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Question Number 35897    Answers: 0   Comments: 2

log_x^(1/2 ) 64=3. What is x?

$${log}_{{x}^{\mathrm{1}/\mathrm{2}\:} } \mathrm{64}=\mathrm{3}.\:{What}\:{is}\:{x}? \\ $$

Question Number 35895    Answers: 2   Comments: 1

Question Number 35892    Answers: 2   Comments: 2

Question Number 35886    Answers: 1   Comments: 0

Question Number 35882    Answers: 0   Comments: 4

An open pipe 40cm long and a closed pipe 32cm long,with the same diameter,sound the same fundamental frequency in unison. Find the end correction of these pipes.

$${An}\:{open}\:{pipe}\:\mathrm{40}{cm}\:{long}\:{and}\:{a} \\ $$$${closed}\:{pipe}\:\mathrm{32}{cm}\:{long},{with}\:{the} \\ $$$${same}\:{diameter},{sound}\:{the}\:{same} \\ $$$${fundamental}\:{frequency}\:{in}\:{unison}. \\ $$$${Find}\:{the}\:{end}\:{correction}\:{of}\:{these} \\ $$$${pipes}. \\ $$

Question Number 35881    Answers: 1   Comments: 1

A train travelling at 30m/s , approaching a stationary observer emits a tone at a frequency of 480Hz. If the speed of sound is 330m/s, determine the frequency of the note the observer hears.

$${A}\:{train}\:{travelling}\:{at}\:\mathrm{30}{m}/{s}\:, \\ $$$${approaching}\:{a}\:{stationary}\:{observer} \\ $$$${emits}\:{a}\:{tone}\:{at}\:{a}\:{frequency}\:{of}\:\mathrm{480}{Hz}. \\ $$$${If}\:{the}\:{speed}\:{of}\:{sound}\:{is}\:\mathrm{330}{m}/{s}, \\ $$$${determine}\:{the}\:{frequency}\:{of}\:{the} \\ $$$${note}\:{the}\:{observer}\:{hears}. \\ $$

Question Number 35880    Answers: 0   Comments: 0

Determine the intensity level of 12 people′s speech,if each person speaks with an intensity level of 60dB

$${Determine}\:{the}\:{intensity}\:{level}\:{of} \\ $$$$\mathrm{12}\:{people}'{s}\:{speech},{if}\:{each}\:{person} \\ $$$${speaks}\:{with}\:{an}\:{intensity}\:{level}\:{of} \\ $$$$\mathrm{60}{dB} \\ $$

Question Number 35879    Answers: 1   Comments: 0

Find the velocity of the source along the line joining the source to a stationary observer if,as a result of the motion,the frequency of the note heard is a)increased in the ratio 18:15 b)decreased in the ratio 15:18

$${Find}\:{the}\:{velocity}\:{of}\:{the}\:{source}\: \\ $$$${along}\:{the}\:{line}\:{joining}\:{the}\:{source} \\ $$$${to}\:{a}\:{stationary}\:{observer}\:{if},{as}\:{a} \\ $$$${result}\:{of}\:{the}\:{motion},{the}\:{frequency} \\ $$$${of}\:{the}\:{note}\:{heard}\:{is} \\ $$$$\left.{a}\right){increased}\:{in}\:{the}\:{ratio}\:\mathrm{18}:\mathrm{15} \\ $$$$\left.{b}\right){decreased}\:{in}\:{the}\:{ratio}\:\mathrm{15}:\mathrm{18} \\ $$

Question Number 35878    Answers: 1   Comments: 1

A sonometer wire 60cm long,under a tension of 120N,vibrates in unison with a turning fork of frequency 512Hz.The wire is then shortened by 2cm,the tension remaining the same.If the fork and wire are sounded together,how many beats per second are heard? What decrease in tension will be required to restore the frequency of vibration of the wire to the original value?

