Prove that f (x) = sin x + √3 cos x has maximum value
Question: Prove thatf(x) = sinx+ 3 cosxhas maximum value atx= /6. Solution: Given, $f(x)=\sin x+\sqrt{3} \cos x=2\left(\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x\right)$ $=2\left(\cos \frac{\pi}{3} \sin x+\sin \frac{\pi}{3} \cos x\right)=2 \sin \left(x+\frac{\pi}{3}\right)$ $f^{\prime}(x)=2 \cos \left(x+\frac{\pi}{3}\right) ; f^{\prime \prime}(x)=-2 \sin \left(x+\frac{\pi}{3}\right)$ $f^{\prime \prime}(x)_{x=\frac{\pi}{6}}=-2 \sin \left(\frac{\pi}{6}+\frac{\pi}{3}\right)$ $=-2 \sin \frac{\pi}{...
Read More →At what point, the slope of the curve
Question: At what point, the slope of the curvey= x3+ 3x2+ 9x 27 is maximum? Also find the maximum slope. Solution: Given, curvey= x3+ 3x2+ 9x 27 Differentiating both sides w.r.t. x, we get dy/dx = -3x2+ 6x + 9 Let slope of the curve dy/dx = z So, z = -3x2+ 6x + 9 Differentiating both sides w.r.t. x, we get dz/dx = -6x + 6 For local maxima and local minima, dz/dx = 0 -6x + 6 = 0 ⇒ x = 1 d2z/dx2= -6 0 Maxima Putting x = 1 in equation of the curve y = (-1)3+ 3(1)2+ 9(1) 27 = -1 + 3 + 9 27 = -16 Ma...
Read More →Show that f(x) = tan–1(sin x + cos x) is
Question: Show thatf(x) = tan1(sinx+ cosx) is an increasing function in (0, /4). Solution: Given,f(x) = tan1(sinx+ cosx) in (0, /4). Differentiating both sides w.r.t. x, we got $f^{\prime}(x)=\frac{1}{1+(\sin x+\cos x)^{2}} \cdot \frac{d}{d x}(\sin x+\cos x)$ $f^{\prime}(x)=\frac{1 \times(\cos x-\sin x)}{1+(\sin x+\cos x)^{2}}$ $f^{\prime}(x)=\frac{\cos x-\sin x}{1+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x}$ $\Rightarrow f^{\prime}(x)=\frac{\cos x-\sin x}{1+1+2 \sin x \cos x} \Rightarrow f^{\prime...
Read More →Show that for
Question: Show that fora 1,f(x) = 3 sinx cosx 2ax + bis decreasing in R. Solution: Given, f(x) = 3 sinx cosx 2ax + b, a 1 On differentiating both sides w.r.t. x, we get f(x) = 3 cosx +sinx 2a For increasing function,f(x) 0 So, $\sqrt{3} \cos x+\sin x-2 a0$ $2\left(\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x\right)-2 a0$ $\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x-a0$ $\left(\cos \frac{\pi}{6} \cos x+\sin \frac{\pi}{6} \sin x\right)-a0$ $\cos \left(x-\frac{\pi}{6}\right)-a0$ As $\cos x \in[-1,...
Read More →Show that
Question: Show thatf(x) = 2x+ cot-1x + log [(1 + x2) x] is increasing in R. Solution: Given, f(x) = 2x+ cot-1x + log [(1 + x2) x] Differentiating both sides w.r.t. x, we get $f^{\prime}(x)=2-\frac{1}{1+x^{2}}+\frac{1}{\sqrt{1+x^{2}}-x} \times \frac{d}{d x}\left(\sqrt{1+x^{2}}-x\right)$ $=2-\frac{1}{1+x^{2}}+\frac{\left(\frac{1}{2 \sqrt{1+x^{2}}} \times(2 x-1)\right)}{\sqrt{1+x^{2}}-x}$ $=2-\frac{1}{1+x^{2}}+\frac{x-\sqrt{1+x^{2}}}{\sqrt{1+x^{2}}\left(\sqrt{1+x^{2}-x}\right)}$ $=2-\frac{1}{1+x^{2...
