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financial accounting information for decisions
Questions and Answers of
Financial Accounting Information For Decisions
A curve has equation y = ln(x2 – 2)/x2 – 2.Find the approximate change in y as x increases from √3 to √3 + p, where p is small.
The point A, where x = 0, lies on the curve y = ln(4x2 + 3)/x-1. The normal to the curve at A meets the x-axis at the point B.i. Find the equation of this normal.ii. Find the area of the triangle
Variables x and y are connected by the equation y = 3 + 2x – 5e-x.Find the approximate change in y as x increases from ln 2 to ln 2 + p, where p is small.
A curve has equation y = A sin x + B cos 2x.The curve has a gradient of 5√3 when x = (π/6) and has a gradient of 6 + 2/√2 when x = π/4.Find the value of A and the value of B.
i. Given that y = tan 2x/x, find dy/dx.ii. Hence find the equation of the normal to the curve y = tan2x/x at the point where x = π/8.
Variables x and y are connected by the equation y = ln x/x2 + 3.Find the approximate change in y as x increases from 1 to 1 + p, where p is small.
A curve has equation y = A sin x + B sin 2x.The curve passes through the point P(π/2, 3) and has a gradient of 3√2/2 when x = π/4.Find the value of A and the value of B.
A curve has equation y = 2x sin x + π/3. The curve passes through the point P(π/2, a).a. Find, in terms of π, the value of a.b. Using your value of a, find the equation of the normal to the curve
Variables x and y are connected by the equation y = 3 + ln(2x – 5)Find the approximate change in y as x increases from 4 to 4 + p, where p is small.
Find dy/dx for each of the following.a. ey = sin 3xb. ey = 3 cos 2x
A curve has equation x = ½ [ey(3x + 7) + 1].Find the value of dy/dx when x = 1.
The diagram shows part of the curve y = ln(x + 1) – ln x. The tangent to the curve at the point P(1, ln 2) meets the x-axis at A and the y-axis at B. The normal to the curve at P meets the x-axis
Variables x and y are connected by the equation y = sin 2x.Find the approximate increase in y as x increases from π/8 to π/8 + p, where p is small.
a. By writing sec x as 1/cos x, find d/dx(sec x).b. By writing cosec x as a/sin x, find d/dx (cosec x).c. By writing cot x as cos x/sin x, find d/dx(cot x).
Find dy/dx for each of the following.a. ey = 4x2 – 1b. ey = 5x3 – 2xc. ey = (x + 3) (x – 4)
Variables x and y are such that y = e2x + e-2x.a. Find dy/dx.b. By using the substitution u = e2x, find the value of y when dy/dx = 3.c. Given that x is decreasing at the rate of 0.5 units s-1, find
A curve has equation y = x ex.The tangent to the curve at the point P(1. e) meets the y-axis at the point A.The normal to the curve at P meets the x-axis at the point B.Find the area of triangle OAB,
Find the gradient of the tangent toa. y = 2x cos 3x when x = π/3b. y = 2 – cos x/3 tan x when x = π/4.
Find dy/dx for each of the following.a. y = log3 xb. y = log2 x2c. y = log4 (5x – 1)
A curve has equation y = xex.a. Find, in terms of e, the coordinates of the stationary point on this curve and determine its nature.b. Find, in terms of e, the equation of the normal to the curve at
Given that y = x2/cos 4x, finda. dy/dx,b. The approximate change in y when x increases from π/4 to π/4 + p, where p is small.
A curve has equation y = 5 – e2x.The curve crosses the x-axis at A and the y-axis at B.a. Find the coordinates of A and B.b. The normal to the curve at B meets the x-axis at the point C.Find the
Differentiate with respect to x.a. ecos xb. ecos 5xc. etan xd. e(sin x + cos x)e. ex sin xf. ex cos1/2xg. ex(cos x + sin x)h. x2ecos xi. ln(sin x)j. x2 ln(cos x)k. sin 3x/e2x-1l. x sin x/ex
Use the laws logarithms to help differentiate these expressions with respect to x.a.b. ln 1/(2x - 5)c. ln[x(x - 5)4]d. ln(2x + 1/x -1)e. ln(2 - x/x2)f. ln [x(x + 1)/x + 2]g. ln [2x + 3/(x-5)(x+1)]h.