$${A}\:{sonometer}\:{wire}\:\mathrm{60}{cm}\:{long},{under} \\ $$$${a}\:{tension}\:{of}\:\mathrm{120}{N},{vibrates}\:{in} \\ $$$${unison}\:{with}\:{a}\:{turning}\:{fork}\:{of} \\ $$$${frequency}\:\mathrm{512}{Hz}.{The}\:{wire}\:{is}\:{then} \\ $$$${shortened}\:{by}\:\mathrm{2}{cm},{the}\:{tension} \\ $$$${remaining}\:{the}\:{same}.{If}\:{the}\:{fork} \\ $$$${and}\:{wire}\:{are}\:{sounded}\:{together},{how} \\ $$$${many}\:{beats}\:{per}\:{second}\:{are}\:{heard}? \\ $$$${What}\:{decrease}\:{in}\:{tension}\:{will}\:{be} \\ $$$${required}\:{to}\:{restore}\:{the}\:{frequency} \\ $$$${of}\:{vibration}\:{of}\:{the}\:{wire}\:{to}\:{the} \\ $$$${original}\:{value}? \\ $$

Question Number 35877    Answers: 1   Comments: 0

If cos A=(3/4) , then 32 sin ((A/2)) sin (((5A)/2))=

$$\mathrm{If}\:\:\mathrm{cos}\:{A}=\frac{\mathrm{3}}{\mathrm{4}}\:,\:\mathrm{then}\:\mathrm{32}\:\mathrm{sin}\:\left(\frac{{A}}{\mathrm{2}}\right)\:\mathrm{sin}\:\left(\frac{\mathrm{5}{A}}{\mathrm{2}}\right)= \\ $$

Question Number 35876    Answers: 0   Comments: 3

please help why is it not advisable to use a small angle of incidence when performing experiment on refraction in a rectangular prism????

$${please}\:{help} \\ $$$$ \\ $$$${why}\:{is}\:{it}\:{not}\:{advisable}\:{to}\:{use}\:{a} \\ $$$${small}\:{angle}\:{of}\:{incidence}\:{when} \\ $$$${performing}\:{experiment}\:{on}\:{refraction} \\ $$$${in}\:{a}\:{rectangular}\:{prism}???? \\ $$

