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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

Question Number 35789    Answers: 0   Comments: 1

Question Number 35791    Answers: 1   Comments: 0

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.

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

Question Number 35771    Answers: 0   Comments: 2

Question Number 35769    Answers: 2   Comments: 0

Question Number 35768    Answers: 0   Comments: 0

Question Number 35766    Answers: 0   Comments: 1

let f(x)=arctan(1+x) and f_n (x)= e^(−nx) 1) define fof_n (x) and f_n of(x) 2) study the nature of the series Σ_(n=0) ^(+∞) fof_n (x) and Σ_(n=0) ^∞ f_n of(x) .

$${let}\:{f}\left({x}\right)={arctan}\left(\mathrm{1}+{x}\right)\:\:{and}\:\:{f}_{{n}} \left({x}\right)=\:{e}^{−{nx}} \\ $$$$\left.\mathrm{1}\right)\:{define}\:{fof}_{{n}} \left({x}\right)\:{and}\:{f}_{{n}} {of}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{nature}\:{of}\:{the}\:{series}\:\sum_{{n}=\mathrm{0}} ^{+\infty} {fof}_{{n}} \left({x}\right) \\ $$$${and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{f}_{{n}} {of}\left({x}\right)\:. \\ $$

Question Number 35763    Answers: 1   Comments: 0

Question Number 35743    Answers: 1   Comments: 1

find x log(log2+log_2 (x+1))=0

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}} \\ $$$$\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{log}}\mathrm{2}+\boldsymbol{\mathrm{log}}_{\mathrm{2}} \left(\boldsymbol{{x}}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$

Question Number 35739    Answers: 1   Comments: 2

Question Number 35738    Answers: 2   Comments: 0

Given that ∫_0 ^k x^2 = 16 find the value of k

$$\:{Given}\:{that}\:\:\int_{\mathrm{0}} ^{{k}} {x}^{\mathrm{2}} =\:\mathrm{16} \\ $$$${find}\:{the}\:{value}\:{of}\:{k} \\ $$

Question Number 35737    Answers: 0   Comments: 4

Question Number 35732    Answers: 1   Comments: 1

Question Number 35729    Answers: 1   Comments: 1

find the value of I =∫_0 ^π (dx/(2−cosx))

$${find}\:{the}\:{value}\:{of}\:\:\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{2}−{cosx}} \\ $$

Question Number 35833    Answers: 1   Comments: 1

fond lim_(n→+∞) n^a {ln(1+e^(−n) ) −e^(−n) } with a>0

$${fond}\:{lim}_{{n}\rightarrow+\infty} \:\:{n}^{{a}} \:\:\left\{{ln}\left(\mathrm{1}+{e}^{−{n}} \right)\:−{e}^{−{n}} \right\}\:{with}\:{a}>\mathrm{0} \\ $$

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