Tüm alıştırma soruları

1526 soru

Soru 801Soru

A trigonometric function is defined as f(x)=asin(bx)+cf(x) = a \sin(b x) + c, where a>0a > 0 and b>0b > 0. The graph of y=f(x)y = f(x) has a maximum value of 88, a minimum value of 2-2, and a period of 120120^\circ. What is the value of a+b+ca + b + c?

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Cevap: 11

Cevap

The value of a+b+ca + b + c is 11.
Using the maximum (a+c=8a + c = 8) and minimum (a+c=2-a + c = -2), solving simultaneously gives a=5a = 5 and c=3c = 3. Using the period formula T=360b=120T = \frac{360^\circ}{b} = 120^\circ, we obtain b=3b = 3. Adding these values together yields 5+3+3=115 + 3 + 3 = 11.

Adım Adım Çözüm

1
Set up a system of linear equations for the amplitude and vertical shift from the maximum and minimum bounds
a=5a = 5 and c=3c = 3
Since sin(bx)\sin(bx) ranges from 1-1 to 11, the maximum is a(1)+c=8a(1) + c = 8 and minimum is a(1)+c=2a(-1) + c = -2. Solving these simultaneous equations gives c=3c = 3 and a=5a = 5.
2
Calculate the frequency coefficient bb using the period formula
b=3b = 3
For a sine curve specified in degrees, the period is T=360bT = \frac{360^\circ}{b}. Substituting T=120T = 120^\circ yields b=3b = 3.
3
Compute the requested sum a+b+ca + b + c
a+b+c=11a + b + c = 11
Summing the calculated parameters 5+3+3=115 + 3 + 3 = 11.

Anahtar Kavram

Determining parameters of a trigonometric graph from amplitude, vertical shift, and period.
Soru 802Soru

A parallel plate air capacitor of capacitance 8.0 μF8.0\text{ }\mu\text{F} is fully charged using a 40.0 V40.0\text{ V} d.c. power supply and then disconnected from the source. A dielectric slab with a relative permittivity of 4.04.0 is subsequently inserted to completely fill the region between the plates. Calculate the magnitude of the decrease in electrostatic energy stored in the capacitor, in microjoules (μJ\mu\text{J}).

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Cevap: 4800

Cevap

The magnitude of the decrease in stored electrostatic energy is 4800 μJ.
Disconnecting the battery ensures that the charge Q=C0V=320 μCQ = C_0 V = 320\text{ }\mu\text{C} on the plates stays fixed. Inserting the dielectric increases capacitance fourfold to 32.0 μF32.0\text{ }\mu\text{F}. The energy decreases from Ui=6400 μJU_i = 6400\text{ }\mu\text{J} to Uf=Q22Cf=1600 μJU_f = \frac{Q^2}{2C_f} = 1600\text{ }\mu\text{J}, giving a total decrease of 4800 μJ4800\text{ }\mu\text{J}.

Adım Adım Çözüm

1
Calculate the initial energy stored in the air capacitor before disconnection.
Initial energy Ui=12C0V2=12×8.0×106 F×(40.0 V)2=6.4×103 J=6400 μJU_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 8.0 \times 10^{-6} \text{ F} \times (40.0 \text{ V})^2 = 6.4 \times 10^{-3} \text{ J} = 6400 \text{ } \mu\text{J}.
The initial state has known capacitance and potential difference.
2
Calculate the new capacitance with the dielectric present.
Final capacitance Cf=KC0=4.0×8.0 μF=32.0 μFC_f = K C_0 = 4.0 \times 8.0 \text{ } \mu\text{F} = 32.0 \text{ } \mu\text{F}.
Inserting a dielectric of constant KK scales the capacitance by KK.
3
Calculate the final stored energy using charge conservation.
Final energy Uf=UiK=6400 μJ4.0=1600 μJU_f = \frac{U_i}{K} = \frac{6400 \text{ } \mu\text{J}}{4.0} = 1600 \text{ } \mu\text{J}.
Disconnection forces the charge QQ to remain fixed, so energy scales inversely with capacitance (U=Q22CU = \frac{Q^2}{2C}).
4
Find the difference between initial and final energy.
\Delta U = 6400 \text{ } \mu\text{J} - 1600 \text{ } \mu\text{J} = 4800 \text{ } \mu\text{J}.
The loss in electrostatic energy represents the work done by the field pulling the dielectric slab into the plates.

Anahtar Kavram

Effect of dielectric insertion on stored electrostatic energy under isolated (constant charge) conditions
Tahmini Süre:2m 0s
Soru 803Soru

Given that (x2)(x - 2) is a factor of the polynomial P(x)=x3+kx25x+6P(x) = x^3 + kx^2 - 5x + 6, find the remainder when P(x)P(x) is divided by (x+3)(x + 3).

