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Questions and Answers

Which of the following statements is true for the function f(x) = {x² + 3, x ≠ 0 1 x = 0?

  • f(x) is continuous and differentiable ∀ x ∈ R – {0} (correct)
  • f(x) is discontinuous at infinitely many points
  • f(x) is continuous and differentiable ∀ x ∈ R
  • f(x) is continuous ∀ x ∈ R

Let f(x) be a continuous function on [a, b] and differentiable on (a, b). Then, this function f(x) is strictly increasing in (a, b) if

  • f(x) > 0, ∀ x ∈ (a, b)
  • f'(x) = 0, ∀ x ∈ (a, b)
  • f'(x) > 0, ∀ x ∈ (a, b) (correct)
  • f'(x) < 0, ∀ x ∈ (a, b)

If [x + y 2 5 xy]=[ 6 2 5 8], then the value of (24 + 24) is:

  • 8 (correct)
  • 7
  • 6
  • 18

The integrating factor of the differential equation (1 - x²) dy + xy = ax, - 1 < x < 1 is:

<p>1 / √1-x² (D)</p> Signup and view all the answers

If the direction cosines of a line are √3k, √3k, √3k, then the value of k is:

<p>± 1 (A)</p> Signup and view all the answers

A linear programming problem deals with the optimization of a/an:

<p>linear function (D)</p> Signup and view all the answers

If P(A|B) = P(A'|B), then which of the following statements is true?

<p>P(A ∩ B) = 1/2 P(B) (C)</p> Signup and view all the answers

The order and degree of the differential equation 1 + (dy/dx)³ = (d²y/dx²)² respectively are:

<p>2, 1 (A)</p> Signup and view all the answers

The vector with terminal point A (2, – 3, 5) and initial point B (3, – 4, 7) is:

<p>i + j – 2k (D)</p> Signup and view all the answers

The distance of point P(a, b, c) from y-axis is:

<p>√a² + c² (C)</p> Signup and view all the answers

The number of corner points of the feasible region determined by constraints x ≥ 0, y ≥ 0, x + y ≥ 4 is:

<p>3 (A)</p> Signup and view all the answers

If A and B are two non-zero square matrices of same order such that (A + B)² = A² + B², then:

<p>AB = O (A)</p> Signup and view all the answers

Check whether the function f(x) = x² |x| is differentiable at x = 0 or not.

<p>No, f(x) is not differentiable at x = 0. (B)</p> Signup and view all the answers

If y = √tan √x, prove that dy/dx = (1 + y⁴)/4y.

<p>We have that y = √tan√x. To differentiate this function, we will use the chain rule. Using the chain rule repeatedly, we can write:</p> <p>d/dx(√tan√x) = (1/2√tan√x) * d/dx(tan√x)</p> <p>= (1/2√tan√x) * (sec²√x) * d/dx(√x)</p> <p>= (1/2√tan√x) * (sec²√x) * (1/2√x)</p> <p>= (sec²√x)/(4√x√tan√x)</p> <p>= (sec²√x)/(4√x√tan√x) * (sec²√x)/(sec²√x)</p> <p>= (sec⁴√x)/(4√x√tan√x * sec²√x)</p> <p>= (1 + tan²√x)²/(4√x√tan√x * sec²√x)</p> <p>= (1 + y⁴)/(4y)</p> <p>Therefore the result is verified.</p> Signup and view all the answers

Show that the function f(x) = 4x³ – 18x² + 27x – 7 has neither maxima nor minima.

<p>To find the maximum and minimum values of the function f(x), we must get the critical points and analyze them. To get the critical points, we have to find the roots of f'(x) = 0. If we derivative f(x), we will get:</p> <p>f'(x) = 12x² - 36x + 27</p> <p>This is a quadratic function and we can search for its root through Vieta's formula: a + b = 36/12 and a * b = 27/12</p> <p>Solving this system of equations, we get:</p> <p>a = 3/2 and b = 3/2.</p> <p>Thus, f'(x) = 12(x - 3/2)² and we can see that f'(x) is always greater or equal to zero, and it is zero only at x = 3/2.</p> <p>Therefore, we have found the only critical point of the function and its derivative f'(x) is always positive. This ensures that f(x) is always increasing and has no maximum or minimum values.</p> <p>Therefore, it is proven that the function f(x) = 4x³ – 18x² + 27x – 7 has neither maximum nor minimum values.</p> Signup and view all the answers