Question Number 35873    Answers: 1   Comments: 0

Question Number 35872    Answers: 1   Comments: 1

Question Number 35871    Answers: 0   Comments: 0

new idea (and solution) to questions 35178 & 35195 triangle: ABC; a=BC, b=CA, c=AB; α=∠CAB, β=∠ABC, γ=∠BCA d=(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))) put it as this: A= ((0),(0) ), B= ((c),(0) ), C= ((((−a^2 +b^2 +c^2 )/(2c))),((d/(2c))) ) circumcircle: center=M_1 = (((c/2)),((((a^2 +b^2 −c^2 )c)/(2d))) ) radius=R=((abc)/d) (calculated by intersection of circles with centers A, B, C or of symmetry−axes of AB and AC) 2 circles touching b, c and circumcircle, one from inside, the other from outside: center=M_2 lies on y=kx with k=tan (α/2) M_2 = ((x),((xtan (α/2))) ) radius=r_1 =R−∣M_1 M_2 ∣=xtan (α/2) (inside) r_2 =∣M_1 M_2 ∣−R=xtan (α/2) (outside) (obviously any circle with center M_2 (x) and touching the x−axis has radius xtan (α/2) and also obviously the touching point of 2 circles is located on the line connecting their centers) 1. ∣M_1 M_2 ∣=R−xtan (α/2) M_1 M_2 =(R−xtan (α/2))^2 2. ∣M_1 M_2 ∣=R+xtan (α/2) M_1 M_2 =(R+xtan (α/2))^2 tan (α/2)=((sin (α/2))/(cos (α/2)))=((√((1−cos α)/2))/(√((1+cos α)/2)))=(√((1−cos α)/(1+cos α)))= [cos α=((−a^2 +b^2 +c^2 )/(2bc))] =(√((a^2 −b^2 +2bc−c^2 )/(−a^2 +b^2 +2bc+c^2 )))=(√(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))) M_1 M_2 =(m_1 −m_2 )^2 +(n_1 −n_2 )^2 = =((c/2)−x)^2 +((((a^2 +b^2 −c^2 )c)/(2d))−xtan (α/2))^2 = [after some transformation work] =((4bc)/((a+b+c)(−a+b+c)))x^2 −((2bc(b+c))/((a+b+c)(−a+b+c)))x+((a^2 b^2 c^2 )/d^2 ) [((a^2 b^2 c^2 )/d^2 )=R^2 ] (R±xtan (α/2))^2 =x^2 tan^2 (α/2)±2Rxtan (α/2)+R^2 = =(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))x^2 ±((2abc)/((a+b+c)(−a+b+c)))x+R^2 so we have ((4bc)/((a+b+c)(−a+b+c)))x^2 −((2bc(b+c))/((a+b+c)(−a+b+c)))x= =(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)))x^2 ±((2abc)/((a+b+c)(−a+b+c)))x which leads to x_3 =0 (as I explained before, the point A can be seen as a circle with radius 0 still meeting the requirements) x_1 =((2bc)/(a+b+c)) ⇒ r_1 =2bc(√(((a+b−c)(a−b+c))/((a+b+c)^3 (−a+b+c)))) x_2 =((2bc)/(−a+b+c)) ⇒ r_2 =2bc(√(((a+b−c)(a−b+c))/((a+b+c)(−a+b+c)^3 ))) for the circles corresponding with the points B and C just interchange {a, b, c} with {b, c, a} and {c, a, b}

