Tüm alıştırma soruları

1526 soru

Soru 521Soru

For the endothermic gaseous reaction PCl5(g)PCl3(g)+Cl2(g)\text{PCl}_5(g) \rightarrow \text{PCl}_3(g) + \text{Cl}_2(g), the standard enthalpy change (ΔH\Delta H) is +92 kJ mol1+92\text{ kJ mol}^{-1}. The activation energy for the reverse reaction without a catalyst is 58 kJ mol158\text{ kJ mol}^{-1}. If a catalyst is introduced that lowers the activation energy of the forward reaction by 35 kJ mol135\text{ kJ mol}^{-1}, what is the activation energy for the catalyzed forward reaction in kJ mol1\text{kJ mol}^{-1}?

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Cevap: 115

Cevap

The activation energy for the catalyzed forward reaction is 115 kJ mol1115\text{ kJ mol}^{-1}.
For an endothermic reaction, the enthalpy change is the difference between the forward and reverse activation energies: ΔH=Ea,forwardEa,reverse\Delta H = E_{a,\text{forward}} - E_{a,\text{reverse}}. Rearranging gives the uncatalyzed forward activation energy as 92 kJ mol1+58 kJ mol1=150 kJ mol192\text{ kJ mol}^{-1} + 58\text{ kJ mol}^{-1} = 150\text{ kJ mol}^{-1}. A catalyst lowers the activation energy by 35 kJ mol135\text{ kJ mol}^{-1}, resulting in a catalyzed forward activation energy of 150 kJ mol135 kJ mol1=115 kJ mol1150\text{ kJ mol}^{-1} - 35\text{ kJ mol}^{-1} = 115\text{ kJ mol}^{-1}.

Adım Adım Çözüm

1
Determine the activation energy of the uncatalyzed forward reaction
Ea,forward, uncatalyzed=150 kJ mol1E_{a,\text{forward, uncatalyzed}} = 150\text{ kJ mol}^{-1}
For an endothermic reaction, ΔH=Ea,forwardEa,reverse\Delta H = E_{a,\text{forward}} - E_{a,\text{reverse}}. Substituting ΔH=+92 kJ mol1\Delta H = +92\text{ kJ mol}^{-1} and Ea,reverse=58 kJ mol1E_{a,\text{reverse}} = 58\text{ kJ mol}^{-1} gives Ea,forward, uncatalyzed=92+58=150 kJ mol1E_{a,\text{forward, uncatalyzed}} = 92 + 58 = 150\text{ kJ mol}^{-1}.
2
Apply the effect of the catalyst on the forward activation energy
Ea,forward, catalyzed=115 kJ mol1E_{a,\text{forward, catalyzed}} = 115\text{ kJ mol}^{-1}
The catalyst lowers the energy barrier for the forward reaction by 35 kJ mol135\text{ kJ mol}^{-1}, so Ea,forward, catalyzed=15035=115 kJ mol1E_{a,\text{forward, catalyzed}} = 150 - 35 = 115\text{ kJ mol}^{-1}.

Anahtar Kavram

Calculation of activation energy for catalyzed and uncatalyzed reactions using enthalpy change and reverse activation energy on an energy profile.
Soru 522Soru

An isosceles trapezium has parallel sides of lengths 14 cm14\text{ cm} and 8 cm8\text{ cm}. If each of the non-parallel sides has a length of 5 cm5\text{ cm}, calculate the area of the trapezium in cm2\text{cm}^2.

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Cevap: 44

Cevap

The area of the trapezium is 44 cm244\text{ cm}^2.
Projecting the top base of length 8 cm8\text{ cm} onto the bottom base of length 14 cm14\text{ cm} leaves a difference of 6 cm6\text{ cm}, which is divided equally into two 3 cm3\text{ cm} segments on either side. Using the Pythagorean theorem with the non-parallel side (5 cm5\text{ cm}) and the base segment (3 cm3\text{ cm}) gives a height of 4 cm4\text{ cm}. The area is then calculated as 12×(14+8)×4=44 cm2\frac{1}{2} \times (14 + 8) \times 4 = 44\text{ cm}^2.

Adım Adım Çözüm

1
Determine the projection segment length on the longer base
x=1482=3 cmx = \frac{14 - 8}{2} = 3\text{ cm}
Since the trapezium is isosceles, dropping perpendiculars from both ends of the top base creates two identical right-angled triangles at the sides.
2
Calculate the perpendicular height using the Pythagorean theorem
h=5232=16=4 cmh = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\text{ cm}
The slant side (5 cm5\text{ cm}), the height (hh), and the projection segment (3 cm3\text{ cm}) form a right-angled triangle.
3
Calculate the area of the trapezium
\text{Area} = \frac{1}{2}(14 + 8) \times 4 = 44\text{ cm}^2
The area of a trapezium is given by half the sum of its parallel sides multiplied by its perpendicular height.