A curve has equation y = 5e2x – 4x – 3.The tangent to the curve at the point (0, 2) meets the x-axis at the point A. Find the coordinates of A.
A curve has equation y = e1/2x + 1.The curve crosses the y-axis at P.The normal to the curve at P meets the x-axis at Q.Find the coordinates of Q.
Variable x and y related by the equation y = 10 – 4 sin2 x, where 0 ≤ x ≤ π/2.Given that x is increasing at a rate of 0.2 radians per second, find the corresponding rate of change of y when y
A curve has equation y = x sin 2x for 0 ≤ x ≤ π radians.a. Find the equation of the normal to the curve at the point P(π/4, π/4).b. The normal at P intersects the x-axis at Q and the y-axis at
Differentiate with respect to x.a. x sin xb. 2 sin 2x cos 3xc. x2 tan xd. x tan3 (x/2)e. 5/cos 3xf. x/cos xg. tan x/xh. sin x/2 + cos xi. sin x/3x – 1j. 1/sin3 2xk. 3x/sin 2xl. sin x + cos x/ sin x
A curve has equation y = x2 ln 3x.Find the value of dy/dx and d2y/dx2 at the point where x = 2.
Find the equation of the tangent toa. y = 5/e2x + 3 at x = 0b.c. y = x2(1 + ex) at x = 1. y = Ve2* +1 at x = In 5 %3D
Find dy/dx whena. y = cos 2x sin (x/3),b. y = tan x/1 + ln x.
Differentiate with respect to x.a. sin3 xb. 5 cox2 (3x)c. sin2 x – 2 cos xd. (3 – cos x)4e. 2 sin3(2x + π/6)f. 3 cos4 x + 2 tan2(2x – π/4)
Differentiate with respect to x.a. x ln xb. 2x2 ln xc. (x – 1) ln xd. 5x ln x2e. x2 ln (ln x)f. ln 2x/xg. 4/ ln xh. ln(2x + 1)/x2i. ln(x3 – 1)/2x + 3
Differentiate with respect to x.a. xexb. x2e2xc. 3xe-xd. √x exe. ex/xf. e2x/√xg. ex+1/ex-1h. xe2x – e2x/2i. x2ex – 5/ex + 1
a. Find the equation of the tangent to the curve y = x3 – ln x at the point on the curve where x = 1.b. Show that the tangent bisects the line joining the points (-2, 16) and (12, 2).
A curve has equation y = 3 sin (2x + π/2).Find the equation of the normal to the curve at the point on the curve where x = π/4.
Differentiate with respect to x.a. 2 + sin xb. 2 sin x + 3 cos xc. 2 cos x – tan xd. 3 sin 2xe. 4 tan 5xf. 2 cos 3x – sin 2xg. tan(3x + 2)h. sin(2x + π/3)i. 2 cos(3x – π/6)
Differentiate with respect to x.a. ln 5xb. ln 12xc. ln (2x + 3)d. 2 + in(I – x2)e. ln(3x + 1)2f. ln √x + 2g. ln(2 – 5x)4h. 2x + ln(4/x)i. 5 – ln 3/(2 – 3x)j. ln (ln x)k. ln (√x + 1)2l.
Differentiate with respect to x.a. e7xb. e3xc. 3e5xd. 2e-4xe. 6e-x/2f. e3x+1g, ex2+1h. 5x – 3e√xi. 2 + 1/e3xj. 2(3 – e2x)k. ex + e-x/2l. 5(x2 + ex2)
Relative to an origin O, the position vectors of points P, Q and R are -6i + 8j, -4i + 2j and 5i + 5j respectively.a. Find the magnitude of:i. PQ(vector)ii. PR(vector)iii. QR(vector).b. Show that
Relative to an origin O, the position vector of P is 8i + 3j and the position vector of Q is – 12i – 7j. R lies on the x-axis and OR(vector) = OP(vector) + µOQ(vector).Find OR(vector).
