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financial accounting information for decisions
Questions and Answers of
Financial Accounting Information For Decisions
a. Find the area of the region enclosed by the curve y = 12/x2, the x-axis and the lines x = 1 and x = 4.b. The line x = p divides the region in part a into two equal parts. Find the value of p.
A curve is such that dy/dx = 2x3 + 6/x2.Given that the curve passes through the point (- 1, 10) find the equation of the curve.
Sketch the following pairs of curves and find the area enclosed between their graphs for x ≥ 0.a. y = x2 and y = x(2 – x)b. y = x3 and y = 4x – 3x2
A curve is such that dy/dx = 2(kx – 1)5 where k is a constant.Given that the curve passes through the points (0, 1) and (1, 8) find the equation of the curve.
A curve is such that d2y/dx2 = 8e-2x.Given that dy/dx = 2 when x = 0 and that the curve passes through the point (1, 2/c2), find the equation of the curve.
A curve is such that d2y/dx2 = 45 cos 3x + 2 sin x.Given that dy/dx = - 2 when x = 0 and that the curve passes through the point (π, -1), find the equation of the curve.
a. Find the value of the constant A such that 6x-5/2x-3 = 3 + A/2x -3.b. Hence show that 6x - 5 7 dx 6+2 ln 3 s 2x-3
a. Find d/dx(xe2x – e2x/2).b. Hence find ʃxe2x dx.
Given thatFind the value of k. 2In 7, 3 2 1. 3x - 1
a. Find d/dx (x2 ln x).b. Hence evaluate ʃ1e 4x ln x dx.
a. Show that d/dx (xex – ex) xex.b. Use your result from part a to evaluate the area of the shaded region. y = x e 2
A curve is such that dy/dx = (2 - √x)2/√x.Given that the curve passes through the point (9, 14) find the equation of the curve.
Find the shaded area enclosed by the curve y = 3√x the line y = 10 – x and the x-axis. y = 3Jx y = 10 – x ol
The point P(3/2, 5) lies on the curve for which dy/dx = 2e3-2x.The point Q(1, k) also lies on the curve.a. Find the value of k.The normal to the curve at the points P and Q intersect at the point
A curve is such that dy/dx = k cos 3x – 4, where k is a constant.At the point (π, 2) the gradient of the curve is – 10.a. Find the value of k.b. Find the equation of the curve.
a. Differentiate 4x3 ln(2x + 1) with respect to x.b. i. Given that y = 2x/√x+2, show that dy/dx = x+4/(√x+2)3.ii. Hence findiii. Hence evaluate 5x + 20 dx. (Jä + 2 )*
a. Show that d/dx(sin x/1 – cos x) can be written in the form k/cos x – 1, and state the value of k.b. Hence find ʃ 5/cos x -1 dx.
a. Given that 4x/2x + 3 = 2 + A/2x + 3, find the value of the constant A.b. Hence show that 4x dx 2-3 ln 2x + 3 3
a. Given that y = x sin 3x, find dy/dx.b. Hence evaluate ʃ0π/6 x cos 3x dx.
The tangent to the curve y = 6x – x2 at the point (2, 8) cuts the x-axis at the point P.a. Find the coordinates of P.b. Find the area of the shaded region. (2, 8) y = 6x – x 2 10 6 х
The point (π/2, 5) lies on the curve for which dy/dx = 4 sin(2x – π/2).a. Find the equation of the curve.b. Find the equation of the normal to the curve at the point where x = π/3.
i. Show that d/dx(e4x/4 – xe4x) = pxe4x, where p is an integer to be found.ii. Hence find the exact value of ʃ0ln2 xe4x dx, giving your answer in the form a ln 2 + b/c, where a, b and c are
a. Given that y = (x + 8)√x – 4, show that dy/dx = kx/√x – 4, and state the value of k.b. Hence find ʃ x/√x – 4 dx.
a. Find the quotient and remainder when 4x2 + 4x is divided by 2x + 1.b. Hence show that 4x2 + 4x dx = 2 -In 3. o 2x +1 2.
a. Show that d/dx(x ln x) = 1 + ln x.b. Use your result from part a to evaluate the area of the shaded region. y = In x
A curve is such that d2y/dx2 = 12x – 12.The gradient of the curve at the point (2, 9) is 8.a. Express y in terms of x.b. Show that the gradient of the curve is never less than 2.
The curve y = √2x + 1 meets the y-axis at the point P.The tangent at the point Q(12, 5) to this curve meets the y-axis at the point R.Find the area of the shaded region POR. Q(12, 5). y=/2x + 1 R P
The point P(π/3, 3) lies on the curve for which dy/dx = 3 cos(3x – π/2).The point Q(π/2, k) also lies on the curve.a. Find the value of k.The tangents to the curve at the points P and Q
Find the value of1. 6x-8x dx. 2х -1
The diagram shows part of the curve y = 3x – x3/2 and the lines y = 3x and 2y = 27 – 3x. The curve and the line y = 3x meet the x-axis at O and the curve and the line 2y = 27 – 3x meet the
a. Given that y = 1/x2 – 7, show that dy/dx = kx/(x2 – 7)2, and state the value of k.b. Hence finds 4x dx. (x² – 7)*
a. Show that d/dx(x cos x) = cos x – x sin x.b. Use your result from part a to evaluate the area of the shaded region. y = x sin x
A curve is such that dy/dx = kx – 5 where k is a constant.The gradient of the normal to the curve at the point (2, -1) is – 1/3. Find the equation of the curve.
