JEE Main Question Paper with Solution 2019 April 9th Shift 2 - Afternoon

A

$ 6.67 \times 10^{-3} \; Wb$

B

$ 6.67 \times 10^{-4} \; Wb$

C

$ 3.67 \times 10^{-4} \; Wb$

D

$ 3.67 \times 10^{-3} \; Wb$

Solution

$\phi_{q} = \frac{\mu_{0} i_{1}R^{2}}{2\left(R^{2} +x^{2}\right)^{\frac{3}{2}}} \times\pi r^{2} = 10^{-3} $
$ \phi_{p} = \frac{\mu_{0}i_{2}r^{2}}{2\left(r^{2}+x^{2}\right)^{\frac{3}{2}}} \times\pi R^{2}$
$ \frac{\phi_{P}}{\phi_{Q} } = \frac{i_{2}}{i_{1}} . \frac{\left(R^{2} +x^{2}\right)^{\frac{3}{2}}}{\left(r^{2} +x^{2}\right)^{\frac{3}{2}}} = \frac{\phi_{P}}{10^{-3}} $
$ \frac{2}{3} = \frac{\phi_{P}}{10^{-3}} $
$ \phi_{P} = 6.67 \times10^{-4} $

A

$\frac{12}{5} \Omega $

B

$\frac{5}{3} \Omega $

C

$\frac{5}{2} \Omega $

D

$\frac{7}{2} \Omega $

Solution

$R =\frac{\rho\ell^{2}}{A \ell D} d = \frac{\rho d \ell^{2}}{m}$
$ R \propto \ell^{2}$
$ R = 12\Omega $ (new resistance of wire)
$ R_{1} = 2 \Omega R_{2} = 10 \Omega $
$ R_{eq} = \frac{10 \times2}{10+2} = \frac{5}{3} \Omega$

A

0.2 ohm

B

0.002 ohm

C

0.02 ohm

D

0.5 ohm

Solution

$S(0.5 - 0.002) = 50 \times 0.002$
$S = \frac{50 \times 0.002}{(0.5 - 0.002)} = \frac{0.1}{0.498} = 0.2 $

A

$a + \frac{b^2}{4c}$

B

$a + \frac{b^2}{c}$

C

$a + \frac{b^2}{2c}$

D

$a + \frac{b^2}{3c}$

Solution

$x =at +bt^{2} -ct^{3}$
$ v = \frac{dx}{dt}=a+2bt-3ct^{2} $
$ a= \frac{dv}{dt} =2b-6ct =0 \Rightarrow t = \frac{b}{3c} $
$ v_{\left(at t= \frac{b}{3c}\right)} = a +2b \left(\frac{b}{3c}\right)-3c\left(\frac{b}{3c}\right) $
$ =a +\frac{b^{2}}{3c} .$

A

$\sqrt{2} $

B

$\frac{4}{3}$

C

$\sqrt{3} $

D

$\frac{3}{2}$

Solution

$\frac{1}{f_{1}} = \frac{1}{2} \times\frac{2}{18} = \frac{1}{18} $
$ \frac{1}{f_{2}} = \frac{\left(\mu_{1} -1\right)}{-18} $
when $\mu_1$ is filled between lens and mirror
$ P= \frac{2}{18} - \frac{2}{18} \left(\mu_{1} - 1\right) = \frac{2-2\mu_{1} +2}{18} $
$ =F_{m} =- \left(\frac{18}{2-\mu_{1}}\right) $
$ 2=6-3\mu_{1} $
$ 3\mu_{1} = 4 $
$ \mu_{1} =4/3 $

A

$10^{-3}$

B

$10^{-1}$

C

$10^{-4}$

D

$10^{-2}$

Solution

$\tau =\vec{M} \times\vec{B} $
$ C \theta =i NAB $
$ 10^{-6} \times\frac{\pi}{180} = 10^{-3} \times10^{-4} \times175 \times B $
$ B =10^{-3} $ Tesla.

A

B

C

D

Solution

$\phi_{\text{outer}} \left(\mu_{0} nKte^{-\alpha t}\right) 4\pi R^{2}$
$ \varepsilon =\frac{-d\phi}{dt} =-Ce^{-\alpha t} \left[1-\alpha t\right]$
$ i_{\text{induced}} = \frac{-Ce^{-\alpha t} \left[1-at\right]}{\left(\text{Resistance}\right)} $
At $t=0 \,\,\,\,\,i_{\text{induced}} =-ve $

A

$10^{-3} K $

B

$10^{-4} K $

C

$10^{-1} K $

D

$10^{-5} K $

Solution

By law of conservation of energy
$\frac{1}{2} kx^2 = (m_1s_1+ m_2 s_2) \Delta T$
$\Delta T = \frac{16 \times 10^{-2}}{4384} = 3.65 \times 10^{-5}$ .

A

$\lambda_x + \lambda_y$

B

$\frac{\lambda_a \lambda_y}{\lambda_x + \lambda_y}$

C

$\frac{\lambda_x \lambda_y}{| \lambda_y - \lambda_x | }$

D

$\lambda_x - \lambda_y$

Solution

By momentum conservation
$P_x - P_y = P_P$
$\frac{h}{\lambda_{x}} - \frac{h}{\lambda_{y}} = \frac{h}{\lambda_{p}} $
$ \lambda_{p} = \frac{\lambda_{x}\lambda_{y}}{\left|\lambda_{y} -\lambda_{x}\right|} $

A

5 : 3

B

2 : 1

C

4 : 3

D

3 : 1

Solution

Magnification is $2$
If image is real, $x_1 = \frac{3f}{2}$
If image is virtual, $x_2 = \frac{f}{2}$
$\frac{x_1}{x_2} = 3: 1 $

