CUET UG Applied Mathematics Booster Test 2 - Applications of Derivatives -

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CUET UG Applied Mathematics Booster Test 2 - Applications of Derivatives

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Applications of Derivatives

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QUESTION 1 OF 20

Applying the second derivative test to find the local extrema of the polynomial f(x) = 4x^3/3 + 6x^2 + 8x + 7, evaluating the second derivative at the critical point x = -2 yields f''(-2) = -4. This mathematically confirms that x = -2 is strictly a:

Answer Review

1 Applying the second derivative test to find the local extrema of the polynomial f(x) = 4x^3/3 + 6x^2 + 8x + 7, evaluating the second derivative at the critical point x = -2 yields f''(-2) = -4. This mathematically confirms that x = -2 is strictly a:

A Point of local minimum

B Point of local maximum

C Point of inflexion

D Point of absolute discontinuity

💡 Short Explanation

Second derivative test decides concavity Negative value implies concave down Concave down at critical point → local maximum

✅ Correct Answer Explanation

At a critical point: If f''(c) < 0 → function is concave downward Concave downward at x = -2 implies peak behavior Hence x = -2 is a local maximum point.

❓ Why Other Options Are Incorrect

A → requires f''(c) > 0
C → inflexion requires sign change of concavity
D → discontinuity is unrelated to derivatives

🎯 Strategy

Used: Extreme Word Filter

Application: Spotting the negative evaluation distinguishes local peaks from minima, rapidly eliminating unrelated discontinuous or inflexion-based mathematical behaviors.

Final Logic: f''(c) < 0 ⇒ concave down ⇒ local maximum

🧠 Memory Trick

"Minus second derivative = peak"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Second derivative test

2 Match the critical points of f(x) = x^3 - 3x^2 + 3x + 5 to their evaluated extremum properties.

List I	List II
1. f'(1)	a. Strictly > 0
2. f''(1)	b. Exactly 0
3. Left neighbourhood of 1, f'(x)	c. Strictly > 0 (maintains positive sign)
4. Right neighbourhood of 1, f'(x)	d. Evaluates to 0, implying test failure

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

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

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

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

💡 Short Explanation

f'(x) = 3(x−1)² No sign change at x=1 Indicates stationary inflexion behavior

✅ Correct Answer Explanation

At x=1: derivative does not change sign → (II) f''(1)=0 → neutral second derivative behavior Thus correct matching follows A-II, B-IV, C-I, D-III.

❓ Why Other Options Are Incorrect

A → incorrect mapping of sign change
B → mismatches derivative behavior
D → swaps extremum interpretation

🎯 Strategy

Used: Option Grouping

Application: Grouping the universally positive neighborhood derivative behaviors quickly isolates the single correct structural pairing, preventing mapping confusion entirely.

Final Logic: derivative no sign change ⇒ correct classification

🧠 Memory Trick

"No sign change = no extremum"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Higher order stationary point

3 Which of the following strict mathematical statements regarding absolute extrema on closed vs open intervals are universally TRUE?
1) A function f defined in a closed interval may have a local extremum value even at the boundary points.
2) Not every critical point inside an open interval is definitively a point of local extremum.
3) A function f defined in an open interval will have a local extremum value exclusively at a critical point.
4) Absolute extrema cannot exist on any unbounded interval (–∞, ).

A 1, 2, 3

B 1, 2

C 2, 3, 4

D 1, 3

💡 Short Explanation

Closed interval allows boundary extrema Critical points may or may not be extrema Open intervals do not guarantee extrema

✅ Correct Answer Explanation

1 true: boundary points can be extrema 2 true: critical point not always extremum 3 true: extrema occur at critical points or boundaries 4 false: extrema can exist on unbounded intervals (e.g., parabola minimum)

❓ Why Other Options Are Incorrect

4 → incorrect universal claim

🎯 Strategy

Used: Elimination

Application: Eliminating the clearly false unbounded claim immediately rules out incorrect combinations, perfectly isolating the correct universally true mathematical statements.

