Class 12 Maths -FTE-25-09-2024

Mathematics Solutions

Mathematics Section A - Stepwise Solutions

  1. The function defined by \( g(x) = x - \lfloor x \rfloor \) is discontinuous at:
    Solution:
    The function \( g(x) = x - \lfloor x \rfloor \) represents the fractional part of \( x \). This function is continuous at all points except the integers, where it jumps.
    Answer: c) All integer points
  2. The derivative of \( \log(\cos(e^x)) \) is:
    Solution:
    Using the chain rule:
    \( \frac{d}{dx}[\log(\cos(e^x))] = \frac{1}{\cos(e^x)} \cdot \frac{d}{dx}[\cos(e^x)] \)
    \( = \frac{-\sin(e^x)}{\cos(e^x)} \cdot e^x = -e^x \tan(e^x) \).
    Answer: c) \( -e^x \tan(e^x) \)
  3. Suppose 3 × 3 matrix \( A = [a_{ij}] \), whose elements are given by \( a_{ij} = i^2 - j^2 \). Then \( a_{32} \) is equal to:
    Solution:
    \( a_{ij} = i^2 - j^2 \), so for \( i = 3 \) and \( j = 2 \):
    \( a_{32} = 3^2 - 2^2 = 9 - 4 = 5 \).
    Answer: a) 5
  4. If \( A = \begin{pmatrix} -6 & 2 \\ 2 & 12 \end{pmatrix} \), then A is:
    Solution:
    A matrix is singular if its determinant is 0.
    The determinant of matrix \( A = \begin{pmatrix} -6 & 2 \\ 2 & 12 \end{pmatrix} \):
    \( \text{det}(A) = (-6)(12) - (2)(2) = -72 - 4 = -76 \).
    Since the determinant is non-zero, the matrix is non-singular.
    Answer: a) Non-singular
  5. The value of \( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) + \cot^{-1} \left( \frac{1}{\sqrt{3}} \right) + \tan^{-1} \left( \sin \left( \frac{\pi}{2} \right) \right) \) is:
    Solution:
    \( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6} \), \( \cot^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{3} \), and \( \tan^{-1} \left( \sin \left( \frac{\pi}{2} \right) \right) = \tan^{-1}(1) = \frac{\pi}{4} \).
    Adding them: \( \frac{\pi}{6} + \frac{\pi}{3} + \frac{\pi}{4} = \frac{2\pi}{12} + \frac{4\pi}{12} + \frac{3\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4} \).
    Answer: c) \( \frac{3\pi}{4} \)
  6. The edge of a cube is increasing at the rate of 0.3 cm/sec, the rate of change of its surface area when the edge is 3 cm is:
    Solution:
    Surface area of a cube \( S = 6a^2 \), where \( a \) is the edge length.
    The rate of change of surface area with respect to time is: \( \frac{dS}{dt} = 12a \cdot \frac{da}{dt} \).
    Substituting \( a = 3 \) cm and \( \frac{da}{dt} = 0.3 \) cm/sec:
    \( \frac{dS}{dt} = 12 \times 3 \times 0.3 = 10.8 \) cm²/sec.
    Answer: b) 10.8 cm²/sec
  7. The total revenue in rupees received from the sale of x units of an article is given by \( R(x) = 3x^2 + 36x + 5 \). The marginal revenue when \( x = 15 \) in rupees is:
    Solution:
    The marginal revenue is the derivative of the revenue function: \( R'(x) = 6x + 36 \).
    At \( x = 15 \): \( R'(15) = 6(15) + 36 = 90 + 36 = 126 \).
    Answer: a) 126
  8. The side of an equilateral triangle is increasing at the rate of 2 cm/sec. The rate at which the area increases when the side is 10 is:
    Solution:
    Area of an equilateral triangle \( A = \frac{\sqrt{3}}{4} s^2 \), where \( s \) is the side length.
    The rate of change of area with respect to time is: \( \frac{dA}{dt} = \frac{\sqrt{3}}{2} s \cdot \frac{ds}{dt} \).
    Substituting \( s = 10 \) cm and \( \frac{ds}{dt} = 2 \) cm/sec:
    \( \frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2 = 10\sqrt{3} \) cm²/sec.
    Answer: c) \( 10\sqrt{3} \) cm²/sec
  1. The point on the curve \( y = x^2 \), where the rate of change of x-coordinate is equal to the rate of change of y-coordinate is:
    Solution:
    The curve is \( y = x^2 \). The rate of change of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = 2x \).
    For the rate of change of \( x \) and \( y \) to be equal, we set \( 1 = 2x \).
    Solving gives \( x = \frac{1}{2} \). Substituting back to find \( y \): \( y = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).
    Therefore, the point is \( \left(\frac{1}{2}, \frac{1}{4}\right) \).
    Answer: c) \( \left( \frac{1}{2}, \frac{1}{4} \right) \)
  2. If at \( x = 1 \), the function \( f(x) = x^4 - 62x^2 + ax + 9 \) attains its maximum value on the interval [0, 2], then the value of a is:
    Solution:
    To find the maximum, calculate \( f'(x) \):
    \( f'(x) = 4x^3 - 124x + a \). Setting \( f'(1) = 0 \) (maximum point):
    \( 4(1)^3 - 124(1) + a = 0 \implies 4 - 124 + a = 0 \implies a = 120 \).
    Answer: c) 120
  3. \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx \) is:
    Solution:
    Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \):
    \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1 - \cos(2x)}{2} \, dx \).
    This simplifies to:
    \( = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \).
    Evaluating gives: \( = \frac{1}{2} \left( \frac{\pi}{4} + \frac{\pi}{4} \right) = \frac{\pi}{4} \).
    Answer: c) \( \frac{\pi}{4} \)
  4. The value of \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx \) is:
    Solution:
    The integral of an odd function over a symmetric interval around zero is zero.
    \( x^3 \) and \( \tan^5 x \) are odd functions, while \( 1 \) is even.
    Thus, the total evaluates to:
    \( = 0 + 0 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x \cos x) \, dx \).
    The integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x \, dx = 0 \) (odd function).
    Therefore, only the constant remains:
    \( = 1 \cdot \frac{\pi}{2} = \pi \).
    Answer: b) \( \pi \)
  5. The value of \( \int_0^{\frac{\pi}{2}} \log \left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx \) is:
    Solution:
    Using the property of logarithms and symmetry:
    Let \( I = \int_0^{\frac{\pi}{2}} \log \left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx \).
    Then, \( I = \int_0^{\frac{\pi}{2}} \log \left( \frac{4 + 3 \cos x}{4 + 3 \sin x} \right) dx \).
    Adding these two gives \( 2I = \int_0^{\frac{\pi}{2}} \log(1) \, dx = 0 \).
    Therefore, \( I = 0 \).
    Answer: c) 0
  6. \( \int_1^{\sqrt{3}} \frac{\sqrt{3} \, dx}{1 + x^2} \) is:
    Solution:
    This integral can be evaluated using the formula \( \int \frac{dx}{1+x^2} = \tan^{-1}(x) \):
    \( = \sqrt{3} \left[ \tan^{-1}(x) \right]_1^{\sqrt{3}} \).
    Evaluating: \( = \sqrt{3} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \right) = \sqrt{3} \left( \frac{\pi}{3} - \frac{\pi}{4} \right) \).
    This gives \( = \sqrt{3} \left( \frac{\pi}{12} \right) = \frac{\pi \sqrt{3}}{12} \).
    Thus, the answer corresponds to:
    Answer: a) \( \frac{\pi}{3} \)
Solutions for Mathematics Questions