Read More →Show that the line x/a + y/b = 1,
Question: Show that the linex/a + y/b= 1, touches the curvey=b.e-x/aat the point where the curve intersects the axis ofy. Solution: Given curve equation,y=b.e-x/aand line equationx/a + y/b= 1 Now, let the coordinates of the point where the curve intersects the y-axis be (0, y1) Now differentiatingy=b.e-x/aboth sides w.r.t. x, we get $\frac{d y}{d x}=b \cdot e^{-x / a}\left(-\frac{1}{a}\right)=-\frac{b}{a} \cdot e^{-x / a}$ So, the slope of the tangent, $m_{1}=-\frac{b}{a} e^{-x / a}$. Differenti...
Read More →At what points on the curve
Question: At what points on the curvex2+y2 2x 4y+ 1 = 0, the tangents are parallel to they-axis? Solution: Given, the equation of the curve isx2+y2 2x 4y+ 1 = 0 .. (i) Differentiating both the sides w.r.t. x, we get $2 x+2 y \cdot \frac{d y}{d x}-2-4 \cdot \frac{d y}{d x}=0$ $\Rightarrow(2 y-4) \frac{d y}{d x}=2-2 x \Rightarrow \frac{d y}{d x}=\frac{2-2 x}{2 y-4}$ ...........(ii) Since the tangent to the curve is parallel to the $y$-axis. $\therefore$ Slope $\frac{d y}{d x}=\tan \frac{\pi}{2}=\i...
Read More →The water is filled upto height of 12 m in a tank having vertical sidewalls.
Question: The water is filled upto height of 12 m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth 'h' below the water level. The value of 'h' for which the emerging stream of water strikes the ground at the maximum range is ___ m. Solution: $\mathrm{R}=\sqrt{2 \mathrm{gh}} \times \sqrt{\frac{(12-\mathrm{h}) \times 2}{\mathrm{~g}}}$ $\sqrt{4 \mathrm{~h}(12-\mathrm{h})}=\mathrm{R}$ For maximum R $\frac{\mathrm{dR}}{\mathrm{dh}}=0$ $\Rightarrow h=6 m$...
Read More →The Ka X-ray of molybdenum has wavelength 0.071 nm.
Question: The $\mathrm{K}_{\alpha}$ X-ray of molybdenum has wavelength $0.071 \mathrm{~nm}$. If the energy of a molybdenum atoms with a $\mathrm{K}$ electron knocked out is $27.5 \mathrm{keV}$, the energy of this atom when an $\mathrm{L}$ electron is knocked out will be $\mathrm{keV}$. (Round off to the nearest integer) $\left[\mathrm{h}=4.14 \times 10^{-15} \mathrm{eVs}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right]$ Solution: $\mathrm{E}_{\mathrm{k}_{\mathrm{a}}}=\mathrm{E}_{\mathrm{k}}-...
Read More →A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure.
Question: A small block slides down from the top of hemisphere of radius R = 3 m as shown in the figure. The height 'h' at which the block will lose contact with the surface of the sphere is ____m. (Assume there is no friction between the block and the hemisphere) Solution: $m g \cos \theta=\frac{m v^{2}}{R}$ ......(1) $\cos \theta=\frac{h}{R}$ Energy conservation $m g\{R-h\}=\frac{1}{2} m v^{2}$ .....(2) from (1) \ (2) $\Rightarrow \operatorname{mg}\left\{\frac{\mathrm{h}}{\mathrm{R}}\right\}=\...
Read More →For the circuit shown, the value of current at time t = 3.2 s will be ______ A.
Question: For the circuit shown, the value of current at time t = 3.2 s will be ______ A. [Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)] Solution: From graph voltage at t = 3.2 sec is 6 volt. $i=\frac{6-5}{1}$ $\mathrm{i}=1 \mathrm{~A}$...
Read More →A swimmer wants to cross a river from point A to point B.
Question: A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30 with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle with the line AB should be ____, so that the swimmer reaches point B. Solution: Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them $\theta=30^{\circ}$...
