Application of Integrals JEE Mathematics Worksheets Set 01

Advanced Subjective Questions

Question. A polynomial function f(x) satisfies the condition f(x + 1) = f(x) + 2x + 1. Find f(x) if f(0) = 1. Find also the equations of the pair of tangents from the origin on the curve y = f(x) and compute the area enclosed by the curve and the pair of tangents.
Answer: f(x + 1) = f(x) + 2x + 1 ; f(0) = 1
put x = 0
f(1) = f(0) + 1 = 2
put x = 1
f(2) = f(1) + 3 = 5
put x = 2
f(3) = f(2) + 5 = 10
\( \Rightarrow f(x) = 1 + x^2 \)
Let the pair of tangents
y = mx
mx = 1 + \( x^2 \Rightarrow x^2 - mx + 1 = 0 \)
D = 0 \( \Rightarrow m = \pm 2 \)
pair of tangent y = \( \pm 2x \)
Area = \( 2 \int_{0}^{1} (1 + x^2 - 2x) dx = \frac{2}{3} \)

Question. Find the equation of the line passing through the origin and dividing the curvilinear triangle with vertex at the origin, bounded by the curves \( y = 2x - x^2 \), y = 0 and x = 1 into two parts of equal area.
Answer: Area bounded by \( y = 2x - x^2 \), y = 0 & x = 1
\( A = \int_{0}^{1} (2x - x^2) dx = \frac{2}{3} \)
Let line is y = mx which divides the area into two equal parts so area of \( \Delta OAB = \frac{1}{2} \left( \frac{2}{3} \right) = \frac{1}{3} \)
\( \Rightarrow \frac{1}{2} \times 1 \times m = \frac{1}{3} \Rightarrow m = \frac{2}{3} \)
so line is \( y = \frac{2}{3}x \)

Question. Consider the curve \( y = x^n \) where n > 1 in the \( 1^{st} \) quadrant. If the area bounded by the curve, the x-axis and the tangent line to the graph of \( y = x^n \) at the point (1, 1) is maximum then find the value of n.
Answer: Equation of tangent at (1, 1)
\( y - 1 = n(x - 1) \Rightarrow y = nx - n + 1 \) ....(i)
area bounded by the curve, the tangent & the x-axis is
\( A = \left| \int_{0}^{1} (nx - n + 1 - x^n) dx \right| = \left| \frac{n^2 - n}{2(n + 1)} \right| \)
Now A to be maximum, \( \frac{dA}{dn} = 0 \Rightarrow \) at \( n = \sqrt{2} + 1 \)

Question. Consider the collection of all curve of the form \( y = a - bx^2 \) that pass through the point (2, 1), where 'a' and 'b' are positive constants. Determine the value of 'a' and 'b' that will minimize the area of the region bounded by \( y = a - bx^2 \) and x-axis. Also find the minimum area.
Answer: at (2, 1) \( \Rightarrow a = 1 + 4b \)
area bounded by curve & x-axis
\( A = 2 \left[ \int_{0}^{\sqrt{a/b}} ((1 + 4b) - bx^2) dx \right] \)
for area to be minimum \( \frac{dA}{db} = 0 \)
\( \Rightarrow b = \frac{1}{8} \) & \( a = \frac{3}{2} \)
\( A_{min} = 4\sqrt{3} \) sq. units.

Question. In the adjacent figure, graphs of two functions y = f(x) and y = sinx are given. y = sinx intersects, y = f(x) at A(a, f(a)); B(\( \pi \), 0) and C(\( 2\pi \), 0). \( A_i \) (i = 1, 2, 3) is the area bounded by the curves y = f(x) and y = sinx between x = 0 and x = a; i = 1, between x = a and x = \( \pi \); i = 2, between x = \( \pi \) and x = \( 2\pi \); i = 3. If \( A_1 = 1 - \sin a + (a - 1) \cos a \), determine the function f(x). Hence determine 'a' and \( A_1 \). Also calculate \( A_2 \) and \( A_3 \).
Answer: From the figure it is clear that
\( \int_{0}^{a} (\sin x - f(x)) dx = 1 - \sin a + (a - 1)\cos a \)
differentiate w.r.t. a
\( \sin a - f(a) = - \cos a + \cos a - (a - 1) \sin a \)
\( \Rightarrow \sin a - f(a) = - a \sin a + \sin a \)
\( \Rightarrow f(a) = a \sin a \Rightarrow f(x) = x \sin x \)
The points where f(x) & sinx intersect are x sinx = sinx \( \Rightarrow \sin x = 0 \text{ or } x = 1 \).
We can say that a = 1
\( A_1 = \int_{0}^{1} (\sin x - x \sin x) dx = (1 - \sin 1) \)
\( A_2 = \int_{1}^{\pi} (f(x) - \sin x) dx = \int_{1}^{\pi} (x \sin x - \sin x) dx = (\pi - 1 - \sin 1) \)
\( A_3 = \left| \int_{\pi}^{2\pi} (\sin x - x \sin x) dx \right| = (3\pi - 2) \)

