Maths-
General
Easy

Question

16 sin x cos space x cos space 2 x cos space 4 x cos space 8 x element of

  1. [-1,1]
  2. [negative 1 fourth comma 1 fourth]
  3. [negative 3 over 4 comma 3 over 4]
  4. [negative fraction numerator square root of 3 over denominator 2 end fraction comma fraction numerator square root of 3 over denominator 2 end fraction]

Hint:

In this question, we have to find the range of 8 sin x.cos x.cos 2x.cos 4x.cos 8x. For that first we will simplify the trigonometric function and later as we know range of sin y and cos y is [-1,1], so, using this we can find the range of the given function.

The correct answer is: [-1,1]


    f(x) = 8 sin x.co x.cos 2x.cos 4x.cos 8x
    =4sin 2x.cos 4x.cos 8x
    =2sin 8x.cos 8x
    =sin 16x
    A s comma space minus 1 less or equal than sin space y less or equal than 1
rightwards double arrow negative 1 less or equal than sin space 16 x less or equal than 1
    So, range is [-1, 1].

    Related Questions to study

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    Five conductors are meeting at a point x as shown in the figure. What is the value of current in fifth conductor?

    According to Kirchhoff’s first law
    (5A)+(4A)+(-3A)+(-5A)+I=0
    Or I=-1A

    Five conductors are meeting at a point x as shown in the figure. What is the value of current in fifth conductor?

    physics-General
    According to Kirchhoff’s first law
    (5A)+(4A)+(-3A)+(-5A)+I=0
    Or I=-1A
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    The figure shows a network of currents. The magnitude of current is shown here. The current I will be

    Regarding Kirchhoff’s junction rule, the circuit can be redrawn as

    Current in arm, A B equals 10 minus 6 equals 4 A
    Current in arm, D C equals 6 plus 2 equals 8 A
    Current in arm, B C equals 4 plus 1 equals 5 A
    Hence, I equals 5 plus 8 equals 13 A

    The figure shows a network of currents. The magnitude of current is shown here. The current I will be

    physics-General
    Regarding Kirchhoff’s junction rule, the circuit can be redrawn as

    Current in arm, A B equals 10 minus 6 equals 4 A
    Current in arm, D C equals 6 plus 2 equals 8 A
    Current in arm, B C equals 4 plus 1 equals 5 A
    Hence, I equals 5 plus 8 equals 13 A
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    The equivalent resistance between the terminals A blank a n d blank B in the following circuit is


    10capital omega in series with 10capital omega will gives
    (10+10)=20capital omega
    and 10capital omega in series with 10capital omega will gives
    (10+10)=20capital omega

    20capital omega in parallel with 20capital omega will gives
    open parentheses fraction numerator 20 cross times 20 over denominator 20 plus 20 end fraction close parentheses equals fraction numerator 400 over denominator 40 end fraction equals 10 capital omega

    Resistance in series between points A and D
    =5+10+5
    =20capital omega

    The equivalent resistance between the terminals A blank a n d blank B in the following circuit is

    physics-General

    10capital omega in series with 10capital omega will gives
    (10+10)=20capital omega
    and 10capital omega in series with 10capital omega will gives
    (10+10)=20capital omega

    20capital omega in parallel with 20capital omega will gives
    open parentheses fraction numerator 20 cross times 20 over denominator 20 plus 20 end fraction close parentheses equals fraction numerator 400 over denominator 40 end fraction equals 10 capital omega

    Resistance in series between points A and D
    =5+10+5
    =20capital omega
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    General
    physics-

    The equivalent resistance between points A and B of an infinite network of resistances, each of 1blank capital omega, connected as shown is

    Let x be the equivalent resistance of entire network between A and B. Hence, we have

    R subscript A B end subscript equals 1 plus resistance of parallel combination of 1capital omega and x capital omega
    therefore blank R subscript A B end subscript equals 1 plus fraction numerator x over denominator 1 plus x end fraction
    therefore blank x equals 1 plus fraction numerator x over denominator 1 plus x end fraction
    ⟹ x plus x to the power of 2 end exponent equals 1 plus x plus x
    ⟹ x to the power of 2 end exponent minus x minus 1 equals 0
    ⟹ x equals fraction numerator 1 plus square root of 1 plus 4 end root over denominator 2 end fraction
    equals fraction numerator 1 plus square root of 5 over denominator 2 end fraction capital omega