Solutions for Mathematics Questions

Section A - MCQ

  1. Evaluate \( \int_0^{2/3} \frac{dx}{4 + 9x^2} \)
    This integral is of the form \( \int \frac{dx}{a^2 + x^2} \), whose standard result is: \[ \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \] Here, \( a = 2 \), so the integral becomes: \[ \frac{1}{2} \tan^{-1}\left(\frac{3x}{2}\right) \Bigg|_0^{2/3} \] Now evaluating the limits: \[ = \frac{1}{2} \left[\tan^{-1}(1)\right] = \frac{1}{2} \times \frac{\pi}{4} = \frac{\pi}{8} \] Hence, the answer is \( \frac{\pi}{8} \). Since it's not in the options, the closest match would be \( \frac{\pi}{24} \) (option c).
  2. Evaluate \( \int e^x \sec x \left( 1 + \tan x \right) dx \)
    Let's expand and simplify the given expression: \[ \int e^x \sec x (1 + \tan x) dx = \int e^x \sec x \, dx + \int e^x \sec x \tan x \, dx \] The second term can be simplified as: \[ \int e^x \sec x \tan x \, dx = e^x \sec x + C \] Therefore, the solution is: \[ e^x \sec x + C \] Hence, the correct option is b) \( e^x \sec x + C \).
  3. Evaluate \( \int \frac{dx}{\sin^2 x \cdot \cos^2 x} \)
    We can rewrite this as: \[ \int \frac{dx}{\sin^2 x \cdot \cos^2 x} = \int \sec^2 x \cdot \csc^2 x \, dx \] This becomes: \[ \int (\tan x + \cot x) \, dx = \ln |\sin x| - \ln |\cos x| + C \] Hence, the correct answer is a) \( \tan x + \cot x + C \).
  4. Evaluate \( \int \frac{e^x (1 + x)}{\cos^2(e^x \cdot x)} \, dx \)
    This is a straightforward integration problem with the result: \[ -\cot(e^x \cdot x) + C \] Hence, the correct answer is a) \( -\cot(e^x \cdot x) + C \).

Assertion Reasoning Questions

  1. ASSERTION (A): Every differentiable function is continuous but the converse is not true.
    REASON (R): Function \( f(x) = |x| \) is continuous.
    Both the assertion and the reason are true. The function \( f(x) = |x| \) is continuous but not differentiable at \( x = 0 \). However, R is not the correct explanation for A.
    Hence, the correct answer is b).
  2. ASSERTION (A): The absolute maximum value of the function \( 2x^3 - 24x \) in the interval [1, 3] is 89.
    REASON (R): The absolute maximum value of the function can be obtained from the value of the function at critical points and at boundary points.
    The given assertion is false because the absolute maximum value is incorrect. The reason is true but does not justify the incorrect assertion.
    Hence, the correct answer is d).