Question. Show that the area bounded by the curve \( y = \frac{\ln x - c}{x} \), the x-axis and the vertical line through the maximum point of the curve is independent of the constant c.
Answer: \( y = \frac{\ln x - c}{x} \)
\( \frac{dy}{dx} = 0 \Rightarrow x = e^{c + 1} \)
y = 0 \( \Rightarrow x = e^c \)
Area = \( \int_{e^c}^{e^{c + 1}} \left( \frac{\ln x - c}{x} \right) dx = \frac{1}{2} \)

Question. For what value of 'a' is the area of the figure bounded by the lines, \( y = \frac{1}{x} \), \( y = \frac{1}{2x - 1} \), x = 2 and x = a equal to \( \ln \frac{4}{\sqrt{5}} \)?
Answer: \( A = \int_{1}^{a} \left( \frac{1}{x} - \frac{1}{2x - 1} \right) dx = \ln \frac{4}{\sqrt{5}} \)
a = 8 or \( \frac{2}{5} (6 - \sqrt{2}) \)

Question. Compute the area of the loop of the curve \( y^2 = x^2 \left[ \frac{1 + x}{1 - x} \right] \).
Answer: Area = \( 2 \left| \int_{-1}^{0} x \sqrt{\frac{1 + x}{1 - x}} dx \right| = 2 - \frac{\pi}{2} \)

Question. For the curve \( f(x) = \frac{1}{1 + x^2} \), let two points on it are \( A (\alpha, f(\alpha)) \), \( B \left( - \frac{1}{\alpha}, f \left( - \frac{1}{\alpha} \right) \right) \) (\( \alpha > 0 \)). Find the minimum area bounded by the line segments OA, OB and f(x), where 'O' is the origin.
Answer: \( y = \frac{1}{1 + x^2} \)
\( A = \int_{-1/\alpha}^{0} \left( \frac{1}{1 + x^2} - \left( \frac{\alpha^3}{1 + \alpha^2} \right) x \right) dx + \int_{0}^{\alpha} \left( \frac{1}{1 + x^2} - \frac{1}{\alpha(1 - \alpha^2)} x \right) dx \)
for area to be minimum \( \frac{dA}{d\alpha} = 0 \) & \( A = \frac{\pi - 1}{2} \)

Question. Let 'c' be the constant number such that c > 1. If the least area of the figure given by the line passing through the point (1, c) with gradient 'm' and the parabola \( y = x^2 \) is 36 sq. units find the value of (\( c^2 + m^2 \)).
Answer: Equation of line
y - c = m (x - 1) \( \Rightarrow \) y = mx + c - m ........(i)
solving line & parabola
\( x = \frac{m \pm \sqrt{m^2 - 4m + 4c}}{2} \begin{matrix} \nearrow x_1 \\ \searrow x_2 \end{matrix} (\text{let}) \)
\( A = \left| \int_{x_1}^{x_2} (x^2 - (mx + c - m)) dx \right| = 36 \)