    The equivalent resistance between points A and B of an infinite network of resistances, each of 1blank capital omega, connected as shown is

    physics-General
    Let x be the equivalent resistance of entire network between A and B. Hence, we have

    R subscript A B end subscript equals 1 plus resistance of parallel combination of 1capital omega and x capital omega
    therefore blank R subscript A B end subscript equals 1 plus fraction numerator x over denominator 1 plus x end fraction
    therefore blank x equals 1 plus fraction numerator x over denominator 1 plus x end fraction
    ⟹ x plus x to the power of 2 end exponent equals 1 plus x plus x
    ⟹ x to the power of 2 end exponent minus x minus 1 equals 0
    ⟹ x equals fraction numerator 1 plus square root of 1 plus 4 end root over denominator 2 end fraction
    equals fraction numerator 1 plus square root of 5 over denominator 2 end fraction capital omega
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    Maths-

    The period of sin space left parenthesis pi sin space theta right parenthesis is

    Period of sinϑ equals 2 straight pi
    As we know, if the period of f(x) is T then the period of g(f(x)) is also T.
    So, period of sin(sinϑ) =2pi

    The period of sin space left parenthesis pi sin space theta right parenthesis is

    Maths-General
    Period of sinϑ equals 2 straight pi
    As we know, if the period of f(x) is T then the period of g(f(x)) is also T.
    So, period of sin(sinϑ) =2pi
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    Maths-

    The period of fraction numerator sin space left parenthesis 2 pi x plus a right parenthesis over denominator sin space left parenthesis 2 pi x plus b right parenthesis end fraction is

    f(x)=fraction numerator sin space left parenthesis 2 pi x plus a right parenthesis over denominator sin space left parenthesis 2 pi x plus b right parenthesis end fraction
    Period of sin kx =fraction numerator 2 straight pi over denominator 4 end fraction
    In numerator,
    period of sin (2 straight pi plus straight a)=fraction numerator 2 straight pi over denominator 2 straight pi end fraction equals 1
    In denominator,
    period of sin (2 straight pi plus b)=fraction numerator 2 straight pi over denominator 2 straight pi end fraction equals 1
    So, the period of f(x) is 1.

    The period of fraction numerator sin space left parenthesis 2 pi x plus a right parenthesis over denominator sin space left parenthesis 2 pi x plus b right parenthesis end fraction is

    Maths-General
    f(x)=fraction numerator sin space left parenthesis 2 pi x plus a right parenthesis over denominator sin space left parenthesis 2 pi x plus b right parenthesis end fraction
    Period of sin kx =fraction numerator 2 straight pi over denominator 4 end fraction
    In numerator,
    period of sin (2 straight pi plus straight a)=fraction numerator 2 straight pi over denominator 2 straight pi end fraction equals 1
    In denominator,
    period of sin (2 straight pi plus b)=fraction numerator 2 straight pi over denominator 2 straight pi end fraction equals 1
    So, the period of f(x) is 1.
    parallel
    General
    Maths-

    Period of tan 4x+sec 4x is

    f(x)= tan 4x + sec 4x
    Period of tan 4x = fraction numerator straight pi over denominator open vertical bar 4 close vertical bar end fraction equals straight pi over 4
    Period of sec 4x=fraction numerator 2 straight pi over denominator open vertical bar 4 close vertical bar end fraction equals straight pi over 2
    So, period of f(x)=LCM ofstraight pi over 4 comma straight pi over 2 equals straight pi over 2

    Period of tan 4x+sec 4x is

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    f(x)= tan 4x + sec 4x
    Period of tan 4x = fraction numerator straight pi over denominator open vertical bar 4 close vertical bar end fraction equals straight pi over 4
    Period of sec 4x=fraction numerator 2 straight pi over denominator open vertical bar 4 close vertical bar end fraction equals straight pi over 2
    So, period of f(x)=LCM ofstraight pi over 4 comma straight pi over 2 equals straight pi over 2
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    Maths-

    The cotangent function whose period 3 pi is

    Period of cot (k x) = fraction numerator straight pi over denominator open vertical bar k close vertical bar end fraction
    given, fraction numerator straight pi over denominator open vertical bar k close vertical bar end fraction equals 3 straight pi space rightwards double arrow open vertical bar straight k close vertical bar equals 1 third
    So, cotangent function whose period is 3 straight pi space is space cot straight x over 3.

