JEE Main Full Mock Test: Practice Test 03
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- $\frac{m v_0^3}{2\sqrt{2} g}$
- $\frac{m v_0^3}{4\sqrt{2} g}$
- $\frac{m v_0^3}{\sqrt{2} g}$
- $\frac{m v_0^3}{2 g}$
Answer: (B) $\frac{m v_0^3}{4\sqrt{2} g}$
At the highest point, velocity is horizontal: $v_x = v_0 \cos 45^\circ = \frac{v_0}{\sqrt{2}}$. Maximum height $H = \frac{v_0^2 \sin^2 45^\circ}{2g} = \frac{v_0^2}{4g}$. Angular momentum about point of projection $L = m v_x H = m \left(\frac{v_0}{\sqrt{2}}\right) \left(\frac{v_0^2}{4g}\right) = \frac{m v_0^3}{4\sqrt{2} g}$.
- $W_1 < W_2$
- $W_1 = W_2$
- $W_1 > W_2$
- $W_1 = 2 W_2$
Answer: (C) $W_1 > W_2$
In a $P-V$ diagram, the isothermal curve lies above the adiabatic curve during expansion. Area under $P-V$ curve represents work done. Thus $W_{\text{isothermal}} > W_{\text{adiabatic}}$, i.e., $W_1 > W_2$.
- $\frac{1}{2} \frac{\varepsilon_0 A V^2}{d} (K - 1)$
- $\frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(1 - \frac{1}{K}\right)$
- $\frac{\varepsilon_0 A V^2}{d} \left(1 - \frac{1}{K}\right)$
- $\frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(1 + \frac{1}{K}\right)$
Answer: (B) $\frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(1 - \frac{1}{K}\right)$
Initial energy $U_i = \frac{1}{2} C_0 V^2$. When battery is disconnected, charge $Q$ is constant. Final capacitance $C_f = K C_0$. Final energy $U_f = \frac{Q^2}{2 C_f} = \frac{U_i}{K}$. Work done magnitude $= U_i - U_f = \frac{1}{2} \frac{\varepsilon_0 A V^2}{d} \left(1 - \frac{1}{K}\right)$.
- $\frac{T}{4}$
- $\frac{T}{6}$
- $\frac{T}{12}$
- $\frac{T}{8}$
Answer: (D) $\frac{T}{8}$
Equation of motion: $x = A \sin(\omega t)$. For $x = \frac{A}{\sqrt{2}}$, $\sin(\omega t) = \frac{1}{\sqrt{2}} \implies \omega t = \frac{\pi}{4} \implies \frac{2\pi}{T} t = \frac{\pi}{4} \implies t = \frac{T}{8}$.
- $R$
- $\frac{R}{\sqrt{2}}$
- $\sqrt{2} R$
- $\frac{R}{2}$
Answer: (B) $\frac{R}{\sqrt{2}}$
Electric field along axis of ring $E = \frac{k Q x}{(R^2 + x^2)^{3/2}}$. Differentiating $E$ w.r.t $x$ and equating to $0$ gives $x = \frac{R}{\sqrt{2}}$.
- $100\text{ rad/s}$
- $1000\text{ rad/s}$
- $316\text{ rad/s}$
- $500\text{ rad/s}$
Answer: (C) $316\text{ rad/s}$
Resonant angular frequency $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{100 \times 10^{-3} \times 100 \times 10^{-6}}} = \sqrt{10^5} \approx 316.2\text{ rad/s}$.
- $\frac{r_1}{r_2}$
- $\frac{r_2}{r_1}$
- $\left(\frac{r_2}{r_1}\right)^2$
- $\sqrt{\frac{r_2}{r_1}}$
Answer: (B) $\frac{r_2}{r_1}$
By conservation of angular momentum: $m v_1 r_1 = m v_2 r_2 \implies \frac{v_1}{v_2} = \frac{r_2}{r_1}$.
- $0.123\text{ \AA}$
- $12.27\text{ \AA}$
- $1.227\text{ \AA}$
- $2.45\text{ \AA}$
Answer: (C) $1.227\text{ \AA}$
$\lambda = \frac{12.27}{\sqrt{V}}\text{ \AA} = \frac{12.27}{\sqrt{100}}\text{ \AA} = 1.227\text{ \AA}$.