Question. Let \( A_n \) be the area bounded by the curve \( y = (\tan x)^n \) and the lines x = 0, y = 0 and x = \( \pi/4 \). Prove that for n > 2, \( A_n + A_{n - 2} = 1/ (n - 1) \) and deduce that \( 1/(2n + 2) < A_n < 1/(2n - 2) \).
Answer: \( A_n = \int_{0}^{\pi/4} (\tan x)^n dx \)
\( 0 < \tan x < 1 \) when \( 0 < x < \pi/4 \)
\( 0 < (\tan x)^{n + 1} < (\tan x)^n, n \in \mathbb{N} \)
\( \Rightarrow \int_{0}^{\pi/4} (\tan x)^{n + 1} dx < \int_{0}^{\pi/4} (\tan x)^n dx \)
\( \Rightarrow A_{n + 1} < A_n \) for n > 2
\( A_n + A_{n + 2} = \int_{0}^{\pi/4} [(\tan x)^n + (\tan x)^{n + 2}] dx \)
\( = \int_{0}^{\pi/4} (\tan x)^n (1 + \tan^2 x) dx = \frac{1}{n + 1} \)
since \( A_{n + 2} < A_{n + 1} < A_n \)
so \( A_n + A_{n + 2} < 2A_n \Rightarrow \frac{1}{n + 1} < 2A_n \Rightarrow \frac{1}{2n + 2} < A_n \) .....(1)
Also for n > 2 ; \( A_n + A_n < A_n + A_{n - 2} = \frac{1}{n - 1} \Rightarrow 2A_n < \frac{1}{n - 1} \Rightarrow A_n < \frac{1}{2n - 2} \) .....(2)
combining (1) & (2) use get \( \frac{1}{2n + 2} < A_n < \frac{1}{2n - 2} \).

Question. If f(x) is monotonic in (a, b) then prove that the area bounded by the ordinates at x = a; x = b; y = f(x) and y = f(c), \( c \in (a, b) \) is minimum when \( c = \frac{a + b}{2} \). Hence if the area bounded by the graph of \( f(x) = \frac{x^3}{3} - x^2 + a \), the straight lines x = 0, x = 2 and the x-axis is minimum then find the value of 'a'.
Answer: Proof. \( A = \int_{a}^{c} (f(c) - f(x)) dx + \int_{c}^{b} (f(x) - f(c)) dx \)
\( = f(c) [c - a + c - b] - \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx \)
differentiating w.r.t. 'c'
\( \frac{dA}{dc} = f(c).2 + (2c - a - b). f'(c) - f(c) - f(b) . 0 - f(c) \)
\( \Rightarrow (2c - a - b) (f'(c)) = 0 \Rightarrow 2c - a - b = 0 \)
\( c = \frac{a + b}{2} \)
\( f'(x) = x^2 - 2x + a = 0 \)
\( \frac{2 + a}{2} = - a \Rightarrow a = \frac{2}{3} \)

Question. Consider the two curves \( C_1 : y = 1 + \cos x \) and \( C_2 : y = 1 + \cos (x - \alpha) \) for \( \alpha \in (0, \pi/2) \); x \( \in [0, \pi] \). Find the value of \( \alpha \), for which the area of the figure bounded by the curves \( C_1, C_2 \) and x = 0 is same as that of the figure bounded by \( C_2 \), y = 1 and x = \( \pi \). For this value of \( \alpha \), find the ratio in which the line y = 1 divides the area of the figure by the curves \( C_1, C_2 \) and x = \( \pi \).
Answer: \( 1 + \cos x = 1 + \cos (x - \alpha) \Rightarrow x = \frac{\alpha}{2} \)
Now \( \alpha/2 \)
\( \int_{0}^{\alpha/2} ((1 + \cos x) - (1 + \cos(x - \alpha))) dx = - \int_{\frac{\pi}{2} + \alpha}^{\pi} (1 - (1 + \cos(\pi - \alpha))) dx \)
\( \Rightarrow (\sin x - \sin(x - \alpha)) \Big|_{0}^{\alpha/2} = \sin(x - \alpha) \Big|_{\pi}^{\frac{\pi}{2} + \alpha} \)
\( \Rightarrow 2 \sin \frac{\alpha}{2} - \sin \alpha = 1 - \sin \alpha \Rightarrow \alpha = \frac{\pi}{3} \)
Area bounded by \( c_1, c_2 \) & y = 0
\( = \int_{0}^{\pi/6} (c_1 - c_2) dx + \int_{\pi/6}^{\pi} (c_2 - c_1) dx = 2 \) sq. units

Question. Find the whole area included between the curve \( x^2y^2 = a^2(y^2 - x^2) \) and its asymptotes (asymptotes are the lines which meet the curve at infinity).
Answer: \( x^2y^2 = a^2y^2 - a^2x^2 \)
\( y^2 = \frac{a^2x^2}{a^2 - x^2} \)
Area = \( 4 \int_{0}^{a} \frac{ax}{\sqrt{a^2 - x^2}} dx = 4a^2 \)

Question. For what values of a \( \in [0, 1] \) does the area of the figure bounded by the graph of the function y = f(x) and the straight lines x = 0, x = 1 and y = f(a) is at a minimum and for what values it is at a maximum if \( f(x) = \sqrt{1 - x^2} \). Find also the maximum and the minimum areas.
Answer: \( A = \int_{0}^{\sqrt{1 - a^2}} (\sqrt{1 - y^2}) dy \)
\( \frac{dA}{da} = 0, a = \frac{1}{2} \)
so \( A_{min} = \frac{3\sqrt{3} - \pi}{12} \) & maxima at a = 0.