Relative to an origin O, the position vector of A is -6i + 4j and the position vector of B is 18i + 6j. C lies on the y-axis and OC(vector) = OA(vector) + λOB(vector). Find OC(vector).
Relative to an origin O, the position vectors of A, B and C are -2i + 7j, 2i – j and 6i + λj respectively.a. Find the value of λ when AC = 17.b. Find the value of λ when is a straight line.c.
Relative to an origin O, the position vectors of points A, B and C are -5i – 11j, 23i – 4j and λ(I – 3j) respectively.Given that C les on the line AB, find the value of λ.
Relative to an origin O, the position vector of A isAnd the position vector of B isThe point A, B and C are such that BC(vector) = 2AB(vector). Find the position vector of C. 3
Relative to an origin O, the position vector of A is 3i – 2j and the position vector of B is 15i + 7j.a. Find AB(vector).The point C lies on AB such that AC(vector) = 1/3 AB(vector).b. Find the
Relative to an origin O, the position vector of A is 6i + 6j and the position vector of B is 12i – 2j.a. Find AB(vector).The point C lies on AB such that AC(vector) = ¾ AB(vector).b. Find the
At time t = 0, boat P leaves the origin and travels with velocity (3i + 4j) km h-1. Also at time t = 0, boat Q leaves the point with position vector (-10i + 17j) km and travels with velocity (5i +
At 12:00 hours two boats, A and B, have position vectors (-10i + 40j) km and (70i + 10j) km and are moving with velocities (20i + 10j) km h-1 and (-10i + 30j) km h-1 respectively.a. Find the position
Find λ and µ such that λa + µb = c.a = 5i – 6j, b = - I + 2j and c = - 13i + 18j.
a. The vectors p and q are such that p = 11i – 24j and q = 2i + αj.i. Find the value of each of the constants α and β such that p + 2q = (α + β)I – 20j.ii. Using the values of α and β
At 15:00 hours, a submarine departs from point A and travels a distance of 120 km to a point B.The position vector, r km, of the submarine relative to an origin O, t hours after 15:00 is given bya.
OA(vector) = a and OB(vector) = b.OA: AE = 1: 3 and AB: BC = 1: 2.OB = BDa. Find, in terms of a and/or b,i. OE(vector)ii. OD(vector)iii. OC(vector).b. Find, in terms of a and/or b,a. CE(vector)ii.
Relative to an origin O, the position vector of A isAnd the position vector of B isa. Find:i. |OA(vector)|ii. |OB(vector)|iii. |AB(vector)|.The point A, B and C lie on a straight line such that
P = 7i – 2j and q = I + µj.Find λ and µ such that λp + q = 36i – 13j.
The figure shows points A, B and C with position vectors a, b and c respectively, relative to an origin O. The point P lies on AB such that AP: AB = 3: 4. The point Q lies on OC such that OQ:QC = 2:
At 12:00 hours, a boat sails from a point P.The position vector, r km, of the boat relative to an origin O, t hours after 12:00 is given bya. Write down the position vector of the point P.b. Write
O, A, B and C are four points such thatOA(vector) = 7a – 5b, OB(vector) = 2a + 5b and OC(vector) = 2a + 13b.a. Findi. AC(vector)ii. AB(vector).b. Use your answers to part a to explain why B lies on
Relative to an origin O, the position vector of A isAnd the position vectors of B isa. Find AB(vector).The points A, B and C lie on a straight line such that AC(vector) = 2AB(vector).b. Find the
P = 9i + 12j, q = 3i – 3j and r = 7i + jFinda. |p + q|b. |p + q + r|.