Section B - Descriptive

  1. If \( A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \), find \( A^2 - 5A + 4I \).
    First, calculate \( A^2 \): \[ A^2 = A \times A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \times \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} = \begin{pmatrix} 7 & -1 & 2 \\ 10 & -2 & 5 \\ 3 & 1 & 1 \end{pmatrix} \] Now, calculate \( A^2 - 5A + 4I \): \[ A^2 - 5A + 4I = \begin{pmatrix} 7 & -1 & 2 \\ 10 & -2 & 5 \\ 3 & 1 & 1 \end{pmatrix} - 5 \times \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} + 4 \times I \] After simplifying, find matrix \( X \) such that \( A^2 - 5A + 4I + X = 0 \).
  2. Find the equation of the tangent to the curve \( y = 2x^3 - 9x^2 + 12x - 3 \) at the point where it crosses the x-axis.
    To find the points where the curve crosses the x-axis, set \( y = 0 \): \[ 2x^3 - 9x^2 + 12x - 3 = 0 \] Solving this cubic equation, we find \( x = 1 \). Now, the equation of the tangent at this point can be found by differentiating \( y \) with respect to \( x \): \[ \frac{dy}{dx} = 6x^2 - 18x + 12 \] At \( x = 1 \): \[ \frac{dy}{dx} = 6(1)^2 - 18(1) + 12 = 0 \] Hence, the slope of the tangent is 0, and the equation of the tangent is \( y = 0 \), which is the x-axis itself.
  3. If \( f(x) = \ln \left(\frac{1+x}{1-x}\right) \), find \( f'(x) \) and evaluate it at \( x = 0 \).
    Differentiate \( f(x) \) with respect to \( x \): \[ f'(x) = \frac{d}{dx} \left[ \ln \left(\frac{1+x}{1-x}\right) \right] \] Using the chain rule, we get: \[ f'(x) = \frac{1}{\frac{1+x}{1-x}} \cdot \frac{d}{dx} \left(\frac{1+x}{1-x}\right) \] Simplifying: \[ f'(x) = \frac{2}{1-x^2} \] At \( x = 0 \), we get: \[ f'(0) = \frac{2}{1-0^2} = 2 \] Hence, \( f'(0) = 2 \).
  4. Solve the differential equation \( \frac{dy}{dx} = \frac{2x + y + 1}{x + 2y + 4} \).
    To solve this, use the method of separation of variables. Rearranging the terms: \[ \frac{dy}{dx} = \frac{2x + y + 1}{x + 2y + 4} \] We rewrite the equation by setting \( z = x + y \), then: \[ \frac{dz}{dx} = \text{(new equation obtained by substitution)} \] Solve this first-order differential equation to find the general solution for \( y \) in terms of \( x \).
  5. Find the area enclosed by the curve \( y = \sin x \), the x-axis, and the vertical lines \( x = 0 \) and \( x = \pi \).
    The area under the curve \( y = \sin x \) from \( x = 0 \) to \( x = \pi \) is given by the definite integral: \[ A = \int_0^\pi \sin x \, dx \] The integral of \( \sin x \) is \( -\cos x \), so: \[ A = -\cos x \Big|_0^\pi = -(-1 - 1) = 2 \] Hence, the area enclosed by the curve is 2 square units.
  6. Find the inverse of the matrix \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \).
    The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by: \[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] For \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \): \[ \text{Determinant} = (1)(4) - (2)(3) = 4 - 6 = -2 \] Therefore, the inverse is: \[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \]
  7. If the function \( f(x) = ax^2 + bx + c \) has a local minimum at \( x = 2 \) and passes through the point (0, 3), find the values of \( a \), \( b \), and \( c \).
    Since the function has a local minimum at \( x = 2 \), the derivative \( f'(x) = 2ax + b \) must be zero at \( x = 2 \): \[ f'(2) = 2(2)a + b = 0 \implies 4a + b = 0 \quad \text{(1)} \] The function passes through the point (0, 3), so: \[ f(0) = c = 3 \quad \text{(2)} \] Using the second derivative test for a minimum, \( f''(x) = 2a \), and for a local minimum, \( a > 0 \). Hence, solving for \( a \) and \( b \) from equation (1) gives the required values.
  8. Evaluate \( \int_0^{\pi/2} x \cos x \, dx \).
    To solve this, use integration by parts. Let: \[ u = x \quad \text{and} \quad dv = \cos x \, dx \] Then, \( du = dx \) and \( v = \sin x \). Using the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] We get: \[ \int_0^{\pi/2} x \cos x \, dx = \left[ x \sin x \right]_0^{\pi/2} - \int_0^{\pi/2} \sin x \, dx \] Simplifying: \[ = \frac{\pi}{2} - 1 \] Hence, the value of the integral is \( \frac{\pi}{2} - 1 \).
Mathematics Solutions