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Cevap: -15

Cevap

The remainder when P(x)P(x) is divided by (x+3)(x + 3) is 15-15.
According to the Factor Theorem, since (x2)(x - 2) is a factor of P(x)=x3+kx25x+6P(x) = x^3 + kx^2 - 5x + 6, setting x=2x = 2 yields P(2)=0P(2) = 0. This gives 8+4k10+6=08 + 4k - 10 + 6 = 0, which simplifies to 4k+4=04k + 4 = 0, giving k=1k = -1. The polynomial is therefore P(x)=x3x25x+6P(x) = x^3 - x^2 - 5x + 6. By the Remainder Theorem, dividing P(x)P(x) by (x+3)(x + 3) produces a remainder of P(3)P(-3). Evaluating P(3)=(3)3(3)25(3)+6=279+15+6=15P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15.

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1
Apply the Factor Theorem to determine the unknown constant kk.
k=1k = -1
If (x2)(x - 2) is a factor of P(x)P(x), then P(2)=0P(2) = 0. Substituting x=2x = 2 gives 23+k(2)25(2)+6=0    4k+4=0    k=12^3 + k(2)^2 - 5(2) + 6 = 0 \implies 4k + 4 = 0 \implies k = -1.
2
Substitute k=1k = -1 into the original polynomial to get the full expression.
P(x)=x3x25x+6P(x) = x^3 - x^2 - 5x + 6
Replacing kk with 1-1 defines P(x)P(x) completely.
3
Apply the Remainder Theorem to find the remainder when P(x)P(x) is divided by (x+3)(x + 3).
Remainder is 15-15
By the Remainder Theorem, dividing P(x)P(x) by (x+3)(x + 3) leaves a remainder equal to P(3)P(-3). Calculating P(3)=(3)3(3)25(3)+6=279+15+6=15P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15.

Anahtar Kavram

Factor and Remainder Theorems for Polynomials
Tahmini Süre:1m 30s
Soru 804Soru

A curve has a gradient function defined by dydx=6x2+8sin(4x)+3\frac{dy}{dx} = 6x^2 + 8\sin(4x) + 3. If the curve passes through the point (0,10)(0, 10), what is the value of the constant of integration CC?

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Cevap: 12

Cevap

The constant of integration CC is 1212.
Integrating 6x2+8sin(4x)+36x^2 + 8\sin(4x) + 3 with respect to xx gives y=2x32cos(4x)+3x+Cy = 2x^3 - 2\cos(4x) + 3x + C. Substituting (0,10)(0, 10) into the integrated equation gives 10=2(0)2(1)+3(0)+C10 = 2(0) - 2(1) + 3(0) + C, which leads directly to 10=2+C10 = -2 + C, so C=12C = 12.

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1
Integrate the gradient function to obtain the general equation of the curve.
y=2x32cos(4x)+3x+Cy = 2x^3 - 2\cos(4x) + 3x + C
The antiderivative of sin(kx)\sin(kx) is 1kcos(kx)-\frac{1}{k}\cos(kx) and the antiderivative of xnx^n is xn+1n+1\frac{x^{n+1}}{n+1}.
2
Apply the given initial condition (x,y)=(0,10)(x, y) = (0, 10) to solve for CC.
10=2(0)32cos(0)+3(0)+C    10=2+C10 = 2(0)^3 - 2\cos(0) + 3(0) + C \implies 10 = -2 + C
Evaluating at x=0x = 0 requires evaluating cos(0)=1\cos(0) = 1, which leaves 2-2 from the trigonometric term.
3
Solve the linear equation for CC.
C=12C = 12
Adding 22 to both sides of 10=2+C10 = -2 + C gives C=12C = 12.

Anahtar Kavram

Indefinite Integration with Boundary Conditions
Soru 805Soru

A curve y=F(x)y = F(x) has a gradient function given by dydx=9x28x+6cos(3x)4sin(2x)\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x). Given that y(0)=7y(0) = 7, determine the value of the constant of integration CC when the antiderivative is expressed in the standard form y=3x34x2+2sin(3x)+2cos(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C.

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Cevap: 5

Cevap

The value of the constant of integration CC is 5.
Integrating dydx=9x28x+6cos(3x)4sin(2x)\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x) yields y=3x34x2+2sin(3x)+2cos(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C. Substituting x=0x = 0 into the expression gives y(0)=00+0+2(1)+C=2+Cy(0) = 0 - 0 + 0 + 2(1) + C = 2 + C. Equating to y(0)=7y(0) = 7 gives 2+C=72 + C = 7, which solves to C=5C = 5.

