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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

Question Number 20013 Answers: 0 Comments: 4

Show that the equation (1/(x − a)) + (1/(x − b)) + (1/(x − c)) = 0 can have a pair of equal roots if a = b = c.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\frac{\mathrm{1}}{{x}\:−\:{a}}\:+\:\frac{\mathrm{1}}{{x}\:−\:{b}} \\ $$$$+\:\frac{\mathrm{1}}{{x}\:−\:{c}}\:=\:\mathrm{0}\:\mathrm{can}\:\mathrm{have}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{equal} \\ $$$$\mathrm{roots}\:\mathrm{if}\:{a}\:=\:{b}\:=\:{c}. \\ $$

Question Number 20131 Answers: 1 Comments: 4

In rectangle ABCD,AB=8, BC=20.P is a point on AD so that ∠BPC=90°.If r_1 ,r_2 ,r_3 are the radii of the incircles of APB, BPC, and CPD. find r_1 +r_2 +r_3

$${In}\:{rectangle}\:{ABCD},{AB}=\mathrm{8}, \\ $$$${BC}=\mathrm{20}.{P}\:{is}\:{a}\:{point}\:{on}\:{AD}\:{so} \\ $$$${that}\:\angle{BPC}=\mathrm{90}°.{If}\:{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ,{r}_{\mathrm{3}} \:{are}\:{the} \\ $$$${radii}\:{of}\:{the}\:{incircles}\:{of}\:{APB}, \\ $$$${BPC},\:{and}\:{CPD}.\:{find}\:{r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} \\ $$

Question Number 20001 Answers: 1 Comments: 0

The number of the roots of the quadratic equation 8sec^2 θ − 6secθ + 1 = 0 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic} \\ $$$$\mathrm{equation}\:\mathrm{8sec}^{\mathrm{2}} \theta\:−\:\mathrm{6sec}\theta\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{is} \\ $$

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