The diagram shows the graphs of y = 2 + cos 2x and y = 1 + 2 cos 2x for 0 ≤ x ≤ π.Find the area of the shaded region. y = 2 + cos 2x TT x y = 1+ 2 cos 2x
a. Find d/dx (2x3 ln x).b. Hence find ʃx2 ln x dx.
The diagram shows the graphs of y = 6/x + 3 and y = 4 – x. Find the area of the shaded region. y = x + 3 y = 4 - x
a. Differentiate x cos x with respect to x.b. Hence find ʃx sin x dx.
a. Given that y = e2x (sin 2x + cos 2x), show that dy/dx = 4e2x cos 2x.b. Hence find ʃe2x cos 2x dx.
a. Find d/dx(x2√2x – 7).b. Hence finds 5x2 -14x + 3 dx. V2x - 7
Find.a. ʃ 8/x dxb. ʃ 5/x dxc. ʃ 1/2x dxd. ʃ 5/3x dxe. ʃ 1/3x+2 dxf. ʃ 1/1 – 8x dxg. ʃ 7/2x -1 dxh. ʃ 4/2 – 3x dxi. ʃ 5/2(5x – 1) dx
a. Given that y = x + 5/√(2x – 1, show that dy/dx = x – 6/√2x – 1)3.b. Hence finds x – 6 dx. V(2x – 1)
a. Given that y = ex2, find dy/dx.b. Use your answer to part a to find ʃxex2 dx.c. Hence evaluate ʃ20 xex2 dx.
Evaluate.a.b.c.d.e.f.g.h.i.j.k.l. 7x° dx
a. Given that y = (x + 1) √2x – 1, find dy/dx,b. Hence evaluates dx. 2x 1
Find the area of the region enclosed by the curve y = 1 + cos x and the line y = 1. y = 1+ cos x -y = 1
Find the area of each shaded region.a.b.c.d.e.f. y = x 3– 6x 2 + 9x 3
Find y in terms of x for each of the following.a. dy/dx = 12x4b. dy/dx = 5x8c. dy/dx = 7x3d. dy/dx = 4/x3e. dy/dx = 1/2x2f. dy/dx = 3/√x
Find each of the following.a. ʃ4x7 dxb. ʃ12x5 dxc. ʃ2x-3 dxd. ʃ 4/x2 dxe. f. 3 dx
Find:a.b.c.d.e.f.g.h.i. |(x + 2)° dx
Find:a. ʃe5x dxb. ʃe9x dxc. ʃe1/2x dxd. ʃe-2x dxe. ʃ4ex dxf. ʃ2e4x dxg. ʃe7x+4 dxh. ʃe5-2x dxi. ʃ1/3 e6x-1 dx
Find:a.b.c.d.e.f.g.h.i. 7x +1+ dx
A curve is such that dy/dx = 4x + 1/(x + 1)2 for x > 0. The curve passes through the point (1/2. 5/6).a. Find the equation of the curve.b. Find the equation of the normal to the curve at the point
a. Differentiate (3x2 – 1)5 with respect to x.b. Hence finds fx( 3x² - 1)* dx.
Evaluate.a.b.c.d.e.f. (2x + 1)° dx
a. Given that y = x√3x2 + 4, find dy/dx.b. Hence evaluates 3x? dx. V3x? + 4 0.
Find the area of each shaded region.a.b.c.d. ソラx(x-2)
Find y in terms of x for each of the following.a. dy/dx = 7x6 + 2x4 + 3b. dy/dx = 2x5 – 3x3 + 5xc. dy/dx = 3/x4 – 15/x2 + xd. dy/dx = 18/x10 + 6/x7 -2
Find the area of the region bounded by the curve y = 2 + 3x – x2, the line x = 2 and the line y = 2. y= 2+3x- x?
Find each of the following.a.b.c.d.e.f. (x + 2)(x + 5) dx
A curve is such that dy/dx = (4x + 1)4.Given that the curve passes through the point (0, - 1.95) find the equation of the curve.
Find:a.b.c.d.e.f. Je* (5 - e*) dx
Finda.b.c.d.e.f. (1- sin x) dx
A curve is such that dy/dx = 5/2x – 1 for x > 0.5.Given that the curve passes through the point (1, 3), find the equation of the curve.
a. Find ʃ(1 – 6/x2) dx.b. Hence find the value of the positive constant k for which 3k dx = 2.
a. Differentiate x ln x with respect to x.b. hence finds ʃ ln x dx.