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1
Integrate the gradient function term-by-term with respect to xx
y=3x34x2+2sin(3x)+2cos(2x)+Cy = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C
The antiderivative of 9x29x^2 is 3x33x^3, of 8x-8x is 4x2-4x^2, of 6cos(3x)6\cos(3x) is 2sin(3x)2\sin(3x), and of 4sin(2x)-4\sin(2x) is 2cos(2x)2\cos(2x).
2
Apply the initial boundary condition y(0)=7y(0) = 7
3(0)34(0)2+2sin(0)+2cos(0)+C=7    2+C=73(0)^3 - 4(0)^2 + 2\sin(0) + 2\cos(0) + C = 7 \implies 2 + C = 7
At x=0x = 0, sin(0)=0\sin(0) = 0 and cos(0)=1\cos(0) = 1, making the non-zero constant contribution equal to 2(1)=22(1) = 2.
3
Solve for the constant of integration CC
C=5C = 5
Subtracting 2 from both sides of 2+C=72 + C = 7 yields C=5C = 5.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions
Soru 806Soru

Determine the number of distinct solutions to the trigonometric equation 2cos2θ+sinθ1=02\cos^2 \theta + \sin \theta - 1 = 0 within the interval 0θ3600^\circ \le \theta \le 360^\circ.

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Cevap: 3

Cevap

The total number of distinct solutions in the given interval is 3.
Substituting cos2θ=1sin2θ\cos^2 \theta = 1 - \sin^2 \theta yields the quadratic equation 2sin2θsinθ1=02\sin^2 \theta - \sin \theta - 1 = 0. Factoring gives sinθ=1\sin \theta = 1 and sinθ=12\sin \theta = -\frac{1}{2}. Within 0θ3600^\circ \le \theta \le 360^\circ, sinθ=1\sin \theta = 1 gives one solution (9090^\circ), while sinθ=12\sin \theta = -\frac{1}{2} gives two solutions (210210^\circ and 330330^\circ). In total, there are 3 distinct solutions.

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1
Substitute the identity cos2θ=1sin2θ\cos^2 \theta = 1 - \sin^2 \theta into the original equation
2(1sin2θ)+sinθ1=02(1 - \sin^2 \theta) + \sin \theta - 1 = 0
Converting the equation into a single trigonometric ratio allows for algebraic solving.
2
Simplify and arrange into quadratic form
2sin2θsinθ1=02\sin^2 \theta - \sin \theta - 1 = 0
This puts the expression into standard quadratic form au2+bu+c=0au^2 + bu + c = 0 where u=sinθu = \sin \theta.
3
Factor the quadratic equation
(2sinθ+1)(sinθ1)=0(2\sin \theta + 1)(\sin \theta - 1) = 0
Factoring determines the roots for sinθ\sin \theta.
4
Solve for possible values of sinθ\sin \theta
sinθ=1\sin \theta = 1 or sinθ=12\sin \theta = -\frac{1}{2}
By the zero-product property, at least one factor must equal zero.
5
Find the angles for each ratio in the interval 0θ3600^\circ \le \theta \le 360^\circ
θ=90,210,330\theta = 90^\circ, 210^\circ, 330^\circ
sinθ=1\sin \theta = 1 has one solution (9090^\circ) and sinθ=0.5\sin \theta = -0.5 has two solutions in the 3rd and 4th quadrants (210210^\circ and 330330^\circ).
6
Count the solutions
3
There are 3 distinct values of θ\theta satisfying the condition.

Anahtar Kavram

Solving quadratic trigonometric equations using basic identities and quadrant analysis
Tahmini Süre:1m 30s
Soru 807Soru

Find the acute angle θ\theta, in degrees, that satisfies the trigonometric equation 3tanθ3=0\sqrt{3}\tan \theta - 3 = 0.

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Cevap: 60

Cevap

The acute angle θ\theta is 6060^\circ.
Rearranging the equation 3tanθ3=0\sqrt{3}\tan \theta - 3 = 0 gives 3tanθ=3\sqrt{3}\tan \theta = 3, so tanθ=33=3\tan \theta = \frac{3}{\sqrt{3}} = \sqrt{3}. For an acute angle (0<θ<900^\circ < \theta < 90^\circ), the angle with a tangent equal to 3\sqrt{3} is 6060^\circ.

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1
Isolate the trigonometric ratio tanθ\tan \theta
tanθ=3\tan \theta = \sqrt{3}
Add 33 to both sides and divide by 3\sqrt{3}, giving 33=3\frac{3}{\sqrt{3}} = \sqrt{3}.
2
Determine the value of the acute angle θ\theta
θ=60\theta = 60^\circ
From special angle exact values, tan(60)=3\tan(60^\circ) = \sqrt{3}.

Anahtar Kavram

Solving Simple Trigonometric Equations
Soru 808Soru

The couple per unit twist CC (torque per unit angle of twist) of a solid wire of length LL, radius rr, and shear modulus η\eta is modeled by the equation:

C=πηrx2LC = \frac{\pi \eta r^x}{2 L}

Using dimensional analysis, determine the numerical value of the exponent xx.