Question. Let \( C_1 \) and \( C_2 \) be two curves passing through the origin as shown in the figure. A curve C is said to "bisect the area" the region between \( C_1 \) and \( C_2 \), if for each point P of C, the two shaded regions A and B shown in the figure have equal areas. Determine the upper curve \( C_2 \), given that the bisecting curve C has the equation \( y = x^2 \) and that the lower curve \( C_1 \) has the equation \( y = x^2/2 \).
Answer: Let a point P(t, \( t^2 \)) is on the curve c.
y cordinate of Q is also \( t^2 \) & R = (t, \( t^2/2 \)) & Q = (x, \( t^2 \))
Area of OPQ = Ara of OPR
\( \Rightarrow \int_{0}^{t^2} (\sqrt{y} - f(y)) dy = \int_{0}^{t} (x^2 - x^2/2) dx \)
differntiating both sides w.r.t. t
\( f(t) = \frac{16}{9} t^2 \Rightarrow f(x) = \frac{16}{9} x^2 \)

Question. Given \( f(x) = \int_{0}^{x} e^t (\ln\sec t - \sec^2 t) dt \); \( g(x) = -2e^x \tan x \). Find the area bounded by the curves y = f(x) and y = g(x) between the ordinates x = 0 and x = \( \pi/3 \).
Answer: \( f(x) = \int_{0}^{x} e^t (\ln \sec t - \sec^2 t) dt \)
\( = \left[ e^t \ln \sec t \right]_{0}^{x} - \int_{0}^{x} e^t \tan t dt - \int_{0}^{x} e^t \sec^2 t dt \)
\( = e^x \ln \sec x - \int_{0}^{x} e^t (\tan t + \sec^2 t) dt \)
\( = e^x \ln \sec x - e^x \tan x \)
Area = \( \int_{0}^{\pi/3} (f(x) - g(x)) dx \)
\( = \int_{0}^{\pi/3} (e^x \ln \sec x + e^x \tan x) dx \)
\( = \left[ e^x \ln \sec x \right]_{0}^{\pi/3} \)
\( = e^{\pi/3} \ln 2 \)

Subjective Questions

Question. Find the area bounded on the right by the line x + y = 2, on the left by the parabola \( y = x^2 \) and below by the x-axis.
Answer: \( A = \int_{0}^{1} x^2 dx + \frac{1}{2} = \frac{5}{6} \)

Question. Find the value of c for which the area of the figure bounded by the curves y = sin 2x, the straight lines x = \( \pi/6 \), x = c and the abscissa axis is equal to 1/2.
Answer: \( \int_{\pi/6}^{c} \sin 2x dx = \frac{1}{2} \)
on solving c = \( -\frac{\pi}{6} \) or \( \frac{\pi}{3} \)

Question. The tangent to the parabola \( y = x^2 \) has been drawn so that the abscissa \( x_0 \) of the point of tangency belongs to the interval [1, 2]. Find \( x_0 \) for which the triangle bounded by the tangent, the axis of ordinates and the straight line \( y = x_0^2 \) has the greatest area.
Answer: Equation of tangent at \( (x_0, x_0^2) \)
\( 2x x_0 - y = x_0^2 \) .......(i)
line : \( y = x_0^2 \) .......(ii)
axis of ordinates y-axis .......(iii)
on solving (i) & (ii) & (i) & (iii)
area of triangle
\( A = x_0^3 \)
It will be greatest for \( x_0 = 2 \)
so \( A = 8 \)

Question. Compute the area of the region bounded by the curves y = e.x. \( \ln x \) and \( y = \ln x / (e.x) \) where \( \ln e = 1 \).
Answer: Intersecting points are \( \frac{1}{e} \) & 1
so \( A = \int_{1/e}^{1} \left( \frac{\ln x}{ex} - ex \ln x \right) dx = \frac{e^2 - 5}{4e} \) sq. units.