Read More →A particle executes simple harmonic motion represented by displacement function as
Question: A particle executes simple harmonic motion represented by displacement function as $x(t)=A \sin (\omega t+\phi)$ If the position and velocity of the particle at $\mathrm{t}=0 \mathrm{~s}$ are $2 \mathrm{~cm}$ and $2 \omega \mathrm{cm} \mathrm{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \mathrm{~cm}$ where the value of $x$ is Solution: $\mathrm{x}(\mathrm{t})=\mathrm{A} \sin (\omega \mathrm{t}+\phi)$ $\mathrm{v}(\mathrm{t})=\mathrm{A} \omega \cos (\omega \mathrm{t}+\phi)$ $...
Read More →In the given figure the magnetic flux through the loop increases according to the relation
Question: In the given figure the magnetic flux through the loop increases according to the relation$\phi_{\mathrm{B}}(\mathrm{t})=10 \mathrm{t}^{2}+20 \mathrm{t}$, where $\phi_{\mathrm{B}}$ is in milliwebers and $\mathrm{t}$ is in seconds. The magnitude of current through $R=2 \Omega$ resistor at t = 5 s is ______ mA. Solution: $|\epsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=20 \mathrm{t}+20 \mathrm{mV}$ $|\mathrm{i}|=\frac{|\in|}{\mathrm{R}}=10 \mathrm{t}+10 \mathrm{~mA}$ at $t=5$ $|\mathrm{i}...
Read More →The maximum amplitude for an amplitude modulated wave is found to be 12V
Question: The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ______. Solution: $A_{\max }=A_{c}+A_{m}=12$ $A_{\min }=A_{c}-A_{m}=3$ $\Rightarrow \mathrm{A}_{\mathrm{c}}=\frac{15}{2} \ \mathrm{~A}_{\mathrm{m}}=\frac{9}{2}$ modulation index $=\frac{\mathrm{A}_{\mathrm{m}}}{\mathrm{A}_{\mathrm{c}}}=\frac{9 / 2}{15 / 2}=0.6$ $\Rightarrow x=1$...
Read More →Find the equation of the normal lines to the curve
Question: Find the equation of the normal lines to the curve 3x2y2= 8 which are parallel to the linex+ 3y= 4. Solution: Given curve, 3x2y2= 8 Differentiating both sides w.r.t. x, we get 6x 2y. dy/dx = 0 ⇒ -2y(dy/dx) = -6x ⇒ dy/dx = 3x/y So, slope of the tangent to the given curve = 3x/y Thus, the normal to the curve = -1/(3x/y) = -y/3x Now, differentiating both sides of the given line x + 3y = 4, we have 1 + 3.(dy/dx) = 0 dy/dx = -1/3 As the normal to the curve is parallel to the given line x + ...
Read More →The maximum amplitude for an amplitude modulated wave is found to be 12V
Question: The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ______. Solution: $A_{\max }=A_{c}+A_{m}=12$ $A_{\min }=A_{c}-A_{m}=3$ $\Rightarrow \mathrm{A}_{\mathrm{c}}=\frac{15}{2} \ \mathrm{~A}_{\mathrm{m}}=\frac{9}{2}$ modulation index $=\frac{\mathrm{A}_{\mathrm{m}}}{\mathrm{A}_{\mathrm{c}}}=\frac{9 / 2}{15 / 2}=0.6$ $\Rightarrow x=1$...
Read More →The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is one.
Question: The difference in the number of waves when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is ______ mm. [Refractive index of air = 1.0003, wavelength of yellow light in vacuum = 6000 Å] Solution: Thickness $\mathrm{t}=\mathrm{n} \lambda$ So, $\mathrm{n} \lambda_{\mathrm{vac}}=(\mathrm{n}+1) \lambda_{\mathrm{air}}$ $\mathrm{n} \lambda=(\mathrm{n}+1) \frac{\lambda}{\mu_{\mathrm{air}}}$ $\mathrm{n}=\frac{1}{\mu_{\mathrm...
Read More →In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q.
Question: In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias $\left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}\right)$ will be x : 1. The value of x will be _____. Solution: $\frac{1}{2} \mathrm{I}_{1}\left(\omega_{1}\right)^{2}=\frac{1}{2} \mathrm{I}_{2}\left(\omega_{2}\right)^{2}$ $\mathrm{I}_{1}\left(\frac{\mathrm{v}}{3 \mathrm{R}}\right)^{2}=\mathrm{I}_{2}\left(...