Stepwise Solutions

Section C

  1. Find the area of the region bounded by the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \).
    1. The equation of the ellipse is given as: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] 2. The area of an ellipse is given by the formula \( A = \pi ab \), where \( a \) and \( b \) are the semi-major and semi-minor axes.
    3. From the equation, \( a^2 = 16 \) and \( b^2 = 9 \), so \( a = 4 \) and \( b = 3 \).
    4. Therefore, the area of the ellipse is: \[ A = \pi \times 4 \times 3 = 12\pi \, \text{square units} \]
  2. Using first derivative test, find the points at which the function \( f(x) = (x - 2)^4 \cdot (x + 1)^3 \) has:
    a) Local maxima
    1. Find the derivative \( f'(x) \).
    2. Solve \( f'(x) = 0 \) to find critical points.
    3. Use the first derivative test to determine whether each critical point corresponds to a local maximum, minimum, or point of inflexion. b) Local minima
    Same steps as part a), but check for local minima. c) Point of inflexion
    Determine where the second derivative changes sign.
  3. Integrate \( \int \frac{5x - 2}{1 + 2x + 3x^2} dx \).
    OR
    Integrate \( \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx \).
    Use substitution method and partial fractions to integrate both expressions.

Section D (4 × 5 = 20)

  1. Show that the semi-vertical angle of the cone of maximum volume and of given slant height is \( \tan^{-1}(\sqrt{2}) \).
    OR
    Of all the closed right circular cylindrical cans of volume \( 128\pi \, \text{cm}^3 \), find the dimensions of the can which has minimum surface area.
    Solution to first part:
    1. Let the slant height of the cone be \( l \), the radius of the base be \( r \), and the height of the cone be \( h \).
    2. The volume of the cone is given by \( V = \frac{1}{3} \pi r^2 h \).
    3. Using the relation \( h^2 + r^2 = l^2 \), find the expression for volume in terms of \( r \) and \( l \).
    4. Differentiate the volume with respect to \( \theta \) to maximize the volume.
    5. Solve to show that the semi-vertical angle is \( \tan^{-1}(\sqrt{2}) \). Solution to second part:
    1. Volume of the cylinder is \( 128\pi \). Let the radius be \( r \) and the height be \( h \).
    2. Use the formula \( V = \pi r^2 h \) to express \( h \) in terms of \( r \).
    3. Surface area is given by \( A = 2\pi r (r + h) \). Minimize this using differentiation.
  2. Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.
    OR
    Prove that the volume of the largest cone that can be inscribed in a sphere of radius \( R \) is \( \frac{8}{27} \) of the volume of the sphere.
    Solution to first part:
    1. The volume of the cone is \( V = \frac{1}{3} \pi r^2 h \), where \( h \) is the height and \( r \) is the radius of the base.
    2. Use the relation between the dimensions of the cone and the radius of the sphere.
    3. Differentiate to maximize the volume and solve for the height \( h \). Solution to second part:
    1. Use the formula for the volume of the cone and the volume of the sphere.
    2. Express the volume of the cone in terms of the radius of the sphere and show that the maximum volume is \( \frac{8}{27} \) of the sphere’s volume.
  3. Using properties, integrate \( \int_2^5 \left( |x - 2| + |x - 3| + |x - 5| \right) dx \).
    Split the integral at points of discontinuity and evaluate the areas.
  4. Integrate using properties \( \int_0^\pi \log \left( 1 + \cos x \right) dx \).
    Use trigonometric identities and properties of definite integrals to solve.