The position vectors of points A and B relative to an origin O are a and b respectively. The point P is such that OP(vectors) = µOA(vectors). The point Q is such that OQ(vectors) = λOB(vectors).
At 12:00 hours, a tanker sails from a point P with position vector (5i + 12j) km relative to an origin O. The tanker sails south-east with a speed of 12√2 km h-1.a. Find the velocity vector of the
OA(vector) = a, OB(vector) = b and O is the origin.OX = λOA(vector) and OY(vector) = µOB(vector).a. i. Find Bx in terms of λ, a and b.ii. Find AY(vector) in terms of µ, a and b.b. 5
Relative to an origin O, the position vector of A is 4i – 2j and the position vector of B is λi + 2j.The unit vector in the direction of AB(vector) is 0.3i + 0.4j. Find the value of λ.
P = 8i – 6j, q = -2i + 3j and r = 10iFinda. 2qb. 2p + qc. 1/2p – 3rd. 1/2r – p – q.
Relative to an origin O, points A, B and C have position vectorsRespectively. All distance are measured in kilometers. A man drives at a constant speed directly from A to B in 20 minutes.i. Calculate
At 12:00 hours, a ship leaves a point Q with position vector (10i + 38j) km relative to an origin O. The ship to an origin O. The ship travels with velocity (6i – 8j) km h-1.a. Find the speed of
OA(vector) = a, OB(vector) = b, BX(vector) = 3/5 BA(vector) ANDOY(vector) = 5/7 OA(vector).OP(vector) = λOX(vector) AND BP(vector) = µBY(vector).a. Find OP(vector) in terms of λ, a and b.b.
Relative to an origin O, the position vector of P is – 2i – 4j and the position vector of Q is 8i + 20j.a. Find PQ(vector).b. Find ||PQ(vector)|.c. Find the unit vector in the direction of
Find the unit vector in the direction of each of these vectors.a. 6i + 8jb. 5i + 12jc. -4i – 3jd. 8i – 15je. 3i + 3j
a. The four points O, A, B and C are such that OA(vectors) = 5a, OB(vectors) = 15b, OC(vectors) = 24b = 3a. Show that B lies on the line AC.b. Relative to an origin O, the position vector of the
A particle starts at a point P with position vector (-20i + 60j)m relative to an origin O.The particle travels with velocity (12i 16j) ms-1.a. Find the speed of the particle.b. Find the position
OA(vector) = a, OB(vector) = b.M is the midpoint of AB and OY(vector) = ¾ OA(vector).OX(vector) = λOM(vector) and BX(vector) = µBY(vector).a. Find in terms of a and b,i. AB(vector)ii.
Relative to an origin O, the position vector of A is -7i – 7j and the position vector of B is 9i + 5j.The point C lies on AB such that AC(vector) = 3 CB(vector).a. Find AB(vector).b. Find the unit
The vector OA(vector) has a magnitude of 25 units and is parallel to the vector -3i + 4j.The vector OB(vector) has a magnitude of 26 units and is parallel to the vector 12i + 5j.Find:a. OA(vector)b.
The position vectors of the points A and B relative to an origin O are -2i + 17j and 6i + 2j respectively.i. Find the vector AB(vectors).ii. Find the unit vector in the direction of AB(vectors).iii.
The vector PQ(Vector) has a magnitude of 39 units is parallel to the vector 12j – 5j.Find PQ(Vector).
a. A car travels north-east with a speed of 18√2 km h-1. Find the velocity vector of the car.b. A boat sails on a bearing of 030° with a speed of 20 km h-1. Find the velocity vector of the boat.c.
OA(vector) = a, OB(vector) = b.B is the midpoint of OD and AC(vector) = 2/3 OA(vector).AX(vector) = λAD(vector) and BX(vector) = µBC(vector).a. Find OX(vector) in terms of λ, a and b.b. Find
The vector AB(Vector) has magnitude of 20 units and is parallel to the vector 4i + 3j.Find AB(Vector).
