CUET UG Mathematics Booster Test 1 - Increasing and Decreasing Functions -

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CUET UG Mathematics Booster Test 1 - Increasing and Decreasing Functions

⏱️ 30 Minutes

❓ 20 Questions

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CUET UG Mathematics Booster Test 1 - Increasing and Decreasing Functions

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QUESTION 1 OF 20

Which of the following conditions on a graph unequivocally indicates a strictly increasing function over an interval?

Answer Review

1 Which of the following conditions on a graph unequivocally indicates a strictly increasing function over an interval?

A The tangent is horizontal at all points

B The slope of the tangent is positive at all points

C The function has a maximum point

D The graph intersects the x-axis

💡 Short Explanation

Positive slope means rising graph. Function values increase with x. This is the derivative test for increasing functions.

✅ Correct Answer Explanation

A strictly increasing function has positive derivative throughout the interval. Graphically, this means the tangent slope remains positive at every point. Option B directly expresses this condition. A horizontal tangent indicates constant behavior, while having a maximum point or crossing the x-axis does not guarantee monotonic increase.

❓ Why Other Options Are Incorrect

Option A → Horizontal tangents imply slope zero, not strict increase.
Option C → A maximum point often indicates a change from increasing to decreasing.
Option D → Crossing the x-axis does not determine monotonicity.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Relate graph behavior to derivative sign.

Final Logic: Positive tangent slope everywhere ⇒ strictly increasing.

🧠 Memory Trick

Positive Slope = Increasing Function

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Increasing Functions

2 (Vectors) If the displacement vector of a particle is given by \(\vec{r}(t)=(x(t),y(t))\), the path is considered decreasing if the rate of change of the y-component with respect to t is:

A Negative

B Zero

C Positive

D Undefined

💡 Short Explanation

Decreasing means values reduce. Negative derivative indicates reduction. y-component falls as time increases.

✅ Correct Answer Explanation

The rate of change of the y-component is \(dy/dt\). If \(dy/dt<0\), then y decreases as time increases. Therefore the motion is decreasing in the y-direction. Option A correctly represents this condition. The remaining options do not indicate decreasing behavior.

❓ Why Other Options Are Incorrect

Option B → Indicates constant y-value.
Option C → Represents increasing y-values.
Option D → Does not describe a valid decreasing rate.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret derivative sign physically.

Final Logic: Negative rate ⇒ decreasing component.

🧠 Memory Trick

Negative Rate = Downward Change

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Rate of Change

3 (Area) If f is strictly increasing on \(\left[a\ ,\ b\right]\), then the area under \(f^{'}(x)\)from a to b is:

A Negative

B Zero

C Undefined

D Positive

💡 Short Explanation

Increasing functions have positive derivative. Integral of positive quantity is positive. Net change is positive.

✅ Correct Answer Explanation

For a strictly increasing function, \(f^{'}(x)>0\) on the interval. Therefore, \(\int_{a}^{b}\,f^{'}(x) dx=f(b)-f(a)\) and since \(f(b)>f(a)\), the result is positive. Thus Option D is correct. The other options contradict increasing behavior.

❓ Why Other Options Are Incorrect

Option A → Negative area would imply decreasing behavior.
Option B → Zero area implies no net change.
Option C → The integral is well-defined.

🎯 Strategy

Used: Elimination

Application: Use the relationship between derivative sign and net change.

Final Logic: Increasing ⇒ positive derivative ⇒ positive integral.

🧠 Memory Trick

Increase ⇒ Positive Change

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Fundamental Theorem of Calculus

4 (Integral) If f is decreasing on \(\left[a\ ,\ b\right]\), then \(\int_{a}^{b}\,f^{'}(x) dx\) is:

A Positive

B Zero

C Negative

D Infinite

💡 Short Explanation

Decreasing functions have negative derivative. Net change is negative. Integral equals function difference.

✅ Correct Answer Explanation

By the Fundamental Theorem of Calculus, \(\int_{a}^{b}\,f^{'}(x) dx=f(b)-f(a)\) For a decreasing function, \(f(b)<f(a)\). Hence \(f(b)-f(a)<0\). Therefore the integral is negative, making Option C correct.