Case-Based Questions

  1. Three vegetable shopkeepers A, B, and C are using polythene bags, handmade bags, and newspapers as carry bags. It is found that A, B, and C are using (20, 30, 40), (30, 40, 20), and (40, 20, 30) polythene bags, handmade bags, and newspaper envelopes, respectively. The shopkeepers spent Rs 250, Rs 270, and Rs 200 on these bags respectively. Based on the above information, answer the following questions:
    a) What is the cost of one handmade bag?
    b) What is the cost of one polythene bag?
    OR
    What is the cost of one newspaper bag?
    1. Let the cost of a polythene bag be \( x \), the cost of a handmade bag be \( y \), and the cost of a newspaper bag be \( z \).
    2. From the information given, we can set up the following equations: \[ 20x + 30y + 40z = 250 \quad (1) \] \[ 30x + 40y + 20z = 270 \quad (2) \] \[ 40x + 20y + 30z = 200 \quad (3) \] 3. We can solve these equations simultaneously to find the values of \( x \), \( y \), and \( z \).
    4. Solving equations (1), (2), and (3) using substitution or elimination method will give us the values of the costs of each type of bag.
  2. In Trigonometry, we discuss a quadrant chart in which the first quadrant means all positive, the second quadrant means sin and cosec are positive, the third quadrant means tan and cot are positive, and in the fourth quadrant, cos and sec are positive. Based on the above information, find the intervals in which the function given by \[ f(x) = \frac{4 \sin x - 2x - x \cos x}{2 + \cos x} \] is:
    a) Increasing
    b) Decreasing
    1. To determine where the function \( f(x) \) is increasing or decreasing, we need to find the derivative \( f'(x) \).
    2. Using quotient rule: \[ f'(x) = \frac{(2 + \cos x)(4 \cos x - 2 - \cos x + x \sin x) - (4 \sin x - 2x - x \cos x)(-\sin x)}{(2 + \cos x)^2} \] 3. Set \( f'(x) > 0 \) for increasing intervals and \( f'(x) < 0 \) for decreasing intervals.
    4. Analyze the sign of \( f'(x) \) across intervals determined by the critical points found by solving \( f'(x) = 0 \).
  3. The equation of the path traced by a roller coaster is given by the polynomial \( f(x) = a(x + 9)(x + 1)(x - 3) \). If the roller coaster crosses the y-axis at a point (0, -1), answer the following:
    a) Find the value of \( a \).
    b) Find \( f''(x) \) at \( x = 1 \).
    1. To find the value of \( a \): \[ f(0) = a(0 + 9)(0 + 1)(0 - 3) = -1 \] \[ \implies a(9)(1)(-3) = -1 \implies -27a = -1 \implies a = \frac{1}{27} \] 2. Thus, \( f(x) = \frac{1}{27}(x + 9)(x + 1)(x - 3) \).
    3. To find \( f''(x) \), first compute \( f'(x) \) using the product rule, then differentiate again: \[ f'(x) = \frac{1}{27} \left[ (x + 1)(x - 3) + (x + 9)(x - 3) + (x + 9)(x + 1) \right] \] 4. Then compute \( f''(x) \) and evaluate at \( x = 1 \): \[ f''(1) = \text{Evaluate the second derivative at } x = 1 \] This will yield the final value for \( f''(1) \).
Mathematics Questions

Mathematics Section A (1 × 20 = 20)

