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Question Number 20851    Answers: 2   Comments: 0

if (x+y)^(m+n) =x^m y^n ,show that (dy/dx)=(y/x)

$${if}\:\left({x}+{y}\right)^{{m}+{n}} ={x}^{{m}} {y}^{{n}} ,{show}\:{that}\:\frac{{dy}}{{dx}}=\frac{{y}}{{x}} \\ $$

Question Number 20850    Answers: 1   Comments: 0

given that y=log(√((1+cos^2 x)/(1−e^(2x) ))),find (dy/dx)

$${given}\:{that}\:{y}={log}\sqrt{\frac{\mathrm{1}+{cos}^{\mathrm{2}} {x}}{\mathrm{1}−{e}^{\mathrm{2}{x}} }},{find} \\ $$$$\frac{{dy}}{{dx}} \\ $$

Question Number 20849    Answers: 1   Comments: 0

The time period,T of pendulum of length l is given by T=2Π(√(l/g))where l amd g are constant.Find the approximate percentage increase in time T, when the length of pendulum increases by 4%

$${The}\:{time}\:{period},{T}\:\:{of}\:{pendulum}\:{of} \\ $$$${length}\:{l}\:{is}\:{given}\:{by}\:{T}=\mathrm{2}\Pi\sqrt{\frac{{l}}{{g}}}{where} \\ $$$$\:{l}\:{amd}\:{g}\:{are}\:{constant}.{Find}\:{the}\: \\ $$$${approximate}\:{percentage}\:{increase}\: \\ $$$${in}\:{time}\:{T},\:{when}\:{the}\:{length}\:{of}\: \\ $$$${pendulum}\:{increases}\:{by}\:\mathrm{4\%} \\ $$

Question Number 20847    Answers: 1   Comments: 0

the rectangle is known to be twice as long as its wide.if the width is measured as 20 ±0.2cm. find the area in the form of (A±b)

$${the}\:{rectangle}\:{is}\:{known}\:{to}\:{be}\:{twice} \\ $$$${as}\:{long}\:{as}\:{its}\:{wide}.{if}\:{the}\:{width}\:{is} \\ $$$${measured}\:{as}\:\mathrm{20}\:\pm\mathrm{0}.\mathrm{2}{cm}. \\ $$$${find}\:{the}\:{area}\:{in}\:{the}\:{form}\:{of}\:\left({A}\pm{b}\right) \\ $$$$ \\ $$

Question Number 20867    Answers: 1   Comments: 0

The number of irrational roots of the equation (x − 1)(x − 2)(3x − 2)(3x + 1) = 21 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{irrational}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\left({x}\:−\:\mathrm{1}\right)\left({x}\:−\:\mathrm{2}\right)\left(\mathrm{3}{x}\:−\:\mathrm{2}\right)\left(\mathrm{3}{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{21}\:\mathrm{is} \\ $$

Question Number 20842    Answers: 1   Comments: 0

Acceleration of a particle which is at rest at x = 0 is a^→ = (4 − 2x) i^∧ . Select the correct alternative(s). (a) Maximum speed of the particle is 4 units (b) Particle further comes to rest at x = 4 (c) Particle oscillates about x = 2 (d) Particle will continuously accelerate along the x-axis.

$$\mathrm{Acceleration}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{which}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{rest}\:\mathrm{at}\:{x}\:=\:\mathrm{0}\:\mathrm{is}\:\overset{\rightarrow} {{a}}\:=\:\left(\mathrm{4}\:−\:\mathrm{2}{x}\right)\:\overset{\wedge} {{i}}.\:\mathrm{Select} \\ $$$$\mathrm{the}\:\mathrm{correct}\:\mathrm{alternative}\left(\mathrm{s}\right). \\ $$$$\left({a}\right)\:\mathrm{Maximum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{4}\:\mathrm{units} \\ $$$$\left({b}\right)\:\mathrm{Particle}\:\mathrm{further}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{at} \\ $$$${x}\:=\:\mathrm{4} \\ $$$$\left({c}\right)\:\mathrm{Particle}\:\mathrm{oscillates}\:\mathrm{about}\:{x}\:=\:\mathrm{2} \\ $$$$\left({d}\right)\:\mathrm{Particle}\:\mathrm{will}\:\mathrm{continuously}\:\mathrm{accelerate} \\ $$$$\mathrm{along}\:\mathrm{the}\:{x}-\mathrm{axis}. \\ $$

Question Number 20836    Answers: 0   Comments: 0

Question Number 20832    Answers: 2   Comments: 2

Given that y=log(√((1−cos^2 x)/(1−e^(2x) ))) find (dy/dx)

$${Given}\:{that}\:{y}={log}\sqrt{\frac{\mathrm{1}−{cos}^{\mathrm{2}} {x}}{\mathrm{1}−{e}^{\mathrm{2}{x}} }} \\ $$$${find}\:\frac{{dy}}{{dx}} \\ $$