❓ Why Other Options Are Incorrect

Option A → Positive value implies increase.
Option B → Would indicate no change.
Option D → The integral remains finite.

🎯 Strategy

Used: Elimination

Application: Connect decreasing behavior with derivative sign.

Final Logic: Decreasing ⇒ negative net change.

🧠 Memory Trick

Decrease ⇒ Negative Integral

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Net Change Interpretation

5 (Case/Numerical) A function is given by \(f(x)=C\), where C is constant. What is the rate of change?

A 1

B 0

C C

D Undefined

💡 Short Explanation

Constant functions do not vary. Derivative measures change. No change means zero derivative.

✅ Correct Answer Explanation

For a constant function, \(f(x)=C\) its derivative is \(f^{'}(x)=0\) for every x. Since the function value never changes, the rate of change is zero. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Represents a constant slope, not a constant function.
Option C → The derivative is independent of C.
Option D → The derivative exists and equals zero.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recall the derivative rule for constants.

Final Logic: Constant function ⇒ zero rate of change.

🧠 Memory Trick

Constant ⇒ Derivative Zero

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Derivative of Constant Function

6 (Assertion–Reason Format)
Assertion (A): \(f(x)=x\) is a strictly monotonic function on \(R\).
Reason (R): Its derivative is constant and equal to 1.

A Both false

B Both true, R explains A

C A false, R true

D A true, R false

💡 Short Explanation

Derivative equals 1 everywhere. Positive derivative implies increasing function. Reason explains assertion directly.

✅ Correct Answer Explanation

For \(f(x)=x\), \(f^{'}(x)=1\) which is positive for all real x. A positive derivative throughout the domain implies strict increase. Hence the function is strictly monotonic. Both statements are true, and the reason correctly explains the assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are true.
Option C → Assertion is true.
Option D → Reason correctly explains the assertion.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Connect derivative sign with monotonicity.

Final Logic: Positive derivative everywhere ⇒ strictly increasing.

🧠 Memory Trick

f(x)=x Always Rises

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Monotonic Functions

7 If a function is strictly increasing at a point \(x_{0}\), then there exists an open interval around \(x_{0}\)such that for all \(x_{1}<x_{2}\)in that interval:

A \(f(x_{1})<f(x_{2})\)

B \(f(x_{1})>f(x_{2})\)

C \(f(x_{1})=f(x_{2})\)

D \(f^{'}(x_{0})=0\)

💡 Short Explanation

Increasing functions preserve order. Larger inputs give larger outputs. This is the local definition.

✅ Correct Answer Explanation

A function is strictly increasing if \(x_{1}<x_{2}\Rightarrow f(x_{1})<f(x_{2})\) within a neighborhood of the point. This condition directly describes increasing behavior. Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Defines decreasing behavior.
Option C → Defines constant behavior.
Option D → Zero derivative does not guarantee increase.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Match the formal definition.

Final Logic: Input order preserved ⇒ increasing.

🧠 Memory Trick

Increase Preserves Inequality

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Increasing Functions

8 (Multiple Correct) If a function is strictly decreasing at a point, which are true?
I. The tangent slope is negative
II. Function value decreases as x increases
III. \(f^{'}(x)=0\)

A I, II, and III

B I and II only

C I only

D II and III only

💡 Short Explanation

Negative slope indicates decrease. Function values fall as x increases. Zero derivative does not indicate decrease.

✅ Correct Answer Explanation

For a strictly decreasing function, the derivative is negative and function values decrease as x increases. Therefore Statements I and II are true. Statement III is false because \(f^{'}(x)=0\) corresponds to a critical point, not necessarily decreasing behavior. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Includes false Statement III.
Option C → Omits Statement II.
Option D → Includes incorrect Statement III.

🎯 Strategy

Used: Option Grouping

Application: Evaluate each statement independently.

Final Logic: Only I and II characterize decreasing behavior.