- $20\text{ cm}$
- $\frac{40}{3}\text{ cm}$
- $40\text{ cm}$
- $\frac{80}{3}\text{ cm}$
Answer: (D) $\frac{80}{3}\text{ cm}$
Magnification $m = -3 = \frac{v}{u} \implies v = -3u$. Lens formula: $\frac{1}{20} = \frac{1}{-3u} - \frac{1}{u} = -\frac{4}{3u} \implies 3u = -80 \implies |u| = \frac{80}{3}\text{ cm}$.
- $2 \times 10^{-5}\text{ T}$
- $4 \times 10^{-5}\text{ T}$
- $1 \times 10^{-5}\text{ T}$
- $8 \times 10^{-5}\text{ T}$
Answer: (B) $4 \times 10^{-5}\text{ T}$
$B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(10)}{2\pi (0.05)} = 4 \times 10^{-5}\text{ T}$.
- $25\%$
- $75\%$
- $50\%$
- $90\%$
Answer: (C) $50\%$
$T_1 = 600\text{ K}, T_2 = 300\text{ K}$. Efficiency $\eta = 1 - \frac{300}{600} = 0.5 = 50\%$.
- $2 R$
- $4 R$
- $R / 2$
- $R / 4$
Answer: (B) $4 R$
Since volume is constant, stretching to length $2L$ reduces area to $A/2$. New resistance $R' = \rho \frac{2L}{A/2} = 4R$.
- $330\text{ m/s}$
- $400\text{ m/s}$
- $404\text{ m/s}$
- $360\text{ m/s}$
Answer: (C) $404\text{ m/s}$
Beat frequency $= v \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) = v \left(1 - \frac{1}{1.01}\right) = 4 \implies v = 404\text{ m/s}$.
- $25\text{ J}$
- $12\text{ J}$
- $16\text{ J}$
- $9\text{ J}$
Answer: (D) $9\text{ J}$
$W = \vec{F} \cdot \vec{d} = (3)(3) + (4)(0) + (0)(4) = 9\text{ J}$.
- $[M^1 L^2 T^{-2} A^{-1}]$
- $[M^1 L^1 T^{-2} A^{-1}]$
- $[M^1 L^2 T^{-1} A^{-2}]$
- $[M^1 L^0 T^{-2} A^{-1}]$
Answer: (A) $[M^1 L^2 T^{-2} A^{-1}]$
$[\Phi] = [B][A] = [M T^{-2} A^{-1}][L^2] = [M^1 L^2 T^{-2} A^{-1}]$.
- Diffusion of majority charge carriers
- Drift of minority charge carriers
- Drift of majority charge carriers
- Diffusion of minority charge carriers
Answer: (B) Drift of minority charge carriers
Reverse current in p-n junction is caused by the drift of minority charge carriers across the depletion layer under reverse bias.
- $16\text{ J}$
- $12\text{ J}$
- $24\text{ J}$
- $8\text{ J}$
Answer: (C) $24\text{ J}$
Total $\text{KE} = \frac{1}{2} m v^2 \left(1 + \frac{k^2}{R^2}\right) = \frac{1}{2}(2)(16)(1 + 0.5) = 24\text{ J}$.
- $I_0$
- $I_0 / 2$
- $I_0 / 4$
- $I_0 / \sqrt{2}$
Answer: (B) $I_0 / 2$
When unpolarized light passes through a linear polarizer, the transmitted intensity is half of the initial intensity: $I = I_0 / 2$.
- $\sqrt{2} + 1$
- $1$
- $\sqrt{2} - 1$
- $\frac{1}{\sqrt{2}}$
Answer: (C) $\sqrt{2} - 1$
Time to top $t = u/g$. Time from top to $H/2$ is $t_2 = u/(\sqrt{2}g)$. Upward time to $H/2$ is $t_1 = t - t_2$. Ratio $t_1 / t_2 = \sqrt{2} - 1$.
- $\text{Fe}^{56}$
- $\text{U}^{235}$
- $\text{He}^4$
- $\text{O}^{16}$
Answer: (A) $\text{Fe}^{56}$
$\text{Fe}^{56}$ has maximum binding energy per nucleon (about $8.8\text{ MeV}$).