Evaluate.a.b.c.d.e.f.g.h.i. e2x dx 0.
a. Given that y = 1/x2 + 5, find dy/dx.b. Hence evaluates 4x dx. (x² + 5 )*
Find the total shaded region.1. у%3 (x+2)(х+ 1) (х-1) -2 -1
Find y in terms of x for each of the following.a. dy/dx = 3x(x – 2)b. dy/dx = x2(4x2 – 3)c. dy/dx = (x + 2√x)2d. dy/dx = x(x – 3) (x + 4)e. dy/dx = x5 – 3x/2x3f. dy/dx = (2x – 3) (x –
Find the area of the region bounded by the curve y = 3x2 + 2, the line y = 14 and the y-axis. y = 3x2 + 2 %3D 14 Ol
Find each of the following.a.b.c.d.e.f. x2 - 5 dx .2
A curve is such that dy/dx = √2x + 1.Given that the curve passes through the point (4, 11) find the equation of the curve.
Find:a.b.c. 2e* + dx
A curve is such that dy/dx = cos x – sin x.Given that the curve passes through the point (π/2, 3), find the equation of the curve.
A curve is such that dy/dx = 2x + 5/x for x > 0.Given that the curve passes through the point (e, e2), find the equation of the curve.
The diagram shows part of the curve of y = 9x2 – x3, which meets the x-axis at the origin O and the point A. The line y – 2x + 18 = 0 passes through A and meets the y-axis at the point B.a. Show
a. Show that d/dx(ln x/x) = 1 – ln x/x2.b. Hence finds Inx dx.
Evaluate.a.b.c.d.e.f. | sin x dx
A curve is such that dy/dx = 3x2 – 4x + 1.Given that the curve passes through the point (0, 5) find the equation of the curve.
Find the area of the shaded region.1. y 2 sin 2x +3 cos x RIN
A curve is such that dy/dx = 1/√10-x.Given that the curve passes through the point (6, 1), find the equation of the curve.
A curve is such that dy/dx = 2e2x + e-x.Given that the curve passes through the point (0, 4), find the equation of the curve.
A curve is such that dy/dx = 1 – 4 cos 2x.Given that the curve passes through the point (π/4, 1), find the equation of the curve.
A curve is such that dy/dx = 1/x+e for x > -e.Given that the curve passes through the point (e, 2 + ln 2), find the equation of the curve.
The diagram shows part of the curve y = 2 sin 3x. The normal to the curve y = 2 sin 3x at the point where x = π/9 meets the y-axis at the point P.a. Find the coordinates of P.b. Find the area of the
a. Given that y = x√x2 – 4, find dy/dx.b. Hence finds x2 – 2 dx. Vx² - 4 .2
Evaluate.a.b.c.d.e.f. 2 dx 1 3x +1
a. Show that d/dx (x/cos x) = cos x + x sin x/cos2 x.b. Hence evaluate. 4 COS x + x sin x dx. 5 cos? x
Find the area enclosed by the curve y = 6/√x, the x-axis and the lines x = 4 and x = 9.
A curve is such that dy/dx = 6x(x – 1).Given that the curve passes through the point (1, - 5) find the equation of the curve.
The diagram shows an isosceles triangle PQR inscribed in a circle, centre O, radius r cm.PR = QR and angle ORP = θ radians.Triangle PQR has an area of A cm2.a. Show that A = r2 sin 2θ + r2 sin 2θ
The diagram shows a semi-circle with diameter EF of length 12 cm.Angle GEF = θ radians and the shaded region has an area of A cm2.a. Show that A = 36θ + 18 sin 2θ.b. Given that θ is increasing at
A curve has equation y = x2ex.The curve has a minimum point at P and a maximum point at Q.a. Find the coordinates of P and the coordinates of Q.b. The tangent to the curve at the point A(1, e) meets
A curve has equation y = x ln x.The curve crosses the x-axis at the point A and has a minimum point at B.Find the coordinates of A and the coordinates of B.
A curve has equation y = Ae2x + Be-2x.The gradient of the tangent at the point (0, 10) is – 12.a. Find the value of A and the value of B.b. Find the coordinates of the turning point on the curve
The diagram shows a circle, centre O, radius r cm. The points A and B lie on circle such that angle AOB = 2θ radians.i. Find, in terms of r and θ, an expression for the length of the chord AB.ii.
Find the coordinates of the stationary points on these curves and determine their nature.a. y = 4 sin x + 3 cos x for 0 ≤ x ≤ π/2b. y = 6 cos x/2 + 8 sin x/2 for 0 ≤ x ≤ 2πc. y = 5 sin
A curve has equation y = x/x2 + 1.i. Find the coordinates of the stationary points on the curve.ii. Show that d2y/dx2 = px3 + qx/(x2 + 1)3, where p and q are integers to be found, and determine the
Find the coordinates of the stationary points on these curves and determine their nature.a. y = xex/2b. y = x2e2xc. y = ex – 7x + 2d. y = 5e2x – 10x – 1e. y = (x2 – 8)e-xf. y = x2 ln xg. y =
Variables x and y are such that y = (x -3) ln(2x2 + 1).i. Find the value of dy/dx when x = 2.ii. Hence find the approximate change in y when x changes from 2 to 2.03.
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