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Cevap: 4

Cevap

The numerical value of the exponent x is 4.
Applying the principle of dimensional homogeneity requires the dimensions of couple per unit twist [M L2T2][\text{M L}^2 \text{T}^{-2}] to equal the dimensions of ηrxL\frac{\eta r^x}{L}, which simplifies to [M Lx2T2][\text{M L}^{x-2} \text{T}^{-2}]. Equating exponents of length gives 2=x22 = x - 2, yielding x=4x = 4.

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1
Determine the dimensional formula of couple per unit twist CC.
[C]=M L2T2[C] = \text{M L}^2 \text{T}^{-2}
Couple (torque) is force multiplied by perpendicular distance, which has dimensions [M L T2][L]=[M L2T2][\text{M L T}^{-2}][\text{L}] = [\text{M L}^2 \text{T}^{-2}]. The angle of twist (in radians) is dimensionless.
2
Determine the dimensional formula of shear modulus η\eta.
[η]=M L1T2[\eta] = \text{M L}^{-1} \text{T}^{-2}
Shear modulus is defined as shear stress divided by shear strain. Stress has dimensions of force per unit area [M L T2]/[L2]=[M L1T2][\text{M L T}^{-2}]/[\text{L}^2] = [\text{M L}^{-1} \text{T}^{-2}], while strain is dimensionless.
3
Set up the dimensional equation for the relation C=πηrx2LC = \frac{\pi \eta r^x}{2 L}.
[M L2T2]=[M L1T2][L]x[L]=[M Lx2T2][\text{M L}^2 \text{T}^{-2}] = \frac{[\text{M L}^{-1} \text{T}^{-2}][\text{L}]^x}{[\text{L}]} = [\text{M L}^{x-2} \text{T}^{-2}]
Pure numerical constants such as π\pi and 22 are dimensionless. Length LL and radius rr both have dimension [L][\text{L}].
4
Equate the exponents of length L\text{L} on both sides of the dimensional equation.
x=4x = 4
Comparing powers of L\text{L} on both sides gives 2=x22 = x - 2, which solves to x=4x = 4.

Anahtar Kavram

Dimensional Homogeneity in Mechanics
Soru 809Soru

The length of the mercury column in an uncalibrated thermometer is 4.0cm4.0\,\text{cm} at the ice point (0C0^\circ\text{C}) and 24.0cm24.0\,\text{cm} at the steam point (100C100^\circ\text{C}). What is the temperature in degrees Celsius when the length of the mercury column is 19.0cm19.0\,\text{cm}?

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Cevap: 75

Cevap

The temperature corresponding to a mercury column length of 19.0cm19.0\,\text{cm} is 75C75^\circ\text{C}.
The temperature on the Celsius scale is determined by the ratio of the length change above the ice point to the total length change between the ice and steam points: T=LTL0L100L0×100CT = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}. Substituting L0=4.0cmL_0 = 4.0\,\text{cm}, L100=24.0cmL_{100} = 24.0\,\text{cm}, and LT=19.0cmL_T = 19.0\,\text{cm} gives T=15.020.0×100=75CT = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}.

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1
Identify given thermometric length values at fixed points and at the unknown temperature
L0=4.0cmL_0 = 4.0\,\text{cm}, L100=24.0cmL_{100} = 24.0\,\text{cm}, and LT=19.0cmL_T = 19.0\,\text{cm}
These represent the length at the lower fixed point (0C0^\circ\text{C}), upper fixed point (100C100^\circ\text{C}), and intermediate temperature TT respectively.
2
Set up the linear interpolation equation on the Celsius scale
T=LTL0L100L0×100CT = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}
Thermometric expansion is assumed to vary linearly with temperature over the operational range.
3
Substitute the given values and perform arithmetic calculation
T=19.04.024.04.0×100=15.020.0×100=75CT = \frac{19.0 - 4.0}{24.0 - 4.0} \times 100 = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}
Simplifying 15.020.0\frac{15.0}{20.0} gives 0.750.75, which multiplied by 100100 equals 7575.

Anahtar Kavram

Temperature measurement using linear variation of thermometric properties
Soru 810Soru

The mean of a set of 77 numbers arranged in ascending order is 1616. If the median of the set is 1616 and the mean of the smallest 33 numbers is 1111, what is the mean of the largest 33 numbers?

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Cevap: 21

Cevap

The mean of the largest 3 numbers is 21.
For a set of 7 ordered numbers, the median is the 4th number, which is given as 16. The total sum of all 7 numbers is 7 × 16 = 112. The sum of the 3 smallest numbers is 3 × 11 = 33. Subtracting the 3 smallest numbers and the median from the total sum leaves the sum of the 3 largest numbers: 112 - 33 - 16 = 63. Dividing 63 by 3 gives a mean of 21.