Question Number 20831    Answers: 1   Comments: 0

if x(√(1+y)) + y(√(1+x))=0 prove that (dy/dx)=−(1+x)^(−2)

$${if}\:{x}\sqrt{\mathrm{1}+{y}}\:+\:{y}\sqrt{\mathrm{1}+{x}}=\mathrm{0}\:{prove}\:{that}\: \\ $$$$\frac{{dy}}{{dx}}=−\left(\mathrm{1}+{x}\right)^{−\mathrm{2}} \\ $$

Question Number 20827    Answers: 1   Comments: 0

find cube root of 12167 by divison method.

$${find}\:{cube}\:{root}\:{of}\:\mathrm{12167}\:{by}\:{divison}\:{method}. \\ $$

Question Number 20825    Answers: 0   Comments: 1

Question Number 20823    Answers: 1   Comments: 0

Question Number 20819    Answers: 1   Comments: 1

Question Number 20810    Answers: 0   Comments: 3

What is the number of triples (a, b, c) of positive integers such that (i) a < b < c < 10 and (ii) a, b, c, 10 form the sides of a quadrilateral?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{triples}\:\left({a},\:{b},\:{c}\right)\:\mathrm{of} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left(\mathrm{i}\right)\:{a}\:<\:{b}\:<\:{c}\:<\:\mathrm{10}\:\mathrm{and}\:\left(\mathrm{ii}\right)\:{a},\:{b},\:{c},\:\mathrm{10}\:\mathrm{form} \\ $$$$\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{quadrilateral}? \\ $$

Question Number 20800    Answers: 2   Comments: 1

If Re(((z + 4)/(2z − 1))) = (1/2), then z is represented by a point lying on (1) A circle (2) An ellipse (3) A straight line (4) No real locus

$$\mathrm{If}\:\mathrm{Re}\left(\frac{{z}\:+\:\mathrm{4}}{\mathrm{2}{z}\:−\:\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{then}\:{z}\:\mathrm{is}\:\mathrm{represented} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{point}\:\mathrm{lying}\:\mathrm{on} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{A}\:\mathrm{circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{An}\:\mathrm{ellipse} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{A}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{No}\:\mathrm{real}\:\mathrm{locus} \\ $$

Question Number 20799    Answers: 1   Comments: 0

If z^2 + z∣z∣ + ∣z∣^2 = 0, then locus of z is

$$\mathrm{If}\:{z}^{\mathrm{2}} \:+\:{z}\mid{z}\mid\:+\:\mid{z}\mid^{\mathrm{2}} \:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is} \\ $$

Question Number 20798    Answers: 0   Comments: 0

{ ((4x−y=11)),((2y+5x=5)),((x−2z=9)) :} z+y−x=?

$$\begin{cases}{\mathrm{4}\boldsymbol{{x}}−\boldsymbol{{y}}=\mathrm{11}}\\{\mathrm{2}\boldsymbol{{y}}+\mathrm{5}\boldsymbol{{x}}=\mathrm{5}}\\{\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{z}}=\mathrm{9}}\end{cases}\:\:\:\:\boldsymbol{{z}}+\boldsymbol{{y}}−\boldsymbol{{x}}=? \\ $$

Question Number 20786    Answers: 2   Comments: 0

Two particles A and B start from the same position along the circular path of radius 0.5 m with a speed v_A = 1 ms^(−1) and v_B = 1.2 ms^(−1) in opposite direction. Determine the time before they collide.

$$\mathrm{Two}\:\mathrm{particles}\:{A}\:\mathrm{and}\:{B}\:\mathrm{start}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{position}\:\mathrm{along}\:\mathrm{the}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{0}.\mathrm{5}\:\mathrm{m}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:{v}_{{A}} \:=\:\mathrm{1}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:{v}_{{B}} \:=\:\mathrm{1}.\mathrm{2}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{opposite}\:\mathrm{direction}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{time}\:\mathrm{before}\:\mathrm{they}\:\mathrm{collide}. \\ $$

Question Number 20785    Answers: 1   Comments: 0

If in a ΔABC, tan(A/2), tan(B/2) and tan(C/2) are in HP, then the minimum value of cot(B/2) is greater than (1) 2(√3) (2) ((√3)/2) (3) (√3) (4) (1/(√3))

$$\mathrm{If}\:\mathrm{in}\:\mathrm{a}\:\Delta{ABC},\:\mathrm{tan}\frac{{A}}{\mathrm{2}},\:\mathrm{tan}\frac{{B}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{tan}\frac{{C}}{\mathrm{2}} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{HP},\:\mathrm{then}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cot}\frac{{B}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$