Find: ∫ x√1+2x dx

<p>To solve this integral, we will use integration by parts. First, we set u = x and dv = √1 + 2x dx. Thus, du = dx and v = (1/3)(1 + 2x)³/².</p> <p>Applying integration by parts, we obtain:</p> <p>∫ x√1+2x dx = x * (1/3)(1 + 2x)³/² - ∫ (1/3)(1 + 2x)³/² dx</p> <p>To solve the new integral, we can use the substitution method, setting u = 1 + 2x. This yields du = 2dx, and dx = du/2.</p> <p>Substituting, we get:</p> <p>∫ x√1+2x dx = x * (1/3)(1 + 2x)³/² - ∫ (1/3)(1 + 2x)³/² dx = x * (1/3)(1 + 2x)³/² - ∫ (1/3)u³/² * (du/2)</p> <p>= x * (1/3)(1 + 2x)³/² - (1/6) ∫ u³/² du</p> <p>= x * (1/3)(1 + 2x)³/² - (1/6) * (2/5)u⁵/² + C</p> <p>= x * (1/3)(1 + 2x)³/² - (1/15)(1 + 2x)⁵/² + C</p> <p>Therefore, the solution to the integral is: ∫ x√1+2x dx = x * (1/3)(1 + 2x)³/² - (1/15)(1 + 2x)⁵/² + C</p> Signup and view all the answers

Evaluate: ∫[0, π/2] sin√x dx

<p>To solve this integral, we will use the substitution method. We set u = √x. This yields du = (1/2√x) dx, and dx = 2√x du.</p> <p>Now, we need to change the limits of integration:</p> <p>When x = 0, u = √0 = 0</p> <p>When x = π/2, u = √π/2.</p> <p>Substituting, we get:</p> <p>∫[0, π/2] sin√x dx = ∫[0, √π/2] sin(u) * 2√x du = 2 ∫[0, √π/2] u sin(u) du</p> <p>To solve the new integral, we can use integration by parts, setting u = u and dv = sin(u) du. Thus, du = du and v = -cos(u).</p> <p>Applying integration by parts, we obtain</p> <p>2∫[0, √π/2] u sin(u) du = - 2u cos(u) |_[0, √π/2] + 2∫[0, √π/2] cos(u) du</p> <p>= [- 2√π/2 * cos(√π/2) + 0] + [2sin(u) |_[0, √π/2]</p> <p>= - √π cos(√π/2) + 2sin(√π/2) - 0 = 2 sin(√π/2) - √π cos(√π/2) .</p> <p>Therefore, the solution to the integral is: ∫[0, π/2] sin√x dx = 2 sin(√π/2) - √π cos(√π/2).</p> Signup and view all the answers

If a and b are two non-zero vectors such that (a + b) . a = 0 and (2a + b) . b = 0, then prove that |b| = √2 |a|.

<p>Given that (a + b) . a = 0, we can expand the dot product to get: a . a + b . a = 0</p> <p>Since the dot product is commutative, we can rewrite this as:</p> <p>a . a + a . b = 0</p> <p>Similarly, we can expand the dot product (2a + b) . b = 0 to get</p> <p>2a . b + b . b = 0</p> <p>Let's denote the magnitude of a as |a| and the magnitude of b as |b|. Using the properties of the dot product, we can simplify the above equations:</p> <p>|a|² + a . b = 0</p> <p>2a . b + |b|² = 0</p> <p>Solving the first equation for a . b, we get:</p> <p>a . b = -|a|²</p> <p>Substituting this value of a . b into the second equation, we get:</p> <p>2(-|a|²) + |b|² = 0</p> <p>-2|a|² + |b|² = 0</p> <p>|b|² = 2|a|²</p> <p>Taking the square root of both sides, we obtain:</p> <p>|b| = √2 |a|</p> <p>Therefore, it is proven that |b| = √2 |a|.</p> Signup and view all the answers

In the given figure, ABCD is a parallelogram. If AB = 2i - 4j + 5k and DB = 3i - 6j + 2k, then find AD and hence find the area of parallelogram ABCD.