- $\sqrt{3}$
- $\frac{1}{\sqrt{3}}$
- $\frac{1}{2}$
- $\frac{\sqrt{3}}{2}$
Answer: (B) $\frac{1}{\sqrt{3}}$
Constant velocity $\implies mg \sin 30^\circ = \mu_k mg \cos 30^\circ \implies \mu_k = \tan 30^\circ = \frac{1}{\sqrt{3}}$.
- $\frac{2T}{r}$
- $\frac{T}{r}$
- $\frac{4T}{r}$
- $\frac{8T}{r}$
Answer: (C) $\frac{4T}{r}$
A soap bubble has two surfaces, so $\Delta P = \frac{4T}{r}$.
- $180^\circ$
- $90^\circ$
- $45^\circ$
- $0^\circ$
Answer: (D) $0^\circ$
$U = -p E \cos\theta$. Minimum value occurs at $\theta = 0^\circ$ where $U = -pE$.
- $2 \times 10^{-7}\text{ T}$
- $1.8 \times 10^{10}\text{ T}$
- $2 \times 10^{-6}\text{ T}$
- $6 \times 10^{-7}\text{ T}$
Answer: (A) $2 \times 10^{-7}\text{ T}$
$B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7}\text{ T}$.
- $2.75 \times 10^4\text{ J/m}^3$
- $1.375 \times 10^4\text{ J/m}^3$
- $5.5 \times 10^4\text{ J/m}^3$
- $1.1 \times 10^4\text{ J/m}^3$
Answer: (B) $1.375 \times 10^4\text{ J/m}^3$
Energy density $= \frac{1}{2} Y (\text{strain})^2 = \frac{1}{2} (1.1 \times 10^{11}) (0.5 \times 10^{-3})^2 = 1.375 \times 10^4\text{ J/m}^3$.
- $\text{NH}_3$
- $\text{H}_2\text{O}$
- $\text{BF}_3$
- $\text{SO}_2$
Answer: (C) $\text{BF}_3$
$\text{BF}_3$ has symmetrical trigonal planar geometry ($sp^2$), cancelling bond dipoles to give zero overall dipole moment.
- $6$ and $+3$
- $4$ and $+2$
- $6$ and $+2$
- $4$ and $+3$
Answer: (A) $6$ and $+3$
'en' is bidentate $\implies \text{Coordination Number} = 2(2) + 2 = 6$. Oxidation state: $x + 2(0) - 2 = +1 \implies x = +3$.
- Methylamine
- Dimethylamine
- Trimethylamine
- Aniline
Answer: (B) Dimethylamine
For methyl amines in aqueous medium, order of basicity is $2^\circ > 1^\circ > 3^\circ > \text{Aniline}$. Dimethylamine is most basic.
- $5$
- $3$
- $1$
- $0$
Answer: (C) $1$
$\text{Fe}^{3+} = 3d^5$. $\text{CN}^-$ is strong field, pairs electrons to $t_{2g}^5 e_g^0$, leaving 1 unpaired electron.
- $\text{Na}_2\text{CO}_3$
- $\text{K}_2\text{CO}_3$
- $\text{Cs}_2\text{CO}_3$
- $\text{Li}_2\text{CO}_3$
Answer: (D) $\text{Li}_2\text{CO}_3$
$\text{Li}^+$ has high polarising power due to small size, decomposing $\text{Li}_2\text{CO}_3$ easily upon heating.
- $3.87\text{ BM}$
- $4.90\text{ BM}$
- $2.84\text{ BM}$
- $1.73\text{ BM}$
Answer: (A) $3.87\text{ BM}$
$\text{Cr}^{3+} = [\text{Ar}] 3d^3 \implies n = 3$. $\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87\text{ BM}$.
- Acetone
- Glucose
- Acetophenone
- Acetic acid
Answer: (B) Glucose
Glucose is a reducing sugar containing an aldehyde group in open-chain form, giving positive Tollens' test.
- $\text{HF} > \text{HCl} > \text{HBr} > \text{HI}$
- $\text{HCl} > \text{HBr} > \text{HI} > \text{HF}$
- $\text{HI} > \text{HBr} > \text{HCl} > \text{HF}$
- $\text{HBr} > \text{HI} > \text{HCl} > \text{HF}$
Answer: (C) $\text{HI} > \text{HBr} > \text{HCl} > \text{HF}$
Bond strength decreases down halogen group: $\text{HI} > \text{HBr} > \text{HCl} > \text{HF}$.