A

$1.5 \times 10^{-7} \; rad$

B

$2.0 \times 10^{-7} \; rad$

C

$3.0 \times 10^{-7} \; rad$

D

$4.5 \times 10^{-7} \; rad$

Solution

Limit of resolution $= \frac{1.22\lambda}{d} $
$= \frac{1.22 \times600 \times 10^{-9}}{250 \times10^{-2} } $
$ = 2.9 \times10^{-7} \text{rad} $

A

2s

B

5s

C

2.5 s

D

3s

Solution

Given moment of inertia 'I' = 1.5 kgm$^2$
Angular Acc. "a" = $20\, Rad/s^2$
$KE =\frac{1}{2} I\omega^{2} $
$ 1200 = \frac{1}{2} 1.5 \times\omega^{2} $
$ \omega^{2} = \frac{1200\times2}{1.5} = 1600 $
$ \omega = 40 \text{rad}/s^{2} $
$ \omega =\omega_{0} +\alpha t $
$ 40=0+20t $
$ t=2\sec$

A

$OR$

B

$AND$

C

$NOR$

D

$NAND$

Solution

Method 1
Truth table can be formed as

A	B	Equivalent
0	0	0
0	1	1
1	0	1
1	1	1

Hence the Equivalent is "OR" gate.
Method 2

A

$15 \times 10^{-8} \; N$

B

$35 \times 10^{-8} \; N$

C

$10 \times 10^{-8} \; N$

D

$20 \times 10^{-8} \; N$

Solution

Force on the surface (25% reflecting and rest absorbing)
$F = \frac{25}{100} \left(\frac{2I}{C} \right)+ \frac{75}{100} \left(\frac{I}{C}\right) = \frac{125}{100}\left(\frac{I}{C}\right) $
$ =\frac{125}{100} \times\left(\frac{50}{3\times10^{8}} \right) = 20.83 \times10^{-8} N $

A

$37 \; cm^2$

B

$37.0 \; cm^2$

C

$37.03 \; cm^2$

D

$37.030 \; cm^2$

Solution

Total Area = $A_1 + A_2 + ...... A_7$
= $A + A + ......... 7$ times
= $37.03\, cm^2$.
Addition of $7$ terms all having $2$ terms beyond decimal, so final answer must have $2$ terms beyond decimal (as per rules of significant digits.)

A

proportional to carrier frequency

B

inversely proportional to modulation frequency

C

inversely proportional to carrier frequency

D

independent of both carrier and modulation frequency

Solution

The physical size of antenna of reciver and transmitter both inversely proportional to carrier frequency.

A

$E \propto \frac{1}{D}$

B

$E \propto \frac{1}{D^3}$

C

$E \propto \frac{1}{D^2}$

D

$E \propto \frac{1}{D^4}$

Solution

Electric field at $p = 2E_{1} \cos\theta_{1} -2E_{1}\cos\theta_{2} $
$ = \frac{2Kq}{\left(d^{2}+D^{2}\right)} \times\frac{D}{\left(d^{2}+D^{2}\right)^{1/2}} - \frac{2Kq}{\left[\left(2d\right)^{2}+D^{2}\right]} \times\frac{D}{\left[\left(2d\right)^{2}+D^{2}\right]^{1/2}} $
$ =2KqD \left[\left(d^{2}+D^{2}\right)^{-\frac{3}{2}} - \left(4d^{2} +D^{2}\right)^{-\frac{3}{2}} \right] $
$ = \frac{2KaD}{D^{3}} \left[\left(1- \frac{d^{2}}{D^{2}}\right)^{-3/2} -\left(1+ \frac{4d^{2}}{D^{2}}\right)^{-3/2}\right] $
Applying binomial approximation $ \because d < < D $
$ = \frac{2KqD}{D^{3}} \left[ 1- \frac{3}{2} \frac{d^{2}}{D^{2}} - \left(1- \frac{3\times4d^{2}}{2D^{2}}\right)\right] $
$ = \frac{2KqD}{D^{3}} \left[\frac{12}{2} \frac{d^{3}}{D^{2}} - \frac{3}{2} \frac{d^{2}}{D^{2}}\right] $
$ = \frac{9kqd^{2}}{D^{4}} $

A

A has one vibrational mode and B has two

B

Both A and B have a vibrational mode each

C

A is rigid but B has a vibrational mode

D

A has a vibrational mode but B has none

Solution

For A
$R = C_p - C_v = 7$
$C_{V} = \frac{fR}{2} = 22 \Rightarrow f= \frac{44}{7} = 6.3 $
$ f \simeq6 \begin{matrix} \to &5 \left(\text{Rotation}+\text{Translatioal}\right)\\ \to &1 \left(\text{Vibration}\right)\end{matrix} $
For B
$ R = C_{P} -C_{V} =9 $
$ C_{V} = \frac{fR}{2} = 21 \Rightarrow f= \frac{42}{9} $
$ f \simeq 5 \begin{matrix} \to &5 \left(\text{Rotation}+\text{Translatioal}\right)\\ \to &0 \left(\text{Vibration}\right)\end{matrix}$

A

40

B

100

C

25

D

50

Solution

$\vec{r} = 15t^{2} \hat{i} +\left(4-20 t^{2}\right)\hat{j} $
$ \vec{v} = \frac{d\vec{r}}{dt} =30 t \hat{i} + \left(-40t\right)\hat{j} $
$ \vec{a} = \frac{d \vec{v}}{dt} =30 \hat{i} - 40 \hat{j}$
$ \left|\vec{a}\right| = 50 m/s^{2} $

A

$T/R^2$ is a constant

B

$TR$ is a constant

C

$T^2/R^3$ is a constant

D

$T/R$ is a constant

Solution

$ m = \int^{R}_{0} \rho4\pi r^{2}dr $
$ m = 4\pi KR$
$ v \propto \sqrt{4\pi K} $
$ \frac{T}{R} = \frac{2\pi}{\sqrt{4\pi K}}$