Final Logic: only 1,2,3 satisfy all conditions

🧠 Memory Trick

"Endpoints + critical points = extrema check"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Absolute maxima/minima

4 Identify the incorrect analytical statement regarding the absolute extrema of the linear function f(x) = 2x + 5 evaluated strictly on the open interval (-2, 4).

A The function is strictly monotonically increasing.

B The function has no defined critical points where f'(x) = 0.

C The absolute maximum value strictly evaluates to 13.

D The function has neither an absolute maximum nor an absolute minimum value on this open interval.

💡 Short Explanation

Linear function increases on interval Open interval does not include endpoints Hence no max/min attained

✅ Correct Answer Explanation

Even though limit at x→4 gives 13, it is not attained inside open interval → so no absolute maximum exists.

❓ Why Other Options Are Incorrect

A → true
B → true
D → true

🎯 Strategy

Used: Extreme Word Filter

Application: Highligting the absolute terminology instantly flags the functionally unattained maximum on the open interval, perfectly resolving boundary condition confusion.

Final Logic: open interval ⇒ endpoint value not attained

🧠 Memory Trick

"Open interval = no endpoint value"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Monotonic functions

5 The graph of the fractional power function f(x) = x^(2/3) possesses a critical point behaviour at x = 0 where f'(0) is algebraically not defined. Geometrically, this specific non-differentiable point is strictly classified as a:

A Smooth turning point

B Point of inflexion

C Cusp

D Vertical asymptote

💡 Short Explanation

Derivative undefined at 0 Sharp point behavior Not smooth

✅ Correct Answer Explanation

At x=0: derivative undefined curve has sharp point with steep change Thus classified as cusp.

❓ Why Other Options Are Incorrect

A → not smooth
B → concavity change not applicable
D → function defined at 0

🎯 Strategy

Used: Contextual Matching

Application: Properly matching the algebraically undefined mathematical derivative specifically with its sharply pointed geometric behavior immediately excludes smoothly turning alternatives.

Final Logic: non-smooth sharp point ⇒ cusp

🧠 Memory Trick

"Sharp tip = cusp"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Non-differentiable points

6 A smooth turning curve like f(x) = |x| + 3 possesses a critical point at x = 0. Since the derivative f'(0) is fundamentally undefined due to differing left and right limits, the geometric feature explicitly present at (0,3) is a:

A Cusp

B Corner

C Horizontal tangent

D Inflexion

💡 Short Explanation

Left and right derivatives differ Sharp bend at origin Continuous but not differentiable

✅ Correct Answer Explanation

At x=0: slope from left = -1 slope from right = +1 → forms a corner point.

❓ Why Other Options Are Incorrect

A → cusp involves infinite slope
C → no horizontal tangent
D → no concavity change

🎯 Strategy

Used: Elimination

Application: Rapidly eliminating infinite cusp and smooth inflexion attributes confirms a corner point by directly addressing differing constant finite slopes.

Final Logic: piecewise slope mismatch ⇒ corner

🧠 Memory Trick

"Different slopes = corner"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Absolute value function

7 In rigorous optimization using derivatives, if the second derivative test yields f''(c) = 0 at a confirmed critical point 'c', the absolute mandatory next step is to:

A Local max

B Local min

C First derivative test

D Discontinuous

💡 Short Explanation

Second derivative test inconclusive Need sign analysis Check f'(x) behavior

✅ Correct Answer Explanation

When f''(c)=0: second derivative test fails must use first derivative sign change test

❓ Why Other Options Are Incorrect

A → not justified
B → not justified
D → unrelated

🎯 Strategy

Used: Elimination

Application: Systematically removing completely unjustified extremum conclusions for a zero derivative perfectly highlights the strictly mandatory fallback to first derivative testing.

Final Logic: f''=0 ⇒ switch to first derivative test

🧠 Memory Trick

"Zero second derivative → change test"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Second derivative test

8 During objective function analysis for the wire cut into a square and a circle (Total length = 40), the combined area A(x) features a second derivative A''(x). Evaluated analytically, A''(x) strictly equals:

A 1/8 + 1/(2π)

B 1/4 + 1/π

C 1/16 + 1/(4π)

D 1/2 + π

💡 Short Explanation

Differentiate area twice Apply constraint expression Simplify constants

✅ Correct Answer Explanation

From A(x), differentiation yields constant second derivative terms: A''(x)=1/8 + 1/(2π)

❓ Why Other Options Are Incorrect

B → incorrect scaling
C → incorrect coefficient
D → dimensionally invalid

🎯 Strategy

Used: Substitution

Application: Substituting precise constraint values strictly into the combined algebraic equation quickly verifies the exact constant curvature without secondary calculation confusion.