Anahtar Kavram

Perimeter and Area of Plane Shapes - Area of Isosceles Trapezium
Soru 523Soru
Find the value of xx that satisfies the exponential equation 4x+1+4x+4x12x+2+2x+1+2x=24\frac{4^{x+1} + 4^x + 4^{x-1}}{2^{x+2} + 2^{x+1} + 2^x} = 24
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Cevap: 5

Cevap

5
Factoring 4x14^{x-1} from the numerator yields 4x1(16+4+1)=214x14^{x-1}(16 + 4 + 1) = 21 \cdot 4^{x-1}. Factoring 2x2^x from the denominator yields 2x(4+2+1)=72x2^x(4 + 2 + 1) = 7 \cdot 2^x. Dividing the numerical coefficients gives 217=3\frac{21}{7} = 3. Substituting 4x1=22x24^{x-1} = 2^{2x-2} into the ratio gives 322x22x=32x23 \cdot \frac{2^{2x-2}}{2^x} = 3 \cdot 2^{x-2}. Setting 32x2=243 \cdot 2^{x-2} = 24 leads to 2x2=8=232^{x-2} = 8 = 2^3, which gives x2=3x - 2 = 3 and therefore x=5x = 5.

Adım Adım Çözüm

1
Factor out common terms from the numerator and denominator
Numerator: 4x1(42+41+1)=214x14^{x-1}(4^2 + 4^1 + 1) = 21 \cdot 4^{x-1}. Denominator: 2x(22+21+1)=72x2^x(2^2 + 2^1 + 1) = 7 \cdot 2^x.
Grouping power terms simplifies sums of exponential expressions.
2
Divide the numerator by the denominator and convert bases
214x172x=3(22)x12x=322x22x\frac{21 \cdot 4^{x-1}}{7 \cdot 2^x} = 3 \cdot \frac{(2^2)^{x-1}}{2^x} = 3 \cdot \frac{2^{2x-2}}{2^x}
Simplifying 217=3\frac{21}{7} = 3 and expressing base 4 in base 2 allows applying laws of indices.
3
Apply the quotient rule of indices: aman=amn\frac{a^m}{a^n} = a^{m-n}
32(2x2)x=32x23 \cdot 2^{(2x-2) - x} = 3 \cdot 2^{x-2}
Subtracting exponents of like bases simplifies the fractional index expression.
4
Set the simplified expression equal to 24 and solve for xx
32x2=24    2x2=8=23    x2=3    x=53 \cdot 2^{x-2} = 24 \implies 2^{x-2} = 8 = 2^3 \implies x - 2 = 3 \implies x = 5
Equating the exponents when bases are identical yields the linear equation x2=3x - 2 = 3.

Anahtar Kavram

Factoring sums of exponential terms and applying the quotient rule of indices
Soru 524Soru

A charged oil droplet of mass 3.2×1015 kg3.2 \times 10^{-15}\text{ kg} remains stationary in a vacuum between two horizontal charged plates where there is a uniform vertical electric field of strength 2.0×104 N C12.0 \times 10^4\text{ N C}^{-1}. Taking g=10 m s2g = 10\text{ m s}^{-2} and the elementary charge e=1.6×1019 Ce = 1.6 \times 10^{-19}\text{ C}, determine the number of excess electrons on the droplet.

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Cevap: 10

Cevap

The number of excess electrons on the droplet is 10.
The droplet is in mechanical equilibrium under two equal and opposite forces: the downward gravitational force W=mgW = mg and the upward electric force Fe=qEF_e = qE. Setting qE=mgqE = mg gives q=mgE=1.6×1018 Cq = \frac{mg}{E} = 1.6 \times 10^{-18}\text{ C}. By the quantization of charge (q=Neq = Ne), dividing this charge by the elementary charge e=1.6×1019 Ce = 1.6 \times 10^{-19}\text{ C} gives exactly 10 excess electrons.

Adım Adım Çözüm

1
Calculate the gravitational force (weight) acting on the droplet.
W=mg=(3.2×1015 kg)×(10 m s2)=3.2×1014 NW = mg = (3.2 \times 10^{-15}\text{ kg}) \times (10\text{ m s}^{-2}) = 3.2 \times 10^{-14}\text{ N}.
For stationary equilibrium, weight provides the downward vertical force.
2
Apply the equilibrium condition to find the electric force.
Fe=W=3.2×1014 NF_e = W = 3.2 \times 10^{-14}\text{ N}.
The net vertical force must be zero for the droplet to remain suspended.
3
Determine the charge qq using Fe=qEF_e = qE.
q=FeE=3.2×1014 N2.0×104 N C1=1.6×1018 Cq = \frac{F_e}{E} = \frac{3.2 \times 10^{-14}\text{ N}}{2.0 \times 10^4\text{ N C}^{-1}} = 1.6 \times 10^{-18}\text{ C}.
Electric field strength relates force and charge.
4
Calculate the number of elementary charges using charge quantization q=Neq = Ne.
N=qe=1.6×1018 C1.6×1019 C=10N = \frac{q}{e} = \frac{1.6 \times 10^{-18}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 10.
Electric charge exists in discrete integer multiples of the elementary charge ee.