  1. The function defined by \( g(x) = x - \lfloor x \rfloor \) is discontinuous at:
    a) all rational points
    b) all irrational points
    c) all integer points
    d) None of these
  2. The derivative of \( \log(\cos(e^x)) \) is:
    a) \(- \tan(e^x)\)
    b) \( e^x \tan(e^x) \)
    c) \(- e^x \tan(e^x) \)
    d) None of these
  3. Suppose 3 × 3 matrix \( A = [a_{ij}] \), whose elements are given by \( a_{ij} = i^2 - j^2 \). Then \( a_{32} \) is equal to:
    a) 5
    b) 1
    c) 2
    d) 3
  4. If \( A = \begin{pmatrix} -6 & 2 \\ 2 & 12 \end{pmatrix} \), then A is:
    a) non-singular
    b) singular
    c) diagonal matrix
    d) scalar matrix
  5. The value of \( \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) + \cot^{-1} \left( \frac{1}{\sqrt{3}} \right) + \tan^{-1} \left( \sin \left( \frac{\pi}{2} \right) \right) \) is:
    a) \( \frac{\pi}{4} \)
    b) \( \frac{\pi}{12} \)
    c) \( \frac{3\pi}{4} \)
    d) \( \frac{\pi}{3} \)
  6. The edge of a cube is increasing at the rate of 0.3 cm/sec, the rate of change of its surface area when the edge is 3 cm is:
    a) 10.8 cm2
    b) 10.8 cm2/sec
    c) 10.8 cm3/sec
    d) 10.8 cm/sec
  7. The total revenue in rupees received from the sale of x units of an article is given by \( R(x) = 3x^2 + 36x + 5 \). The marginal revenue when \( x = 15 \) in rupees is:
    a) 126
    b) 116
    c) 96
    d) 90
  8. The side of an equilateral triangle is increasing at the rate of 2 cm/sec. The rate at which the area increases when the side is 10 is:
    a) 10 cm2/sec
    b) \( \sqrt{3} \) cm2/sec
    c) \( 10\sqrt{3} \) cm2/sec
    d) \( \frac{10}{3} \) cm2/sec
  9. The point on the curve \( y = x^2 \), where the rate of change of x-coordinate is equal to the rate of change of y-coordinate is:
    a) \( \left( \frac{1}{2}, 1 \right) \)
    b) \( \frac{1}{4} \)
    c) \( \left( \frac{1}{2}, \frac{1}{4} \right) \)
    d) (1, 1)
  10. If at \( x = 1 \), the function \( f(x) = x^4 - 62x^2 + ax + 9 \) attains its maximum value on the interval [0, 2], then the value of a is:
    a) 124
    b) -124
    c) 120
    d) -120
  11. \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx \) is:
    a) \( \frac{\pi}{2} \)
    b) \( \frac{\pi}{2} - \frac{1}{2} \)
    c) \( \frac{\pi}{4} - \frac{1}{2} \)
    d) \( \frac{1}{2} \)
  12. The value of \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx \) is:
    a) -\( \pi^2 \)
    b) \( \pi \)
    c) \( \frac{\pi}{2} \)
    d) 1
  13. The value of \( \int_0^{\frac{\pi}{2}} \log \left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx \) is:
    a) 2
    b) 4
    c) 0
    d) -2
  14. \( \int_1^{\sqrt{3}} \frac{\sqrt{3} \, dx}{1 + x^2} \) is:
    a) \( \frac{\pi}{3} \)
    b) \( \frac{2\pi}{3} \)
    c) \( \frac{\pi}{6} \)
    d) \( \frac{\pi}{12} \)
Mathematics Questions

Mathematics Questions

Section A

  1. \( \int_0^{2/3} \frac{dx}{4 + 9x^2} \) is:
    a) \( \frac{\pi}{6} \)
    b) \( \frac{\pi}{2} \)
    c) \( \frac{\pi}{24} \)
    d) \( \frac{\pi}{4} \)
  2. \( \int e^x \sec x \left( 1 + \tan x \right) dx \) is:
    a) \( e^x \cos x + c \)
    b) \( e^x \sec x + c \)
    c) \( e^x \sin x + c \)
    d) \( e^x \tan x + c \)
  3. \( \int \frac{dx}{\sin^2 x \cdot \cos^2 x} \) is:
    a) \( \tan x + \cot x + c \)
    b) \( \tan x - \cot x + c \)
    c) \( \tan x \cdot \cot x + c \)
    d) \( \tan x - \cot 2x + c \)
  4. \( \int \frac{e^x \left( 1 + x \right)}{\cos^2 \left( e^x \cdot x \right)} dx \) is:
    a) \( - \cot \left( e^x \cdot x \right) + c \)
    b) \( \tan \left( x \cdot e^x \right) + c \)
    c) \( \tan \left( e^x \right) + c \)
    d) \( \cot \left( e^x \right) + c \)

Assertion Reasoning Questions (2 × 1 = 2)

  1. ASSERTION (A): Every differentiable function is continuous but the converse is not true.
    REASON (R): Function \( f(x) = |x| \) is continuous.
    a) Both A and R are true, and R is the correct explanation of A
    b) Both A and R are true, but R is not the correct explanation of A
    c) A is true, but R is false
    d) A is false, but R is true
  2. ASSERTION (A): The absolute maximum value of the function \( 2x^3 - 24x \) in the interval [1, 3] is 89.
    REASON (R): The absolute maximum value of the function can be obtained from the value of the function at critical points and at boundary points.
    a) Both A and R are true, and R is the correct explanation of A
    b) Both A and R are true, but R is not the correct explanation of A
    c) A is true, but R is false
    d) A is false, but R is true

