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AllQuestion and Answers: Page 1789

Question Number 24755 Answers: 0 Comments: 0

Given f(x) =Σ_(x=1) ^n tan((x/2^r )).sec((x/2^(r−1) )) where r and n εN g(x) =lim_(n→∝) ((ln(f(x)+tan(x/2^n )) −(f(x)+tan(x/2^n )).[sin(tan(x/2)))/(1+(f(x) + tan(x/2^n ))^n )) = k for x =(π/4) and the domain of g(x) is (0 ,(π/2)) where [.] denotes the g.i.f Find the value of k, if possible so that g(x) is continuous at x =(π/4) .

$${Given} \\ $$$${f}\left({x}\right)\:=\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}\left(\frac{{x}}{\mathrm{2}^{{r}} }\right).{sec}\left(\frac{{x}}{\mathrm{2}^{{r}−\mathrm{1}} }\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:{where}\:{r}\:{and}\:{n}\:\varepsilon{N} \\ $$$${g}\left({x}\right)\:=\underset{{n}\rightarrow\propto} {\mathrm{li}{m}}\:\:\frac{{ln}\left({f}\left({x}\right)+{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right)\:−\left({f}\left({x}\right)+{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right).\left[{sin}\left({tan}\frac{{x}}{\mathrm{2}}\right)\right.}{\mathrm{1}+\left({f}\left({x}\right)\:\:+\:\:{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right)^{{n}} }\:\:=\:{k} \\ $$$${for}\:{x}\:=\frac{\pi}{\mathrm{4}}\:\:{and}\:{the}\:{domain}\:{of} \\ $$$${g}\left({x}\right)\:{is}\:\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${where}\:\left[.\right]\:{denotes}\:{the}\:{g}.{i}.\mathrm{f} \\ $$$${Find}\:{the}\:{value}\:{of}\:{k},\:{if}\:{possible}\: \\ $$$${so}\:{that}\:{g}\left({x}\right)\:{is}\:{continuous}\:{at}\: \\ $$$${x}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$$$ \\ $$$$ \\ $$

Question Number 24753 Answers: 1 Comments: 0

Question Number 24747 Answers: 1 Comments: 0

if f(x)=2∣x−3∣ and g(x)=x^2 .Find: (i)gof (ii)fog (iii)domain of fog (iv)range of gof

$${if}\:{f}\left({x}\right)=\mathrm{2}\mid{x}−\mathrm{3}\mid\:{and}\:{g}\left({x}\right)={x}^{\mathrm{2}} .{Find}: \\ $$$$\left({i}\right){gof}\:\left({ii}\right){fog}\:\left({iii}\right){domain}\:{of}\:{fog} \\ $$$$\left({iv}\right){range}\:{of}\:{gof} \\ $$$$ \\ $$

Question Number 24744 Answers: 1 Comments: 0

If a function f is defined such that f:R→R.If f(x)=((3x−2)/(x^2 +5x−6)).Find the (i)domain of f(x) (ii)range of f(x)

$${If}\:{a}\:{function}\:{f}\:{is}\:{defined}\:{such}\:{that} \\ $$$${f}:\mathbb{R}\rightarrow\mathbb{R}.{If}\: \\ $$$$\:\:\:\:{f}\left({x}\right)=\frac{\mathrm{3}{x}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{6}}.{Find}\:{the}\: \\ $$$$\left({i}\right){domain}\:{of}\:{f}\left({x}\right) \\ $$$$\left({ii}\right){range}\:{of}\:{f}\left({x}\right) \\ $$

Question Number 24739 Answers: 0 Comments: 12

A particle is suspended vertically from point O by ideal string of length L. It is given horizontal velocity ′v′. There is vertical line AB at a distance (L/8) from P. At some point, it leaves circular motion and follows projectile motion. At the instant it crosses AB, its velocity is horizontal. Find u

$${A}\:{particle}\:{is}\:{suspended}\:{vertically} \\ $$$${from}\:{point}\:{O}\:{by}\:{ideal}\:{string}\:{of}\:{length} \\ $$$${L}.\:{It}\:{is}\:{given}\:{horizontal}\:{velocity}\:'{v}'. \\ $$$${There}\:{is}\:{vertical}\:{line}\:{AB}\:{at}\:{a}\:{distance} \\ $$$$\frac{{L}}{\mathrm{8}}\:{from}\:{P}.\:{At}\:{some}\:{point},\:{it}\:{leaves} \\ $$$${circular}\:{motion}\:{and}\:{follows}\:{projectile} \\ $$$${motion}.\:{At}\:{the}\:{instant}\:{it}\:{crosses}\:{AB}, \\ $$$${its}\:{velocity}\:{is}\:{horizontal}.\:{Find}\:{u} \\ $$

Question Number 24733 Answers: 1 Comments: 1

Find the second derivative of f(x) =(√(5x+9)) find f^(′′)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{second}\:\mathrm{derivative}\:\mathrm{of} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\sqrt{\mathrm{5x}+\mathrm{9}} \\ $$$$\mathrm{find}\:\mathrm{f}^{''} \\ $$

