On treatment of 100 mL of 0.1 M solution of CoCl_{3} × 6H_{2}O with excess AgNO_{3}; 1.2 × 10^{22} ions are precipitated. The complex is

Solution:

Millimoles of AgNO_{3 = }

Millimoles of CoCl_{3}·6H_{2}O = 0.1 × 100 = 10

∴ Each mole of CoCl_{3}·6H_{2}O gives two chloride ions.

∴ [Co(H_{2}O)_{5}Cl]Cl_{2}-H_{2}O

QUESTION: 56

The major product obtained in the following reaction is

Solution:

DIBAL — H reduces esters and carboxylic acids into aldehydes

QUESTION: 57

A water sample has ppm level concentration of following anions

F^{-} = 10; SO_{4}^{2-} = 100; NO_{3}^{-} = 50

The anion/anions that make/makes the water sample unsuitable for drinking is/are

Solution:

Permissible limit of F^{–} in drinking water is upto 1 ppm. Excess concentration of F^{–}

> 10 ppm causes decay of bones.

QUESTION: 58

1 gram of a carbonate (M_{2}CO_{3}) on treatment with excess HCl produces 0.01186 mole of CO_{2}. The molar mass of M_{2}CO_{3} in g mol^{-1} is

Solution:

M_{2}CO_{3} + 2HCl → 2MCl + H_{2}O + CO_{2}

QUESTION: 59

Given

Among the following, the strongest reducing agent is

Solution:

Positive E° is for Cr, hence it is strongest reducing agent.

QUESTION: 60

The group having isoelectronic species is

Solution:

Mg^{2+}, Na^{+}, O^{2–} and F^{– }all have 10 electrons each.

QUESTION: 61

Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point

Solution:

(k^{2} - 7k+ 10) + 4k^{2} + 20k = ± 56

5k^{2} +13k+ 10 = ± 56

= 2, –4.6

Equation of AD,

x = 2 ...(i)

Also equation of BE,

2y -4=x - 5

x - 2y-1 = 0 ...(ii)

Solving (i) & (ii), 2y = 1

y = 1/2

Orthocentre is

QUESTION: 62

If, for a positive integer n, the quadratic equation, x(x + 1) + (x + 1) (x + 2) + .....

+ (x + n -1 ) (x + n) = 10n has two consecutive integral solutions, then n is equal to

Solution:

QUESTION: 63

The function defined as is

Solution:

f '(x) changes sign in different intervals.

∴ Not injective.

yx^{2} - x+ y = 0

For y ≠ 0

For, y = 0 ⇒ x = 0

∴ Part of range

∴ Surjective but not injective.

QUESTION: 64

The following statement (p → q) → [(~ p → q) → q] is

Solution:

QUESTION: 65

If S is the set of distinct values of b for which the following system of linear equations

x +y+ z = 1

x +ay+ z = 1

ax + by+ z = 0

has no solution, then S is

Solution:

⇒ –(1 – a)^{2} = 0

⇒ a = 1

For a = 1

Eq. (1) & (2) are identical i.e.,x + y + z = 1

To have no solution with x + by + z = 0

b = 1

QUESTION: 66

The area (in sq. units) of the region {(x, y) : x ≥ 0, x + y ≤ 3, x^{2} ≤ 4 y and y ≤ 1 + √x} is

Solution:

Area of shaded region

= 5/2 sq. unit

QUESTION: 67

For any three positive real numbers a, b and c, 9(25a^{2} +b^{2} )+ 25(c^{2} - 3ac ) = 15b(3a + c ). Then

Solution:

9(25a^{2} + b^{2}) + 25 (c^{2} - 3ac) = 15b (3a + c)

⇒ (15a)^{2} + (3b)^{2} + (5c)^{2} - 45ab - 15bc - 75ac = 0

⇒ (15a - 3b)^{2} + (3b - 5c)^{2} + (15a - 5c)^{2} = 0

It is possible when

15a - 3b= 0 and 3b -5c= 0 and 15a - 5c= 0

15a = 3b= 5c

∴ b, c, a are in A.P.

QUESTION: 68

A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is

Solution:

QUESTION: 69

The normal to the curve y (x -2)( x- 3) = x + 6 at the point where the curve intersects the y-axis passes through the point

Solution:

y (x -2)( x- 3) = x + 6

At y-axis, x = 0, y = 1

Now, on differentiation.

Now slope of normal = –1

Equation of normal y – 1 = –1(x – 0)

y + x – 1 = 0 ... (i)

Line (i) passes through (1/2,1/2)

QUESTION: 70

A hyperbola passes through the point P(√2,√3) and has foci at (±2, 0). Then the tangent to this hyperbola at P also passes through the point

Solution:

a^{2} +b^{2}= 4

⇒ b^{2} = 3

∴ a^{2}= 1

∴ Tangent at

Clearly it passes through (2√2, 3√3 )

QUESTION: 71

Let a, b, c ∈ R. If f(x) = ax^{2} + bx + c is such that a + b + c = 3 and

then is equal to

Solution:

As, f (x + y)= f (x) + f (y) + xy

Given, f (1) = 3

Putting, x =y= 1 ⇒ f (2) = 2f (1)+ 1 = 7

Similarly, x = 1,y= 2 ⇒ f (3) =f (1)+ f (2) + 2 = 12

= 3 + 7 + 12 + 18 + ... = S (let)

Now, S = 3+ 7 + 12 + 18 + ... +t_{n}

Again, S = 3+7 + 12 + ... +t_{n -1} + t_{n}

We get, t_{n} = 3+ 4 + 5 + ... terms

QUESTION: 72

Let and be a vector such that and the angle between be 30°. Then is equal to

Solution:

QUESTION: 73

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β then tan β is

Solution:

QUESTION: 74

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is

Solution:

2r + θr =20 ... (i)

A to be maximum

Hence for r = 5, A is maximum

Now, 10 + θ·5 = 20 ⇒ θ = 2 (radian)

QUESTION: 75

The integral is equal to

Solution:

QUESTION: 76

If ( 2 + sinx) dy/dx + (y + 1)cos x = 0 and y(0) = 1, then y(π/2) is equal to

Solution:

ln| y + 1 |+ ln (2 + sinx ) = lnC

(y + 1) (2+ sinx ) = C

Put x = 0, y = 1

(1 + 1)× 2 = C ⇒ C = 4

Now, ( y +1) (2+ sinx ) = 4

( y +1) (2+ 1) = 4

QUESTION: 77

Let In = tan^{n} xdx,(n > 1). If I_{4} +I_{6}= a tan^{5} x + bx^{5} + C, where C is a constant of integration, then the ordered pair (a, b) is equal to

Solution:

Let tanx = t

sec^{2}x dx = dt

QUESTION: 78

Let ω be a complex number such that 2ω + 1 = z where z =√-3 . If

then k is equal to

Solution:

2ω + 1 = z , z = √3i

C_{1} → C_{1} + C_{2 }+ C_{3}

= 3 (ω^{2} - ω^{4})

= -3√3i

= –3z

∴ k = –z

QUESTION: 79

The value of

(^{21}C_{1} - ^{10}C_{1}) + (^{21}C_{2 }- ^{10}C_{2}) + (^{21}C_{3} - ^{10}C_{3}) + (^{21}C_{4} - ^{10}C_{4}) +... + (^{21}C_{10} - ^{10}C_{10}) is

Solution:

^{21}C_{1} + ^{21}C_{2} +... + ^{21}C_{10} = 1/2 {^{21}C_{0} + ^{21}C_{1} +... + ^{21}C_{21}}-1

= 2^{20} – 1

(^{10}C_{1} + ^{10}C_{2} +... + ^{10}C_{10}) = 2^{10} -1

∴ Required sum = (2^{20} – 1) – (2^{10} – 1)

= 2^{20} – 2^{10}

QUESTION: 80

equals

Solution:

= 1/16.

QUESTION: 81

If 5 (tan^{2}x – cos^{2}x) = 2cos 2x + 9, then the value of cos 4x is

Solution:

5 tan^{2}x = 9 cos^{2}x + 7

5 sec^{2}x – 5 = 9 cos^{2}x + 7

Let cos^{2}x = t

cos4x = 2 cos^{2} 2x – 1

QUESTION: 82

If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line, x/1 = y/4 = z/5 is Q, then PQ is equal to

Solution:

Equation of PQ,

Let M be (λ+1, 4λ- 2, 5λ+ 3)

As it lies on 2x + 3y – 4z + 22 = 0

λ = 1

For Q, λ = 2

Distance PQ

=

QUESTION: 83

The distance of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the lines and is

Solution:

Let the plane be

a(x - 1)+ b( y + 1) + c(z +1) = 0

It is perpendicular to the given lines

a – 2b + 3c = 0

Solving, a : b : c = 5 : 7 : 3

∴ The plane is 5x + 7y + 3z + 5 = 0

Distance of (1, 3, –7) from this plane = 10/√83

QUESTION: 84

If for x∈ the derivative of tan^{ -1} is √x × g(x) , then g(x) equals

Solution:

QUESTION: 85

The radius of a circle, having minimum area, which touches the curve y = 4 – x^{2} and the lines, y = |x| is

Solution:

x^{2} =-(y -4)

Let a point on the parabola

Equation of normal at P is

It passes through centre of circle, say (0, k)

(Length of perpendicular from (0, k) to y = x)

Equation of circle is

It passes through point P

t^{4} +t^{2} (8k- 28) + 8k^{2} - 128k + 256 = 0

For t = ⇒ k^{2} - 16k+ 32 = 0

k = 8±4√2

(14 - 4k )^{2}+ (14 - 4k ) (8k - 28) + 8k^{2} -128k + 256 = 0

2k^{2} +4k-15 = 0

From (iii) & (iv),

But from options, r = 4(√2-1)

QUESTION: 86

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is

Solution:

n = 10

p(Probability of drawing a green ball) = 15/25

QUESTION: 87

The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x = – 4, then

the equation of the normal to it at is

Solution:

x = –4

-a= -4 x e

a = 2

Now, b^{2} =a^{2} (1- e^{2} ) = 3

Equation to ellipse

Equation of normal is

QUESTION: 88

If two different numbers are taken from the set {0, 1, 2, 3, ......, 10}; then the probability that their sum as well as absolute difference are both multiple of 4, is

Solution:

Total number of ways = ^{11}C_{2} = 55

Favourable ways are

(0, 4), (0, 8), (4, 8), (2, 6), (2, 10), (6, 10)

Probability = 6/55

QUESTION: 89

For three events A, B and C, P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P (Exactly one of C or A occurs) = 1/4 and P(All the three events occur simultaneously) = 1/16 Then the probability that at least one of the events occurs, is

Solution:

QUESTION: 90

,then adj (3A^{2 }+ 12A) is equal to

Solution:

= (2 – 2λ- λ + λ^{2} ) - 12

f (λ)= λ^{2} - 3λ - 10

∵ A satisfies f (λ )

∴ A^{2} – 3A –10I = 0

A^{2} – 3A = 10I

3A^{2} – 9A = 30I

3A^{2} + 12A = 30I + 21A

### April 2017 - Sports

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- JEE Main 2017 Question Paper with Solutions (2nd April)
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