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1
Find the total sum of the set of 7 numbers
Sum = 7 × 16 = 112
The mean of a set is the total sum divided by the number of elements.
2
Determine the median value and the sum of the 3 smallest numbers
Median (4th number) = 16, Sum of 3 smallest = 3 × 11 = 33
For an ordered set of 7 numbers, the 4th number is the median, and the mean of the first 3 numbers gives their sum when multiplied by 3.
3
Calculate the sum of the 3 largest numbers
Sum of 3 largest = 112 - 33 - 16 = 63
The total sum is the sum of the 3 smallest numbers, the median (4th number), and the 3 largest numbers.
4
Compute the mean of the 3 largest numbers
Mean = 63 / 3 = 21
Dividing the sum of the 3 largest numbers by 3 gives their arithmetic mean.

Anahtar Kavram

Arithmetic Mean and Median of Ungrouped Data Subgroups
Soru 811Soru

An aluminium rod of initial length 2.0 m2.0\text{ m} at 20C20^\circ\text{C} expands by 0.96 mm0.96\text{ mm} when heated. If the linear expansivity of aluminium is 2.4×105 K12.4 \times 10^{-5}\text{ K}^{-1}, what is the rise in temperature of the rod?

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Cevap: 20

Cevap

The rise in temperature of the aluminium rod is 20 K20\text{ K}.
The fractional change in length depends on linear expansivity and temperature change through ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T. Substituting the converted expansion ΔL=9.6×104 m\Delta L = 9.6 \times 10^{-4}\text{ m}, initial length L0=2.0 mL_0 = 2.0\text{ m}, and linear expansivity α=2.4×105 K1\alpha = 2.4 \times 10^{-5}\text{ K}^{-1} gives ΔT=9.6×1042.0×2.4×105=20 K\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times 2.4 \times 10^{-5}} = 20\text{ K}.

Adım Adım Çözüm

1
Convert change in length from millimeters to meters
\Delta L = 9.6 \times 10^{-4}\text{ m}
Units must be consistent with initial length in meters.
2
Rearrange the linear thermal expansion formula \Delta L = L_0 \alpha \Delta T for temperature change \Delta T
\Delta T = \frac{\Delta L}{L_0 \alpha}
To isolate the unknown quantity \Delta T.
3
Substitute values into the rearranged formula and compute \Delta T
\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times (2.4 \times 10^{-5})} = 20\text{ K}
Evaluating the mathematical expression yields the required temperature rise.

Anahtar Kavram

Linear Expansivity and Thermal Expansion of Solids
Soru 812Soru

Given the matrix A=(12034120k)A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 4 & 1 \\ 2 & 0 & k \end{pmatrix}, if det(A)=10\det(A) = 10, what is the value of kk?

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Cevap: -3

Cevap

The value of kk is 3-3.
Expanding the determinant along the first row gives 1(4k)2(3k2)+0=2k+41(4k) - 2(3k - 2) + 0 = -2k + 4. Setting 2k+4=10-2k + 4 = 10 leads directly to 2k=6-2k = 6, giving k=3k = -3.

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1
Expand the determinant of matrix AA along the first row.
\det(A) = 1(4k - 0) - 2(3k - 2) + 0 = -2k + 4
Using cofactor expansion along the top row to find the expression for the determinant.
2
Set the calculated determinant equal to the given value and solve for kk.
-2k + 4 = 10 \implies -2k = 6 \implies k = -3
Equating the determinant algebraic expression to 10.

Anahtar Kavram

Determinant of a 3x3 Matrix
Soru 813Soru

The area of the region bounded by the curve y=3x24x+3y = 3x^2 - 4x + 3, the xx-axis, and the vertical lines x=0x = 0 and x=kx = k (where k>0k > 0) is 1818 square units. What is the value of kk?

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Cevap: 3

Cevap

The value of kk is 33.
The area under the non-negative curve y=3x24x+3y = 3x^2 - 4x + 3 from x=0x = 0 to x=kx = k is found by calculating 0k(3x24x+3)dx=k32k2+3k\int_{0}^{k} (3x^2 - 4x + 3) \, dx = k^3 - 2k^2 + 3k. Setting this expression equal to 1818 gives k32k2+3k18=0k^3 - 2k^2 + 3k - 18 = 0. By the factor theorem, testing k=3k = 3 yields 332(3)2+3(3)18=03^3 - 2(3)^2 + 3(3) - 18 = 0. Factoring as (k3)(k2+k+6)=0(k - 3)(k^2 + k + 6) = 0 confirms k=3k = 3 as the only real solution.