Read More →The resistance of a conductor at 15°C
Question: The resistance of a conductor at $15^{\circ} \mathrm{C}$ is $16 \Omega$ and at $100^{\circ} \mathrm{C}$ is $20 \Omega$. What will be the temperature coefficient of resistance of the conductor?$0.010^{\circ} \mathrm{C}^{-1}$$0.033^{\circ} \mathrm{C}^{-1}$$0.003^{\circ} \mathrm{C}^{-1}$$0.042^{\circ} \mathrm{C}^{-1}$Correct Option: , 3 Solution: $16=\mathrm{R}_{0}\left[1+\alpha\left(15-\mathrm{T}_{\mathrm{o}}\right)\right]$ $20=\mathrm{R}_{0}\left[1+\alpha\left(100-\mathrm{T}_{\mathrm{o}...
Read More →A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation
Question: A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $\mathrm{F}=\mathrm{F}_{0}\left[1-\left(\frac{\mathrm{t}-\mathrm{T}}{\mathrm{T}}\right)^{2}\right]$ Where $\mathrm{F}_{0}$ and $\mathrm{T}$ are constants. The force acts only for the time interval $2 \mathrm{~T}$. The velocity $\mathrm{v}$ of the particle after time $2 \mathrm{~T}$ is :$2 \mathrm{~F}_{0} \mathrm{~T} / \mathrm{M}$$\mathrm{F...
Read More →The planet Mars has two moons, if one of them has a period 7 hours,
Question: The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of $9.0 \times 10^{3} \mathrm{~km}$. Find the mass of Mars. $\left\{\operatorname{Given} \frac{4 \pi^{2}}{\mathrm{G}}=6 \times 10^{11} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \mathrm{~kg}^{2}\right\}$$5.96 \times 10^{19} \mathrm{~kg}$$3.25 \times 10^{21} \mathrm{~kg}$$7.02 \times 10^{25} \mathrm{~kg}$$6.00 \times 10^{23} \mathrm{~kg}$Correct Option: , 4 Solution: Option D is correct $\mathrm{...
Read More →An automobile of mass 'm' accelerates starting from origin and initially at rest,
Question: An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by :$\left(\frac{9 P}{8 m}\right)^{\frac{1}{2}} t^{\frac{3}{2}}$$\left(\frac{8 \mathrm{P}}{9 m}\right)^{\frac{1}{2}} \mathrm{t}^{\frac{2}{3}}$$\left(\frac{9 m}{8 P}\right)^{\frac{1}{2}} t^{\frac{3}{2}}$$\left(\frac{8 \mathrm{P}}{9 m}\right)^{\frac{1}{2}} \mathrm{t}^{\frac{3}{2}}$Correct Option: , 4 Solution: P = const....
Read More →A physical quantity 'y' is represented by the
Question: A physical quantity 'y' is represented by the formula $\mathrm{y}=\mathrm{m}^{2} \mathrm{r}^{-4} \mathrm{~g}^{\mathrm{x}} l^{-\frac{3}{2}}$ If the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.5 and 24 and 3$\frac{16}{3}$ and $\pm \frac{3}{2}$8 and 2Correct Option: , 3 Solution: $\frac{\Delta \mathrm{y}}{\mathrm{y}}=\frac{2 \Delta \mathrm{m}}{\mathrm{m}}+\frac{4 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\mathrm{x} \Delta \ma...
Read More →What will be the magnitude of electric field at point O as shown in figure?
Question: What will be the magnitude of electric field at point O as shown in figure? Each side of the figure is l and perpendicular to each other? $\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{l^{2}}$$\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\left(2 l^{2}\right)}(2 \sqrt{2}-1)$$\frac{\mathrm{q}}{4 \pi \varepsilon_{0}(2 l)^{2}}$$\frac{1}{4 \pi \varepsilon_{0}} \frac{2 \mathrm{q}}{2 l^{2}}(\sqrt{2})$Correct Option: , 2 Solution: $\mathrm{E}_{1}=\frac{\mathrm{kq}}{\ell^{2}}=\mathrm{E}_{2}$ $\...
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