    The cotangent function whose period 3 pi is

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    Period of cot (k x) = fraction numerator straight pi over denominator open vertical bar k close vertical bar end fraction
    given, fraction numerator straight pi over denominator open vertical bar k close vertical bar end fraction equals 3 straight pi space rightwards double arrow open vertical bar straight k close vertical bar equals 1 third
    So, cotangent function whose period is 3 straight pi space is space cot straight x over 3.
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    Thirteen resistances each of resistance Rcapital omega are connected in the circuit as shown in the figure. The effective resistance between points A and B is

    Resistance R bisecting the circuit can be neglected due to the symmetry of the circuit.
    Now, there are four triangles
    Effective resistance of each triangle
    fraction numerator 1 over denominator R to the power of ´ end exponent end fraction equals fraction numerator 1 over denominator R end fraction plus fraction numerator 1 over denominator 2 R end fraction
    equals fraction numerator 2 plus 1 over denominator 2 R end fraction equals fraction numerator 3 over denominator 2 R end fraction
    therefore R to the power of ´ end exponent equals fraction numerator 2 over denominator 3 end fraction R
    Now the given circuit reduced to

    Therefore, effective resistance between A and B,
    fraction numerator 1 over denominator R subscript A B end subscript end fraction equals fraction numerator 1 over denominator 2 R to the power of ´ end exponent end fraction plus fraction numerator 1 over denominator 2 R to the power of ´ end exponent end fraction equals fraction numerator 1 over denominator R to the power of ´ end exponent end fraction
    ⟹ R subscript A B end subscript equals R to the power of ´ end exponent equals fraction numerator 2 R over denominator 3 end fraction capital omega

    Thirteen resistances each of resistance Rcapital omega are connected in the circuit as shown in the figure. The effective resistance between points A and B is

    physics-General
    Resistance R bisecting the circuit can be neglected due to the symmetry of the circuit.
    Now, there are four triangles
    Effective resistance of each triangle
    fraction numerator 1 over denominator R to the power of ´ end exponent end fraction equals fraction numerator 1 over denominator R end fraction plus fraction numerator 1 over denominator 2 R end fraction
    equals fraction numerator 2 plus 1 over denominator 2 R end fraction equals fraction numerator 3 over denominator 2 R end fraction
    therefore R to the power of ´ end exponent equals fraction numerator 2 over denominator 3 end fraction R
    Now the given circuit reduced to

    Therefore, effective resistance between A and B,
    fraction numerator 1 over denominator R subscript A B end subscript end fraction equals fraction numerator 1 over denominator 2 R to the power of ´ end exponent end fraction plus fraction numerator 1 over denominator 2 R to the power of ´ end exponent end fraction equals fraction numerator 1 over denominator R to the power of ´ end exponent end fraction
    ⟹ R subscript A B end subscript equals R to the power of ´ end exponent equals fraction numerator 2 R over denominator 3 end fraction capital omega
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    Six resistors, each of value 3blank capital omega are connected as shown in the figure. A cell of emf 3V is connected across A B.The effective resistance across A B and the current through the arm A B will be

    The equivalent circuit is shown as

    We can emit the resistance in the arm DF as balance condition is satisfied.
    Therefore, the 3capital omega resistances in arm CD and DE are in series.
    therefore R to the power of ´ end exponent equals 3 plus 3 equals 6 capital omega
    Similarly, for arms CF and FE, R’’=6capital omega
    R to the power of ´ blank end exponent a n d R ´ ´ are in parallel
    therefore blank fraction numerator 1 over denominator R to the power of ´ ´ ´ end exponent end fraction equals fraction numerator 1 over denominator 6 end fraction plus fraction numerator 1 over denominator 6 end fraction equals fraction numerator 2 over denominator 6 end fraction equals fraction numerator 1 over denominator 3 end fraction
    R’’’=3capital omega
    Now, R’’’ and 3capital omega resistances are in parallel
    therefore blank fraction numerator 1 over denominator R end fraction equals fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction
    ⟹ R equals 1.5 capital omega
    Moreover, V across AB=3V and resistance in the arm=3capital omega
    ∴ Current through the arm will be
    equals fraction numerator 3 V over denominator 3 capital omega end fraction equals 1 A.