Question. A figure is bounded by the curves \( y = \left| \sqrt{2} \sin \frac{\pi x}{4} \right| \), \( y = 0 \), \( x = 2 \) and \( x = 4 \). At what angles to the positive x-axis straight lines must be drawn through (4, 0) so that these lines partition the figure into three parts of the same size.
Answer: \( A = \left| \int_{2}^{4} \sqrt{2} \sin \frac{\pi x}{4} dx \right| = \frac{4\sqrt{2}}{\pi} \) sq. units.
let the line is \( y = m(x - 4) \)
\( A = \left( \frac{4\sqrt{2}}{\pi} \right) \times \frac{1}{3} = \int_{2}^{4} \left( \sqrt{2} \sin \frac{\pi x}{4} - mx + 4m \right) dx \)
similarly for \( m_2 \)

Question. Find the area of the region bounded by the curves, \( y = \log_e x \), \( y = \sin^4 \pi x \) and x = 0.
Answer: [Graph provided in the detailed solution]

Question. Find the area bounded by the curves \( y = \sqrt{1 - x^2} \) and \( y = x^3 - x \). Also find the ratio in which the y-axis divided this area.
Answer: \( A = \left| \int_{-1}^{1} \left( \sqrt{1 - x^2} - x^3 + x \right) dx \right| \)

Question. If the area enclosed by the parabolas \( y = a - x^2 \) and \( y = x^2 \) is \( 18\sqrt{2} \) sq. units. Find the value of 'a'.
Answer: \( A = 2 \int_{0}^{\sqrt{a/2}} (2x^2 - a) dx = 18\sqrt{2} \)
\( \Rightarrow a^{3/2} = 9 \times 3 \Rightarrow a = 9 \)

Question. The line 3x + 2y = 13 divides the area enclosed by the curve, \( 9x^2 + 4y^2 - 18x - 16y - 11 = 0 \) into two parts. Find the ratio of the larger area to the smaller area.
Answer: point of intersection of line & curve = (3, 2) & (1, 5).

Question. Find the area of the region enclosed by the curve \( y = x^4 - 2x^2 \) and \( y = 2x^2 \).
Answer: Intersecting points = 0, \( \pm 2 \)
so area = \( 2 \int_{0}^{2} (2x^2 - x^4 + 2x^2) dx = \frac{128}{15} \)

Question. Find the values of m (m > 0) for which the area bounded by the line y = mx + 2 and \( x = 2y - y^2 \) is, (i) 9/2 square units and (ii) minimum. Also find the minimum area.
Answer: (a) Area = \( \frac{1}{2} \cdot 2 \times \left( \frac{-1 - 2m}{m^2} \right) + \int_{0}^{1} (2y - y^2) dy = \frac{3}{2} \)
(b) If area is minimum
\( \frac{dA}{dm} = 0 \)
\( \Rightarrow m = \infty \)
& \( A_{min} = \frac{4}{3} \).

Question. Consider two curves \( C_1 : y = \frac{1}{x} \) and \( C_2 : y = \ln x \) on the xy plane. Let \( D_1 \) denotes the region surrounded by \( C_1, C_2 \) and the line x = 1 and \( D_2 \) denotes the region surrounded by \( C_1, C_2 \) and the line x = a. If \( D_1 = D_2 \). Find the value of 'a'.
Answer: [Graph provided in the detailed solution]

Question. Find the area enclosed between the curves : \( y = \log_e (x + e) \), \( x = \log_e (1/y) \) and the x-axis.
Answer: \( x = \log_e (1/y) \Rightarrow y = e^{-x} \)
\( A = \int_{1-e}^{0} \log (x + e) dx + \int_{0}^{\infty} e^{-x} dx = 2 \)

Question. Find the value (s) of the parameter 'a' (a > 0) for each of which the area of the figure bounded by the straight line, \( y = \frac{a^2 - ax}{1 + a^4} \) and the parabola \( y = \frac{x^2 + 2ax + 3a^2}{1 + a^4} \) is the greatest.
Answer: Intersecting points of the curve is \( x = -a \) & \( x = -2a \)
so \( A = \int_{-2a}^{-a} \left( \frac{a^2 - ax}{1 + a^4} - \left( \frac{x^2 + 2ax + 3a^2}{1 + a^4} \right) \right) dx \)
\( \frac{dA}{da} = 0 \Rightarrow a = 3^{1/4} \)