A

$v / (2 \sqrt{2} )$

B

$2 \sqrt{2} v$

C

$ \sqrt{2} v$

D

$ v/ \sqrt{2} $

Solution

Linear momentum conservation
$m 2v + 2m v = m \times 0 + m \frac{v'}{\sqrt{2}} \times 2 $
$v' = 2 \sqrt{2} v $

A

0.5

B

0.7

C

0.6

D

0.8

Solution

In $1^{st}$ situation
$V_{b} \rho_{b}g = V_{s}\rho_{w}g $
$ \frac{V_{s}}{V_{b}} =\frac{\rho_{b}}{\rho_{w}} = \frac{4}{5} $ .....(i)
here $V_b$ is volume of block
$V_s$ is submerged volume of block
$\rho_b$ is density of block
$\rho_w$ is density of water
& Let $\rho_o$ is density of oil
finally in equilibrium condition
$ V_{b} \rho_{b} g = \frac{V_{b}}{2} \rho_{o} g + \frac{V_{b}}{2} \rho_{w}g $
$2\rho_{b} = \rho_{0} + \rho_{w} $
$\Rightarrow \frac{\rho_{o}}{\rho_{w}} = \frac{3}{5} = 0.6 $

A

2250 Hz

B

2060 Hz

C

2150 Hz

D

2300 Hz

Solution

$f =\frac{v+v_{0}}{v-v_{s}} f_{0} (v_{0} \& v_{s} $ is taken $\oplus$ when approaching each other)
$2000 = \frac{340+\left(-20\right)}{340-\left(-20\right)} f_{0 }$
$ f_{0} = 2250 $ Hz.

A

$\frac{2v^2}{7g}$

B

$\frac{v^2}{g}$

C

$\frac{2v^2}{5g}$

D

$\frac{v^2}{2g}$

Solution

Applying Linear momentum conservation
$mv = (m + M) v_c$
$v_c = \frac{v}{5}$
applying work energy theorem
$ - mgh = \frac{1}{2} (m + M) v_c^2 - \frac{1}{2} mv^2$
solve , $h = \frac{2v^2}{5g}$

A

$6.04 \,eV$

B

$13.60 \,eV$

C

$54.40 \,eV$

D

$48.36 \,eV$

Solution

$E = -13.6 \frac{z^{2}}{n^{2}} eV$
As He+ is 1st excited state
$\therefore z = 2, n = 2$
$E = -13.6\, eV$
As total energy of $He^+$ in 1st excited state is $-13.6\, eV$, ionisation energy should be $+13.6\,eV$.

A

$\frac{\theta_{2} + \theta_{1}}{2} $

B

$ \frac{\theta_{1}}{10} + \frac{9\theta_{2}}{10}$

C

$ \frac{\theta_{1}}{3} + \frac{2\theta_{2}}{3}$

D

$ \frac{\theta_{1}}{6} + \frac{5\theta_{2}}{6}$

Solution

Let the temperature of interface be " $\theta$ "
$i_1 = i_2$ {Steady state conduction}
$\frac{3KA\left(\theta_{2}-\theta\right)}{d} = \frac{KA\left(\theta - \theta_{1}\right)}{3d} $
$\theta = \frac{9 \theta_{2}}{10} + \frac{\theta_{1}}{10} $

A

$\frac{M\omega_0}{M +3 m} $

B

$\frac{M\omega_0}{M + m} $

C

$\frac{M\omega_0}{M + 2 m} $

D

$\frac{M\omega_0}{M + 6 m} $

Solution

Applying angular momentum conservation, about axis of rotation
$L_i = L_f$
$\frac{ML^2}{12} \omega_0 = \left( \frac{ML^2}{12} + m \left( \frac{L}{2} \right)^2 \times 2 \right) \omega $
$\Rightarrow \; \omega = \frac{M \omega_0}{M + 6m}$

A

$\frac{V}{K + n}$

B

V

C

$\frac{(n + 1) V}{(K + n)}$

D

$\frac{nV}{K + n}$

Solution

Total charge of the system = $CV + nCV$
$= (n + 1)CV$
After the insertion of dielectric of constant $K$
New potential (common)
$V_C = \frac{total \; charge}{total \; capacitance}$
$ = \frac{( n + 1) CV}{KC + nC} = \frac{(n + 1) V}{K + n}$

A

$10^{-5} \Omega m$

B

$10^{-6} \Omega m$

C

$10^{-7} \Omega m$

D

$10^{- 8} \Omega m$

Solution

$\rho = \frac{m}{ne^2 \tau}$
$ = 1.67 \times 10^{-8} \Omega m $

A

320 m/s, 120 Hz

B

180 m/s, 80 Hz

C

180 m/s, 120 Hz

D

320 m/s, 80 Hz

Solution

$3\left(\frac{v}{2\ell}\right) =240 $
$ 3 \frac{v}{2\times2} = 240 $
$v = 320\, m/s$
fundamental frequency $ = \frac{v}{2\ell} = \frac{320}{2\times2} = 80 $

A

(B) < (C) < (D) < (A)

B

(A) < (B) < (D) < (C)

C

(B) < (A) < (D) < (C)

D

(B) < (C) < (A) < (D)

Solution

$S_N1$ Reactivity order
Order $C > D > A > B$

A

bauxite

B

siderite

C

calamine

D

malachite

Solution

1. Bauxite- $AlO _{ x }( OH )_{3-2 x }$ where $0 < x < 1$
2. Siderite $- FeCO _3$
3. Calamine $- ZnCO _3$
4. Malachite $- CuCO _3 \cdot Cu ( OH )_2$

A

8

B

12

C

- 12

D

-8

Solution

$\Delta U = q + w$
$q = - 2\,kJ, W = 10\,kJ$
$\Delta U = 8\,kJ$

A

quartz

B

kieselguhr

C

cristobalite

D

tridymite

Solution

Kieselguhr is amorphous form of silica, it's a fact

A

B

C

D

A

The electron can be found at a distance $2a_0$ from the nucleus

B

The probability density of finding the electron is maximum at the nucleus.