Question Number 20777    Answers: 0   Comments: 6

In the figure shown, mass ′m′ is placed on the inclined surface of a wedge of mass M. All the surfaces are smooth. Find the acceleration of the wedge.

$${In}\:{the}\:{figure}\:{shown},\:{mass}\:'{m}'\:{is} \\ $$$${placed}\:{on}\:{the}\:{inclined}\:{surface}\:{of}\:{a} \\ $$$${wedge}\:{of}\:{mass}\:{M}.\:{All}\:{the}\:{surfaces} \\ $$$${are}\:{smooth}.\:{Find}\:{the}\:{acceleration}\:{of} \\ $$$${the}\:{wedge}. \\ $$

Question Number 20764    Answers: 1   Comments: 0

A small bead is slipped on a horizontal rod of length l. The rod starts moving with a horizontal acceleration a in a direction making an angle α with the length of the rod. Assuming that initially the bead is in the middle of the rod, find the time elapsed before the bead leaves the rod. Coefficient of friction between the bead and the rod is μ. (Neglect gravity).

$${A}\:{small}\:{bead}\:{is}\:{slipped}\:{on}\:{a}\:{horizontal} \\ $$$${rod}\:{of}\:{length}\:{l}.\:{The}\:{rod}\:{starts}\:{moving} \\ $$$${with}\:{a}\:{horizontal}\:{acceleration}\:{a}\:{in}\:{a} \\ $$$${direction}\:{making}\:{an}\:{angle}\:\alpha\:{with}\:{the} \\ $$$${length}\:{of}\:{the}\:{rod}.\:{Assuming}\:{that} \\ $$$${initially}\:{the}\:{bead}\:{is}\:{in}\:{the}\:{middle}\:{of} \\ $$$${the}\:{rod},\:{find}\:{the}\:{time}\:{elapsed}\:{before} \\ $$$${the}\:{bead}\:{leaves}\:{the}\:{rod}.\:{Coefficient}\:{of} \\ $$$${friction}\:{between}\:{the}\:{bead}\:{and}\:{the}\:{rod} \\ $$$${is}\:\mu.\:\left({Neglect}\:{gravity}\right). \\ $$

Question Number 20745    Answers: 1   Comments: 3

Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the Figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed (1/8) rotation, (ii) their range is less than half the disc radius, and (iii) ω remains constant throughout. Then (a) P lands in the shaded region and Q in the unshaded region (b) P lands in the unshaded region and Q in the shaded region (c) Both P and Q land in the unshaded region (d) Both P and Q land in the shaded region

$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{disc}\:\mathrm{rotating}\:\mathrm{in}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{plane}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{angular}\:\mathrm{speed}\:\omega \\ $$$$\mathrm{about}\:\mathrm{its}\:\mathrm{centre}\:{O}.\:\mathrm{The}\:\mathrm{disc}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{shaded}\:\mathrm{region}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diameter} \\ $$$$\mathrm{and}\:\mathrm{an}\:\mathrm{unshaded}\:\mathrm{region}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Figure}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{disc}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{orientation}\:\mathrm{as}\:\mathrm{shown},\:\mathrm{two} \\ $$$$\mathrm{pebbles}\:{P}\:\mathrm{and}\:{Q}\:\mathrm{are}\:\mathrm{simultaneously} \\ $$$$\mathrm{projected}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{towards}\:{R}.\:\mathrm{The} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{projection}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:{y}-{z} \\ $$$$\mathrm{plane}\:\mathrm{and}\:\mathrm{is}\:\mathrm{same}\:\mathrm{for}\:\mathrm{both}\:\mathrm{pebbles}\:\mathrm{with} \\ $$$$\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{disc}.\:\mathrm{Assume}\:\mathrm{that}\:\left({i}\right)\:\mathrm{they} \\ $$$$\mathrm{land}\:\mathrm{back}\:\mathrm{on}\:\mathrm{the}\:\mathrm{disc}\:\mathrm{before}\:\mathrm{the}\:\mathrm{disc}\:\mathrm{has} \\ $$$$\mathrm{completed}\:\frac{\mathrm{1}}{\mathrm{8}}\:\mathrm{rotation},\:\left({ii}\right)\:\mathrm{their}\:\mathrm{range} \\ $$$$\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{half}\:\mathrm{the}\:\mathrm{disc}\:\mathrm{radius},\:\mathrm{and} \\ $$$$\left({iii}\right)\:\omega\:\mathrm{remains}\:\mathrm{constant}\:\mathrm{throughout}. \\ $$$$\mathrm{Then} \\ $$$$\left({a}\right)\:{P}\:\mathrm{lands}\:\mathrm{in}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{region}\:\mathrm{and}\:{Q} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{unshaded}\:\mathrm{region} \\ $$$$\left({b}\right)\:{P}\:\mathrm{lands}\:\mathrm{in}\:\mathrm{the}\:\mathrm{unshaded}\:\mathrm{region}\:\mathrm{and} \\ $$$${Q}\:\mathrm{in}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{region} \\ $$$$\left({c}\right)\:\mathrm{Both}\:{P}\:\mathrm{and}\:{Q}\:\mathrm{land}\:\mathrm{in}\:\mathrm{the}\:\mathrm{unshaded} \\ $$$$\mathrm{region} \\ $$$$\left({d}\right)\:\mathrm{Both}\:{P}\:\mathrm{and}\:{Q}\:\mathrm{land}\:\mathrm{in}\:\mathrm{the}\:\mathrm{shaded} \\ $$$$\mathrm{region} \\ $$