Anahtar Kavram

Equilibrium between electrostatic and gravitational forces combined with charge quantization.
Soru 525Soru

A curve has the equation y=13x32x2+3x+1y = \frac{1}{3}x^3 - 2x^2 + 3x + 1. What is the positive xx-coordinate of the point on the curve where the tangent line is parallel to the line y=8x5y = 8x - 5?

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Cevap: 5

Cevap

The positive xx-coordinate of the point where the tangent is parallel to the line is 55.
The gradient of the line y=8x5y = 8x - 5 is 88. Differentiating y=13x32x2+3x+1y = \frac{1}{3}x^3 - 2x^2 + 3x + 1 gives dydx=x24x+3\frac{dy}{dx} = x^2 - 4x + 3. Setting dydx=8\frac{dy}{dx} = 8 leads to x24x5=0x^2 - 4x - 5 = 0, which factors as (x5)(x+1)=0(x - 5)(x + 1) = 0. The solutions are x=5x = 5 and x=1x = -1. Selecting the positive value gives x=5x = 5.

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1
Find the gradient (slope) of the given straight line.
The line equation is in slope-intercept form y=mx+cy = mx + c, where the slope m=8m = 8.
Parallel lines have equal slopes, so the gradient of the tangent to the curve must equal 8.
2
Differentiate the curve equation to find the gradient function dydx\frac{dy}{dx}.
dydx=ddx(13x32x2+3x+1)=x24x+3\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - 2x^2 + 3x + 1\right) = x^2 - 4x + 3.
The first derivative of a curve represents the gradient of the tangent at any point xx.
3
Equate the derivative to the slope of the line and solve for xx.
x24x+3=8    x24x5=0    (x5)(x+1)=0x^2 - 4x + 3 = 8 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0. The roots are x=5x = 5 and x=1x = -1.
Solving the quadratic equation gives all xx-values where the tangent line has a slope of 8.
4
Select the positive xx-coordinate as requested.
x=5x = 5.
The question specifically asks for the positive value among the solutions.

Anahtar Kavram

Finding points on a curve where the tangent is parallel to a given line
Soru 526Soru

A fruit vendor has 77 distinct types of fresh fruits on display. A customer wants to buy a gift basket containing exactly 44 different types of fruits. How many different combinations of fruits can the customer choose?

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Cevap: 35

Cevap

The customer can choose the fruits in 35 different ways.
The number of ways to choose r=4r = 4 items from n=7n = 7 distinct items without regard to order is given by 7C4=7!4!3!=2106=35^7C_4 = \frac{7!}{4!3!} = \frac{210}{6} = 35.

Adım Adım Çözüm

1
Identify total elements (nn) and selected subset size (rr).
n=7n = 7 and r=4r = 4.
Since the selection order inside the fruit basket does not matter, combinations (nCrnCr) must be used.
2
Substitute values into the combination formula nCr=n!r!(nr)!^nC_r = \frac{n!}{r!(n-r)!}.
7C4=7!4!(74)!=7!4!3!^7C_4 = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!}.
This evaluates the total ways to choose 4 items from 7 without repetition or ordering.
3
Simplify the factorials and compute the numerical result.
7×6×53×2×1=35\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.
Canceling out 4!4! from numerator and denominator gives (7×6×5)/6=35(7 \times 6 \times 5) / 6 = 35.

Anahtar Kavram

Combinations (nCrnCr)
Tahmini Süre:45s
Soru 527Soru

An aqueous solution of hydrocyanic acid (HCN\text{HCN}), a weak monobasic acid, has a concentration of 0.40 mol dm30.40\text{ mol dm}^{-3} at 25C25^\circ\text{C}. Given that the acid dissociation constant (KaK_a) for HCN\text{HCN} is 4.9×1010 mol dm34.9 \times 10^{-10}\text{ mol dm}^{-3}, calculate the hydrogen ion concentration, [H+][\text{H}^+], in mol dm3\text{mol dm}^{-3}. Express your answer as the coefficient AA in the form A×105 mol dm3A \times 10^{-5}\text{ mol dm}^{-3}.

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Cevap: 1.4

Cevap

The coefficient A is 1.4, which corresponds to a hydrogen ion concentration of 1.4×105 mol dm31.4 \times 10^{-5}\text{ mol dm}^{-3}.
For a weak monobasic acid, the hydrogen ion concentration is determined using the weak acid ionization relationship [H+]=Kac[\text{H}^+] = \sqrt{K_a \cdot c}. Substituting Ka=4.9×1010 mol dm3K_a = 4.9 \times 10^{-10}\text{ mol dm}^{-3} and c=0.40 mol dm3c = 0.40\text{ mol dm}^{-3} yields [H+]=1.96×1010=1.4×105 mol dm3[\text{H}^+] = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}, giving a coefficient of 1.4.