C

The magnitude of potential energy is double that of its kinetic energy on an average.

D

The total energy of the electron is maximum when it is at a distance $a^0$ from the nucleus.

A

6 and 8

B

8 and 6

C

8 and 8

D

6 and 6

Solution

Towards common transition element and inner transition metal ion given ligand can have maximum denticities of 6 and 8 respectively.

A

(I), (II) and (III)

B

(II) and (III) only

C

(I) and (III) only

D

(I) and (II) only

Solution

All statements are correct
$B_2O_3 \rightarrow$ acidic
$Al_2O_3 \; \& \; Ga_2O_3$ are amphoteric oxides of In & Tl are basic

A

$O_2$

B

$NO$

C

$B_2$

D

$CO$

Solution

$O_2,NO,B_2$ are paramagnetic according to M.O.T. where as $CO$ is diamagnetic.

A

Lys-Asp

B

Ser-Lys

C

Gln-Asp

D

Asp-Gln

Solution

Lysine has $-NH_2$ group hence gives $\oplus ve$ carbyl amine test and serine has $-OH$ group hence gives $\oplus ve$ serric ammonium nitrate test

A

Only the reason is correct.

B

Both the assertion and reason are correct and the reason is the correct explanation for the assertion.

C

Only the assertion is correct.

D

Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion.

Solution

Assertion is correct as Haemetite ore is used for extraction of Fe.
Haemetite is an oxide ore so reason is incorrect

A

2.0 pm

B

2.0 nm

C

1.0 pm

D

0.1 nm

Solution

Millimoles $= 10 \times 10^{-3} = 10^{-2} $
Moles = $10^{-5}$
No. of molecules = $6 \times 10^{23} \times 10^{-5} = 6 \times 10^{+18 }$ surface area occupied by one molecule
$= \frac{0.24}{6\times10^{18} } = 0.04 \times10^{-18 } cm^{2} $
$ 4 \times10^{-20} =a^{2}$
$ a = 2\times10^{-10}cm =2pm$

A

C is the thermodynamically stable product.

B

Formation of A and B from C has highest enthalpy of activation.

C

D is kinetically stable product.

D

Activation enthalpy to form C is $5kJ \; mol^{-1}$ less than that to form D.

Solution

A + B $\to$ C + D
Activation enthalpy for $C = 20 - 5 = 15\,kJ/mol$
Activation enthalpy for $D = 15 - 5 = 10\,kJ/mol$

A

Ne

B

Ar

C

Xe

D

Kr

Solution

As $b \uparrow \; \Rightarrow $ slope $\uparrow$
Hence, Xe, will have highest slope

A

0.20

B

0.05

C

0.10

D

0.15

Solution

0.1 eq. of $Ni^{+2}$ will be discharged.
No. of eq = (No of moles) $\times$ (n-factor)
0.1 = (No. of moles) $\times$ 2
No. of moles of Ni = $\frac{0.1}{2} = 0.05 $

A

Acetone as substrate and methanol in stoichiometric amount

B

Propanal as substrate and methanol in stoichiometric amount

C

Acetone as substrate and methanol in excess

D

Propanal as substrate and methanol in excess

Solution

Acetone as substrate is less rective than propanal towards neucleophilic addition.

A

chain and dimeric

B

chain and chain

C

dimeric and dimeric

D

dimeric and chain

Solution

$BeCl_2$ exist as $(BeCl_2)_{n}$ polymeric chain in solid form, while $BeCl_2$ exist as dimer $(BeCl_2)_2$ in vapour phase.

A

B

C

D

Solution

PE diagram for $S_N1$
$S_N1$ is two step reaction where in step (1) formation of carbocation is RDS

A

B

C

D

A

berkelium (Bk) and californium (Cf)

B

nobelium (No) and lawrencium (Lr)

C

actinium (Ac) and thorium (Th)

D

neptunium (Np) and plutonium (Pu)

Solution

Np and Pu show maximum no. of oxidations states starting from +3 to +7 all oxidation states.

A

0.12 K

B

0.36 K

C

0.18 K

D

0.24 K

Solution

$K_f$ = $4\, K-kg/mol$
$m = 0.03\, mol/kg$
$i = 3$
$\Delta T_f = iK_f \times m$
$ \Delta T_f = 3 \times 4 \times 0.03 = 0.3 6K$

A

Neurotransmitter

B

Antidepressant

C

Antihistamine

D

Antacid

Solution

Nor adrenaline is a neutro transmitter and it belongs to catecholamine family that fanctions in brain & body as a hermone & neutro transmitter.

A

$C_6H_5SO_2Cl$

B

$C_6H_5COCl$

C

$SOCl_2$

D

$(COCl)_2$

Solution

[Benzene Sulphonyl chloride]

A

troposphere

B

mesosphere

C

stratosphere

D

thermosphere

Solution

It's a fact, the layer of atmosphere between 10km to 50km above sea level is called as stratosphere.