$$\mathrm{new}\:\mathrm{idea}\:\left(\mathrm{and}\:\mathrm{solution}\right)\:\mathrm{to}\:\mathrm{questions}\:\mathrm{35178}\:\&\:\:\mathrm{35195} \\ $$$$ \\ $$$$\mathrm{triangle}: \\ $$$${ABC};\:{a}={BC},\:{b}={CA},\:{c}={AB}; \\ $$$$\alpha=\angle{CAB},\:\beta=\angle{ABC},\:\gamma=\angle{BCA} \\ $$$${d}=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$$ \\ $$$$\mathrm{put}\:\mathrm{it}\:\mathrm{as}\:\mathrm{this}: \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix},\:{B}=\begin{pmatrix}{{c}}\\{\mathrm{0}}\end{pmatrix},\:{C}=\begin{pmatrix}{\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{c}}}\\{\frac{{d}}{\mathrm{2}{c}}}\end{pmatrix} \\ $$$$ \\ $$$$\mathrm{circumcircle}: \\ $$$$\mathrm{center}={M}_{\mathrm{1}} =\begin{pmatrix}{\frac{{c}}{\mathrm{2}}}\\{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){c}}{\mathrm{2}{d}}}\end{pmatrix} \\ $$$$\mathrm{radius}={R}=\frac{{abc}}{{d}} \\ $$$$\left(\mathrm{calculated}\:\mathrm{by}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{circles}\:\mathrm{with}\right. \\ $$$$\mathrm{centers}\:{A},\:{B},\:{C}\:\mathrm{or}\:\mathrm{of}\:\mathrm{symmetry}−\mathrm{axes}\:\mathrm{of} \\ $$$$\left.{AB}\:\mathrm{and}\:{AC}\right) \\ $$$$ \\ $$$$\mathrm{2}\:\mathrm{circles}\:\mathrm{touching}\:{b},\:{c}\:\mathrm{and}\:\mathrm{circumcircle}, \\ $$$$\mathrm{one}\:\mathrm{from}\:\mathrm{inside},\:\mathrm{the}\:\mathrm{other}\:\mathrm{from}\:\mathrm{outside}: \\ $$$$\mathrm{center}={M}_{\mathrm{2}} \:\mathrm{lies}\:\mathrm{on}\:{y}={kx}\:\mathrm{with}\:{k}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$${M}_{\mathrm{2}} =\begin{pmatrix}{{x}}\\{{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{radius}={r}_{\mathrm{1}} ={R}−\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid={x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\left(\mathrm{inside}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{r}_{\mathrm{2}} =\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid−{R}={x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\left(\mathrm{outside}\right) \\ $$$$ \\ $$$$\left(\mathrm{obviously}\:\mathrm{any}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:{M}_{\mathrm{2}} \left({x}\right)\right. \\ $$$$\mathrm{and}\:\mathrm{touching}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{has}\:\mathrm{radius}\:{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{also}\:\mathrm{obviously}\:\mathrm{the}\:\mathrm{touching}\:\mathrm{point}\:\mathrm{of} \\ $$$$\mathrm{2}\:\mathrm{circles}\:\mathrm{is}\:\mathrm{located}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:\mathrm{connecting} \\ $$$$\left.\mathrm{their}\:\mathrm{centers}\right) \\ $$$$ \\ $$$$\mathrm{1}.\:\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid={R}−{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:{M}_{\mathrm{1}} {M}_{\mathrm{2}} =\left({R}−{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}.\:\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid={R}+{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:{M}_{\mathrm{1}} {M}_{\mathrm{2}} =\left({R}+{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}}=\frac{\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{\mathrm{2}}}}{\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\alpha}{\mathrm{2}}}}=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{cos}\:\alpha}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{cos}\:\alpha=\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{bc}}\right] \\ $$$$=\sqrt{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{bc}−{c}^{\mathrm{2}} }{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{bc}+{c}^{\mathrm{2}} }}=\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}} \\ $$$$ \\ $$$${M}_{\mathrm{1}} {M}_{\mathrm{2}} =\left({m}_{\mathrm{1}} −{m}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({n}_{\mathrm{1}} −{n}_{\mathrm{2}} \right)^{\mathrm{2}} = \\ $$$$=\left(\frac{{c}}{\mathrm{2}}−{x}\right)^{\mathrm{2}} +\left(\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){c}}{\mathrm{2}{d}}−{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} = \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left[\mathrm{after}\:\mathrm{some}\:\mathrm{transformation}\:\mathrm{work}\right] \\ $$$$=\frac{\mathrm{4}{bc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} −\frac{\mathrm{2}{bc}\left({b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}+\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{d}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{{d}^{\mathrm{2}} }={R}^{\mathrm{2}} \right] \\ $$$$ \\ $$$$\left({R}\pm{x}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:\frac{\alpha}{\mathrm{2}}\pm\mathrm{2}{Rx}\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}+{R}^{\mathrm{2}} = \\ $$$$=\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} \pm\frac{\mathrm{2}{abc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}+{R}^{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\frac{\mathrm{4}{bc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} −\frac{\mathrm{2}{bc}\left({b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}= \\ $$$$\:\:\:\:\:=\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x}^{\mathrm{2}} \pm\frac{\mathrm{2}{abc}}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)}{x} \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}_{\mathrm{3}} =\mathrm{0}\:\left(\mathrm{as}\:\mathrm{I}\:\mathrm{explained}\:\mathrm{before},\:\mathrm{the}\:\mathrm{point}\:{A}\:\mathrm{can}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{be}\:\mathrm{seen}\:\mathrm{as}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{0}\:\mathrm{still} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{meeting}\:\mathrm{the}\:\mathrm{requirements}\right) \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{2}{bc}}{{a}+{b}+{c}}\:\Rightarrow\:{r}_{\mathrm{1}} =\mathrm{2}{bc}\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)^{\mathrm{3}} \left(−{a}+{b}+{c}\right)}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}{bc}}{−{a}+{b}+{c}}\:\Rightarrow\:{r}_{\mathrm{2}} =\mathrm{2}{bc}\sqrt{\frac{\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)}{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)^{\mathrm{3}} }} \\ $$$$ \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{circles}\:\mathrm{corresponding}\:\mathrm{with}\:\mathrm{the}\:\mathrm{points} \\ $$$${B}\:\mathrm{and}\:{C}\:\mathrm{just}\:\mathrm{interchange}\:\left\{{a},\:{b},\:{c}\right\}\:\mathrm{with} \\ $$$$\left\{{b},\:{c},\:{a}\right\}\:\mathrm{and}\:\left\{{c},\:{a},\:{b}\right\} \\ $$