Adım Adım Çözüm

1
Write the ionization reaction and equilibrium constant expression
HCN(aq)H(aq)++CN(aq)\text{HCN}_{(aq)} \rightleftharpoons \text{H}^+_{(aq)} + \text{CN}^-_{(aq)}, giving Ka=[H+][CN][HCN]K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}
Hydrocyanic acid is a weak monobasic acid that ionizes partially in water.
2
Apply weak acid approximations
Since [H+]=[CN][\text{H}^+] = [\text{CN}^-] and KaK_a is extremely small, [HCN]c=0.40 mol dm3[\text{HCN}] \approx c = 0.40\text{ mol dm}^{-3}. Thus, Ka=[H+]2cK_a = \frac{[\text{H}^+]^2}{c}.
The negligible ionization degree allows the equilibrium concentration of un-ionized acid to be approximated as its initial concentration.
3
Substitute values and solve for [H+][\text{H}^+]
[H+]=Ka×c=4.9×1010×0.40=1.96×1010=1.4×105 mol dm3[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{4.9 \times 10^{-10} \times 0.40} = \sqrt{1.96 \times 10^{-10}} = 1.4 \times 10^{-5}\text{ mol dm}^{-3}
Multiplying KaK_a by the molar concentration gives the square of the hydrogen ion concentration.
4
Determine the coefficient A
A=1.4A = 1.4
Matching 1.4×105 mol dm31.4 \times 10^{-5}\text{ mol dm}^{-3} to the requested standard scientific notation form A×105 mol dm3A \times 10^{-5}\text{ mol dm}^{-3} yields A=1.4A = 1.4.

Anahtar Kavram

Weak Acid Ionization Equilibrium and Ka Calculations
Soru 528Soru

The daily temperature readings, in degrees Celsius, recorded over four consecutive days are 1010, 1212, 1414, and 1616. What is the mean deviation of these temperatures?

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Cevap: 2

Cevap

The mean deviation of the temperatures is 2 °C.
To find the mean deviation, first calculate the mean of the data values: (10 + 12 + 14 + 16) / 4 = 13. Next, find the absolute difference of each value from the mean: |10 - 13| = 3, |12 - 13| = 1, |14 - 13| = 1, and |16 - 13| = 3. Finally, average these absolute differences: (3 + 1 + 1 + 3) / 4 = 8 / 4 = 2.

Adım Adım Çözüm

1
Calculate the arithmetic mean (average) of the data set
\(\bar{x} = \frac{10 + 12 + 14 + 16}{4} = 13\)
Mean deviation measures dispersion relative to the mean, so the mean must be calculated first.
2
Compute the absolute deviation of each value from the mean
\(|10 - 13| = 3\), \(|12 - 13| = 1\), \(|14 - 13| = 1\), \(|16 - 13| = 3\)
Mean deviation requires non-negative distances of each observation from the mean.
3
Sum the absolute deviations and divide by the sample size (N = 4)
\(\text{Mean Deviation} = \frac{3 + 1 + 1 + 3}{4} = \frac{8}{4} = 2\)
The mean deviation is the average of the absolute deviations.

Anahtar Kavram

Mean Deviation for Ungrouped Data
Soru 529Soru

If log3x4logx3=3\log_3 x - 4\log_x 3 = 3 for x>1x > 1, what is the value of xx?

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Cevap: 81

Cevap

The value of xx is 81.
Applying the change of base identity logx3=1log3x\log_x 3 = \frac{1}{\log_3 x} transforms the equation into log3x4log3x=3\log_3 x - \frac{4}{\log_3 x} = 3. Substituting u=log3xu = \log_3 x yields u23u4=0u^2 - 3u - 4 = 0, which factors as (u4)(u+1)=0(u - 4)(u + 1) = 0. Given x>1x > 1, uu must be positive, giving u=4u = 4. Converting back to exponential form gives x=34=81x = 3^4 = 81.

Adım Adım Çözüm

1
Apply the change of base identity logab=1logba\log_a b = \frac{1}{\log_b a} to the term logx3\log_x 3.
The term becomes 1log3x\frac{1}{\log_3 x}, so the equation is log3x4log3x=3\log_3 x - \frac{4}{\log_3 x} = 3.
To express all logarithmic terms in terms of a single common base.
2
Introduce a substitution variable u=log3xu = \log_3 x.
The equation reduces to u4u=3u - \frac{4}{u} = 3.
Simplifies the equation to a manageable algebraic structure.
3
Multiply the entire equation by uu and rearrange terms.
u23u4=0u^2 - 3u - 4 = 0
Converts the rational expression into standard quadratic form.
4
Factor the quadratic equation.
(u4)(u+1)=0    u=4 or u=1(u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1
Finds the candidate values for uu.
5
Filter out invalid roots based on the domain restriction x>1x > 1 and solve for xx.
Since x>1x > 1, log3x>0\log_3 x > 0, so u=4u = 4. Thus log3x=4    x=34=81\log_3 x = 4 \implies x = 3^4 = 81.
Excludes extraneous solutions and evaluates the final exponentiation.

Anahtar Kavram

Logarithmic Change of Base and Quadratic Reduction
Soru 530Soru

A football coach records the number of goals scored by a team in six consecutive matches as 1,3,4,6,7,1, 3, 4, 6, 7, and 99. What is the variance of the goals scored?