A

lowest dissociation enthalpy

B

strongest van der Waals' interactions

C

strongest hydrogen bonding

D

lowest ionic character

Solution

HF has highest boiling point among hydrogen halides because it has strongest hydrogen bonding

A

1.08

B

1.48

C

1.51

D

1.35

Solution

$\frac{W}{W} \% = 20$
$100\, gm$ solution has $20\, gm\, KI$
$80 \,gm$ solvent has $20\, gm\, KI$
$ m = \frac{\frac{20}{166}}{\frac{80}{1000}} = \frac{20\times1000}{166\times80} = 1.506 \simeq 1.51 \,mol/kg$

A

Formaldehyde

B

Ammonia

C

Methylamine

D

N-Methyl urea

A

(I) and (II) only

B

(I), (II) and (III)

C

(I) and (III) only

D

(II) and (III) only

Solution

Based on NCERT, statement of limitations of VBT, I & III are correct

A

(A)

B

(C)

C

(D)

D

(B)

A

B

C

D

Solution

OH is activating and ortho para directing group towards ESR

A

$ 2 \sqrt{2} + 1 $

B

$ 2 \sqrt{2} -1 $

C

$ \sqrt{2} + 1 $

D

$ 2 \sqrt{2} - 1 $

Solution

$T:y \left(\beta\right) = \frac{1}{2} \left(x+\beta^{2}\right)$
$ 2y \beta=x+\beta^{2} $
$ y=\left(\frac{1}{2\beta}\right) x +\frac{\beta}{2} $
$ m = \frac{1}{2\beta} ; C = \frac{\beta}{2}$
$ \frac{\beta}{2} = \pm \sqrt{\frac{1}{4\beta^{2}} + \frac{1}{2}} $
$ \frac{\beta^{2}}{4} = \frac{1}{4\beta^{2}} + \frac{1}{2} $
$ \frac{\beta^{2}}{4} = \frac{1}{4\beta^{2}} + \frac{1}{2} \frac{\beta^{2}}{4} = \frac{1+2\beta^{2}}{4\beta^{2}}$
$ \Rightarrow \beta^{4} -2\beta^{2} -1 = 0 $
$ \left(\beta^{2} -1\right)^{2} = 2$
$ \beta^{2} -1 = \sqrt{2} $
$ \beta^{2} = \sqrt{2}+1 $

A

190

B

262

C

225

D

157

Solution

$\frac{n\left(n+1\right)}{2} +99 =\left(n-2\right)^{2} $
$ n^{2} + n + 198 = 2\left(n^{2} + 4 - 4n\right)$
$ n^{2} - 9n - 190 = 0 $
$ n^{2} - 19n + 10 - 190 = 0$
$ n\left(n - 19\right) + 10\left(n - 19\right) = 0$
$n = 19$

A

$0$

B

$2 f'(2)$

C

$12 f '(2)$

D

$24 f '(2)$

Solution

$\displaystyle\lim{x\to2} \frac{\int^{f\left(x\right)}_{6} 2t dt }{x-2} $
L Hopital Rule
$ \displaystyle\lim_{x\to2} \frac{2f \left(x\right)f '\left(x\right)}{1} = 2f \left(2\right) f '\left(2\right) 12f '\left(2\right)$

A

$\frac{3}{4}$

B

-4

C

$\frac{1}{2}$

D

$ - \frac{1}{4}$

Solution

$\begin{pmatrix}2&3&-1\\ 1&K&-2\\ 2&-1&1\end{pmatrix} = 0 $
By solving $ K = \frac{9}{2} $
$ 2x+3y-z=0 $ .....(1)
$ x+ \frac{9}{2} y-2z=0 $ ...(2)
$ 2x - y + z = 0 $ ....(3)
$ \left(1\right)-\left(3\right) \Rightarrow 4y - 2z = 0 $
$ 2y = z$ ...(4)
$ \frac{y} {z} = \frac{1}{2} $ ...(5)
put z from eqn. (4) into (1)
$ 2x + 3y - 2y = 0 $
$ 2x + y = 0 $
$ \frac{x}{y} = - \frac{1}{2} $ ....(6)
$ \frac{\left(6\right)}{\left(5\right)} \frac{z}{x} = - 4 $
$ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + K = \frac{1}{2} $

A

(-4, 6)

B

(6, -2)

C

(-6, 4)

D

(4, -2)

Solution

Circle touches internally
$C_1(0, 0); r_1 = 2$
$C_2 : (-3, -4); r_2 = 7$
$C_1C_2 = |r_1 - r_2|$
$S_1 - S_2 = 0 \; \Rightarrow $ eqn. of common tangent
$6x + 8y - 20 = 0$
$3x + 4y = 10$
$(6, -2)$ satisfy it

A

-25

B

25

C

-36

D

-35

Solution

$a - d + a + a + d = 33$ $\Rightarrow $ $a = 11$
$a(a^2 - d^2) = 1155$
$121 - d^2 = 105$
$d^2 = 16 \; \Rightarrow \; d = \pm 4$
If $d = 4$ then Ist term = $7$
If $d = -4$ then Ist term = $15$
$T_{11} = 7 + 40 = 47$

A

$\frac{\pi}{4} - \frac{1}{2} \log_{e}2$

B

$\frac{\pi}{2} - \log_{e}2$

C

$\frac{\pi}{2} - \frac{1}{2} \log_{e}2$

D

$\frac{\pi}{4} - \log_{e}2$

Solution

$I = \int^{1}_{0} x \tan \left(\frac{1}{1+x^{2}\left(x^{2}-1\right)}\right)dx$
$ I = \int^{1}_{0} x \left(\tan^{-1}x^{2} - \tan^{-1}\left(x^{2}-1\right)\right)dx $
$ x^{2} =t \Rightarrow 2xdx =dt $
$ I = \frac{1}{2} \int^{1}_{0} \left(\tan^{-1} t - \tan^{-1} \left(t-1\right)\right)dx $
$ = \frac{1}{2} \int^{1}_{0} \tan^{-1} t dt - \frac{1}{2} \int^{1}_{0} \tan^{-1} \left(t-1\right)dt $
$ = \frac{1}{2} \int^{1}_{0} \tan^{-1} t dt - \frac{1}{2} \int^{1}_{0} \tan^{-1} dt = \int^{1}_{0} \tan^{-1} dt $
$ \tan^{-1} t = \theta \Rightarrow t = \tan\theta $
$ dt = \sec^{2} \theta d\theta$
$ \int^{\pi/4}_{0} \theta .\sec^{2} \theta d \theta $
$ I = \left(\theta. \tan\theta\right)^{\pi/4}_{0} - \int^{\pi/4}_{0} \tan \theta d\theta $
$ = \left(\frac{\pi}{4} - 0\right) - ln \left(\sec\theta\right) ^{\pi/4}_{0} $
$ = \frac{\pi}{4} - \left(\ell n \sqrt{2} - 0\right) $
$ = \frac{\pi}{4} - \frac{1}{2} \ell n2$