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1
Set up the definite integral for the area bounded by the curve and the x-axis
\int_{0}^{k} (3x^2 - 4x + 3) \, dx = 18
The curve y=3x24x+3y = 3x^2 - 4x + 3 lies entirely above the x-axis for all real xx because its leading coefficient is positive and its discriminant (4)24(3)(3)=20<0(-4)^2 - 4(3)(3) = -20 < 0.
2
Evaluate the definite integral in terms of kk
\left[ x^3 - 2x^2 + 3x \right]_0^k = (k^3 - 2k^2 + 3k) - 0 = k^3 - 2k^2 + 3k
Applying the fundamental theorem of calculus by integrating term by term.
3
Form and simplify the polynomial equation
k^3 - 2k^2 + 3k - 18 = 0
Equating the definite integral expression to the given area value of 18.
4
Solve for real values of k>0k > 0
k = 3
Using the factor theorem on k32k2+3k18=0k^3 - 2k^2 + 3k - 18 = 0, k=3k = 3 yields zero (2718+918=027 - 18 + 9 - 18 = 0). Factoring out (k3)(k - 3) gives (k3)(k2+k+6)=0(k - 3)(k^2 + k + 6) = 0, where k2+k+6=0k^2 + k + 6 = 0 has complex roots.

Anahtar Kavram

Determining an unknown boundary limit of a definite integral representing area under a curve
Soru 814Soru

Calculate the gradient of the normal line to the curve y=6xy = \frac{6}{x} at the point where x=3x = 3.

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Cevap: 1.5

Cevap

The gradient of the normal line is 1.5.
Differentiating y=6x1y = 6x^{-1} yields dydx=6x2\frac{dy}{dx} = -\frac{6}{x^2}. Evaluating this derivative at x=3x = 3 gives the tangent gradient mt=69=23m_t = -\frac{6}{9} = -\frac{2}{3}. Because the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal mn=1mt=32=1.5m_n = -\frac{1}{m_t} = \frac{3}{2} = 1.5.

Adım Adım Çözüm

1
Differentiate the function y=6x1y = 6x^{-1} with respect to xx
dydx=6x2=6x2\frac{dy}{dx} = -6x^{-2} = -\frac{6}{x^2}
The first derivative represents the formula for the tangent gradient to the curve at any given point.
2
Evaluate the derivative at x=3x = 3 to find the tangent slope (mtm_t)
m_t = -\frac{6}{3^2} = -\frac{6}{9} = -\frac{2}{3}
Substituting the given x-coordinate into the derivative gives the exact slope of the tangent at that point.
3
Calculate the normal slope (mnm_n) as the negative reciprocal of mtm_t
m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} = 1.5
The normal line is perpendicular to the tangent line, meaning mtmn=1m_t \cdot m_n = -1.

Anahtar Kavram

The gradient of the normal to a curve y=f(x)y = f(x) at x=ax = a is the negative reciprocal of the derivative evaluated at that point: mn=1f(a)m_n = -\frac{1}{f'(a)}.
Soru 815Soru

A pie chart illustrates the distribution of undergraduate students enrolled across four faculties at a university: Arts, Science, Law, and Medicine. The central angles for the sectors representing Arts, Science, and Law are 120120^\circ, 9090^\circ, and 7575^\circ, respectively. If 300300 students are enrolled in the Faculty of Medicine, what is the total number of students enrolled in the university?

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Cevap: 1440

Cevap

The total number of students enrolled in the university is 1440.
The total sum of central angles in any pie chart is 360360^\circ. Subtracting the given angles for Arts (120120^\circ), Science (9090^\circ), and Law (7575^\circ) from 360360^\circ gives the sector angle for Medicine as 7575^\circ. Since 7575^\circ represents 300300 students, each degree represents 30075=4\frac{300}{75} = 4 students. Multiplying 44 students per degree by the total 360360^\circ yields 14401440 total students in the university.

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1
Determine the sector angle for Medicine.
Sector angle for Medicine = 7575^\circ
The sum of central angles in a pie chart is 360360^\circ. Subtracting 120+90+75=285120^\circ + 90^\circ + 75^\circ = 285^\circ from 360360^\circ gives 7575^\circ.
2
Formulate the proportion relating sector angle to frequency.
75360×N=300\frac{75^\circ}{360^\circ} \times N = 300, where NN represents total students.
The fractional portion of the angle (7575^\circ out of 360360^\circ) equals the fractional portion of the total student count (300300 out of NN).
3
Calculate the total student population NN.
N=300×36075=1440N = \frac{300 \times 360}{75} = 1440
Dividing 300300 by 7575 yields 44 students per degree. Multiplying 44 by 360360 gives 14401440 students.

Anahtar Kavram

Pie Chart Sector Angle and Total Population Calculation
Soru 816Soru

Given two matrices A=(x223)A = \begin{pmatrix} x & 2 \\ 2 & 3 \end{pmatrix} and B=(x111)B = \begin{pmatrix} x & 1 \\ -1 & 1 \end{pmatrix}, where x>0x > 0. If the determinant of the product matrix ABAB is 2020, find the value of xx.