Question. For what value of 'a' is the area bounded by the curve \( y = a^2x^2 + ax + 1 \) and the straight line y = 0, x = 0 and x = 1 the least ?
Answer: \( a^2x^2 + ax + 1 \) is clearly positive
\( A = \int_{0}^{1} (a^2 x^2 + ax + 1) dx = \frac{1}{6} \left( 2\left(a + \frac{3}{4}\right)^2 + \frac{39}{8} \right) \)
minimum for \( a = -3/4 \).

Question. Find the positive value of 'a' for which the parabola \( y = x^2 + 1 \) bisects the area of the rectangle with vertices (0, 0), (a, 0), (0, a^2 + 1) and (a, a^2 + 1).
Answer: Area of rectangle = \( a \cdot (a^2 + 1) \)
now area of shaded region.
now \( \int_{1}^{a^2+1} \sqrt{y - 1} dy = \frac{1}{2} [a \cdot (a^2 + 1)] \)
\( \Rightarrow a = \sqrt{3} \)

Question. Compute the area of the curvilinear triangle bounded by the y-axis and the curve, \( y = \tan x \) and \( y = (2/3 \cos x) \).
Answer: on solving \( \tan x = \frac{2}{3} \cos x \)
\( 3 \tan x = 2 \cos x \Rightarrow x = \frac{\pi}{6} \)
so area is = \( \int_{0}^{\pi/6} \left( \frac{2}{3} \cos x - \tan x \right) dx \)

Question. Find the area bounded by the curve \( y = x e^{-x} \); xy = 0 and x = c where c is the x-coordinate of the curve's inflection point.
Answer: \( y = xe^{-x} \)
\( \frac{dy}{dx} = -xe^{-x} + e^{-x} = e^{-x}(1 - x) \)
\( \frac{d^2y}{dx^2} = e^{-x}(-1) + (1 - x)(-e^{-x}) = e^{-x}(x - 2) \)
\( e^{-x}(x - 2) = 0 \Rightarrow x = 2 \quad (\because e^{-x} \neq 0) \)
so c = 2
\( A = \int_{0}^{2} (xe^{-x}) dx = -xe^{-x} \Big|_0^2 + \int_{0}^{2} e^{-x} dx = 1 - 3e^{-2} \) sq. units

Question. Find the value of 'c' for which the area of the figure bounded by the curve, \( y = 8x^2 - x^5 \), the straight lines x = 1 and x = c and the abscissa axis is equal to 16/3.
Answer: \( \int_{1}^{c} (8x^2 - x^5) dx = \frac{16}{3} \)

Question. Compute the area included between the straight lines, x - 3y + 5 = 0; x + 2y + 5 = 0 and the circle \( x^2 + y^2 = 25 \).
Answer: Total area = area of \( \Delta \) + area of sector

Question. Find the area bounded by the curve \( y = x e^{-x^2} \), the x-axis and the line x = c where y(c) is maximum.
Answer: \( y = xe^{-x^2} \)
\( y' = e^{-x^2} - 2x^2 e^{-x^2} = 0 \Rightarrow x = \pm \frac{1}{\sqrt{2}} \)
y is maximum at \( x = \frac{1}{\sqrt{2}} \)
Area = \( \int_{0}^{1/\sqrt{2}} x e^{-x^2} dx = \frac{1}{2} (1 - e^{-1/2}) \)

JEE Problems

Question. For which of the following values of m, is the area of the region bounded by the curve \( y = x - x^2 \) and the line \( y = mx \) equals 9/2 ? 
(a) \( -4 \)
(b) \( -2 \)
(c) 2
(d) 4
Answer: (b) \( -2 \), (d) 4
Solution:
\( y = x - x^2 \); \( y = mx \)
\( mx = x - x^2 \)
\( x^2 = x(1 - m) \) or \( x = 0, 1 - m \)
\( \int (y_1 - y_2) dx = \int (x - x^2 - mx) dx \)
\( = \left[ (1 - m) \frac{x^2}{2} - \frac{x^3}{3} \right]_0^{1-m} = \frac{9}{2} \)
If \( m < 1 \), \( (1 - m)^3 \left[ \frac{1}{2} - \frac{1}{3} \right] = \frac{9}{2} \)
\( (1 - m)^3 = 27 \Rightarrow m = -2 \)
If \( m > 1 \), then \( 1 - m \) will be negative
\( \left[ (1 - m) \frac{x^2}{2} - \frac{x^3}{3} \right]_{1-m}^0 = \frac{9}{2} \)
\( \Rightarrow (1 - m)^3 = -27 \)
\( 1 - m = -3 \Rightarrow m = 4 \)