A

$\frac{1}{36}$

B

$\frac{1}{32}$

C

$\frac{1}{18}$

D

$\frac{1}{16}$

Solution

$\left(\sin10^{\circ} \sin30^{\circ} \sin70^{\circ}\right)\sin30^{\circ} $
$ \frac{1}{4} \left(\sin30^{\circ}\right) = \frac{1}{4} . \frac{1}{4} = \frac{1}{16}$

A

$ 5\,I m (\omega) < 1 $

B

$ 4 \,Im (\omega) > 5 $

C

$ 5 \,Re\, (\omega) > 1 $

D

$ 5 \,Re (\omega) > 4 $

Solution

$\left|z\right| < 1 $
$ 5\omega\left(1-z\right)=5+3z $
$ 5\omega - 5\omega z = 5 +3z $
$ z = \frac{5\omega-5}{3+5\omega}$
$ \left|z \right| = 5 \left|\frac{\omega-1}{3+5\omega} \right|<1 $
$ 5 \left|\omega-1\right|< \left|3+5\omega \right| $
$ 5 \left|\omega-1\right| < 5 \left|\omega+ \frac{3}{5}\right| $
$ \left|\omega-1\right| < 5 \left|\omega - \left( - \frac{3}{5}\right)\right| $

A

964

B

625

C

227

D

232

Solution

$\frac{^{n}C_{r-1}}{^{n}C_{r}} = \frac{2}{15} $
$ \frac{\frac{n!}{\left(r-1\right)!\left(n-r+1\right)!}}{\frac{n!}{r!\left(n-r\right)!}} = \frac{2}{15} $
$ \frac{r}{n-r+1} = \frac{2}{15} $
$ 15r =2n-2r+2 $
$ 17r = 2n+2 $
$ \frac{^{n}C_{r}}{^{n}C_{r+1}} = \frac{15}{70} $
$ \frac{\frac{n!}{r!\left(n-r\right)!}}{\frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}} = \frac{3}{14}$
$ \frac{r+1}{n-r} = \frac{3}{14} $
$ 14r +14 =3n -3r $
$ 3n - 17r =14$
$ \frac{2n-17r =- 2 }{n=16} $
$ 17r=34, r = 2 $
${^{16}C_{1}} , {^{16}C_{2}} , {^{16}C_{3}} $
$ \frac{^{16}C_{1} + ^{16}C_{2} + ^{16}C_{3}}{3 } = \frac{16+120+560}{3} $
$\frac{680+16}{3} = \frac{696}{3} = 232 $

A

$ - \frac{\pi^2}{4\sqrt{3}}$

B

$ - \frac{\pi^2}{2}$

C

$ - \frac{\pi^2}{2\sqrt{3}}$

D

$\frac{\pi^2}{2\sqrt{3}}$

Solution

$\frac{dy}{dx} -y \tan x = 6x \sec x $
$ y\left(\frac{\pi}{3}\right) = 0 ; y \left(\frac{\pi}{6}\right) = 7 $
$ e^{\int pdx} = e^{-\int\tan xdx} = e^{\ell n \cos x} = \cos x $
$ y.\cos x = \int6x \sec x \cos x dx $
$ y.\cos x = \frac{6x^{2}}{2} + C $
$ y = 3x^{2}\sec x + C \sec x $
$ 0=3 . \frac{\pi^{2}}{9} .\left(2\right)+C\left(2\right) $
$ 2C = \frac{-2\pi^{2}}{3} \Rightarrow C = - \frac{\pi^{2}}{3} $
$ y\left(\pi/6\right) = 3. \frac{\pi^{2}}{36} . \left(\frac{2}{\sqrt{3}}\right) + \left(\frac{2}{\sqrt{3}}\right) . \left(- \frac{\pi^{2}}{3}\right) $
$ \Rightarrow y = - \frac{\pi^{2}}{2\sqrt{3}} $

A

$\frac{2}{5}$

B

$\frac{2}{\sqrt{5}}$

C

$\frac{\sqrt{2}}{5}$

D

$\sqrt{\frac{2}{5}}$

Solution

$\left(\frac{-1}{a-1}\right)\left(\frac{-2}{a^{2}} \right) = - 1 $
$ 2= -\left(a^{2}\right)\left(a-1\right) $
$ a^{3} -a^{2} + 2 =0 $
$ \left(a+1\right)\left(a^{2} - 2a +2 \right) = 0 $
$ \therefore a =- 1 $
$\left.\begin{aligned} L_1 : x -2 y + 1 = 0 \\ L_2 : 2x +y - 1 = 0 \end{aligned}\right\rbrace $
$ 0\left(0,0\right) P\left(\frac{1}{5}, \frac{3}{5}\right) $
$ OP = \sqrt{\frac{1}{25} + \frac{9}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}} $

A

$2 / \pi $

B

$1 / 5 \pi $

C

$1/ 10 \pi $

D

$1/ 15 \pi $

Solution

$\tan\theta = \frac{1}{2} = \frac{r}{h} $
$ r = \frac{h}{2} $
$V = \frac{1}{3} \pi r^{2}h $
$ V = \frac{1}{3} \pi. \frac{h^{3}}{4} $
$\frac{dV}{dt} = \frac{\pi}{12} \left(3h\right)^{2} \left(\frac{dh}{dt}\right) $
$ 5 = \frac{\pi}{4} .\left(100\right) \frac{dh}{dt} \Rightarrow \frac{dh}{dt} = \frac{1}{5\pi} $