Question Number 20744    Answers: 1   Comments: 0

Question Number 20739    Answers: 1   Comments: 0

If ∣z_1 ∣ = 2, ∣z_2 ∣ = 3, ∣z_3 ∣ = 4 and ∣2z_1 + 3z_2 + 4z_3 ∣ = 4, then the expression ∣8z_2 z_3 + 27z_3 z_1 + 64z_1 z_2 ∣ equals

$$\mathrm{If}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mathrm{2},\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{3},\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{4}\:\mathrm{and} \\ $$$$\mid\mathrm{2}{z}_{\mathrm{1}} \:+\:\mathrm{3}{z}_{\mathrm{2}} \:+\:\mathrm{4}{z}_{\mathrm{3}} \mid\:=\:\mathrm{4},\:\mathrm{then}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mid\mathrm{8}{z}_{\mathrm{2}} {z}_{\mathrm{3}} \:+\:\mathrm{27}{z}_{\mathrm{3}} {z}_{\mathrm{1}} \:+\:\mathrm{64}{z}_{\mathrm{1}} {z}_{\mathrm{2}} \mid\:\mathrm{equals} \\ $$

Question Number 20733    Answers: 1   Comments: 0

Find the volume of the solid of revolution obtained by revolving area bounded by x = 4 + 6y − 2y^2 , x = −4, x = 0 about the y−axis

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{of}\:\mathrm{revolution} \\ $$$$\mathrm{obtained}\:\mathrm{by}\:\mathrm{revolving}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by} \\ $$$${x}\:=\:\mathrm{4}\:+\:\mathrm{6}{y}\:−\:\mathrm{2}{y}^{\mathrm{2}} ,\:{x}\:=\:−\mathrm{4},\:{x}\:=\:\mathrm{0}\:\mathrm{about} \\ $$$$\mathrm{the}\:{y}−\mathrm{axis} \\ $$

Question Number 20724    Answers: 0   Comments: 3

A sphere is rolling without slipping on a fixed horizontal plane surface. In the Figure, A is a point of contact, B is the centre of the sphere and C is its topmost point. Then (a) v_C ^→ − v_A ^→ = 2(v_B ^→ − v_C ^→ ) (b) v_C ^→ − v_B ^→ = v_B ^→ − v_A ^→ (c) ∣v_C ^→ − v_A ^→ ∣ = 2∣v_B ^→ − v_C ^→ ∣ (d) ∣v_C ^→ − v_A ^→ ∣ = 4∣v_B ^→ ∣

$$\mathrm{A}\:\mathrm{sphere}\:\mathrm{is}\:\mathrm{rolling}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on} \\ $$$$\mathrm{a}\:\mathrm{fixed}\:\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{surface}.\:\mathrm{In}\:\mathrm{the} \\ $$$$\mathrm{Figure},\:\mathrm{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{of}\:\mathrm{contact},\:{B}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{and}\:{C}\:\mathrm{is}\:\mathrm{its}\:\mathrm{topmost} \\ $$$$\mathrm{point}.\:\mathrm{Then} \\ $$$$\left({a}\right)\:\overset{\rightarrow} {{v}}_{\mathrm{C}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{A}} \:=\:\mathrm{2}\left(\overset{\rightarrow} {{v}}_{\mathrm{B}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{C}} \right) \\ $$$$\left({b}\right)\:\overset{\rightarrow} {{v}}_{\mathrm{C}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{B}} \:=\:\overset{\rightarrow} {{v}}_{\mathrm{B}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{A}} \\ $$$$\left({c}\right)\:\mid\overset{\rightarrow} {{v}}_{\mathrm{C}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{A}} \mid\:=\:\mathrm{2}\mid\overset{\rightarrow} {{v}}_{\mathrm{B}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{C}} \mid \\ $$$$\left({d}\right)\:\mid\overset{\rightarrow} {{v}}_{\mathrm{C}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{A}} \mid\:=\:\mathrm{4}\mid\overset{\rightarrow} {{v}}_{\mathrm{B}} \mid \\ $$

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