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Question Number 20138    Answers: 1   Comments: 0

lim_(x→0) ((1−cos ax)/(1−cos bx))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{cos}\:{ax}}{\mathrm{1}−\mathrm{cos}\:{bx}} \\ $$$$ \\ $$

Question Number 20118    Answers: 1   Comments: 0

The quadratic equations x^2 − 6x + a = 0 and x^2 − cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then, find the common root.

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equations}\:{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:{a}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} \:−\:{cx}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\mathrm{have}\:\mathrm{one}\:\mathrm{root}\:\mathrm{in} \\ $$$$\mathrm{common}.\:\mathrm{The}\:\mathrm{other}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{4}\::\:\mathrm{3}.\:\mathrm{Then},\:\mathrm{find}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{root}. \\ $$

Question Number 20116    Answers: 1   Comments: 0

If a and b (≠ 0) are the roots of the equation x^2 + ax + b = 0, then find the least value of x^2 + ax + b (x ∈ R).

$$\mathrm{If}\:{a}\:\mathrm{and}\:{b}\:\left(\neq\:\mathrm{0}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:\left({x}\:\in\:{R}\right). \\ $$

Question Number 20115    Answers: 1   Comments: 0

The value of a for which the equation (1 − a^2 )x^2 + 2ax − 1 = 0 has roots belonging to (0, 1) is

$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\:−\:{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{is} \\ $$

Question Number 20110    Answers: 0   Comments: 1

Question Number 20102    Answers: 2   Comments: 0

Solve the equation: (log _(sin x) cos x)^2 =1

$${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{log}\:_{\mathrm{sin}\:{x}} \mathrm{cos}\:{x}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

Question Number 20091    Answers: 1   Comments: 0

Prove that Σ_(n=0) ^3 tan^2 (((2n + 1)π)/(16)) = 28.

$$\mathrm{Prove}\:\mathrm{that}\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{3}} {\sum}}\mathrm{tan}^{\mathrm{2}} \:\frac{\left(\mathrm{2}{n}\:+\:\mathrm{1}\right)\pi}{\mathrm{16}}\:=\:\mathrm{28}. \\ $$

Question Number 20079    Answers: 0   Comments: 2

Question Number 20068    Answers: 1   Comments: 1

Question Number 20058    Answers: 1   Comments: 0

What is the difference between ∮ and ∫? Where is ∮ used?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{between}\:\oint\:\mathrm{and} \\ $$$$\int?\:\mathrm{Where}\:\mathrm{is}\:\oint\:\mathrm{used}? \\ $$

Question Number 20054    Answers: 1   Comments: 0

If α and β (α < β) are the roots of the equation x^2 + bx + c = 0, where c < 0 < b, then (1) 0 < α < β (2) α < 0 < β < ∣α∣ (3) α < β < 0 (4) α < 0 < ∣α∣ < β

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\left(\alpha\:<\:\beta\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{where} \\ $$$${c}\:<\:\mathrm{0}\:<\:{b},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{0}\:<\:\alpha\:<\:\beta \\ $$$$\left(\mathrm{2}\right)\:\alpha\:<\:\mathrm{0}\:<\:\beta\:<\:\mid\alpha\mid \\ $$$$\left(\mathrm{3}\right)\:\alpha\:<\:\beta\:<\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\alpha\:<\:\mathrm{0}\:<\:\mid\alpha\mid\:<\:\beta \\ $$

Question Number 20053    Answers: 1   Comments: 0

If (4a + c)^2 ≤ 4b^2 then one root of ax^2 + bx + c = 0 lies in (1) (−2, 2) (2) (−1, 1) (3) (−∞, −2) (4) (2, ∞)

$$\mathrm{If}\:\left(\mathrm{4}{a}\:+\:{c}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{4}{b}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{lies}\:\mathrm{in} \\ $$$$\left(\mathrm{1}\right)\:\left(−\mathrm{2},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:\left(−\infty,\:−\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2},\:\infty\right) \\ $$

Question Number 20052    Answers: 1   Comments: 0

If the roots α and β of the equation ax^2 + bx + c = 0 are real and of opposite sign then the roots of the equation α(x − β)^2 + β(x − α)^2 is/are (1) Positive (2) Negative (3) Real and opposite sign (4) Imaginary

$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{real}\:\mathrm{and}\:\mathrm{of}\:\mathrm{opposite} \\ $$$$\mathrm{sign}\:\mathrm{then}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\alpha\left({x}\:−\:\beta\right)^{\mathrm{2}} \:+\:\beta\left({x}\:−\:\alpha\right)^{\mathrm{2}} \:\mathrm{is}/\mathrm{are} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Positive} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Negative} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Real}\:\mathrm{and}\:\mathrm{opposite}\:\mathrm{sign} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Imaginary} \\ $$

Question Number 20049    Answers: 0   Comments: 0

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Question Number 20047    Answers: 1   Comments: 0

Solve for x: ((√(x + 1))/x) + (√(x/(x + 1))) = ((13)/6)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}: \\ $$$$\frac{\sqrt{\mathrm{x}\:+\:\mathrm{1}}}{\mathrm{x}}\:+\:\sqrt{\frac{\mathrm{x}}{\mathrm{x}\:+\:\mathrm{1}}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$

