Limits

01

PYQ 2018

medium

mathematics ID: ap-eamce

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02

PYQ 2019

medium

mathematics ID: ap-eamce

If and then

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03

PYQ 2024

medium

mathematics ID: ap-eamce

Evaluate the following limit:

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04

PYQ 2024

medium

mathematics ID: ap-eamce

.The function is given by: Which of the following is true

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left discontinuous at

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left continuous at

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right discontinuous at

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discontinuous at

05

PYQ 2024

medium

mathematics ID: ap-eamce

A point moves on the curve and its y-coordinate increases at a rate of 6 units per second. When the point is at (2,-1), the rate of change of its x-coordinate is:

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06

PYQ 2024

medium

mathematics ID: ap-eamce

Let . If exists, then the sum of the cubes of the possible values of is: }

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$1$

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$5$

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$7$

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$9$

Official Solution

Correct Option: (3)

$For to exist, the left -hand limit and the right -hand limit must be equal. Left -hand limit: Right -hand limit: For the limit to exist, the left -hand limit must equal the right -hand limit: We can factor the quadratic equation: Thus, the possible values of are and . The sum of the cubes of the possible values of is: Therefore, the sum of the cubes of the possible values of is 7. Final Answer: (3) 7$

07

PYQ 2024

easy

mathematics ID: ap-eamce

$The length of the tangent drawn at the point on the curve is:$

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Official Solution

Correct Option: (2)

$Step 1: Differentiate the given curve equation The given equation of the curve is: Differentiating both sides with respect to : Rearranging for : Step 2: Compute the slope at the given point Let the given point be . To determine and, we use the constraint with . Solving for, we find its corresponding value. Step 3: Use the formula for the length of the tangent The formula for the length of the tangent to a curve at a given point is: Substituting the computed values, we get:$

08

PYQ 2024

medium

mathematics ID: ap-eamce

$If $ y$ is:$

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$Error in$

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$Percentage error in$

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$(error in)$

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$(relative error in)$

Official Solution

Correct Option: (3)

$Step 1: Simplify the expression for . The expression is an infinite geometric series with first term 1 and common ratio . Since the series is infinite, we assume for the series to converge. The sum of the infinite geometric series is . Therefore, . Step 2: Find the relative error in . The relative error in is given by, where is the error in . Taking logarithms on both sides of, we get: . Differentiating both sides, we get: . Therefore, the relative error in is times the error in .$

09

PYQ 2024

medium

mathematics ID: ap-eamce

$Evaluate the limit:$

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Official Solution

Correct Option: (4)

$Step 1: Divide by in the Numerator and Denominator Divide each term in the numerator by : Step 2: Apply the Limit Since as, we get: Step 3: Conclusion Thus, the correct answer is:$

10

PYQ 2024

medium

mathematics ID: ap-eamce

$Evaluate the limit:$

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Official Solution

Correct Option: (1)

$Step 1: Use the Cosine Difference Formula Using the standard trigonometric identity: we apply this to both numerator and denominator: For the numerator: For the denominator: Step 2: Apply the Limit Canceling common terms: Using the small-angle approximation as : Step 3: Conclusion Thus, the correct answer is:$

11

PYQ 2024

easy

mathematics ID: ap-eamce

$Evaluate the limit:$

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Official Solution

Correct Option: (2)

$Step 1: Expanding trigonometric functions Using approximations: Step 2: Evaluating the limit$

12

PYQ 2024

hard

mathematics ID: ap-eamce

$If is continuous at, then}$

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Official Solution

Correct Option: (1)

$Since is continuous at , we must have Thus, we need to find We have \begin{align*} &\lim_{x \to 0} \frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{\sqrt{a + x} - \sqrt{a - x}} &= \lim_{x \to 0} \frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{\sqrt{a + x} - \sqrt{a - x}} \cdot \frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x}} &= \lim_{x \to 0} \frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{2x} \cdot (\sqrt{a + x} + \sqrt{a - x}) &= \lim_{x \to 0} \frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{2x} \cdot 2 \sqrt{a} &= \sqrt{a} \lim_{x \to 0} \frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{x} &= \sqrt{a} \lim_{x \to 0} \frac{\sqrt{a^2 - ax - x^2} - \sqrt{x^2 + ax + a^2}}{x} \cdot \frac{\sqrt{a^2 - ax - x^2} + \sqrt{x^2 + ax + a^2}}{\sqrt{a^2 - ax - x^2} + \sqrt{x^2 + ax + a^2}} &= \sqrt{a} \lim_{x \to 0} \frac{(a^2 - ax - x^2) - (x^2 + ax + a^2)}{x (\sqrt{a^2 - ax - x^2} + \sqrt{x^2 + ax + a^2})} &= \sqrt{a} \lim_{x \to 0} \frac{-2ax - 2x^2}{x (\sqrt{a^2 - ax - x^2} + \sqrt{x^2 + ax + a^2})} &= \sqrt{a} \lim_{x \to 0} \frac{-2a - 2x}{\sqrt{a^2 - ax - x^2} + \sqrt{x^2 + ax + a^2}} &= \sqrt{a} \cdot \frac{-2a}{2 \sqrt{a^2}} &= \sqrt{a} \cdot \frac{-2a}{2a} &= -\sqrt{a}. \end{align*}Therefore, .$