- $\frac{0.693}{k}$
- $\frac{[A]_0}{k}$
- $\frac{1}{k [A]_0}$
- $\frac{[A]_0}{2k}$
Answer: (D) $\frac{[A]_0}{2k}$
For zero order reaction $t_{1/2} = \frac{[A]_0}{2k}$.
- Propanone and Ethanal
- Propanal and Ethanal
- Methanal and Propanone
- Butanone and Methanal
Answer: (A) Propanone and Ethanal
$(\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_3 \xrightarrow{\text{O}_3, \text{Zn/H}_2\text{O}} (\text{CH}_3)_2\text{C=O} + \text{CH}_3\text{CHO}$.
- Cyclooctatetraene
- Cyclopentadienyl anion
- Cyclopentadienyl cation
- Cyclobutadiene
Answer: (B) Cyclopentadienyl anion
Cyclopentadienyl anion is planar and conjugated with $6\,\pi$ electrons ($4n+2$ for $n=1$).
- $0.052\text{ K}$
- $0.104\text{ K}$
- $0.156\text{ K}$
- $0.208\text{ K}$
Answer: (C) $0.156\text{ K}$
$i = 3$ for $\text{BaCl}_2$. $\Delta T_b = i K_b m = 3 \times 0.52 \times 0.1 = 0.156\text{ K}$.
- $+3$
- $+1$
- $+4$
- $+5$
Answer: (D) $+5$
$3(+1) + x + 4(-2) = 0 \implies x = +5$.
- Tropylium cation
- Allyl cation
- tert-Butyl cation
- Benzyl cation
Answer: (A) Tropylium cation
Tropylium cation is aromatic with $6\,\pi$ electrons delocalized over 7 ring carbons, making it exceptionally stable.
- Hydrated $\text{ZnCl}_2$ and dilute $\text{HCl}$
- Anhydrous $\text{ZnCl}_2$ and concentrated $\text{HCl}$
- Anhydrous $\text{AlCl}_3$ and concentrated $\text{HCl}$
- Concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$
Answer: (B) Anhydrous $\text{ZnCl}_2$ and concentrated $\text{HCl}$
Lucas reagent is equimolar anhydrous $\text{ZnCl}_2$ in concentrated $\text{HCl}$.
- 2-Hydroxybutanoic acid
- 3-Hydroxybutyric acid
- 3-Hydroxybutanoic acid
- beta-Hydroxybutyric acid
Answer: (C) 3-Hydroxybutanoic acid
Chain is 4 carbons long with -COOH at position 1 and -OH at position 3: 3-hydroxybutanoic acid.
- $d_{xy}$ and $d_{yz}$
- $d_{xz}$ and $d_{z^2}$
- $d_{xy}$ and $d_{x^2-y^2}$
- $d_{z^2}$ and $d_{x^2-y^2}$
Answer: (D) $d_{z^2}$ and $d_{x^2-y^2}$
The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are axial orbitals with electron lobes lying along coordinate axes.
- $\text{SO}_4^{2-}$
- $\text{H}_2\text{SO}_4$
- $\text{SO}_3$
- $\text{HSO}_3^-$
Answer: (A) $\text{SO}_4^{2-}$
Loss of $H^+$ from $\text{HSO}_4^-$ forms its conjugate base $\text{SO}_4^{2-}$.
- Invertase
- Zymase
- Maltase
- Diastase
Answer: (B) Zymase
Zymase enzyme present in yeast ferment glucose into ethyl alcohol.
- $\text{CH}_3\text{CH}_2\text{Br}$
- $(\text{CH}_3)_2\text{CHBr}$
- $\text{CH}_3\text{Br}$
- $(\text{CH}_3)_3\text{CBr}$
Answer: (C) $\text{CH}_3\text{Br}$
Least sterically hindered methyl halide reacts fastest in $S_{\text{N}}2$.
- Oxygen
- Methane
- Carbon dioxide
- Hydrogen
Answer: (D) Hydrogen
$2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\uparrow$.