Question Number 20042    Answers: 0   Comments: 3

In the situation given, all surfaces are frictionless, pulley is ideal and string is light, F = ((mg)/2) , find the acceleration of block 2.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{situation}\:\mathrm{given},\:\mathrm{all}\:\mathrm{surfaces}\:\mathrm{are} \\ $$$$\mathrm{frictionless},\:\mathrm{pulley}\:\mathrm{is}\:\mathrm{ideal}\:\mathrm{and}\:\mathrm{string}\:\mathrm{is} \\ $$$$\mathrm{light},\:{F}\:=\:\frac{{mg}}{\mathrm{2}}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{acceleration}\:\mathrm{of} \\ $$$$\mathrm{block}\:\mathrm{2}. \\ $$

Question Number 20040    Answers: 0   Comments: 3

The system shown in figure is given an acceleration ′a′ toward left. Assuming all the surfaces to be frictionless, find the force on the sphere by inclined surface.

$$\mathrm{The}\:\mathrm{system}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure}\:\mathrm{is}\:\mathrm{given}\:\mathrm{an} \\ $$$$\mathrm{acceleration}\:'{a}'\:\mathrm{toward}\:\mathrm{left}.\:\mathrm{Assuming} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{surfaces}\:\mathrm{to}\:\mathrm{be}\:\mathrm{frictionless},\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{force}\:\mathrm{on}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{by}\:\mathrm{inclined} \\ $$$$\mathrm{surface}. \\ $$

Question Number 20038    Answers: 1   Comments: 1

In the figure shown, m slides on inclined surface of wedge M. If velocity of wedge at any instant be v, find velocity of m with respect to ground.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown},\:{m}\:\mathrm{slides}\:\mathrm{on} \\ $$$$\mathrm{inclined}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{wedge}\:{M}.\:\mathrm{If}\:\mathrm{velocity} \\ $$$$\mathrm{of}\:\mathrm{wedge}\:\mathrm{at}\:\mathrm{any}\:\mathrm{instant}\:\mathrm{be}\:{v},\:\mathrm{find} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:{m}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{ground}. \\ $$

Question Number 20036    Answers: 1   Comments: 0

Question Number 20031    Answers: 1   Comments: 0

Question Number 20035    Answers: 1   Comments: 1

In the following cases, find out the acceleration of the wedge and the block, if an external force F is applied as shown. (Both pulleys and strings are ideal)

$$\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{cases},\:\mathrm{find}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{and}\:\mathrm{the}\:\mathrm{block}, \\ $$$$\mathrm{if}\:\mathrm{an}\:\mathrm{external}\:\mathrm{force}\:{F}\:\mathrm{is}\:\mathrm{applied}\:\mathrm{as} \\ $$$$\mathrm{shown}.\:\left(\mathrm{Both}\:\mathrm{pulleys}\:\mathrm{and}\:\mathrm{strings}\:\mathrm{are}\right. \\ $$$$\left.\mathrm{ideal}\right) \\ $$

Question Number 20051    Answers: 1   Comments: 0

If x ∈ R then ((x^2 + 2x + a)/(x^2 + 4x + 3a)) can take all real values if (1) a ∈ (0, 2) (2) a ∈ [0, 1] (3) a ∈ [−1, 1] (4) None of these

$$\mathrm{If}\:{x}\:\in\:{R}\:\mathrm{then}\:\frac{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:{a}}{{x}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:+\:\mathrm{3}{a}}\:\mathrm{can}\:\mathrm{take}\:\mathrm{all} \\ $$$$\mathrm{real}\:\mathrm{values}\:\mathrm{if} \\ $$$$\left(\mathrm{1}\right)\:{a}\:\in\:\left(\mathrm{0},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:{a}\:\in\:\left[\mathrm{0},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{3}\right)\:{a}\:\in\:\left[−\mathrm{1},\:\mathrm{1}\right] \\ $$$$\left(\mathrm{4}\right)\:\mathrm{None}\:\mathrm{of}\:\mathrm{these} \\ $$

Question Number 20021    Answers: 1   Comments: 0

A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at the angle β and finds the angle of elevation of the peak of the hill to be γ. Show that the height of the peak above the ground is ((c sin α sin (γ − β))/((sin γ − α))).

$$\mathrm{A}\:\mathrm{person}\:\mathrm{observes}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{a}\:\mathrm{hill}\:\mathrm{from}\:\mathrm{a}\:\mathrm{station}\:\mathrm{to}\:\mathrm{be} \\ $$$$\alpha.\:\mathrm{He}\:\mathrm{walks}\:{c}\:\mathrm{metres}\:\mathrm{along}\:\mathrm{a}\:\mathrm{slope} \\ $$$$\mathrm{inclined}\:\mathrm{at}\:\mathrm{the}\:\mathrm{angle}\:\beta\:\mathrm{and}\:\mathrm{finds}\:\mathrm{the} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\mathrm{elevation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{peak}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{hill}\:\mathrm{to}\:\mathrm{be}\:\gamma.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{height}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{peak}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{is} \\ $$$$\frac{{c}\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\left(\gamma\:−\:\beta\right)}{\left(\mathrm{sin}\:\gamma\:−\:\alpha\right)}. \\ $$