    Six resistors, each of value 3blank capital omega are connected as shown in the figure. A cell of emf 3V is connected across A B.The effective resistance across A B and the current through the arm A B will be

    physics-General
    The equivalent circuit is shown as

    We can emit the resistance in the arm DF as balance condition is satisfied.
    Therefore, the 3capital omega resistances in arm CD and DE are in series.
    therefore R to the power of ´ end exponent equals 3 plus 3 equals 6 capital omega
    Similarly, for arms CF and FE, R’’=6capital omega
    R to the power of ´ blank end exponent a n d R ´ ´ are in parallel
    therefore blank fraction numerator 1 over denominator R to the power of ´ ´ ´ end exponent end fraction equals fraction numerator 1 over denominator 6 end fraction plus fraction numerator 1 over denominator 6 end fraction equals fraction numerator 2 over denominator 6 end fraction equals fraction numerator 1 over denominator 3 end fraction
    R’’’=3capital omega
    Now, R’’’ and 3capital omega resistances are in parallel
    therefore blank fraction numerator 1 over denominator R end fraction equals fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction
    ⟹ R equals 1.5 capital omega
    Moreover, V across AB=3V and resistance in the arm=3capital omega
    ∴ Current through the arm will be
    equals fraction numerator 3 V over denominator 3 capital omega end fraction equals 1 A.
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    In the circuit shown the value of I in ampere is

    We can simplify the network as shown

    So, net resistance,
    R=2.4+1.6=4.0capital omega
    Therefore, current from the battery.
    i equals fraction numerator V over denominator R end fraction equals fraction numerator 4 over denominator 4 end fraction equals 1 A
    Now, from the circuit (b),
    4I’ =6I
    ⟹ I to the power of ´ end exponent equals fraction numerator 3 over denominator 2 end fraction I
    But i=I+I’
    equals I plus fraction numerator 3 over denominator 2 end fraction I equals fraction numerator 5 over denominator 2 end fraction I
    therefore blank 1 equals fraction numerator 5 over denominator 2 end fraction I
    ⟹ I equals fraction numerator 2 over denominator 5 end fraction equals 0.4 A

    In the circuit shown the value of I in ampere is

    physics-General
    We can simplify the network as shown

    So, net resistance,
    R=2.4+1.6=4.0capital omega
    Therefore, current from the battery.
    i equals fraction numerator V over denominator R end fraction equals fraction numerator 4 over denominator 4 end fraction equals 1 A
    Now, from the circuit (b),
    4I’ =6I
    ⟹ I to the power of ´ end exponent equals fraction numerator 3 over denominator 2 end fraction I
    But i=I+I’
    equals I plus fraction numerator 3 over denominator 2 end fraction I equals fraction numerator 5 over denominator 2 end fraction I
    therefore blank 1 equals fraction numerator 5 over denominator 2 end fraction I
    ⟹ I equals fraction numerator 2 over denominator 5 end fraction equals 0.4 A
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    The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement?

    The v minus x equation from the given graph can be written as,
    v equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses blank x plus v subscript 0 end subscript blank open parentheses i close parentheses
    therefore blank a equals fraction numerator d v over denominator d t end fraction equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses fraction numerator d x over denominator d t end fraction equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses blank v
    Substituting v from Eq. (i), we get
    a equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses open square brackets open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses blank x plus v subscript 0 end subscript close square brackets
    blank a equals open parentheses fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses to the power of 2 end exponent blank x minus fraction numerator v subscript 0 end subscript superscript 2 end superscript over denominator x subscript 0 end subscript end fraction
    Thus, a minus x graph is a straight line with positive slope and negative intercept.