Question. If the area bounded by \( y = ax^2 \) and \( x = ay^2 \), \( a > 0 \), is 1, then a equals 
(a) 1
(b) \( \frac{1}{\sqrt{3}} \)
(c) \( \frac{1}{3} \)
(d) \( -\frac{1}{\sqrt{3}} \)
Answer: (b) \( \frac{1}{\sqrt{3}} \)
Solution:
Point of intersections of \( y = ax^2 \) & \( x = ay^2 \) are \( (0, 0) \) & \( \left( \frac{1}{a}, \frac{1}{a} \right) \)
Hence \( \int_0^{1/a} \left( \sqrt{\frac{x}{a}} - ax^2 \right) dx = 1 \Rightarrow a = \frac{1}{\sqrt{3}} \) (as \( a > 0 \))

Question. Match the following 
(i) \( \int_0^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \ln(\sin x)^{\sin x}) dx \)      (A) 1
(ii) Area bounded by \( -4y^2 = x \) and \( x - 1 = -5y^2 \)      (B) 0
(iii) Cosine of the angle of intersection of curves \( y = 3^{x-1} \ln x \) and \( y = x^x - 1 \) is      (C) \( 6\ln 2 \)
                                                                                    (D) 4/3
Answer: (i) - (A), (ii) - (D), (iii) - (A)
Solution:
(i) \( I = \int_0^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \sin x \log(\sin x)) dx \)
\( = \int_0^{\pi/2} d((\sin x)^{\cos x}) = (\sin x)^{\cos x} \Big|_0^{\pi/2} = 1 \)
(ii) Point of intersection of \( -4y^2 = x \) and \( x - 1 = -5y^2 \) is (-4, -1) and (-4, 1)
Hence reqd. area
\( = 2 \left[ \int_0^1 (1 - 5y^2) dy - \int_0^1 (-4y^2) dy \right] = \frac{4}{3} \)
(iii) Point of intersection of \( y = 3^{x-1} \log x \) and \( y = x^x - 1 \) is (1, 0)
Hence \( \frac{dy}{dx} = \frac{3^{x-1}}{x} + 3^{x-1} \log 3 . \log x \)
\( \frac{dy}{dx}\Big|_{(1,0)} = 1 \)
\( y = x^x - 1 \Rightarrow \frac{dy}{dx}\Big|_{(1,0)} = 1 \)
If \( \theta \) is the angle b/w the curves then \( \tan \theta = 0 \)
\( \Rightarrow \cos \theta = 1 \)

Question. The area of the region between the curves \( y = \sqrt{\frac{1 + \sin x}{\cos x}} \) and \( y = \sqrt{\frac{1 - \sin x}{\cos x}} \) bounded by the lines \( x = 0 \) and \( x = \frac{\pi}{4} \) is 
(a) \( \int_0^{\sqrt{2}-1} \frac{t}{(1 + t^2)\sqrt{1 - t^2}} dt \)
(b) \( \int_0^{\sqrt{2}-1} \frac{4t}{(1 + t^2)\sqrt{1 - t^2}} dt \)
(c) \( \int_0^{\sqrt{2}+1} \frac{4t}{(1 + t^2)\sqrt{1 - t^2}} dt \)
(d) \( \int_0^{\sqrt{2}+1} \frac{t}{(1 + t^2)\sqrt{1 - t^2}} dt \)
Answer: (b) \( \int_0^{\sqrt{2}-1} \frac{4t}{(1 + t^2)\sqrt{1 - t^2}} dt \)
Solution:
\( I = \int_0^{\pi/4} \left( \sqrt{\frac{1 + \sin x}{\cos x}} - \sqrt{\frac{1 - \sin x}{\cos x}} \right) dx \)
\( = \int_0^{\pi/4} \left( \sqrt{\frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}}} - \sqrt{\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}}} \right) dx \)
\( = \int_0^{\pi/4} \frac{\left( 1 + \tan \frac{x}{2} \right) - \left( 1 - \tan \frac{x}{2} \right)}{\sqrt{1 - \tan^2 \frac{x}{2}}} dx \)
\( = \int_0^{\pi/4} \frac{2 \tan \frac{x}{2}}{\sqrt{1 - \tan^2 \frac{x}{2}}} dx = \int_0^{\sqrt{2}-1} \frac{4t}{(1 + t^2)\sqrt{1 - t^2}} dt \)
as \( \tan \frac{x}{2} = t \)