🧠 Memory Trick

Negative Slope ⇒ Decrease

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Decreasing Functions

9 (Incorrect Statement) Identify the INCORRECT statement regarding \(f^{'}(x)>0\):

A It implies the function is strictly increasing

B The slope of the tangent is positive

C It proves the function is constant

D It is used in the first derivative test

💡 Short Explanation

Positive derivative means increase. Constant functions have zero derivative. Option C contradicts derivative theory.

✅ Correct Answer Explanation

If \(f^{'}(x)>0\), the function is increasing and the tangent slope is positive. This condition is used in the first derivative test. A constant function requires \(f^{'}(x)=0\), not a positive derivative. Therefore Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Correct consequence of positive derivative.
Option B → Direct interpretation of derivative.
Option D → Standard application in monotonicity tests.

🎯 Strategy

Used: Odd One Out

Application: Identify the statement inconsistent with derivative properties.

Final Logic: Positive derivative cannot imply constant function.

🧠 Memory Trick

Positive ≠ Constant

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: First Derivative Test

10 (Arrange in Order) Determine steps to find intervals where \(f^{'}(x)>0\):
1. Solve the inequality \(f^{'}(x)>0\)
2. Find the derivative \(f^{'}(x)\)
3. Identify the intervals

A 2, 1, 3

B 1, 2, 3

C 3, 2, 1

D 2, 3, 1

💡 Short Explanation

Differentiate first. Solve the sign inequality. Then determine intervals.

✅ Correct Answer Explanation

To find where a function is increasing, first compute the derivative. Next solve the inequality \(f^{'}(x)>0\). Finally identify the intervals satisfying the inequality. Therefore the correct order is 2 → 1 → 3, which corresponds to Option A.

❓ Why Other Options Are Incorrect

Option B → Solves inequality before finding derivative.
Option C → Reverses the logical process.
Option D → Identifies intervals before solving.

🎯 Strategy

Used: Elimination

Application: Follow the standard derivative-test procedure.

Final Logic: Derivative → Inequality → Intervals.

🧠 Memory Trick

D-S-I: Differentiate, Solve, Identify

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Interval Method

11 (Match the Following) Match theorem conditions:

List I	List II
1. f is continuous	a. On open interval
2. f is differentiable	b. Decreasing function
3. \(f^{'}(x)>0\)	c. Increasing function
4. \(f^{'}(x)<0\)	d. On closed interval

A 1-b, 2-c, 3-a, 4-d

B 1-a, 2-b, 3-c, 4-d

C 1-c, 2-a, 3-d, 4-b

D 1-d, 2-a, 3-c, 4-b

💡 Short Explanation

Continuity is considered on closed intervals. Differentiability is checked on open intervals. Derivative sign determines monotonicity.

✅ Correct Answer Explanation

The First Derivative Theorem requires continuity on a closed interval and differentiability on the corresponding open interval. Also, \(f^{'}(x)>0\) implies an increasing function, while \(f^{'}(x)<0\) implies a decreasing function. Therefore the correct matching is 1-d, 2-a, 3-c, 4-b, corresponding to Option D.

❓ Why Other Options Are Incorrect

Option A → Matches continuity and differentiability incorrectly.
Option B → Reverses continuity and differentiability requirements.
Option C → Assigns wrong meanings to derivative signs.

🎯 Strategy

Used: Option Grouping

Application: Match each theorem condition with its standard interpretation.

Final Logic: Continuity→closed interval, Differentiability→open interval.

🧠 Memory Trick

C–Closed, D–Open, +Increase, −Decrease

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: First Derivative Test

12 (Probability) A function is continuous on an interval but not differentiable on the corresponding open interval. The probability that the first derivative test applies is:

A 1

B 0

C 1/2

D Undefined

💡 Short Explanation

First derivative test requires differentiability. Condition is violated here. Therefore applicability is impossible.

✅ Correct Answer Explanation

For the first derivative test to apply, the function must be differentiable on the open interval under consideration. Since the function is not differentiable there, the theorem cannot be used. Hence the probability of applicability is 0, making Option B correct.

❓ Why Other Options Are Incorrect

Option A → The theorem cannot always apply.
Option C → No partial applicability exists in this context.
Option D → The outcome is clearly determined.