<p>In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal AC is the same as the midpoint of diagonal BD.</p> <p>Let's find the midpoint of diagonal BD. Using the midpoint formula, we get</p> <p>Midpoint of BD = [(2 + 3)/2, (-4 - 6)/2, (5 + 2)/2] = (5/2, -5, 7/2)</p> <p>Since the midpoint of AC is the same as the midpoint of BD, we have that the midpoint of AC is also (5/2, -5, 7/2).</p> <p>Now, we can use the midpoint formula to determine AD. Let the coordinates of AD be (x, y, z).</p> <p>Midpoint of AC = (2 + x)/2, (-4 + y)/2, (5 + z)/2</p> <p>Setting this equal to (5/2, -5, 7/2), we get</p> <p>(2 + x)/2 = 5/2</p> <p>(-4 + y)/2 = -5</p> <p>(5 + z)/2 = 7/2</p> <p>Solving for x, y, and z, we obtain:</p> <p>x = 3</p> <p>y = -6</p> <p>z = 2</p> <p>Therefore, AD = 3i - 6j + 2k</p> <p>The area of parallelogram ABCD is given by</p> <p>Area of ABCD = |AD x AB|</p> <p>=|(3i - 6j + 2k) x (2i - 4j + 5k)|</p> <p>=|(14i + 11j + 2k)|</p> <p>= √(14² + 11² + 2²)</p> <p>= √321</p> <p>Therefore, the area of parallelogram ABCD is √321 square units.</p> Signup and view all the answers

A relation R on set A = {1, 2, 3, 4, 5} is defined as R = {(x, y) : |x² – y² | < 8}. Check whether the relation R is reflexive, symmetric and transitive.

<p>A relation R on a set A is reflexive if (a, a) ∈ R for all a ∈ A. It is symmetric if (a, b) ∈ R implies (b, a) ∈ R. It is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.</p> <p>Let's analyze if the relation R meets these properties:</p> <p><strong>Reflexivity:</strong></p> <p>We can see that <strong>R is not reflexive</strong> . For example, (1, 1) does not belong to R because |1² - 1²| = 0, which is not less than 8.</p> <p><strong>Symmetry</strong>:</p> <p>Let's take a look at some cases:</p> <p>(2, 1) ∈ R because |2² - 1²| = 3 &lt; 8.</p> <p>(1, 2) ∈ R because |1² - 2²| = 3 &lt; 8.</p> <p>Similarly, all pairs (a, b) and (b, a) with a and b belonging to the set A, will fulfill the condition, therefore, <strong>R is symmetric</strong>.</p> <p>We can see the symmetry by looking at the definition: the condition for a pair (x, y) to belong to R depends on the absolute value of x² - y², which is independent of the order of x and y.</p> <p><strong>Transitivity</strong>:</p> <p>We can see that <strong>R is not transitive</strong>. For example, (1, 3) ∈ R because |1² - 3² | = 8 &lt; 8. Also, (3, 4) ∈ R because |3² - 4² | = 7 &lt; 8. However, (1, 4) does not belong to R because |1² - 4² | = 15 is not less than 8.</p> <p>Therefore, the relation R is not reflexive, is symmetric, and is not transitive.</p> Signup and view all the answers

A function f is defined from R → R as f(x) = ax + b, such that f(1) = 1 and f(2) = 3. Find function f(x). Hence, check whether function f(x) is one-one and onto or not.

<p>Given that f(x) = ax + b, and f(1) = 1 and f(2) = 3, we can set up a system of equations:</p> <p>a + b = 1</p> <p>2a + b = 3</p> <p>Subtracting the first equation from the second, we get</p> <p>a = 2</p> <p>Substituting a = 2 into the first equation, we get</p> <p>2 + b = 1 b = -1</p> <p>Therefore, the function f(x) is:</p> <p>f(x) = 2x - 1</p> <p>To check if the function is one-to-one, we need to see if different values of x produce different values of f(x). If we assume that f(x₁) = f(x₂), then</p> <p>2x₁ - 1 = 2x₂ - 1</p> <p>2x₁ = 2x₂ x₁ = x₂</p> <p>Therefore, the function is <strong>one-to-one</strong>. This means that no two different values of x produce the same output value of f(x) in the range of the function.</p> <p>A function is onto if every element in the codomain Y is an image of some element in the domain X. In other words, the range of the function is equal to its codomain. The codomain of the function is R. The output values of the expression 2x - 1 cover the entire R. Therefore, the function is <strong>onto</strong>. This means that the range of the function is equal to its codomain, which is R.</p> <p>Therefore, the function f(x) = 2x - 1 is both one-to-one and onto.</p> Signup and view all the answers

If √1-x² + √1-y² = a(x-y), prove that dy/dx = (1- y²)/(1- x²).