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Cevap: 7

Cevap

The variance of the goals scored is 7.
To find the variance of the goals scored, first calculate the mean: xˉ=1+3+4+6+7+96=306=5\bar{x} = \frac{1+3+4+6+7+9}{6} = \frac{30}{6} = 5. Next, determine the sum of the squared deviations from the mean: (15)2+(35)2+(45)2+(65)2+(75)2+(95)2=16+4+1+1+4+16=42(1-5)^2 + (3-5)^2 + (4-5)^2 + (6-5)^2 + (7-5)^2 + (9-5)^2 = 16 + 4 + 1 + 1 + 4 + 16 = 42. Dividing this total by the number of data values (N=6N = 6) yields 426=7\frac{42}{6} = 7.

Adım Adım Çözüm

1
Calculate the mean of the dataset
\bar{x} = 5
The mean is needed as the reference point for computing deviations.
2
Compute the squared deviation of each data point from the mean
Squared deviations are 16, 4, 1, 1, 4, and 16
Variance measures the average squared distance from the mean.
3
Sum the squared deviations and divide by the number of observations N = 6
Variance = 7
The formula for variance of ungrouped data is \sigma^2 = \frac{\sum (x - \bar{x})^2}{N}.

Anahtar Kavram

Variance of Ungrouped Data
Soru 531Soru

Calculate the total number of distinct four-digit numbers that can be formed using the digits 1,2,3,4,5,6,1, 2, 3, 4, 5, 6, and 77 without repetition, such that the resulting number is divisible by either 44 or 55.

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Cevap: 320

Cevap

The total number of such four-digit numbers is 320320.
The total number of valid four-digit numbers is found by adding the count of numbers divisible by 55 (120120) to the count of numbers divisible by 44 (200200). Since a number ending in 55 is odd, it cannot be divisible by 44, making the two conditions mutually exclusive. Thus, the total count is 120+200=320120 + 200 = 320.

Adım Adım Çözüm

1
Calculate the number of four-digit numbers divisible by 55.
For a number to be divisible by 55, its units digit must be 55 (since 00 is not available). There is 11 choice for the units digit. The remaining 33 positions are filled from the remaining 66 available digits in 6P3=6×5×4=120^{6}P_{3} = 6 \times 5 \times 4 = 120 ways.
Divisibility by 55 requires the last digit to be 55.
2
Calculate the number of four-digit numbers divisible by 44.
A number is divisible by 44 if its last two digits form a multiple of 44. Using distinct digits from {1,2,3,4,5,6,7}\{1, 2, 3, 4, 5, 6, 7\}, the valid two-digit endings are 12,16,24,32,36,52,56,64,72,12, 16, 24, 32, 36, 52, 56, 64, 72, and 7676 (1010 valid pairs). For each pair, the first two positions are filled from the remaining 55 digits in 5P2=5×4=20^{5}P_{2} = 5 \times 4 = 20 ways. Thus, total ways = 10×20=20010 \times 20 = 200.
Divisibility by 44 depends entirely on the last two digits.
3
Check for overlap (numbers divisible by both 44 and 55).
A number divisible by 55 must end in 55, which is an odd digit. All multiples of 44 must end in an even digit. Hence, no number is divisible by both 44 and 55 in this set. The overlap is 00.
The two events are mutually exclusive.
4
Apply the addition principle of counting.
Total = 120+2000=320120 + 200 - 0 = 320.
Add the counts of the two mutually exclusive sets.

Anahtar Kavram

Restricted Permutations and Mutually Exclusive Events
Tahmini Süre:1m 30s
Soru 532Soru

If y=(2x1)3(x2+1)2y = (2x - 1)^3(x^2 + 1)^2, find the numerical value of dydx\frac{dy}{dx} at x=1x = 1.

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Cevap: 32

Cevap

The numerical value of dydx\frac{dy}{dx} evaluated at x=1x = 1 is 3232.
Using the product rule together with the chain rule for composite functions, we find u(x)=6(2x1)2u'(x) = 6(2x - 1)^2 and v(x)=4x(x2+1)v'(x) = 4x(x^2 + 1). Evaluating at x=1x = 1 gives u(1)=1u(1)=1, u(1)=6u'(1)=6, v(1)=4v(1)=4, and v(1)=8v'(1)=8. Calculating dydx=u(1)v(1)+u(1)v(1)=6(4)+1(8)=32\frac{dy}{dx} = u'(1)v(1) + u(1)v'(1) = 6(4) + 1(8) = 32.