🎯 Strategy

Used: Elimination

Application: Check whether theorem hypotheses are satisfied.

Final Logic: No differentiability ⇒ no first derivative test.

🧠 Memory Trick

No Derivative = No Derivative Test

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Conditions for Monotonicity

13 (Moving Average/Data) For \(f(x)=x^{2}-4x+6\), the critical point is:

A 0

B 2

C 4

D -2

💡 Short Explanation

Differentiate the function. Set derivative equal to zero. Solve for x.

✅ Correct Answer Explanation

For \(f(x)=x^{2}-4x+6f^{'}(x)=2x-4\) Setting \(f^{'}(x)=0\), \(2x-4=0x=2\) Therefore the critical point occurs at \(x=2\), making Option B correct.

❓ Why Other Options Are Incorrect

Option A → Derivative is −4.
Option C → Derivative is +4.
Option D → Does not satisfy \(f^{'}(x)=0\).

🎯 Strategy

Used: Substitution

Application: Differentiate and solve the resulting equation.

Final Logic: Critical point occurs where derivative equals zero.

🧠 Memory Trick

2x − 4 = 0 ⇒ x = 2

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Critical Points

14 For \(f(x)=x^{3}-3x^{2}+4x\), the interval where the function is strictly decreasing is:

A (0, 2)

B (-∞, 0)

C (2, ∞)

D None

💡 Short Explanation

Find derivative. Check its sign. Determine where it is negative.

✅ Correct Answer Explanation

\(f^{'}(x)=3x^{2}-6x+4=3(x-1)^{2}+1\) Since \(\left(x-1)^{2}\geq 0\right.\), \(f^{'}(x)>0\) for all real x. Therefore the function is strictly increasing everywhere and never decreasing. Hence Option D is correct. The provided answer A is incorrect.

❓ Why Other Options Are Incorrect

Option A → Derivative remains positive on (0,2).
Option B → Function is increasing there.
Option C → Function is also increasing there.

🎯 Strategy

Used: Substitution

Application: Analyze the sign of the derivative.

Final Logic: Positive derivative everywhere ⇒ no decreasing interval.

🧠 Memory Trick

Square + Positive Constant > 0

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Increasing and Decreasing Functions

15 (Graph/Region-based) In which interval is \(f(x)=cos⁡x\) strictly decreasing?

A (-π, 0)

B (π, 2π)

C (0, π/2)

D (0, π)

💡 Short Explanation

Differentiate cos x. Examine the sign of −sin x. Determine where it is negative.

✅ Correct Answer Explanation

For \(f(x)=cos⁡xf^{'}(x)=-sin⁡x\) On \(\left(0\ ,\ \pi \right)\), \(sin⁡x>0\), so \(f^{'}(x)<0\). Therefore the function decreases throughout this interval. Option D is correct. Option C is only part of the complete decreasing interval.

❓ Why Other Options Are Incorrect

Option A → Function increases on this interval.
Option B → Derivative is positive.
Option C → Only a subset of the full decreasing interval.

🎯 Strategy

Used: Elimination

Application: Determine where the derivative is negative.

Final Logic: −sin x < 0 on (0, π).

🧠 Memory Trick

cos Falls from 0 to π

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Trigonometric Monotonicity

16 In which interval is \(f(x)=sin⁡x\) strictly increasing?

A (π, 2π)

B (-π, 0)

C (0, π)

D (π/2, π)

💡 Short Explanation

Differentiate sin x. Increasing where cos x is positive. Compare with the given intervals.

✅ Correct Answer Explanation

\(f^{'}(x)=cos⁡x\) The function is strictly increasing where \(cos⁡x>0\), namely on intervals such as \(\left(-\pi /2,\pi /2\right)\) None of the listed options matches this interval completely. Therefore the provided answer C is incorrect. The question contains flawed options.

❓ Why Other Options Are Incorrect

Option A → cos x is mostly negative.
Option B → Contains regions where cos x changes sign.
Option D → cos x is negative throughout.

🎯 Strategy

Used: Elimination

Application: Check the sign of cos x on each interval.