- Teflon
- Nylon-6
- Buna-S
- Bakelite
Answer: (A) Teflon
Tetrafluoroethene polymerizes to form Polytetrafluoroethylene (Teflon).
- Tetrahedral
- Square planar
- See-saw
- Trigonal bipyramidal
Answer: (B) Square planar
$\text{XeF}_4$ has 4 bond pairs and 2 lone pairs $\implies$ square planar geometry.
- $A > C > B$
- $C > B > A$
- $B > C > A$
- $A > B > C$
Answer: (C) $B > C > A$
More negative $E^\circ$ value corresponds to stronger reducing power: $B (-3.0) > C (-1.2) > A (+0.5)$.
- Presence of excess air
- Presence of oxygen and sulfur dioxide
- Presence of carbon monoxide
- Absence or limited supply of air
Answer: (D) Absence or limited supply of air
Calcination is thermal treatment of ore in absence/limited supply of air.
- $25$
- $125$
- $5$
- $625$
Answer: (A) $25$
$|\text{adj}(A)| = |A|^{n-1} = 5^{3-1} = 25$.
- $\frac{\pi}{2}$
- $\frac{\pi}{4}$
- $\pi$
- $0$
Answer: (B) $\frac{\pi}{4}$
Applying King's property $I + I = \int_0^{\pi/2} 1 dx = \frac{\pi}{2} \implies I = \frac{\pi}{4}$.
- $\sqrt{14}$
- $2\sqrt{14}$
- $\frac{3}{2}\sqrt{14}$
- $\frac{\sqrt{14}}{4}$
Answer: (C) $\frac{3}{2}\sqrt{14}$
Point on line: $(1+r, 2+2r, 3+3r)$. Lies on plane $\implies 1+r - 2-2r + 3+3r = 5 \implies 2r = 3 \implies r = 3/2$. Distance $= |r|\sqrt{1^2+2^2+3^2} = \frac{3}{2}\sqrt{14}$.
- $2$
- $\frac{3}{2}$
- $4$
- $3$
Answer: (D) $3$
$S_\infty = \frac{a}{1 - r} = \frac{1}{1 - 2/3} = 3$.
- $\frac{11!}{2! 2! 2!}$
- $\frac{11!}{2! 2!}$
- $11!$
- $\frac{11!}{4!}$
Answer: (A) $\frac{11!}{2! 2! 2!}$
11 letters with 2 M's, 2 A's, and 2 T's $\implies \frac{11!}{2! 2! 2!}$.
- $y e^{-x} = x + C$
- $y e^x = x + C$
- $y = e^x + C$
- $y = x e^x + C$
Answer: (B) $y e^x = x + C$
$\text{IF} = e^x$. Integrating $y e^x = \int e^x e^{-x} dx = x + C$.
- $p^2 = 4q - 1$
- $p^2 = 4q + 1$
- $q^2 = 4p + 1$
- $p^2 = q + 4$
Answer: (B) $p^2 = 4q + 1$
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 1 \implies p^2 - 4q = 1 \implies p^2 = 4q + 1$.
- $\sqrt{3}$
- $\frac{2}{\sqrt{3}}$
- $\frac{1}{\sqrt{3}}$
- $1$
Answer: (C) $\frac{1}{\sqrt{3}}$
Projection $= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{1 - 1 + 1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.
- $x = 2n\pi \pm \frac{\pi}{3}$
- $x = n\pi + \frac{\pi}{4}$
- $x = n\pi \pm \frac{\pi}{6}$
- $x = n\pi + \frac{\pi}{2}$ or $x = n\pi + (-1)^n \left(-\frac{\pi}{6}\right)$
Answer: (D) $x = n\pi + \frac{\pi}{2}$ or $x = n\pi + (-1)^n \left(-\frac{\pi}{6}\right)$
$\cos x (2\sin x + 1) = 0 \implies \cos x = 0$ or $\sin x = -1/2$. Solutions are $n\pi + \pi/2$ and $n\pi + (-1)^n(-\pi/6)$.
- $\frac{1}{6}$
- $\frac{1}{12}$
- $\frac{5}{36}$
- $\frac{1}{4}$
Answer: (A) $\frac{1}{6}$
Sums > 9 are 10, 11, 12 (6 pairs total: (4,6),(5,5),(6,4),(5,6),(6,5),(6,6)). $P = 6/36 = 1/6$.