Question Number 20019    Answers: 1   Comments: 0

In any triangle ABC, prove that a (cos C − cos B) = 2 (b − c) cos^2 (A/2)

$$\mathrm{In}\:\mathrm{any}\:\mathrm{triangle}\:{ABC},\:\mathrm{prove}\:\mathrm{that} \\ $$$${a}\:\left(\mathrm{cos}\:{C}\:−\:\mathrm{cos}\:{B}\right)\:=\:\mathrm{2}\:\left({b}\:−\:{c}\right)\:\mathrm{cos}^{\mathrm{2}} \:\frac{{A}}{\mathrm{2}} \\ $$

Question Number 20044    Answers: 2   Comments: 0

Question Number 20014    Answers: 0   Comments: 1

A person in lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift′s motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II. List I P. Lift is accelerating vertically up Q. Lift is accelerating vertically down with an acceleration less than the gravitational acceleration R. Lift is moving vertically up with constant speed S. Lift is falling freely List II 1. d = 1.2 m 2. d > 1.2 m 3. d < 1.2 m 4. No water leaks out of the jar

$$\mathrm{A}\:\mathrm{person}\:\mathrm{in}\:\mathrm{lift}\:\mathrm{is}\:\mathrm{holding}\:\mathrm{a}\:\mathrm{water}\:\mathrm{jar}, \\ $$$$\mathrm{which}\:\mathrm{has}\:\mathrm{a}\:\mathrm{small}\:\mathrm{hole}\:\mathrm{at}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{end} \\ $$$$\mathrm{of}\:\mathrm{its}\:\mathrm{side}.\:\mathrm{When}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest},\:\mathrm{the} \\ $$$$\mathrm{water}\:\mathrm{jet}\:\mathrm{coming}\:\mathrm{out}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hole}\:\mathrm{hits} \\ $$$$\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{at}\:\mathrm{a}\:\mathrm{distance}\:{d}\:\mathrm{of} \\ $$$$\mathrm{1}.\mathrm{2}\:\mathrm{m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{person}.\:\mathrm{In}\:\mathrm{the}\:\mathrm{following}, \\ $$$$\mathrm{state}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}'\mathrm{s}\:\mathrm{motion}\:\mathrm{is}\:\mathrm{given}\:\mathrm{in}\:\mathrm{List} \\ $$$$\mathrm{I}\:\mathrm{and}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{where}\:\mathrm{the}\:\mathrm{water}\:\mathrm{jet} \\ $$$$\mathrm{hits}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lift}\:\mathrm{is}\:\mathrm{given}\:\mathrm{in}\:\mathrm{List} \\ $$$$\mathrm{II}.\:\mathrm{Match}\:\mathrm{the}\:\mathrm{statements}\:\mathrm{from}\:\mathrm{List}\:\mathrm{I} \\ $$$$\mathrm{with}\:\mathrm{those}\:\mathrm{in}\:\mathrm{List}\:\mathrm{II}. \\ $$$$\boldsymbol{\mathrm{List}}\:\boldsymbol{\mathrm{I}} \\ $$$$\boldsymbol{\mathrm{P}}.\:\mathrm{Lift}\:\mathrm{is}\:\mathrm{accelerating}\:\mathrm{vertically}\:\mathrm{up} \\ $$$$\boldsymbol{\mathrm{Q}}.\:\mathrm{Lift}\:\mathrm{is}\:\mathrm{accelerating}\:\mathrm{vertically}\:\mathrm{down} \\ $$$$\mathrm{with}\:\mathrm{an}\:\mathrm{acceleration}\:\mathrm{less}\:\mathrm{than}\:\mathrm{the} \\ $$$$\mathrm{gravitational}\:\mathrm{acceleration} \\ $$$$\boldsymbol{\mathrm{R}}.\:\mathrm{Lift}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{vertically}\:\mathrm{up}\:\mathrm{with} \\ $$$$\mathrm{constant}\:\mathrm{speed} \\ $$$$\boldsymbol{\mathrm{S}}.\:\mathrm{Lift}\:\mathrm{is}\:\mathrm{falling}\:\mathrm{freely} \\ $$$$\boldsymbol{\mathrm{List}}\:\boldsymbol{\mathrm{II}} \\ $$$$\mathrm{1}.\:{d}\:=\:\mathrm{1}.\mathrm{2}\:\mathrm{m} \\ $$$$\mathrm{2}.\:{d}\:>\:\mathrm{1}.\mathrm{2}\:\mathrm{m} \\ $$$$\mathrm{3}.\:{d}\:<\:\mathrm{1}.\mathrm{2}\:\mathrm{m} \\ $$$$\mathrm{4}.\:\mathrm{No}\:\mathrm{water}\:\mathrm{leaks}\:\mathrm{out}\:\mathrm{of}\:\mathrm{the}\:\mathrm{jar} \\ $$

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