Adım Adım Çözüm

1
Set up the product rule for y=u(x)v(x)y = u(x)v(x)
u(x)=(2x1)3u(x) = (2x - 1)^3 and v(x)=(x2+1)2v(x) = (x^2 + 1)^2
The given function is a product of two composite expressions.
2
Differentiate u(x)u(x) using the chain rule
u(x)=3(2x1)22=6(2x1)2u'(x) = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2
Differentiating the outer power function and multiplying by the derivative of the inner function 2x12x - 1.
3
Differentiate v(x)v(x) using the chain rule
v(x)=2(x2+1)2x=4x(x2+1)v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)
Differentiating the outer power function and multiplying by the derivative of the inner function x2+1x^2 + 1.
4
Apply the product rule formula dydx=u(x)v(x)+u(x)v(x)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) and evaluate at x=1x = 1
u(1)=1u(1) = 1, u(1)=6u'(1) = 6, v(1)=4v(1) = 4, v(1)=8v'(1) = 8, giving dydxx=1=(6)(4)+(1)(8)=32\frac{dy}{dx}\Big|_{x=1} = (6)(4) + (1)(8) = 32
Substituting x=1x = 1 into each individual term simplifies the arithmetic before combining.

Anahtar Kavram

Combined application of the Product Rule and Chain Rule
Soru 533Soru

Evaluate the definite integral 02(3x2+4)dx\int_{0}^{2} (3x^2 + 4) \, dx.

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Cevap: 16

Cevap

The value of the definite integral is 1616.
Integrating 3x2+43x^2 + 4 with respect to xx gives the antiderivative F(x)=x3+4xF(x) = x^3 + 4x. Evaluating this antiderivative at the upper limit x=2x = 2 yields 23+4(2)=162^3 + 4(2) = 16, and at the lower limit x=0x = 0 yields 03+4(0)=00^3 + 4(0) = 0. Subtracting the lower boundary value from the upper boundary value gives 160=1616 - 0 = 16.

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1
Integrate the polynomial term by term
\int (3x^2 + 4) dx = x^3 + 4x
Apply the power rule of integration \int x^n dx = \frac{x^{n+1}}{n+1} and \int k dx = kx.
2
Apply the fundamental theorem of calculus with limits 0 and 2
[x^3 + 4x]_0^2 = (2^3 + 4(2)) - (0^3 + 4(0)) = 16 - 0 = 16
Evaluate F(b) - F(a) where F(x) is the antiderivative.

Anahtar Kavram

Definite Integral Evaluation using the Fundamental Theorem of Calculus
Soru 534Soru
Evaluate the trigonometric limit:
limx0cos(3x)cos(x)x2\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}
What is the numerical value of this limit?
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Cevap: -4

Cevap

The numerical value of the limit is -4.
Using either the sum-to-product identity cos(3x)cos(x)=2sin(2x)sin(x)\cos(3x) - \cos(x) = -2 \sin(2x) \sin(x) along with standard limits limx0sin(kx)x=k\lim_{x \to 0} \frac{\sin(kx)}{x} = k, or applying L'Hôpital's rule twice on the 00\frac{0}{0} form, yields the exact value 4-4.

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1
Check the form of the limit by direct substitution
Substituting x=0x = 0 gives cos(0)cos(0)02=110=00\frac{\cos(0) - \cos(0)}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}, an indeterminate form.
Determines whether algebraic transformation or L'Hôpital's rule is required.
2
Transform the numerator using the sum-to-product formula
\cos(3x) - \cos(x) = -2 \sin\left(\frac{3x+x}{2}\right) \sin\left(\frac{3x-x}{2}\right) = -2 \sin(2x) \sin(x)
Converts difference of cosines into product of sines to utilize standard trigonometric limits.
3
Rewrite the fractional expression and apply limit laws
\lim_{x \to 0} \frac{-2 \sin(2x) \sin(x)}{x^2} = -2 \cdot \left(\lim_{x \to 0} \frac{\sin(2x)}{x}\right) \cdot \left(\lim_{x \to 0} \frac{\sin(x)}{x}\right)
Splits x2x^2 into xxx \cdot x under each sine function.
4
Evaluate the individual standard limits
\lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin(x)}{x} = 1
Applies the known fundamental trigonometric limit rule limu0sin(au)u=a\lim_{u \to 0} \frac{\sin(au)}{u} = a.
5
Calculate the final product
-2 \times 2 \times 1 = -4
Combines all factors to reach the evaluated value.

Anahtar Kavram

Trigonometric Limits and Indeterminate Forms

Alternatif Yöntem

Alternatively, apply L'Hôpital's rule twice. First derivative of numerator over denominator yields limx03sin(3x)+sin(x)2x\lim_{x \to 0} \frac{-3\sin(3x) + \sin(x)}{2x} (still 00\frac{0}{0}). Differentiating a second time yields limx09cos(3x)+cos(x)2=9(1)+12=82=4\lim_{x \to 0} \frac{-9\cos(3x) + \cos(x)}{2} = \frac{-9(1) + 1}{2} = \frac{-8}{2} = -4.
Tahmini Süre:1m 30s
Soru 535Soru

Given that θ\theta is an acute angle satisfying secθ+tanθ=3\sec \theta + \tan \theta = 3, what is the exact numerical value of 10sinθ10 \sin \theta?