Relative to an origin O, the position vectors of the points A and B are I – 4j and 7i + 20j respectively. The point C lies on AB and is such that AC(vectors) = 2/3 AB(vectors). Find the position
A helicopter flies from a point P with position vector (50i + 100j) km to a point Q.The helicopter flies with a constant velocity of (30i – 40j) km h-1 and takes 2.5 hours to complete the journey.
OP(vector) = a, PY(vector) = 2b, and OQ(vector) = 3b.OX(vector) = λOY(vector) and QX(vector) = µQP(vector).a. Find OX(vector) in terms of λ, a and b.b. Find OX(vector) in terms of µ, a
a. O is the origin, P is the point (1, 5) andb. O is the origin, E is the point (-3, 4) andc. O is the origin, M is the point (4, -2) andFind the position vector of N. PQ = ;) 3 Find OQ.
Find the magnitude of each of these vectors.a. -2ib. 4i + 3jc. 5i – 12jd. -8i – 6je. 7i + 24jf. 15i – 8jg. -4i + 4jh. 5i – 10j
In the diagram OA(vectors) = a, OB(vectors) = b and AP(vectors) = 2/5 AB(vectors).a. Given that PX(vectors) = µOP(vectors), where µ is a constant, express OX(vectors) in terms of µ, a and b.b.
A car travels from a point A with position vector (60i – 40j) km to a point B with position vector (-50i + 18j) km.The car travels with constant velocity and takes 5 hours to complete to
AB(vector) = 5a, DC(vector) = 3a and CB(vector) = b.AX(vector) = λAC(vector) and DX(vector) = µDB(vector).a. Find in terms of a and b,i. AD(vector),ii. DB(vector).b. Find in terms of
Find AB(Vector), in the form ai + bj, for each of the following.a. A(4, 7) and B(3, 4)b. A(0, 6) and B(2, - 4)c. A (3, -3) and B(6, - 2)d. A(7, 0) and B(-5, 3)e. A(-4, -2) and B(-3, 5)f. A(5, -6) and
Write each vector in the form ai + bj.a. AB(Vector)b. AC(Vector)c. AD(Vector)d. AE(Vector)e. BE(Vector)f. DE(Vector)g. EA(Vector)h. DB(Vector)i. DC(Vector) A E D B
Relative to an origin O, the position vectors of the points A and B are 2i – 3j and 11i + 42j respectively.a. Write down an expression AB(vectors).The point C lies on Ab such that AC(vectors) = 1/3
a. Displacement = (21i + 54j) m, time taken = 6 second. Find the velocity.b. Velocity = (5i – 6j)ms-1, time taken = 6 seconds. Find the displacement.c. Velocity = (-4i + 4j) kmh-1, displacement =
OA(vector) = a, OB(vector) = b.R is the midpoint of OA and OP(vector) = 3 OB(vector).AQ(vector) = λAB(vector) and RQ(vector) = µ RP(vector).a. Find OQ(vector) in terms of λ, a and b.b.
A curve has equation y = 4x3 + 3x2 – 6x – 1.a. Show that dy/dx = 0 when x = - 1 and when x = 0.5.b. Find the value of d2y/dx2 when x = - 1 and when x = 0.5.
Find the coordinates of the point on the curve y = 2x2 – x – 1 at which the gradient is 7.
The curve y = 2x3 + ax2 – 12x + 7 has a maximum point at x = -2. Find the value of a.
A cuboid has a total surface area of 400 cm2 and a volume of V cm3.The dimensions of the cuboid are 4x cm by x cm by h cm.a. Express h in terms of V and x.b. Show that V = 160x – 160/5 x3.c. Find
Find the gradient of the curve y = (2x – 5)4 at the point (3, 1).
Find the gradient of the curve y = (x + 2) (x – 5)2 at the points where the curve meets the x-axis.
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