Question Number 35848    Answers: 0   Comments: 1

Find no. of points of discontinuity of : f(x)= cos ∣x∣ + ∣cos x∣ + ∣cos x∣^(2/3) + ∣cos x∣^(5/3) in the interval [−2,3].

$${Find}\:{no}.\:{of}\:{points}\:{of}\:{discontinuity} \\ $$$${of}\:: \\ $$$${f}\left({x}\right)=\:\mathrm{cos}\:\mid{x}\mid\:+\:\mid\mathrm{cos}\:{x}\mid\:+\:\mid\mathrm{cos}\:{x}\mid^{\frac{\mathrm{2}}{\mathrm{3}}} \:+\:\mid\mathrm{cos}\:{x}\mid^{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$$${in}\:{the}\:{interval}\:\left[−\mathrm{2},\mathrm{3}\right]. \\ $$

Question Number 35844    Answers: 3   Comments: 1

Question Number 35832    Answers: 1   Comments: 3

find the value of f(λ) = ∫_(−a) ^a (dx/((λ +_ x^2 )^(3/2) )) λ∈R .

$${find}\:{the}\:{value}\:{of}\:\:{f}\left(\lambda\right)\:=\:\int_{−{a}} ^{{a}} \:\:\:\frac{{dx}}{\left(\lambda\:+_{} {x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\lambda\in{R}\:. \\ $$

Question Number 35828    Answers: 1   Comments: 2

Question Number 35821    Answers: 0   Comments: 4

let f(t) = ∫_0 ^∞ ((e^(−ax) −e^(−bx) )/x^2 ) e^(−tx^2 ) dx with t>0 1) calculate f^′ (t) 2)find a simple form of f(t) 3) find the value of ∫_0 ^∞ ((e^(−2x) −e^(−x) )/x^2 ) e^(−3x^2 ) dx

$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{ax}} \:−{e}^{−{bx}} }{{x}^{\mathrm{2}} }\:{e}^{−{tx}^{\mathrm{2}} } {dx}\:\:\:{with}\:{t}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({t}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\mathrm{2}{x}} \:\:−{e}^{−{x}} }{{x}^{\mathrm{2}} }\:{e}^{−\mathrm{3}{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 35811    Answers: 3   Comments: 6