Comprehension (3 questions together) :
Consider the functions defined implicitly by the equation \( y^3 - 3y + x = 0 \) on various intervals in the real line. If \( x \in (-\infty, -2) \cup (2, \infty) \), the equation implicitly defines a unique real valued differentiable function \( y = f(x) \). If \( x \in (-2, 2) \), the equation implicitly defines a unique real valued differentiable function \( y = g(x) \) satisfying \( g(0) = 0 \).

Question. If \( f(-10\sqrt{2}) = 2\sqrt{2} \), then \( f''(-10\sqrt{2}) = \)
(a) \( \frac{4\sqrt{2}}{7^3 3^2} \)
(b) \( -\frac{4\sqrt{2}}{7^3 3^2} \)
(c) \( \frac{4\sqrt{2}}{7^3 3} \)
(d) \( -\frac{4\sqrt{2}}{7^3 3} \)
Answer: (b) \( -\frac{4\sqrt{2}}{7^3 3^2} \)
Solution:
Differentiating the given equation, we get
\( 3y^2 y' - 3y' + 1 = 0 \)
\( y'(-10\sqrt{2}) = -\frac{1}{21} \)
Differentiation again we get
\( 6y(y')^2 + 3y^2 y'' - 3y'' = 0 \)
\( f''(-10\sqrt{2}) = -\frac{12\sqrt{2}}{(21)^3} = -\frac{4\sqrt{2}}{7^3 3^2} \)

Question. The area of the region bounded by the curve \( y = f(x) \), the x-axis and the lines \( x = a \) and \( x = b \), where \( -\infty < a < b < -2 \), is
(a) \( \int_a^b \frac{x}{3((f(x))^2 - 1)} dx + bf(b) - af(a) \)
(b) \( -\int_a^b \frac{x}{3((f(x))^2 - 1)} dx + bf(b) - af(a) \)
(c) \( \int_a^b \frac{x}{3((f(x))^2 - 1)} dx - bf(b) + af(a) \)
(d) \( -\int_a^b \frac{x}{3((f(x))^2 - 1)} dx - bf(b) + af(a) \)
Answer: (a) \( \int_a^b \frac{x}{3((f(x))^2 - 1)} dx + bf(b) - af(a) \)
Solution:
The reqd. area
\( = \int_a^b f(x) dx = x f(x) \Big|_a^b - \int_a^b x f'(x) dx \)
\( = b f(b) - a f(a) + \int_a^b \frac{x}{3[f(x)^2 - 1]} dx \)

Question. \( \int_{-1}^1 g'(x) dx \) equals
(a) \( 2g(-1) \)
(b) 0
(c) \( -2g(1) \)
(d) \( 2g(1) \)
Answer: (d) \( 2g(1) \)
Solution:
We have \( y' = \frac{1}{3[1 - (f(x))^2]} \) which is even
Hence \( \int_{-1}^1 g'(x) dx = g(1) - g(-1) = 2g(1) \)

Question. Let \( f : [-1, 2] \rightarrow [0, \infty) \) be a continuous function such that \( f(x) = f(1 - x) \) for all \( x \in [-1, 2] \). Let \( R_1 = \int_{-1}^2 x f(x) dx \), and \( R_2 \) be the area of the region bounded by \( y = f(x) \), \( x = -1 \), \( x = 2 \), and the x-axis. Then 
(a) \( R_1 = 2R_2 \)
(b) \( R_1 = 3R_2 \)
(c) \( 2R_1 = R_2 \)
(d) \( 3R_1 = R_2 \)
Answer: (c) \( 2R_1 = R_2 \)

Solution:
\( R_1 = \frac{1}{2} \int_{-1}^2 f(x) dx \)
\( R_2 = \int_{-1}^2 f(x) dx \Rightarrow R_2 = 2R_1 \)