Final Logic: correct differentiation yields A''(x) expression

🧠 Memory Trick

"Second derivative = constant curvature"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Optimization

9 A manufacturer's profit-related condition requires finding the absolute maximum profit. Given P(x) = -9x² + 522x - 2500, evaluating P(29) mathematically yields an absolute maximum weekly profit of:

A ₹2500

B ₹5069

C ₹4962

D ₹7569

💡 Short Explanation

Vertex gives maximum Substitute x=29 Evaluate expression

✅ Correct Answer Explanation

P(29)= -9(841)+522(29)-2500 = -7569+15138-2500 = 5069

❓ Why Other Options Are Incorrect

A → too low
C → miscalculated
D → inflated value

🎯 Strategy

Used: Substitution

Application: Directly substituting the specifically evaluated vertex explicitly yields the precise absolute profit figure, seamlessly bypassing lengthy secondary theoretical quadratic analyses.

Final Logic: plug vertex x into profit function

🧠 Memory Trick

"Vertex gives peak profit"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Profit maximization

10 Based on the marginal equality principle, a firm's Marginal Cost (MC) is proven to fall continuously as output 'x' increases ONLY if the derivative of the Marginal Cost with respect to 'x' (d(MC)/dx) strictly resides in which region?

A >0

B =0

C <0

D Undefined

💡 Short Explanation

MC decreasing means negative slope derivative of MC negative implies downward trend

✅ Correct Answer Explanation

If d(MC)/dx < 0: MC decreases as x increases

❓ Why Other Options Are Incorrect

A → increasing
B → constant
D → undefined behavior

🎯 Strategy

Used: Extreme Word Filter

Application: Isolating the explicitly necessary negative gradient requirement perfectly defines the downward slope, instantly eliminating constant, increasing, or undefined alternatives.

Final Logic: negative derivative ⇒ decreasing function

🧠 Memory Trick

"Negative slope = falling curve"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Marginal cost

11 A printed page must contain 180 sq cm of text with margins. The total area function is A(x) = xy + 5x + 4y + 20. By substituting y = 180/x, solving A'(x) = 0 for the most economical dimension 'x' yields:

A 10 cm

B 12 cm

C 15 cm

D 18 cm

💡 Short Explanation

Substitute constraint y = 180/x Differentiate A(x) Solve A'(x)=0 gives x=12

✅ Correct Answer Explanation

The area function becomes: A(x) = x(180/x) + 5x + 4(180/x) + 20 = 180 + 5x + 720/x + 20 Differentiate: A'(x) = 5 - 720/x² Set A'(x)=0: 5 = 720/x² → x² = 144 → x = 12 Thus, optimal dimension is 12 cm.

❓ Why Other Options Are Incorrect

Option A → does not satisfy derivative condition
Option C → does not satisfy x² = 144
Option D → too large, violates extremum condition

🎯 Strategy

Used: Elimination

Application: Only x satisfying A'(x)=0 is valid; others fail equation check.

Final Logic: Critical point from derivative equation uniquely gives x=12.

🧠 Memory Trick

"720/5 = 144 → √144 = 12"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Optimization

12 For minimum distance/area optimization where a 40m wire is divided, the precise length of the second piece of wire formed strictly into a circle is algebraically evaluated as:

A 40π / (π + 4)

B 160 / (π + 4)

C 40 / (π + 4)

D 160π / (π + 4)

💡 Short Explanation

Form area function Differentiate Solve gives optimal split

✅ Correct Answer Explanation

Let square side be x ⇒ perimeter = 4x Circle wire = 40 − 4x Radius r = (40 − 4x)/(2π) Minimizing total area: A(x) = x² + πr² Solving A'(x)=0 gives: x = 160/(π+4) Thus circle length: 40 − 4x = 40π/(π+4)

❓ Why Other Options Are Incorrect

Option B → corresponds to square part, not circle length
Option C → dimensionally inconsistent
Option D → incorrect scaling factor

🎯 Strategy

Used: Option Grouping

Application: Correctly partitioning the total wire length based on the optimization derivative ensures the circle piece matches the required length.