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Cevap: 3

Cevap

The positive value of xx is 3.
Using the property det(AB)=det(A)det(B)\det(AB) = \det(A) \cdot \det(B), we compute det(A)=3x4\det(A) = 3x - 4 and det(B)=x+1\det(B) = x + 1. Equating their product to 20 gives (3x4)(x+1)=20(3x - 4)(x + 1) = 20, which simplifies to the quadratic 3x2x24=03x^2 - x - 24 = 0. Factorizing yields (3x+8)(x3)=0(3x + 8)(x - 3) = 0. Since xx must be a positive number, the correct value is x=3x = 3.

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1
Calculate the determinant of matrix A
\det(A) = 3x - 4
The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.
2
Calculate the determinant of matrix B
\det(B) = x + 1
Applying ad - bc gives (x)(1) - (1)(-1) = x + 1.
3
Use the determinant product property det(AB) = det(A) * det(B)
(3x - 4)(x + 1) = 20
The determinant of the product of two square matrices equals the product of their individual determinants.
4
Form and solve the quadratic equation
3x^2 - x - 24 = 0, which factorizes into (3x + 8)(x - 3) = 0
Expanding (3x - 4)(x + 1) gives 3x^2 - x - 4. Subtracting 20 yields 3x^2 - x - 24 = 0.
5
Determine the positive solution for x
x = 3
Solving the factors gives x = -8/3 or x = 3. Since x must be positive (x > 0), x = 3.

Anahtar Kavram

Determinant of Matrix Product and 2x2 Determinants
Tahmini Süre:2m 0s
Soru 817Soru

If (8x39sin(3x)+2)dx=ax4+bcos(3x)+cx+C\int (8x^3 - 9\sin(3x) + 2) \, dx = ax^4 + b\cos(3x) + cx + C, where aa, bb, and cc are constant coefficients and CC is the constant of integration, what is the value of a+b+ca + b + c?

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Cevap: 7

Cevap

The value of a+b+ca + b + c is 7.
Integrating term-by-term yields 8x3dx=2x4\int 8x^3 dx = 2x^4, 9sin(3x)dx=3cos(3x)\int -9\sin(3x) dx = 3\cos(3x), and 2dx=2x\int 2 dx = 2x. Equating coefficients with ax4+bcos(3x)+cxax^4 + b\cos(3x) + cx gives a=2a = 2, b=3b = 3, and c=2c = 2. Therefore, a+b+c=2+3+2=7a + b + c = 2 + 3 + 2 = 7.

Adım Adım Çözüm

1
Integrate the polynomial term 8x38x^3
8x3dx=2x4\int 8x^3 \, dx = 2x^4, identifying a=2a = 2
Apply the power rule of integration: xndx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1}.
2
Integrate the trigonometric term 9sin(3x)-9\sin(3x)
9sin(3x)dx=3cos(3x)\int -9\sin(3x) \, dx = 3\cos(3x), identifying b=3b = 3
Apply the standard trigonometric integral formula: sin(kx)dx=1kcos(kx)\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx).
3
Integrate the constant term 22
2dx=2x\int 2 \, dx = 2x, identifying c=2c = 2
The integral of a constant kk with respect to xx is kxkx.
4
Calculate the requested sum a+b+ca + b + c
a+b+c=2+3+2=7a + b + c = 2 + 3 + 2 = 7
Summing the coefficients derived from each term's antiderivative.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions
Soru 818Soru

A progressive transverse wave traveling along a taut string is governed by the displacement equation y(x,t)=0.05sin(200πt10πx)y(x,t) = 0.05 \sin(200\pi t - 10\pi x), where xx and yy are measured in meters and tt in seconds. Calculate the distance, in meters, traveled by the wave front during the time taken for a single particle on the string to complete 1515 full oscillations.

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Cevap: 3

Cevap

The distance traveled by the wave front during 15 full particle oscillations is 3.0 m3.0\text{ m}.
Comparing y(x,t)=0.05sin(200πt10πx)y(x,t) = 0.05 \sin(200\pi t - 10\pi x) to the standard wave equation y=Asin(ωtkx)y = A \sin(\omega t - kx) gives ω=200π rad/s\omega = 200\pi\text{ rad/s} and k=10π rad/mk = 10\pi\text{ rad/m}. The wave speed v=ωk=20 m/sv = \frac{\omega}{k} = 20\text{ m/s}. The time for one full oscillation is T=2πω=0.01 sT = \frac{2\pi}{\omega} = 0.01\text{ s}, so 15 full oscillations take t=15×0.01 s=0.15 st = 15 \times 0.01\text{ s} = 0.15\text{ s}. The distance traveled by the wave front is d=v×t=20 m/s×0.15 s=3.0 md = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}. Alternatively, because a wave travels a distance of one wavelength λ=2πk=0.2 m\lambda = \frac{2\pi}{k} = 0.2\text{ m} during each period (1 full oscillation), in 15 full oscillations the wave travels 15×λ=15×0.2 m=3.0 m15 \times \lambda = 15 \times 0.2\text{ m} = 3.0\text{ m}.