Multiple Choice Questions Rio Mike. Q_1 . The value of 6 + 2 × 4 −15 ÷ 3 is; A) 3 B) 9 C) 4 D) 27 Q_2 . 80 as a product of its prime factor is. A) 2^3 × 5 B) 2 × 5^(3 ) C) 2^2 × 5^2 D) 2^4 × 5 Q_3 . The deteminant of (((6 4)),((3 2)) ) is A) 18 B) 0 C) 24 D) −24 Q_4 . the value of ((1/(16)))^(−(1/2)) is A) −(1/4) B) 4 C) −4 D) (1/2) Q_5 . if −8,m,n,19 are in AP then (m,n) is A) (1,10) B) (2,10) C) (3,13) D) (4,16)o Q_6 . if cos θ = ((12)/(13)) then cot^2 θ is; A) ((169)/(25)) B) ((25)/(169)) C)((169)/(144)) D) ((144)/(169)) Q_7 . the solution set of 3 ≤ 2x−1≤5 is; A) ]x≥−2,x≤(8/3)[ B) [x≥2,x≤−(8/3)[ C) ]x>−2,x≤(3/8)] D) [x≥−2,x≤(8/3)] Q_8 . Which is a factor of f(x)= x^2 −3x + 2 A) (x−1) B) (x−2) C) (x+1) D)(x+3) Q_(9 ) . The sum of the sum of roots and product of roots of the quadratic equation 3x^2 + 6x + 9=0 A) 1 B) −1 C) 32 D) 5 Q_(10) . 2log2−log2 = A) log 8 B) log 6 C) log 3 D) log 2. Q_(11) . Σ_(r=1) ^∞ 3^(2−r) = A) (9/4) B) (9/2) C) ((13)/3) D) (1/2) Q_(12) . The constand term in the binomial expansion of (x^2 + (1/x^2 ))^8 is A) 3^(rd) B) 4^(th) C) 5^(th) D) 6^(th) Q_(13) . cos(180−x) = A) −sinx B) sin x C) cos x D) −cos x Q_(14) . The value of p for which (2,1) , (6,3), and (4,p) are collinear is ; A) 2 B) 1 C)−1 D) −2 Q_(15) . ∫_1 ^2 x^3 dx = A) −sin 7x B) sin 7x C) −7sin7x D) 7sin7x Q_(16) . Given that g : → px −5, the value of p for which g^(−1) (3)=4 is A) (1/2) B) −(1/2) C) −2 D) 2 Q_(17) . The value of θ in the range 0°≤θ≤90° for which sinθ = cos θ is A) 90° B) 60° C) 30° D) 45° Questions