Final Logic: Correct partition of wire gives circle length = 40π/(π+4)

🧠 Memory Trick

"Circle part always carries π in numerator"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Optimization Problems

📄 Passage

In advanced curve analysis, tangents can be drawn to implicit curves from external points not residing on the curve itself. For example, considering the hyperbolic curve 3x^2 - y^2 = 8, we wish to find tangents passing through the external coordinate (4/3, 0). We first assume the point of contact on the curve is (h, k), which inherently must satisfy 3h^2 - k^2 = 8. By differentiating the curve implicitly to find dy/dx, we obtain the tangent slope m = 3x/y. At (h, k), this slope is 3h/k. We then construct the point-slope equation of the tangent and force it to pass through the external point (4/3, 0). This creates a secondary algebraic relation between h and k, allowing us to simultaneously solve for the exact points of contact.

Based strictly on the algebraic passage constraints for behaviour on closed systems, solving the simultaneous equations 3h^2 - k^2 = 8 and 4k = 9h^2 - 3k^2 yields the squared parameter h^2 equal to:

A 44/3

B 44/9

C 8/3

D 16/9

💡 Short Explanation

Substitute the first equation $3h^{2}-k^{2}=8$ into the second. $4k=3\left(3h^{2}-k^{2}\right)$yields 4k = 24, so k = 6. Substitute k=6 back to find $h^{2}=44/3$

✅ Correct Answer Explanation

Solving the given non-linear system requires direct substitution. By factoring the right side of the second equation as $3\left(3h^{2}-k^{2}\right)$, we substitute the first equation's value (8) to get 4k = 24, meaning k = 6. Substituting k back into $3h^{2}-k^{2}=8$ yields $h^{2}=\frac{44}{3} ,$making Option A correct over the originally provided key.

❓ Why Other Options Are Incorrect

Option B -> Represents 44/9, which incorrectly assumes $h=\pm \frac{\sqrt{44}}{3}$ rather than $h=\pm \sqrt{44/3}$during evaluation.
Option C -> Represents 8/3, which mathematically ignores the necessary $k^{2}$ substitution completely.
Option D -> Represents 16/9, which incorrectly evaluates the algebraic constants during isolation.

🎯 Strategy

Used: Substitution

Application: Briefly applying substitution simplifies complex simultaneous equations by evaluating common algebraic groups, avoiding tedious quadratics.

Final Logic: Substituting the known value of $3h^{2}-k^{2}$directly isolates k, leading cleanly to the exact value for $h^{2}$.

🧠 Memory Trick

"Sub and Solve: Group the terms, find the root."

📚 Class XI – Mathematics – Chapter: Conic Sections – Topic: Tangents to Hyperbolas

📄 Passage

In advanced curve analysis, tangents can be drawn to implicit curves from external points not residing on the curve itself. For example, considering the hyperbolic curve $3x^2 - y^2 = 8$, we wish to find tangents passing through the external coordinate (4/3, 0). We first assume the point of contact on the curve is (h, k), which inherently must satisfy $3h^2 - k^2 = 8$. By differentiating the curve implicitly to find dy/dx, we obtain the tangent slope m = 3x/y. At (h, k), this slope is 3h/k. We then construct the point-slope equation of the tangent and force it to pass through the external point (4/3, 0). This creates a secondary algebraic relation between h and k, allowing us to simultaneously solve for the exact points of contact.

During endpoint evaluation of the resulting simultaneous equations, finding the explicit equations of the tangents passing through (4/3, 0) yields:

A $\pm \sqrt{33}x-3y+4=0$

B $\pm \sqrt{11}x-y+4=0$

C $\pm \sqrt{33}x+3y-4=0$

D $x\pm 3y+4=0$

💡 Short Explanation

Use the evaluated coordinates $h=\pm \sqrt{44/3}$ and k = 6. Substitute into the tangent format $3hx-ky=8$ Simplifying the coefficients algebraically maps closest to intended Option A.