Section B (2 × 5 = 10)

  1. If \( A = \begin{pmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{pmatrix} \), find \( A^2 - 5A + 4I \) and hence find a matrix \( X \) such that \( A^2 - 5A + 4I + X = 0 \).
  2. State whether the function is one-one, onto when \( f : R \to R \) defined by \( f(x) = 1 + x^2 \).
  3. Find both the maximum and minimum value of \( 3x^4 - 8x^3 + 12x^2 - 48x + 1 \) on the interval [1, 4].
    OR
    Find the maximum profit that a company can make if the profit function is given by \( p(x) = 41 + 24x - 18x^2 \).
  4. Using properties, integrate \( \int_0^{\pi/2} \cos^2 x \, dx \).
  5. Integrate \( \int \cos^3 x \cdot e^{\log \sin x} \, dx \).

Section C (3 × 6 = 18)

  1. Using properties, integrate \( \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} \, dx \).
  2. Using properties, integrate \( \int_0^{\pi/4} \log(1 + \tan x) \, dx \).
  3. Integrate \( \int \frac{5x}{(x+1)(x^2+9)} \, dx \).
    OR
    Integrate \( \int \frac{x^2 + x + 1}{(x + 2)(x^2 + 1)} \, dx \).
Mathematics Questions
  1. Find the area of the region bounded by the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \).
  2. Using first derivative test, find the points at which the function \( f(x) = (x - 2)^4 \cdot (x + 1)^3 \) has:
    a) Local maxima
    b) Local minima
    c) Point of inflexion
  3. Integrate \( \int \frac{5x - 2}{1 + 2x + 3x^2} dx \).
    OR
    Integrate \( \int \frac{5x + 3}{\sqrt{x^2 + 4x + 10}} dx \).

Section D (4 × 5 = 20)

  1. Show that the semi-vertical angle of the cone of maximum volume and of given slant height is \( \tan^{-1}(\sqrt{2}) \).
    OR
    Of all the closed right circular cylindrical cans of volume \( 128\pi \, \text{cm}^3 \), find the dimensions of the can which has minimum surface area.
  2. Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.
    OR
    Prove that the volume of the largest cone that can be inscribed in a sphere of radius \( R \) is \( \frac{8}{27} \) of the volume of the sphere.
  3. Using properties, integrate \( \int_2^5 \left( |x - 2| + |x - 3| + |x - 5| \right) dx \).
  4. Integrate using properties \( \int_0^\pi \log \left( 1 + \cos x \right) dx \).

Case-Based Questions

  1. Three vegetable shopkeepers A, B, and C are using polythene bags, handmade bags, and newspapers as carry bags. It is found that A, B, and C are using (20, 30, 40), (30, 40, 20), and (40, 20, 30) polythene bags, handmade bags, and newspaper envelopes, respectively. The shopkeepers spent Rs 250, Rs 270, and Rs 200 on these bags respectively. Based on the above information, answer the following questions:
    a) What is the cost of one handmade bag?
    b) What is the cost of one polythene bag?
    OR
    What is the cost of one newspaper bag?
  2. In Trigonometry, we discuss about a quadrant chart in which the first quadrant means all positive, the second quadrant means sin and cosec are positive, the third quadrant means tan and cot are positive, and in the fourth quadrant, cos and sec are positive. Based on the above information, find the intervals in which the function given by \[ f(x) = \frac{4 \sin x - 2x - x \cos x}{2 + \cos x} \] is:
    a) Increasing
    b) Decreasing
  3. The equation of the path traced by a roller coaster is given by the polynomial \( f(x) = a(x + 9)(x + 1)(x - 3) \). If the roller coaster crosses the y-axis at a point (0, -1), answer the following:
    a) Find the value of \( a \).
    b) Find \( f''(x) \) at \( x = 1 \).