- $2$
- $1$
- $1/2$
- $0$
Answer: (B) $1$
Both expressions reduce to $2 \tan^{-1} x$. Thus derivative of $u$ w.r.t $u$ is $1$.
- $\bar{z}$
- $1$
- $z$
- $-z$
Answer: (C) $z$
$\bar{z} = 1/z \implies \frac{1 + z}{1 + 1/z} = z$.
- $\frac{7}{16}$
- $\frac{3}{4}$
- $\frac{\sqrt{5}}{4}$
- $\frac{\sqrt{7}}{4}$
Answer: (D) $\frac{\sqrt{7}}{4}$
$a^2 = 16, b^2 = 9 \implies e = \sqrt{1 - 9/16} = \frac{\sqrt{7}}{4}$.
- $(-2, 0)$
- $(2, 0)$
- $(0, -2)$
- $(0, 2)$
Answer: (A) $(-2, 0)$
For $y^2 = -4ax$, $4a = 8 \implies a = 2$. Focus is $(-2, 0)$.
- $210$
- $252$
- $120$
- $45$
Answer: (B) $252$
$T_{r+1} = {}^{10}C_r x^{10-2r}$. For constant term $10 - 2r = 0 \implies r = 5$. Value $={}^{10}C_5 = 252$.
- $\frac{1}{6}$
- $0$
- $-\frac{1}{6}$
- $-\frac{1}{3}$
Answer: (C) $-\frac{1}{6}$
Using series expansion: $\lim_{x \to 0} \frac{(x - x^3/6 + \dots) - x}{x^3} = -\frac{1}{6}$.
- $25$
- $\sqrt{13}$
- $7$
- $5$
Answer: (D) $5$
$r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 9 - (-12)} = \sqrt{25} = 5$.
- $-\frac{1}{11}$
- $11$
- $-11$
- $\frac{1}{11}$
Answer: (A) $-\frac{1}{11}$
$\frac{dy}{dx} = 3x^2 - 1$. At $x = 2$, slope of tangent $= 11 \implies$ slope of normal $= -1/11$.
- $-2 I$
- $2 I$
- $I$
- $0$
Answer: (B) $2 I$
By Cayley-Hamilton Theorem: $\lambda^2 - 5\lambda - 2 = 0 \implies A^2 - 5A = 2I$.
- $\frac{\pi}{4}$
- $\frac{\pi}{6}$
- $\frac{\pi}{3}$
- $\frac{\pi}{2}$
Answer: (C) $\frac{\pi}{3}$
$\cos\theta = \frac{1(1) + 1(0) + 0(1)}{\sqrt{2}\sqrt{2}} = \frac{1}{2} \implies \theta = \frac{\pi}{3}$.
- $e^x \sec x + C$
- $e^x \sec^2 x + C$
- $e^x \ln(\sec x) + C$
- $e^x \tan x + C$
Answer: (D) $e^x \tan x + C$
Form $\int e^x [f(x) + f'(x)]dx = e^x f(x) + C \implies e^x \tan x + C$.
- $2.4$
- $2.0$
- $1.8$
- $3.0$
Answer: (A) $2.4$
Mean $= 6$. Deviations: $4, 2, 0, 2, 4$. Mean deviation $= 12/5 = 2.4$.
- $x = 1$ only
- $x = 1$ and $x = 2$
- $x = 2$ only
- Everywhere differentiable
Answer: (B) $x = 1$ and $x = 2$
Modulus functions have corners where expressions inside vanish, i.e., $x = 1$ and $x = 2$.
- $0.84$
- $0.76$
- $0.96$
- $0.60$
Answer: (C) $0.96$
$P(A \cap B) = 0.4 \times 0.6 = 0.24 \implies P(A \cup B) = 0.4 + 0.8 - 0.24 = 0.96$.
- $\frac{2\pi}{3}$
- $-\frac{\pi}{3}$
- $\frac{4\pi}{3}$
- $\frac{\pi}{3}$
Answer: (D) $\frac{\pi}{3}$
$\sin(2\pi/3) = \sin(\pi/3)$. Since $\pi/3 \in [-\pi/2, \pi/2]$, principal value is $\frac{\pi}{3}$.