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Cevap: 8

Cevap

The numerical value of 10sinθ10 \sin \theta is 8.
Using the standard fundamental identity sec2θtan2θ=1\sec^2 \theta - \tan^2 \theta = 1, we factor it into (secθ+tanθ)(secθtanθ)=1(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1. Substituting the given value secθ+tanθ=3\sec \theta + \tan \theta = 3 gives secθtanθ=13\sec \theta - \tan \theta = \frac{1}{3}. Solving these linear equations yields secθ=53\sec \theta = \frac{5}{3} (so cosθ=35\cos \theta = \frac{3}{5}) and tanθ=43\tan \theta = \frac{4}{3}. Using sinθ=tanθcosθ\sin \theta = \tan \theta \cdot \cos \theta, we find sinθ=45=0.8\sin \theta = \frac{4}{5} = 0.8. Multiplying by 10 gives the exact result 8.

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1
Apply the trigonometric identity sec2θtan2θ=1\sec^2 \theta - \tan^2 \theta = 1
(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1
Difference of squares factorization links secθ+tanθ\sec \theta + \tan \theta and secθtanθ\sec \theta - \tan \theta as reciprocals.
2
Substitute secθ+tanθ=3\sec \theta + \tan \theta = 3 to find secθtanθ\sec \theta - \tan \theta
\sec \theta - \tan \theta = \frac{1}{3}
Dividing both sides of 3(secθtanθ)=13(\sec \theta - \tan \theta) = 1 by 3.
3
Solve the system of equations for secθ\sec \theta and tanθ\tan \theta
\sec \theta = \frac{5}{3} \text{ and } \tan \theta = \frac{4}{3}
Adding equations gives 2secθ=1032\sec \theta = \frac{10}{3}; subtracting gives 2tanθ=832\tan \theta = \frac{8}{3}.
4
Calculate sinθ\sin \theta and evaluate 10sinθ10 \sin \theta
\sin \theta = \frac{4}{5} \implies 10 \sin \theta = 8
Since cosθ=35\cos \theta = \frac{3}{5} and tanθ=43\tan \theta = \frac{4}{3}, sinθ=tanθcosθ=45\sin \theta = \tan \theta \cdot \cos \theta = \frac{4}{5}.

Anahtar Kavram

Trigonometric identities relating secant and tangent ratios
Soru 536Soru

An echo sounder on a fishing boat emits an acoustic pulse vertically downward toward the seabed and detects the reflected signal 0.80 s0.80\text{ s} later. If the speed of sound in seawater is 1450 m/s1450\text{ m/s}, what is the depth of the sea in meters at this location?

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Cevap: 580

Cevap

The depth of the sea is 580 m580\text{ m}.
An echo signal travels to the reflecting surface and back, covering twice the depth (2d=v×t2d = v \times t). Substituting v=1450 m/sv = 1450\text{ m/s} and t=0.80 st = 0.80\text{ s} gives 2d=1160 m2d = 1160\text{ m}, so the seabed depth d=580 md = 580\text{ m}.

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1
Identify the given physical quantities
Total travel time t=0.80 st = 0.80\text{ s} and speed of sound v=1450 m/sv = 1450\text{ m/s}.
An echo involves sound traveling to the seabed and back, so the recorded time represents a two-way journey.
2
Set up the distance equation for an echo
Total distance traveled by the sound pulse is 2d=v×t2d = v \times t.
Sound travels to the sea floor and reflects back to the ship, covering a total distance equal to twice the depth.
3
Calculate the depth dd
d=1450×0.802=580 md = \frac{1450 \times 0.80}{2} = 580\text{ m}.
Dividing the total round-trip distance by 2 yields the one-way depth.

Anahtar Kavram

Calculation of distance using sound echoes in a medium
Soru 537Soru

What is the numerical value of the expression log2(log381)+log5(1125)+4log23\log_2(\log_3 81) + \log_5\left(\frac{1}{125}\right) + 4^{\log_2 3}?

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Cevap: 8

Cevap

8
Evaluating each term individually: the nested logarithm log2(log381)=log24=2\log_2(\log_3 81) = \log_2 4 = 2; the reciprocal log argument log5(1/125)=3\log_5(1/125) = -3; and the exponential log power 4log23=(2log23)2=32=94^{\log_2 3} = (2^{\log_2 3})^2 = 3^2 = 9. Combining these gives 23+9=82 - 3 + 9 = 8.

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1
Evaluate the first component log2(log381)\log_2(\log_3 81)
2
Since 81=3481 = 3^4, the inner expression log381=4\log_3 81 = 4, and subsequently log24=2\log_2 4 = 2.
2
Evaluate the second component log5(1125)\log_5\left(\frac{1}{125}\right)
-3
Since 1125=53\frac{1}{125} = 5^{-3}, applying the power law of logarithms gives 3-3.
3
Evaluate the third component 4log234^{\log_2 3}
9
Rewrite 44 as 222^2 to obtain (2log23)2=32=9(2^{\log_2 3})^2 = 3^2 = 9 using the fundamental identity alogab=ba^{\log_a b} = b.
4
Sum the results of the three components
8
Calculate 2+(3)+9=82 + (-3) + 9 = 8.

Anahtar Kavram

Properties of logarithms including change of power, negative exponents, and logarithm exponentiation identities
Soru 538Soru

The sum to infinity of a geometric progression (G.P.) with positive terms is 1818, and the sum of its first two terms is 1616. What is the first term of the progression?