13

PYQ 2024

medium

mathematics ID: ap-eamce

$Given the function: If and is continuous for all, then the value of is:$

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Official Solution

Correct Option: (4)

$Step 1: Condition for continuity For to be continuous at, Substituting in the numerator, simplifying, and equating to, we get:$

14

PYQ 2024

medium

mathematics ID: ap-eamce

1

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Official Solution

Correct Option: (2)

$Step 1: Factorizing the numerator and denominator Step 2: Cancel common terms For, canceling, Step 3: Substitute$

15

PYQ 2024

medium

mathematics ID: ap-eamce

$If the equation of the tangent at (2, 3) on y 2 = ax 3 + b is y = 4x - 5, then the value of a 2 + b 2 is:$

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Official Solution

Correct Option: (2)

$Step 1: Use the point (2, 3) on the curve y 2 = ax 3 + b. Since (2, 3) lies on the curve y 2 = ax 3 + b, we have: 3 2 = a(2 3) + b 9 = 8a + b ...(1) Step 2: Differentiate the equation y 2 = ax 3 + b with respect to x. Differentiating both sides with respect to x, we get: 2y (dy/dx) = 3ax 2 dy/dx = (3ax 2) / (2y) Step 3: Find the slope of the tangent at (2, 3). The slope of the tangent at (2, 3) is given by: (dy/dx) | (2, 3) = (3a(2 2)) / (2(3)) = (12a) / 6 = 2a Step 4: Compare the slope with the given tangent equation. The given tangent equation is y = 4x - 5. The slope of this tangent is 4. Therefore, 2a = 4, so a = 2. Step 5: Substitute the value of a in equation (1) to find b. Substitute a = 2 in 9 = 8a + b: 9 = 8(2) + b 9 = 16 + b b = 9 - 16 = -7 Step 6: Calculate a 2 + b 2 . a 2 + b 2 = 2 2 + (-7) 2 = 4 + 49 = 53 Therefore, the value of a 2 + b 2 is 53.$

16

PYQ 2024

medium

mathematics ID: ap-eamce

$If Rolle's theorem is applicable for the function on, then the value of is:$

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Official Solution

Correct Option: (3)

$Step 1: Verify the conditions for Rolle's theorem. The function f(x) = x(x+3)e -x/2 is continuous on [-3, 0] and differentiable on (-3, 0) since it is a product of polynomial and exponential functions. Also, f(-3) = (-3)(-3+3)e -(-3)/2 = (-3)(0)e 3/2 = 0 and f(0) = 0(0+3)e -0/2 = 0(3)e 0 = 0. Since f(-3) = f(0) = 0, Rolle's theorem is applicable. Step 2: Find the derivative of f(x). f(x) = (x 2 + 3x)e -x/2 f'(x) = (2x + 3)e -x/2 + (x 2 + 3x)e -x/2 (-1/2) f'(x) = e -x/2 (2x + 3 - (x 2 + 3x)/2) f'(x) = e -x/2 ((4x + 6 - x 2 - 3x)/2) f'(x) = (e -x/2 /2)(-x 2 + x + 6) Step 3: Set f'(c) = 0 and solve for c. By Rolle's theorem, there exists a c (-3, 0) such that f'(c) = 0. (e -c/2 /2)(-c 2 + c + 6) = 0 Since e -c/2 0, we have -c 2 + c + 6 = 0. c 2 - c - 6 = 0 (c - 3)(c + 2) = 0 c = 3 or c = -2. Since c (-3, 0), we have c = -2. Therefore, the value of c is -2.$

17

PYQ 2024

medium

mathematics ID: ap-eamce

$By considering, the approximate value of is:$

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Official Solution

Correct Option: (4)

$Step 1: Using small-angle approximation We use: Substituting, we compute:$

18

PYQ 2024

medium

mathematics ID: ap-eamce

$Evaluate the limit:$

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4

Official Solution

Correct Option: (3)

$Step 1: Simplifying the fraction inside the limit Divide both numerator and denominator by : As , the terms , , , and tend to zero. Thus, Step 2: Applying Logarithm for Exponential Limit Form We have an indeterminate form , so we take logarithms: Expanding using first-order approximations: Approximating for large , Step 3: Evaluating the Limit Multiplying by , Approximating, For large , the highest degree term dominates: Dividing by , For large , Thus, Exponentiating both sides,$

19

PYQ 2024

medium

mathematics ID: ap-eamce

$Which of the following is true for$

1

$is continuous for all$

2

$has maximum value 2$

3

$has neither minimum nor maximum$

4

$has minimum value 2$

Official Solution

Correct Option: (2)

$We are given the function defined piecewise. Let's examine the continuity of the function. Step 1: Check if the function is continuous at . For to be continuous at, the left-hand and right-hand limits must be equal to the value at . The left-hand limit: The right-hand limit: Since the left-hand and right-hand limits are not equal, the function is not continuous at . Step 2: Analyze the maximum and minimum values. Since takes the value 2 at and the absolute values of the function outside of 0 never exceed 2, the maximum value is 2. Thus, the function has a maximum value of 2.$

High-Yield Trend

Chapter Questions
19 MCQs

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Limits

High-Yield Trend

Chapter Questions 19 MCQs

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Chapter Questions
19 MCQs