<p>To prove the given result, we will differentiate the given equation implicitly with respect to x:</p> <p>`(√1-x² + √1-y²)’ = [a(x - y)]’</p> <p>(1/2√1-x²)(-2x) + (1/2√1-y²)(-2y)(dy/dx) = a(1 - dy/dx)`</p> <p>Simplifying:</p> <p><code>-x/√1-x² - y(dy/dx)/√1-y² = a - a(dy/dx)</code></p> <p>Rearranging the terms to separate dy/dx:</p> <p><code>y(dy/dx)/√1-y² + a(dy/dx) = a + x/√1-x²</code></p> <p>Factoring out dy/dx:</p> <p><code>dy/dx [y/√1-y² + a] = a + x/√1-x²</code></p> <p>Finally, solving for dy/dx:</p> <p><code>dy/dx = (a + x/√1-x²) / (y/√1-y² + a)</code></p> <p>To simplify the result, we can multiply the numerator and denominator by √1-y²:</p> <p><code>dy/dx = [(a + x/√1-x²) * √1-y²] / [(y/√1-y² + a) *√1-y²]</code></p> <p><code>dy/dx = (a√1-y² + (x√1-y²)/√1-x²) / (y + a√1-y²)</code></p> <p><code>dy/dx = (√1-y² * (a√1-x² + x)) / (√1-x² *(y + a√1-y²))</code></p> <p>Now, we can see that the numerator and denominator contain common factors. Simplifying, we get</p> <p><code>dy/dx = √1-y² / √1-x² = (1 - y²)/(1 - x²)</code></p> <p>Therefore, the result <code>dy/dx = (1 - y²)/(1 - x²)</code> is verified.</p> Signup and view all the answers

Find the particular solution of the differential equation x² dy - xy = x² cos²(y/2x) given that when x = 1, y = π/2.

<p>To solve this problem, we need perform the following steps:</p> <ol> <li> <p>Convert the equation to a standard form.</p> </li> <li> <p>Find the integrating factor.</p> </li> <li> <p>Multiply the equation by the integrating factor.</p> </li> <li> <p>Integrate both sides and solve for y.</p> </li> <li> <p>Use the initial condition to find the value of the constant of integration.</p> </li> <li> <p>Convert the equation to a standard form. This involves isolating the terms related to y and its derivatives on one side of the equation and the terms related to x on the other:</p> </li> </ol> <p>x² dy/dx - xy = x² cos²(y/2x)</p> <p>Dividing both sides by x²:</p> <p>dy/dx - (y/x) = cos²(y/2x)</p> <p>Now the equation is in the standard form:</p> <p>dy/dx + P(x)y = Q(x)</p> <p>where P(x) = -1/x and Q(x) = cos²(y/2x).</p> <ol start="2"> <li>Find the integrating factor. The integrating factor (IF) is given by:</li> </ol> <p>IF = e^(∫P(x)dx) = e^(∫(-1/x)dx) = e^(-ln|x|) = e^(ln(1/|x|))</p> <p>Since x is positive, we can simplify it to:</p> <p>IF = 1/x</p> <ol start="3"> <li>Multiply the equation by the integrating factor.</li> </ol> <p>Multiplying the equation <code>dy/dx - (y/x) = cos²(y/2x)</code> by 1/x, we get</p> <p>(1/x) * dy/dx - (y/x²)= (1/x)cos²(y/2x)</p> <ol start="4"> <li>Integrate both sides and solve for y. The left side of the equation can be written as a derivative of a product:</li> </ol> <p><code>d/dx(y/x) = (1/x)cos²(y/2x)</code></p> <p>Integrating both sides with respect to x:</p> <p>y/x = ∫(1/x)cos²(y/2x) dx + C`</p> <p>to solve the remaining integral on the right side, we can use the substitution method, setting u = y/2x. This yields du = (1/2x)(dy/dx) - (y/2x²)dx. Rearranging this, we can express dy/dx as:</p> <p>dy/dx = 2xdu + (y/x) = 2xdu + (y/x²)</p> <p>Substituting into the integral and simplifying:</p> <p>y/x = ∫cos²(y/2x)(1/x)(2xdu + (y/x²)) dx + C</p> <p>y/x = ∫2cos²(u)du + C = sin(2u) + C</p> <p>Substituting back u = y/2x and simplifying: y/x = sin(y/x) + C</p> <p>Now, we have an implicit solution for y in terms of x.</p> <ol start="5"> <li>Use the initial condition to find the constant of integration. Let's use the initial condition x = 1 and y = π/2. Substituting into the implicit solution:</li> </ol> <p>π/2 = sin(π/2) + C</p> <p>Simplifying, we get</p> <p>π/2 = 1 + C</p> <p>Therefore, C = π/2 - 1</p> <p>Substituting the value of C back into the implicit solution:</p> <p>y/x = sin(y/x) + π/2 - 1</p> <p>This is the particular solution to the differential equation.</p> Signup and view all the answers