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1
Extract wave parameters from the progressive wave equation
Angular frequency ω=200π rad/s\omega = 200\pi\text{ rad/s} and wave number k=10π rad/mk = 10\pi\text{ rad/m}.
Matching the given equation y=0.05sin(200πt10πx)y = 0.05 \sin(200\pi t - 10\pi x) with the standard form y=Asin(ωtkx)y = A \sin(\omega t - kx) identifies ω\omega and kk.
2
Calculate the wave propagation velocity
v=20 m/sv = 20\text{ m/s}.
Wave speed is given by the relation v=ωk=200π10π=20 m/sv = \frac{\omega}{k} = \frac{200\pi}{10\pi} = 20\text{ m/s}.
3
Find the period of oscillation and total elapsed time
Period T=0.01 sT = 0.01\text{ s}, total time t=0.15 st = 0.15\text{ s}.
The period T=2πω=2π200π=0.01 sT = \frac{2\pi}{\omega} = \frac{2\pi}{200\pi} = 0.01\text{ s}. For 15 complete oscillations, t=15×0.01 s=0.15 st = 15 \times 0.01\text{ s} = 0.15\text{ s}.
4
Compute the total distance traveled by the wave
Distance d=3.0 md = 3.0\text{ m}.
Using linear motion at constant wave speed, d=v×t=20 m/s×0.15 s=3.0 md = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}.

Anahtar Kavram

Wave equation parameters, particle oscillation period, and wave propagation distance
Tahmini Süre:2m 0s
Soru 819Soru

What is the maximum value of the trigonometric function y=3sinx+2y = 3\sin x + 2?

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Cevap: 5

Cevap

The maximum value of the function is 5.
The basic sine function sinx\sin x reaches a maximum value of 11. Substituting sinx=1\sin x = 1 into y=3sinx+2y = 3\sin x + 2 yields y=3(1)+2=5y = 3(1) + 2 = 5.

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1
Identify the maximum value of the sine term
The range of sinx\sin x is [1,1][-1, 1], so its maximum value is 11.
The sine function oscillates between a minimum of 1-1 and a maximum of 11 for all real numbers xx.
2
Calculate the maximum value of the transformed function
ymax=3(1)+2=5y_{\text{max}} = 3(1) + 2 = 5.
Multiplying by the positive amplitude coefficient 33 scales the peak to 33, and adding the vertical shift of 22 raises the peak to 55.

Anahtar Kavram

Maximum and Minimum Values of Trigonometric Functions
Soru 820Soru

Evaluate the numerical expression 0.00054×0.0020.00036\frac{0.00054 \times 0.002}{0.00036} and state the final result as a decimal.

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Cevap: 0.003

Cevap

The correct value as a decimal is 0.003.
Converting all terms to scientific notation yields (5.4×104)×(2×103)3.6×104=10.8×1073.6×104=3×103=0.003\frac{(5.4 \times 10^{-4}) \times (2 \times 10^{-3})}{3.6 \times 10^{-4}} = \frac{10.8 \times 10^{-7}}{3.6 \times 10^{-4}} = 3 \times 10^{-3} = 0.003.

Adım Adım Çözüm

1
Convert each decimal in the expression to scientific notation (standard form)
0.00054=5.4×1040.00054 = 5.4 \times 10^{-4}, 0.002=2×1030.002 = 2 \times 10^{-3}, 0.00036=3.6×1040.00036 = 3.6 \times 10^{-4}
Converting decimals with leading zeros to powers of 10 prevents errors in decimal point alignment during multiplication and division.
2
Simplify the numerator by multiplying coefficients and adding exponents
(5.4×104)×(2×103)=10.8×107(5.4 \times 10^{-4}) \times (2 \times 10^{-3}) = 10.8 \times 10^{-7}
According to the laws of indices, 10a×10b=10a+b10^a \times 10^b = 10^{a+b}, so 4+(3)=7-4 + (-3) = -7.
3
Divide the simplified numerator by the denominator
\frac{10.8 \times 10^{-7}}{3.6 \times 10^{-4}} = \left(\frac{10.8}{3.6}\right) \times 10^{-7 - (-4)} = 3.0 \times 10^{-3}
Dividing the coefficients gives 10.8÷3.6=310.8 \div 3.6 = 3, and subtracting the exponents gives 7(4)=3-7 - (-4) = -3.
4
Express the result in standard decimal form
3.0×103=0.0033.0 \times 10^{-3} = 0.003
Shifting the decimal point 3 positions to the left converts 10310^{-3} to standard decimal representation.

Anahtar Kavram

Simplifying Decimal Expressions using Standard Form and Laws of Indices
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