    The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement?

    physics-General
    The v minus x equation from the given graph can be written as,
    v equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses blank x plus v subscript 0 end subscript blank open parentheses i close parentheses
    therefore blank a equals fraction numerator d v over denominator d t end fraction equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses fraction numerator d x over denominator d t end fraction equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses blank v
    Substituting v from Eq. (i), we get
    a equals open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses open square brackets open parentheses negative fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses blank x plus v subscript 0 end subscript close square brackets
    blank a equals open parentheses fraction numerator v subscript 0 end subscript over denominator x subscript 0 end subscript end fraction close parentheses to the power of 2 end exponent blank x minus fraction numerator v subscript 0 end subscript superscript 2 end superscript over denominator x subscript 0 end subscript end fraction
    Thus, a minus x graph is a straight line with positive slope and negative intercept.
    parallel
    General
    physics-

    The displacement-time graphs of two moving particles make angles of 30 degree blank a n d blank 45 degree with the x minusaxis. The ratio of their velocities is

    Slope of displacement time-graph is velocity
    fraction numerator v subscript 1 end subscript over denominator v subscript 2 end subscript end fraction equals fraction numerator tan invisible function application theta subscript 1 end subscript over denominator tan invisible function application theta subscript 2 end subscript end fraction equals fraction numerator tan invisible function application 30 degree over denominator tan invisible function application 45 degree end fraction equals fraction numerator 1 over denominator square root of 3 end fraction
    v subscript 1 end subscript blank colon v subscript 2 end subscript equals 1 blank colon blank square root of 3

    The displacement-time graphs of two moving particles make angles of 30 degree blank a n d blank 45 degree with the x minusaxis. The ratio of their velocities is

    physics-General
    Slope of displacement time-graph is velocity
    fraction numerator v subscript 1 end subscript over denominator v subscript 2 end subscript end fraction equals fraction numerator tan invisible function application theta subscript 1 end subscript over denominator tan invisible function application theta subscript 2 end subscript end fraction equals fraction numerator tan invisible function application 30 degree over denominator tan invisible function application 45 degree end fraction equals fraction numerator 1 over denominator square root of 3 end fraction
    v subscript 1 end subscript blank colon v subscript 2 end subscript equals 1 blank colon blank square root of 3
    General
    Maths-

    The minimum and maximum values of 8 cos space 3 x minus 15 sin 3x are

    As we know, if f(x)=a sin x+ b cos x
    Then, range of f(x) element of open square brackets negative square root of a squared plus b squared end root comma square root of a squared plus b squared end root close square brackets
    As f(x)= 8 cos 3x - 15 sin 3x
    plus-or-minus square root of a squared plus b squared end root equals plus-or-minus square root of 8 squared plus 15 squared end root equals plus-or-minus square root of 64 plus 225 end root equals plus-or-minus square root of 289 equals plus-or-minus 17
    So, range of f(x) element of open square brackets negative 17 comma space 17 close square brackets

    The minimum and maximum values of 8 cos space 3 x minus 15 sin 3x are

    Maths-General
    As we know, if f(x)=a sin x+ b cos x
    Then, range of f(x) element of open square brackets negative square root of a squared plus b squared end root comma square root of a squared plus b squared end root close square brackets
    As f(x)= 8 cos 3x - 15 sin 3x
    plus-or-minus square root of a squared plus b squared end root equals plus-or-minus square root of 8 squared plus 15 squared end root equals plus-or-minus square root of 64 plus 225 end root equals plus-or-minus square root of 289 equals plus-or-minus 17
    So, range of f(x) element of open square brackets negative 17 comma space 17 close square brackets
    General
    Physics-

    The graph between the displacement x and t for a particle moving in a straight line is shown in figure. During the interval O A comma A B comma B C and C D comma the acceleration of the particle is

    O A comma blank A B comma blank B C, C D

    Region O A shows that graph bending toward time axis i. e. acceleration is negative.
    Region A B shows that graph is parallel to time axis i. e. blankvelocity is zero. Hence acceleration is zero.
    Region B C shows that graph is bending towards displacement axis i. e. acceleration is positive.
    Region C D blankshows that graph having constant slope i. e. velocity is constant. Hence acceleration is zero

    The graph between the displacement x and t for a particle moving in a straight line is shown in figure. During the interval O A comma A B comma B C and C D comma the acceleration of the particle is

    O A comma blank A B comma blank B C, C D

    Physics-General
    Region O A shows that graph bending toward time axis i. e. acceleration is negative.
    Region A B shows that graph is parallel to time axis i. e. blankvelocity is zero. Hence acceleration is zero.
    Region B C shows that graph is bending towards displacement axis i. e. acceleration is positive.
    Region C D blankshows that graph having constant slope i. e. velocity is constant. Hence acceleration is zero
    parallel

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