$$\:{Multiple}\:{Choice}\:{Questions} \\ $$$$\:\:\:\boldsymbol{{Rio}}\:\boldsymbol{{M}}{ike}. \\ $$$$ \\ $$$${Q}_{\mathrm{1}} .\:{The}\:{value}\:{of}\:\:\mathrm{6}\:+\:\mathrm{2}\:×\:\mathrm{4}\:−\mathrm{15}\:\boldsymbol{\div}\:\mathrm{3} \\ $$$${is}; \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{3}\:\:{B}\right)\:\mathrm{9}\:{C}\right)\:\mathrm{4}\:{D}\right)\:\mathrm{27} \\ $$$$ \\ $$$${Q}_{\mathrm{2}} .\:\mathrm{80}\:{as}\:{a}\:{product}\:{of}\:{its}\:{prime} \\ $$$${factor}\:{is}. \\ $$$$\left.{A}\left.\right)\left.\:\mathrm{2}^{\mathrm{3}} ×\:\mathrm{5}\:{B}\right)\:\mathrm{2}\:×\:\mathrm{5}^{\mathrm{3}\:} \:{C}\right)\:\mathrm{2}^{\mathrm{2}} ×\:\mathrm{5}^{\mathrm{2}} \\ $$$$\left.{D}\right)\:\mathrm{2}^{\mathrm{4}} ×\:\mathrm{5} \\ $$$$ \\ $$$${Q}_{\mathrm{3}} .\:{The}\:{deteminant}\:{of}\:\:\begin{pmatrix}{\mathrm{6}\:\:\:\:\:\:\:\:\:\mathrm{4}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}\:{is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{18}\:\:{B}\right)\:\mathrm{0}\:\:{C}\right)\:\mathrm{24}\:{D}\right)\:−\mathrm{24} \\ $$$$ \\ $$$${Q}_{\mathrm{4}} .\:{the}\:{value}\:{of}\:\left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {is} \\ $$$$\left.{A}\left.\right)\left.\:\left.−\frac{\mathrm{1}}{\mathrm{4}}\:\:\:{B}\right)\:\mathrm{4}\:\:\:{C}\right)\:−\mathrm{4}\:\:\:\:{D}\right)\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${Q}_{\mathrm{5}} .\:{if}\:−\mathrm{8},{m},{n},\mathrm{19}\:{are}\:{in}\:{AP}\:{then}\: \\ $$$$\left({m},{n}\right)\:{is}\: \\ $$$$\left.{A}\left.\right)\left.\:\left.\left(\mathrm{1},\mathrm{10}\right)\:{B}\right)\:\left(\mathrm{2},\mathrm{10}\right)\:{C}\right)\:\left(\mathrm{3},\mathrm{13}\right)\:{D}\right)\:\left(\mathrm{4},\mathrm{16}\right){o} \\ $$$$ \\ $$$${Q}_{\mathrm{6}} .\:{if}\:{cos}\:\theta\:=\:\frac{\mathrm{12}}{\mathrm{13}}\:{then}\:{cot}^{\mathrm{2}} \theta\:{is}; \\ $$$$\left.{A}\left.\right)\left.\:\left.\frac{\mathrm{169}}{\mathrm{25}}\:\:{B}\right)\:\frac{\mathrm{25}}{\mathrm{169}}\:{C}\right)\frac{\mathrm{169}}{\mathrm{144}}\:{D}\right)\:\frac{\mathrm{144}}{\mathrm{169}} \\ $$$$ \\ $$$${Q}_{\mathrm{7}} .\:{the}\:{solution}\:{set}\:{of}\:\mathrm{3}\:\leqslant\:\mathrm{2}{x}−\mathrm{1}\leqslant\mathrm{5} \\ $$$${is};\: \\ $$$$\left.{A}\left.\right)\:\right]{x}\geqslant−\mathrm{2},{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}}\left[\:\:\:{B}\right)\:\left[{x}\geqslant\mathrm{2},{x}\leqslant−\frac{\mathrm{8}}{\mathrm{3}}\left[\right.\right. \\ $$$$\left.{C}\left.\right)\left.\:\left.\right]{x}>−\mathrm{2},{x}\leqslant\frac{\mathrm{3}}{\mathrm{8}}\right]\:\:{D}\right)\:\left[{x}\geqslant−\mathrm{2},{x}\leqslant\frac{\mathrm{8}}{\mathrm{3}}\right] \\ $$$$ \\ $$$${Q}_{\mathrm{8}} .\:{Which}\:{is}\:{a}\:{factor}\:{of}\:{f}\left({x}\right)=\: \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}\:+\:\mathrm{2} \\ $$$$\left.{A}\left.\right)\left.\:\left.\left({x}−\mathrm{1}\right)\:{B}\right)\:\left({x}−\mathrm{2}\right)\:{C}\right)\:\left({x}+\mathrm{1}\right)\:{D}\right)\left({x}+\mathrm{3}\right) \\ $$$$ \\ $$$${Q}_{\mathrm{9}\:} .\:{The}\:{sum}\:{of}\:{the}\:{sum}\:{of}\: \\ $$$${roots}\:{and}\:{product}\:{of}\:{roots}\:{of}\:{the} \\ $$$${quadratic}\:{equation}\:\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{6}{x}\:+\:\mathrm{9}=\mathrm{0} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{1}\:\:\:{B}\right)\:−\mathrm{1}\:\:{C}\right)\:\mathrm{32}\:\:{D}\right)\:\mathrm{5} \\ $$$$ \\ $$$${Q}_{\mathrm{10}} .