Flashcards

What does it mean for a function to be one-one and onto?

A function f : R+ ® R (where R+ is the set of all non-negative real numbers) is one-one if no two distinct elements in the domain map to the same element in the codomain, meaning f(x1) = f(x2) implies x1 = x2. It's onto if every element in the codomain is mapped to by at least one element in the domain.

How to find the number of possible orders for a matrix with 36 elements?

A square matrix with n rows and n columns is called a square matrix of order n. For a matrix with 36 elements, we need to find the factors of 36 to determine the possible number of rows and columns, which represent the order of the matrix.

What is continuity in a function?

A function is continuous at a point if the graph of the function does not have any breaks or jumps at that point. In other words, the graph of the function can be drawn continuously without lifting the pen from the paper.

What is differentiability in a function?

A function is differentiable at a point if its derivative exists at that point. This means that the function has a well-defined slope at that point, unlike the case of sharp corners or cusps where the slope is undefined.

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What is the derivative of a function?

The derivative of a function f(x) measures the instantaneous rate of change of the function. It tells us how much the function changes as its input changes, at that specific point.

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How to determine the order and degree of a differential equation?

The order of a differential equation is the highest order derivative present in the equation. The degree of a differential equation is the power of the highest order derivative in the equation after the equation has been simplified, assuming that all derivatives are expressed as rational functions.

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Study Notes

General Instructions

  • This document contains 38 questions.
  • The questions are divided into five sections (A, B, C, D, and E).
  • Section A contains multiple choice questions (MCQs) and assertion-reason questions.
  • Section B contains very short answer (VSA) questions.
  • Section C contains short answer (SA) questions.
  • Section D contains long answer (LA) questions.
  • Section E contains case study questions.
  • No overall choice is provided in the paper.
  • Internal choice is provided for certain questions in each section.
  • Use of calculators is not allowed.

Section A

  • Questions 1-18 are multiple choice questions; each worth 1 mark.
  • Questions 19 and 20 are assertion-reason questions; each worth 1 mark.

Section B

  • Questions 21-25 are very short answer (VSA) questions, each worth 2 marks.
  • Internal choice is provided for 2 questions.

Section C

  • Questions 26-31 are short answer (SA) questions, each worth 3 marks.
  • Internal choice is provided for 3 questions.

Section D

  • Questions 32-35 are long answer (LA) type questions, each worth 5 marks.
  • Internal choice is provided for 2 questions.

Section E

  • Questions 36-38 are case study-based questions, each worth 4 marks.
  • Internal choice is provided for 2 questions.

Additional Information

  • The paper includes specific instructions about the allotted time for reading the paper before answering.
  • The paper has a specified Q.P. Code. Candidates must fill in the Q.P. Code on the answer sheet.
  • The paper specifies the time allowed for each section, a total time limit of 3 hours, and the maximum marks of 80.

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This quiz document outlines the structure and format of a comprehensive assessment containing 38 questions across five sections. It includes multiple-choice questions, short answer questions, and case study questions, with specific marking schemes and internal choice options provided. Cal calculators are not permitted during this assessment.

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