A

$\frac{5}{2}\left(2+\sqrt{3}\right) $

B

$ 5( \sqrt{3} + 1)$

C

$ 5( 2 + \sqrt{3} )$

D

$ 10 ( \sqrt{3} - 1)$

Solution

$\tan 15^{\circ} = \frac{5}{x}$
$ 2 - \sqrt{13} = \frac{5}{x} $
$ x=5 \left(2+\sqrt{3}\right) $

A

$ 2 \sqrt{34}$

B

$ \sqrt{34}$

C

6

D

$ \sqrt{17}$

Solution

$\overrightarrow{AD} .\left(3\hat{i} +4\hat{k}\right) = 0 $
$ 3\left(3\lambda-3\right)+0+4\left(4\lambda-2\right)=0 $
$\left(9\lambda-9\right)+\left(16\lambda-8\right)=0 $
$25\lambda=17 \Rightarrow \lambda = \frac{17}{25} $
$ \therefore \overrightarrow{AD} = \left(\frac{51}{25} -3\right)\hat{i} +2 \hat{j} + \left(\frac{68}{25} -2\right) \hat{k} $
$= \frac{24}{ 25} \hat{i} +2 \hat{j} +\frac{18}{25} \hat{k} $
$ \left|\overrightarrow{AD}\right| = \sqrt{\frac{576}{625} +4+ \frac{324}{625}} $
$ = \sqrt{\frac{900}{625} + 4} = \sqrt{\frac{3400}{625}} $
$= \sqrt{34} . \frac{10}{25} = \frac{2}{5} \sqrt{34} $
Area of $ \Delta = \frac{1}{2} \times5 \times\frac{2\sqrt{34}}{5} = \sqrt{34} $

A

6

B

2

C

3

D

4

Solution

$A^{T}A =3I_{3} $
$ \begin{pmatrix}0&2x&2x\\ 2y&y&-y\\ 1&-1&1\end{pmatrix}\begin{pmatrix}0&2y&1\\ 2x&y&-1\\ 2x&-y&1\end{pmatrix} = \begin{pmatrix}3&0&0\\ 0&3&0\\ 0&0&3\end{pmatrix} $
$ \begin{pmatrix}8x^{2}&0&0\\ 0&6y^{2}&0\\ 0&0&3\end{pmatrix} = \begin{bmatrix}3&0&0\\ 0&3&0\\ 0&0&3\end{bmatrix} $
$ 8x^{2}= 3 $
$ 6y^{2}=3 $
$ x^{2} =3/8 $
$ y^{2} =1/2 $
$ x = \pm \sqrt{\frac{3}{8}} ; y= \pm \sqrt{\frac{1}{2}} $

A

$4 \pi ( 2 - \sqrt{2})$

B

$8\pi ( 3 - 2 \sqrt{2})$

C

$4 \pi ( 3 + \sqrt{2})$

D

$8 \pi ( 2 - \sqrt{2})$

Solution

Equation of circle is
$\left(x-1\right)^{2} +\left(y-2\right)^{2} +\lambda\left(x-y+1\right)=0$
$ \Rightarrow x^{2} +y^{2} +x\left(\lambda-2\right)+y\left(-4-\lambda\right) +\left(5+\lambda\right) = 0$
As cirlce touches x axis then $ g^{2} -c=0 $
$ \frac{\left(\lambda-2\right)^{2}}{4} = \left(5+\lambda\right)$
$ \lambda^{2} + 4-4\lambda=20 +4\lambda $
$ \lambda^{2} -8\lambda-16 =0 $
$ \lambda = \frac{8\pm \sqrt{128}}{2} $
$ \lambda = 4 \pm4\sqrt{2} $
Radius $ = \left|\frac{\left(-4-\lambda\right)}{2}\right|$
Put $\lambda $ and get least radius.

A

$\frac{2}{5-\pi} $

B

$\frac{2}{\pi - 5 } $

C

$\frac{2}{\pi + 5 } $

D

$\frac{ - 2}{\pi + 5 } $

Solution

$a\left|\pi-5\right|+1 = b \left|5-\pi\right| +3$
$ a\left(5-\pi\right)+1 =b\left(5-\pi\right)+3 $
$ \left(a -b\right)\left(5-\pi\right) = 2 $
$ a-b = \frac{2}{5-\pi}$

A

Both $\displaystyle\lim_{x \to 4-} f(x)$ and $\displaystyle\lim_{x \to 4+} f(x)$ exist but are not equal

B

$\displaystyle\lim_{x\to 4-} f(x)$ exists but $\displaystyle\lim_{x \to 4+} f(x)$ does not exist

C

$\displaystyle\lim_{x\to 4 +} f(x) $ exists but $\displaystyle\lim_{x\to 4-} f(x)$ does not exist

D

$f$ is continuous at $x = 4$

Solution

$f\left(x\right) =\left[x\right] - \left[\frac{x}{4}\right] $
$ \lim_{x\to4+} f\left(x\right) =\lim_{x\to4+} \left(\left(\left[x\right] - \left[\frac{x}{4}\right]\right)\right) = 4-1 =3 $
$ \lim_{x\to4+} f\left(x\right)= \lim_{x\to4-} \left(\left[x\right]- \frac{x}{4}\right) =3-0 =3 $
$ f\left(x\right) = 3 $
$ \therefore $ continuous at x = 4