\:\mathrm{2}{log}\mathrm{2}−{log}\mathrm{2}\:=\: \\ $$$$\left.{A}\left.\right)\left.\:{log}\:\mathrm{8}\:\:\:\:\:\:\:{B}\right)\:{log}\:\mathrm{6}\:\:\:\:{C}\right)\:{log}\:\mathrm{3} \\ $$$$\left.{D}\right)\:{log}\:\mathrm{2}. \\ $$$$ \\ $$$${Q}_{\mathrm{11}} .\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{3}^{\mathrm{2}−{r}} = \\ $$$$\left.{A}\left.\right)\left.\:\left.\frac{\mathrm{9}}{\mathrm{4}}\:\:{B}\right)\:\frac{\mathrm{9}}{\mathrm{2}}\:\:{C}\right)\:\frac{\mathrm{13}}{\mathrm{3}}\:{D}\right)\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$${Q}_{\mathrm{12}} .\:{The}\:{constand}\:{term}\:{in}\:{the}\: \\ $$$${binomial}\:{expansion}\:{of}\:\left({x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{8}} {is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{3}^{{rd}} \:\:{B}\right)\:\mathrm{4}^{{th}} \:{C}\right)\:\mathrm{5}^{{th}} \:{D}\right)\:\mathrm{6}^{{th}} \: \\ $$$$ \\ $$$${Q}_{\mathrm{13}} .\:{cos}\left(\mathrm{180}−{x}\right)\:=\: \\ $$$$\left.{A}\left.\right)\left.\:−{sinx}\:\:\:{B}\right)\:{sin}\:{x}\:\:{C}\right)\:{cos}\:{x}\: \\ $$$$\left.{D}\right)\:−{cos}\:{x} \\ $$$$ \\ $$$${Q}_{\mathrm{14}} .\:{The}\:{value}\:{of}\:{p}\:{for}\:{which}\: \\ $$$$\left(\mathrm{2},\mathrm{1}\right)\:,\:\left(\mathrm{6},\mathrm{3}\right),\:{and}\:\left(\mathrm{4},{p}\right)\:{are}\:{collinear} \\ $$$${is}\:; \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{2}\:\:{B}\right)\:\mathrm{1}\:{C}\right)−\mathrm{1}\:{D}\right)\:−\mathrm{2} \\ $$$$ \\ $$$${Q}_{\mathrm{15}} .\:\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} {dx}\:=\: \\ $$$$\left.{A}\left.\right)\left.\:−{sin}\:\mathrm{7}{x}\:\:\:{B}\right)\:{sin}\:\mathrm{7}{x}\:\:{C}\right)\:−\mathrm{7}{sin}\mathrm{7}{x} \\ $$$$\left.{D}\right)\:\mathrm{7}{sin}\mathrm{7}{x} \\ $$$$ \\ $$$${Q}_{\mathrm{16}} .\:{Given}\:{that}\:{g}\::\:\rightarrow\:{px}\:−\mathrm{5},\:{the}\: \\ $$$${value}\:{of}\:{p}\:{for}\:{which}\:{g}^{−\mathrm{1}} \left(\mathrm{3}\right)=\mathrm{4}\:{is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\frac{\mathrm{1}}{\mathrm{2}}\:{B}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:{C}\right)\:−\mathrm{2}\:\:{D}\right)\:\mathrm{2} \\ $$$$ \\ $$$${Q}_{\mathrm{17}} .\:{The}\:{value}\:{of}\:\theta\:{in}\:{the}\:{range}\: \\ $$$$\mathrm{0}°\leqslant\theta\leqslant\mathrm{90}°\:{for}\:{which}\:{sin}\theta\:=\:{cos}\:\theta\:{is} \\ $$$$\left.{A}\left.\right)\left.\:\left.\mathrm{90}°\:\:{B}\right)\:\mathrm{60}°\:{C}\right)\:\mathrm{30}°\:{D}\right)\:\mathrm{45}°\: \\ $$$$ \\ $$$$\:{Questions}\: \\ $$

Question Number 35798    Answers: 0   Comments: 3

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A 5100 cm^2 trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40 m then find the length of the other parallel side.

$$\mathrm{A}\:\mathrm{5100}\:\mathrm{cm}^{\mathrm{2}} \:\mathrm{trapezium}\:\mathrm{has}\:\mathrm{the}\:\mathrm{perpendicular} \\ $$$$\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{two}\:\mathrm{parallel}\:\mathrm{sides} \\ $$$$\mathrm{60}\:\mathrm{m}.\:\mathrm{If}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallel}\:\mathrm{sides}\:\mathrm{be}\:\mathrm{40}\:\mathrm{m}\: \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{parallel}\:\mathrm{side}. \\ $$

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