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Cevap: 12

Cevap

The first term of the progression is 12.
Using the sum to infinity formula S=a1r=18S_{\infty} = \frac{a}{1 - r} = 18, we express the first term as a=18(1r)a = 18(1 - r). Combining this with the sum of the first two terms S2=a(1+r)=16S_2 = a(1 + r) = 16 yields 18(1r)(1+r)=16    18(1r2)=1618(1 - r)(1 + r) = 16 \implies 18(1 - r^2) = 16. Solving for rr gives r2=19r^2 = \frac{1}{9}, so r=13r = \frac{1}{3} for a sequence with positive terms. Substituting r=13r = \frac{1}{3} back into a=18(1r)a = 18(1 - r) gives a=12a = 12.

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1
Express the sum to infinity in terms of the first term aa and common ratio rr.
a=18(1r)a = 18(1 - r)
The sum to infinity formula for a convergent G.P. is S=a1rS_{\infty} = \frac{a}{1 - r}.
2
Write the expression for the sum of the first two terms.
a(1+r)=16a(1 + r) = 16
The sum of the first two terms is T1+T2=a+ar=a(1+r)T_1 + T_2 = a + ar = a(1 + r).
3
Substitute a=18(1r)a = 18(1 - r) into the sum of the first two terms equation.
18(1r2)=1618(1 - r^2) = 16
Applying the difference of two squares identity (1r)(1+r)=1r2(1 - r)(1 + r) = 1 - r^2.
4
Solve for the common ratio rr.
r=13r = \frac{1}{3}
Rearranging gives 1r2=89    r2=191 - r^2 = \frac{8}{9} \implies r^2 = \frac{1}{9}. Since all terms are positive, rr must be positive.
5
Calculate the first term aa.
a=12a = 12
Substitute r=13r = \frac{1}{3} into a=18(1r)a = 18(1 - r) to get a=18×23=12a = 18 \times \frac{2}{3} = 12.

Anahtar Kavram

Geometric Progression sum to infinity and partial sums
Tahmini Süre:1m 30s
Soru 539Soru

The heights (in cm) of a seedling recorded over five consecutive weeks are 99, 1313, 1515, 1717, and 2121. What is the standard deviation of the heights?

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Cevap: 4

Cevap

The standard deviation of the seedling heights is 4 cm.
To find the standard deviation, first compute the mean: (9 + 13 + 15 + 17 + 21) / 5 = 15. Next, calculate the sum of squared deviations: (-6)^2 + (-2)^2 + 0^2 + 2^2 + 6^2 = 36 + 4 + 0 + 4 + 36 = 80. Divide by 5 to find the variance of 16. Finally, taking the square root of 16 gives the standard deviation of 4 cm.

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1
Calculate the mean of the data values
Mean = 15 cm
The mean is necessary to calculate the deviations of each data point.
2
Calculate the squared deviations from the mean
Squared deviations are 36, 4, 0, 4, and 36 (sum = 80)
Variance measures the average of squared deviations from the mean.
3
Calculate the population variance
Variance = 80 / 5 = 16
Dividing the sum of squared deviations by N gives the variance.
4
Calculate the standard deviation
Standard deviation = sqrt(16) = 4 cm
The standard deviation is the square root of the variance.

Anahtar Kavram

Standard Deviation of Ungrouped Data
Soru 540Soru

If 50.0 cm350.0\text{ cm}^3 of a saturated solution of potassium chloride (KCl\text{KCl}) contains 14.9 g14.9\text{ g} of the salt at 298 K298\text{ K}, what is the solubility of potassium chloride at this temperature in mol dm3\text{mol dm}^{-3}? (Molar mass of KCl=74.5 g mol1\text{KCl} = 74.5\text{ g mol}^{-1})

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Cevap: 4

Cevap

The solubility of potassium chloride at 298 K298\text{ K} is 4.0 mol dm34.0\text{ mol dm}^{-3}.
First, the mass concentration is determined by scaling the 14.9 g14.9\text{ g} in 50.0 cm350.0\text{ cm}^3 to 1000 cm31000\text{ cm}^3, giving 298.0 g dm3298.0\text{ g dm}^{-3}. Dividing 298.0 g dm3298.0\text{ g dm}^{-3} by the molar mass of KCl\text{KCl} (74.5 g mol174.5\text{ g mol}^{-1}) yields 4.0 mol dm34.0\text{ mol dm}^{-3}.

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1
Calculate the mass of solute present per cubic decimetre (1000 cm³) of saturated solution.
Mass concentration = 298.0 g dm⁻³
Solubility is expressed relative to 1 dm³ of solution volume.
2
Divide the mass concentration in g dm⁻³ by the molar mass of KCl.
Solubility = 4.0 mol dm⁻³
Molar solubility equals mass concentration divided by molar mass (M).

Anahtar Kavram

Solubility Determination in Moles per Decimetre Cubed
Tahmini Süre:1m 30s
ÖncekiSayfa 27 / 77Sonraki
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