A

$\sec x - \tan x - \frac{1}{2} $

B

$ x \sec x + \tan x + \frac{1}{2} $

C

$\sec x + x \tan x - \frac{1}{2} $

D

$\sec x + \tan x + \frac{1}{2} $

Solution

$\int e^{\sec x} \left(\sec x \tan x f\left(x\right)+\left(\sec x \tan x+\sec^{2}x \right)dx\right) $
$ =e^{\sec x }f\left(x\right)+C $
Diff. both sides w.r.t. 'x'
$ e^{\sec x} \left(\sec x \tan x f\left(x\right) +\left(\sec x \tan x+\sec^{2}x\right)\right) $
$=e^{\sec x}.\sec x \tan x f\left(x\right)+e^{\sec x} f'\left(x\right) $
$ f'\left(x\right) =\sec^{2}x +\tan x \sec x $
$ \Rightarrow f\left(x\right) =\tan x +\sec x +c $

A

$ 8 \sqrt{3}$

B

$ 4 \sqrt{3}$

C

$10 \sqrt{5}$

D

$8 \sqrt{5}$

Solution

$SOR = \frac{3}{m^{2} +1} \Rightarrow \left(S.O.R\right)_{max} = 3 $
when $m = 0$
$ \alpha+\beta=3 $
$ \alpha\beta = 1 $
$ \left|\alpha^{2} -\beta^{2}\right| =\left| \left|\alpha-\beta\right|\left(\alpha^{2} + \beta^{2} +\alpha\beta\right) \right| $
$ = \left|\sqrt{\left(\alpha-\beta\right)^{2}-\alpha\beta} \left(\left(\alpha+\beta\right)^{2} -\alpha\beta\right)\right| $
$ = \left|\sqrt{9-4} \left(9-1\right)\right| $
$= \sqrt{5} \times8 $

A

12.8

B

13.5

C

13.9

D

13

Solution

Let population = $100$
$n(A) = 25$
$n(B) = 20$
$n (A \cap B) = 8 $
$ n (A \cap \bar{B}) = 17$
$ n (\bar{A} \cap B) = 12$
$\frac{30}{100} \times 17 + \frac{40}{100} \times 12 + \frac{50}{100} \times 8$
$5.1 + 4.8 + 4 = 13.9$

A

$ 63 \sqrt{5}$

B

$ 205 \sqrt{5}$

C

$ 17 / \sqrt{5}$

D

$ 11 / \sqrt{5}$

Solution

$\lambda\left(x+y+z-6\right) +2x+3y+z+5=0 $
$\left(\lambda+2\right)x+\left(\lambda+3\right)y+\left(\lambda+1\right)z+5 -6\lambda=0 $
$ \lambda+1 =0 \Rightarrow \lambda=-1 $
$ P:x +2y +11 =0 =\frac{11}{\sqrt{5}} $

A

F, T, T

B

T, F, F

C

T, T, F

D

F, F, F

Solution

$P \; \Rightarrow (q \vee r) : F$
$P : T q \vee r : F$
$P : T : q : F : r : F$

A

$( 1 , 2 ) \cup (2, \infty)$

B

$( - 1 , 0 ) \cup (1, 2) \cup (3, \infty) $

C

$( - 1 , 0 ) \cup (1, 2) \cup (2, \infty) $

D

$( - 2 , -1 ) \cup ( -1, 0) \cup (2, \infty) $

Solution

$ 4 - x^2 \ne 0 ; x^3 - x > 0 $
$\therefore \; D_f \in (-1 , 0) \cup (1,2) \cup (2, \infty)$

A

915

B

946

C

945

D

916

Solution

$T_{r} = r\left(2r -1\right) $
$ S = \sum2r^{2} -\sum r $
$ S = \frac{2.n\left(n+1\right)\left(2n+1\right)}{6} - \frac{n\left(n+1\right)}{2}$
$ S_{11} = \frac{2}{6} .\left(11\right)\left(12\right)\left(23\right) - \frac{11\left(12\right)}{2} =\left(44\right)\left(23\right)-66 = 946 $

A

7/3

B

9/4

C

7/2

D

8/3

Solution

$\frac{34+x}{2} =35 $
$ x=36 $
$ 42 = \frac{10+22+26+29+34+36+42+67+70+y}{10} $
$ 420-336 =y \Rightarrow y = 84 $
$\frac{y}{x} = \frac{84}{36} = \frac{7}{3} $

A

$\frac{53}{3}$

B

18

C

30

D

16

Solution

$y^{2} = 2x $
$ x-y-4 = 0 $
$ \left(x-4\right)^{2} = 2x $
$ x^{2} +16-8x-2x=0 $
$ x^{2} -10x+16=0 $
$ x=8,2 $
$ y =4,-2 $
$ A = \int^{4}_{-2} \left(y+4 - \frac{y^{2}}{2}\right)dy $
$ = \frac{y^{2}}{2} \bigg|^4_{-2} + 4 y \bigg|^4_{-2} - \frac{y^3}{6} \bigg|^4_{-2} $
$ = ( 8 - 2) + 4 (6) - \frac{1}{6} (64 + 8)$
$ 6+ 24 - 12 = 18 $

A

$\frac{5 \pi}{12}$

B

$\frac{5 \pi}{6}$

C

$\frac{2 \pi}{3}$

D

$\frac{\pi}{4}$

Solution

$\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1 $
$ \frac{1}{4} + \frac{1}{2} + \cos^{2} \gamma = 1$
$ \cos^{2} \gamma = 1 - \frac{3}{4} = \frac{1}{4} $
$ \cos^{2} \gamma = \pm \frac{1}{2} \Rightarrow \gamma = \frac{\pi}{3}$ or $ \frac{2\pi}{3} $

A

72

B

84

C

98

D

56

Solution

$\frac{3\left(a+5\right)}{2} = -1+7 $
$ a+5 = \frac{2\left(6\right)}{3} $
$a = - 1$
sides = $6$ and $14$
$ \Rightarrow A = 84 $