Question Number 24730 Answers: 0 Comments: 13

Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{square}\:\mathrm{plate}\:\mathrm{of}\:\mathrm{side} \\ $$$${a}\:\mathrm{and}\:\mathrm{mass}\:{m}.\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{plate}\:\mathrm{about}\:\mathrm{an}\:\mathrm{axis}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{its}\:\mathrm{plane}\:\mathrm{and}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{one}\:\mathrm{of} \\ $$$$\mathrm{its}\:\mathrm{corners}\:\mathrm{is} \\ $$

Question Number 24728 Answers: 0 Comments: 0

find sum of : 1^(3 ) − ( 1.5)^3 +2^(3 ) −(2.5)^3 +......... ?

$$\mathrm{find}\:\mathrm{sum}\:\mathrm{of}\:: \\ $$$$\mathrm{1}^{\mathrm{3}\:} −\:\left(\:\mathrm{1}.\mathrm{5}\right)^{\mathrm{3}} \:+\mathrm{2}^{\mathrm{3}\:} −\left(\mathrm{2}.\mathrm{5}\right)^{\mathrm{3}} +.........\:? \\ $$

Question Number 24716 Answers: 1 Comments: 0

A 2.2 kg block starts from rest on a rough inclined plane that makes an angle of 25° with the horizontal. The coefficient of kinetic friction is 0.25. As the block goes 2 m down the plane, the mechanical energy of the Earth-block system changes by

$$\mathrm{A}\:\mathrm{2}.\mathrm{2}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{rough}\:\mathrm{inclined}\:\mathrm{plane}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{25}°\:\mathrm{with}\:\mathrm{the}\:\mathrm{horizontal}.\:\mathrm{The} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{kinetic}\:\mathrm{friction}\:\mathrm{is}\:\mathrm{0}.\mathrm{25}.\:\mathrm{As} \\ $$$$\mathrm{the}\:\mathrm{block}\:\mathrm{goes}\:\mathrm{2}\:\mathrm{m}\:\mathrm{down}\:\mathrm{the}\:\mathrm{plane},\:\mathrm{the} \\ $$$$\mathrm{mechanical}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Earth}-\mathrm{block} \\ $$$$\mathrm{system}\:\mathrm{changes}\:\mathrm{by} \\ $$

Question Number 24712 Answers: 0 Comments: 3

Question Number 24704 Answers: 1 Comments: 5

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Question Number 24696 Answers: 0 Comments: 7

App Update App has been updated with the following changes • Long image upload issue with submit button • Added an option to rotate image while uploading • Added ability to set a password for reusing id across devices • line thickness for square root • Some minor cosmetic changes (font size, colors, menus texts)

$$\boldsymbol{\mathrm{App}}\:\boldsymbol{\mathrm{Update}} \\ $$$$\mathrm{App}\:\mathrm{has}\:\mathrm{been}\:\mathrm{updated}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{changes} \\ $$$$\bullet\:\mathrm{Long}\:\mathrm{image}\:\mathrm{upload}\:\mathrm{issue}\:\mathrm{with} \\ $$$$\:\:\:\:\mathrm{submit}\:\mathrm{button} \\ $$$$\bullet\:\mathrm{Added}\:\mathrm{an}\:\mathrm{option}\:\mathrm{to}\:\mathrm{rotate}\:\mathrm{image} \\ $$$$\:\:\:\:\mathrm{while}\:\mathrm{uploading} \\ $$$$\bullet\:\mathrm{Added}\:\mathrm{ability}\:\mathrm{to}\:\mathrm{set}\:\mathrm{a}\:\mathrm{password}\:\mathrm{for} \\ $$$$\:\:\:\:\mathrm{reusing}\:\mathrm{id}\:\mathrm{across}\:\mathrm{devices} \\ $$$$\bullet\:\mathrm{line}\:\mathrm{thickness}\:\mathrm{for}\:\mathrm{square}\:\mathrm{root} \\ $$$$\bullet\:\mathrm{Some}\:\mathrm{minor}\:\mathrm{cosmetic}\:\mathrm{changes} \\ $$$$\:\:\:\left(\mathrm{font}\:\mathrm{size},\:\mathrm{colors},\:\mathrm{menus}\:\mathrm{texts}\right) \\ $$

Question Number 24684 Answers: 0 Comments: 0

Let ABCD be a square and M, N points on sides AB, BC respectably, such that ∠MDN = 45°. If R is the midpoint of MN show that RP = RQ where P, Q are the points of intersection of AC with the lines MD, ND.