Final Logic: Increasing requires cos x > 0.

🧠 Memory Trick

sin′ = cos

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Trigonometric Functions

17 (Integral) If \(f^{'}(x)\)is integrated from a critical point c to b, the result equals:

A \(f(b)-f(c)\)

B 0

C Undefined

D \(f(c)\)

💡 Short Explanation

Apply the Fundamental Theorem of Calculus. Integrating a derivative restores function change. Endpoints determine the result.

✅ Correct Answer Explanation

By the Fundamental Theorem of Calculus, \(\int_{c}^{b}\,f^{'}(x) dx=f(b)-f(c)\) The fact that c is a critical point does not alter this result. Therefore Option A is correct. The remaining options do not represent the net change in the function.

❓ Why Other Options Are Incorrect

Option B → Integral is not necessarily zero.
Option C → The integral is well-defined.
Option D → Omits the upper-end function value.

🎯 Strategy

Used: Substitution

Application: Apply the Fundamental Theorem directly.

Final Logic: Integral of derivative = net change.

🧠 Memory Trick

Derivative Integral = Change

📚 Class XII – NCERT Mathematics – Chapter: Integrals – Topic: Fundamental Theorem of Calculus

18 (Numerical) Using a sign chart for \(f^{'}(x)=2x-4\), the minimum value of slope is:

A 2

B 0

C -4

D 4

💡 Short Explanation

The slope is \(2x-4\). It varies with x. No finite minimum exists.

✅ Correct Answer Explanation

The derivative \(f^{'}(x)=2x-4\) is a linear function. As \(x\rightarrow -\infty\), \(2x-4\rightarrow -\infty\) Therefore the slope has no finite minimum value. Option C is incorrect because −4 is merely one possible slope value. The question is mathematically defective.

❓ Why Other Options Are Incorrect

Option A → Not the minimum value.
Option B → Occurs only at x = 2.
Option C → The slope can be less than −4.

🎯 Strategy

Used: Extreme Word Filter

Application: Examine behavior over the entire domain.

Final Logic: Linear function has no finite minimum slope.

🧠 Memory Trick

Line Extends to −∞

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Sign Charts

19 Over the entire interval \(\left(-\infty ,\infty \right)\), the function is:

A Strictly increasing

B Neither strictly increasing nor strictly decreasing

C Strictly decreasing

D Constant

💡 Short Explanation

Function behavior changes across intervals. It is not monotonic globally. Both increasing and decreasing portions exist.

✅ Correct Answer Explanation

A function that changes from increasing to decreasing or vice versa cannot be classified as strictly increasing or strictly decreasing over its entire domain. Such behavior is called mixed monotonicity. Therefore Option B correctly describes the function over \(\left(-\infty ,\infty \right)\).

❓ Why Other Options Are Incorrect

Option A → Increasing behavior does not persist everywhere.
Option C → Decreasing behavior does not persist everywhere.
Option D → The function changes values.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret global behavior from local intervals.

Final Logic: Mixed behavior ⇒ neither increasing nor decreasing.

🧠 Memory Trick

Up + Down = Neither

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Monotonicity

20 The transition from decreasing to increasing at \(x=2\) represents:

A Constant interval

B Global minimum

C Negative tangent

D Mixed interval behavior

💡 Short Explanation

Derivative changes sign. Function decreases before and increases after. Monotonicity changes at the point.

✅ Correct Answer Explanation

A transition from decreasing to increasing means the derivative changes from negative to positive. This indicates a change in behavior across intervals. While such a point may correspond to a local minimum, the option describing the interval behavior itself is Option D, mixed interval behavior.

❓ Why Other Options Are Incorrect

Option A → The function is not constant.
Option B → Not always guaranteed from the given wording alone.
Option C → A negative tangent describes only one side.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Focus on the phrase "transition from decreasing to increasing."

Final Logic: Change of monotonicity ⇒ mixed interval behavior.

🧠 Memory Trick

Decrease → Increase = Mixed

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: First Derivative Test

CUET UG Mathematics Booster Test 1 - Increasing and Decreasing Functions

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