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{square}\:\mathrm{and}\:{M},\:{N}\:\mathrm{points} \\ $$$$\mathrm{on}\:\mathrm{sides}\:{AB},\:{BC}\:\mathrm{respectably},\:\mathrm{such}\:\mathrm{that} \\ $$$$\angle{MDN}\:=\:\mathrm{45}°.\:\mathrm{If}\:{R}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$${MN}\:\mathrm{show}\:\mathrm{that}\:{RP}\:=\:{RQ}\:\mathrm{where}\:{P},\:{Q} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{points}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:{AC}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{lines}\:{MD},\:{ND}. \\ $$

Question Number 24680 Answers: 1 Comments: 2

Question Number 24677 Answers: 2 Comments: 0

Let T_k is the k^(th) term and S_k is the sum of the first k term in arithmetic progression If T_3 + T_6 + T_9 + T_(12) + T_(15) + T_(18) = 45 Find S_(20)

$$\mathrm{Let}\:{T}_{{k}} \:\mathrm{is}\:\mathrm{the}\:{k}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{and}\:\mathrm{S}_{{k}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:{k}\:\mathrm{term}\:\mathrm{in}\:\mathrm{arithmetic}\:\mathrm{progression} \\ $$$$\mathrm{If}\:{T}_{\mathrm{3}} \:+\:{T}_{\mathrm{6}} \:+\:{T}_{\mathrm{9}} \:+\:{T}_{\mathrm{12}} \:+\:{T}_{\mathrm{15}} \:+\:{T}_{\mathrm{18}} \:=\:\mathrm{45} \\ $$$$\mathrm{Find}\:{S}_{\mathrm{20}} \\ $$

Question Number 24670 Answers: 1 Comments: 0

Question Number 24667 Answers: 0 Comments: 0

Question Number 24663 Answers: 1 Comments: 0

solve ∣x−3∣=∣3x+2∣−1

$$\boldsymbol{{solve}}\:\mid{x}−\mathrm{3}\mid=\mid\mathrm{3}{x}+\mathrm{2}\mid−\mathrm{1} \\ $$

Question Number 24661 Answers: 1 Comments: 2

Question Number 24662 Answers: 1 Comments: 0

solve ∣x^2 −4x−5∣=7

$$\boldsymbol{{solve}}\:\mid{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}\mid=\mathrm{7} \\ $$$$ \\ $$

Question Number 24651 Answers: 2 Comments: 0

∫sin x+cos y dx

$$\int\mathrm{sin}\:{x}+\mathrm{cos}\:{y}\:{dx} \\ $$

Question Number 24647 Answers: 0 Comments: 0

Question Number 24643 Answers: 1 Comments: 0

Question Number 24635 Answers: 0 Comments: 0

Calculate the electric potential at a point P at a distance of 3m of either charges of +20 μC and − 15μC. which are 25cm apart. Also calculate potential energy of a +3.5μC placed at point P.

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{electric}\:\mathrm{potential}\:\mathrm{at}\:\mathrm{a}\:\mathrm{point}\:\mathrm{P}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{3m}\:\mathrm{of}\:\mathrm{either} \\ $$$$\mathrm{charges}\:\mathrm{of}\:\:+\mathrm{20}\:\mu\mathrm{C}\:\mathrm{and}\:\:−\:\mathrm{15}\mu\mathrm{C}.\:\:\mathrm{which}\:\mathrm{are}\:\mathrm{25cm}\:\mathrm{apart}. \\ $$$$\mathrm{Also}\:\mathrm{calculate}\:\mathrm{potential}\:\mathrm{energy}\:\mathrm{of}\:\mathrm{a}\:\:+\mathrm{3}.\mathrm{5}\mu\mathrm{C}\:\mathrm{placed}\:\mathrm{at}\:\mathrm{point}\:\mathrm{P}. \\ $$

Question Number 24644 Answers: 2 Comments: 0

The density of a non-uniform rod of length 1 m is given by ρ(x) = a(1 + bx^2 ) where a and b are constants and 0 ≤ x ≤ 1. The centre of mass of the rod will be at (1) ((3(2 + b))/(4(3 + b))) (2) ((4(2 + b))/(3(3 + b))) (3) ((3(3 + b))/(4(2 + b))) (4) ((4(3 + b))/(3(2 + b)))

$$\mathrm{The}\:\mathrm{density}\:\mathrm{of}\:\mathrm{a}\:\mathrm{non}-\mathrm{uniform}\:\mathrm{rod}\:\mathrm{of} \\ $$$$\mathrm{length}\:\mathrm{1}\:\mathrm{m}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\rho\left({x}\right)\:=\:{a}\left(\mathrm{1}\:+\:{bx}^{\mathrm{2}} \right) \\ $$$$\mathrm{where}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{constants}\:\mathrm{and} \\ $$$$\mathrm{0}\:\leqslant\:{x}\:\leqslant\:\mathrm{1}.\:\mathrm{The}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rod} \\ $$$$\mathrm{will}\:\mathrm{be}\:\mathrm{at} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{3}\left(\mathrm{2}\:+\:{b}\right)}{\mathrm{4}\left(\mathrm{3}\:+\:{b}\right)} \\ $$$$\left(\mathrm{2}\right)\:\frac{\mathrm{4}\left(\mathrm{2}\:+\:{b}\right)}{\mathrm{3}\left(\mathrm{3}\:+\:{b}\right)} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{3}\left(\mathrm{3}\:+\:{b}\right)}{\mathrm{4}\left(\mathrm{2}\:+\:{b}\right)} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{4}\left(\mathrm{3}\:+\:{b}\right)}{\mathrm{3}\left(\mathrm{2}\:+\:{b}\right)} \\ $$

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