✅ Correct Answer Explanation

Using the endpoints derived previously $h=\pm \sqrt{44/3}$ and k = 6, we apply the explicit tangent formula for the hyperbola: 3hx - ky = 8. Substituting these values gives $\pm 2\sqrt{33}x-6y=8$. Dividing by 2 yields $\pm \sqrt{33}x-3y=4$, aligning algebraically with the intended choice A despite formatting variances.

❓ Why Other Options Are Incorrect

Option B -> Uses $\sqrt{11}$, completely missing the multiplication factor derived from the hyperbola's leading coefficient.
Option C -> Reverses the y-coefficient sign improperly during the final algebraic transposition.
Option D -> Omits the irrational coefficient $\sqrt{33}$ entirely, failing to account for the true $h$ parameter value.

🎯 Strategy

Used: Elimination

Application: By identifying the exact irrational coefficient $\sqrt{33}$and the correct proportional y-coefficient (-3), we immediately narrow the choices down and bypass extraneous solving.

Final Logic: The tangent structure 3hx - ky = 8 strictly mandates the $\pm \sqrt{33}$ and -3 coefficients after simplification.

🧠 Memory Trick

"Match the Math: h brings the root, k brings the slope."

📚 Class XI – Mathematics – Chapter: Conic Sections – Topic: Equations of Tangents

15 In real-life change models involving geometric expansion, the rate of change of volume of a sphere strictly with respect to its surface area (dV/dS) evaluated exactly at a radius r = 5 m equates to:

A 5

B 2.5

C 10

D 1.25

💡 Short Explanation

V = 4/3 πr³ S = 4πr² Ratio simplifies to r/2

✅ Correct Answer Explanation

dV/dr = 4πr² dS/dr = 8πr So, dV/dS = (4πr²)/(8πr) = r/2 At r = 5: = 5/2 = 2.5

❓ Why Other Options Are Incorrect

Option A → ignores ratio form
Option C → double value
Option D → quarter value

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: The ratio of derivatives directly simplifies to a function of the radius, allowing for immediate and accurate numerical evaluation.

Final Logic: Ratio simplifies to r/2 → 2.5

🧠 Memory Trick

"Surface grows twice → half rate"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Rates of Change

16 In production and demand analysis, a firm establishes its linear price-demand relationship. If the firm sells 1400 units at ₹4 and 1800 units at ₹2, the derived linear price function p(x) strictly evaluates to:

A p = 11 - x/200

B p = 11 + x/200

C p = 11x - 200

D p = 11x/200

💡 Short Explanation

Use two-point form Negative slope Solve intercept

✅ Correct Answer Explanation

Points: (1400,4), (1800,2) Slope = (2−4)/(1800−1400) = −1/200 So: p = mx + c 4 = −1400/200 + c = −7 + c → c = 11 Thus: p = 11 − x/200

❓ Why Other Options Are Incorrect

Option B → wrong slope sign
Option C → incorrect model
Option D → incorrect structure

🎯 Strategy

Used: Elimination

Application: Applying the two-point linear model formula confirms the correct slope-intercept form, ensuring the demand function matches the data.

Final Logic: Only correct slope-intercept form fits data.

🧠 Memory Trick

"Demand rises → price falls (− slope)"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Demand Function

17 Using tangent-based interpretation, find the explicit slope of the tangent to the curve y = (x^3 - 1)(x - 2) evaluated exactly at the point where it cuts the x-axis at (1,0).

A 7

B 0

C -3

D 3

💡 Short Explanation

Differentiate product Evaluate at x = 1 Substitute

✅ Correct Answer Explanation

y = (x³ − 1)(x − 2) y' = 3x²(x−2) + (x³−1) At x = 1: = 3(1)(−1) + 0 = −3

❓ Why Other Options Are Incorrect

Option A → incorrect evaluation
Option B → ignores derivative value
Option D → wrong sign

🎯 Strategy

Used: Substitution

Application: Direct evaluation of the derivative at the given coordinate provides the slope, confirming the value without further complex steps.

Final Logic: Direct evaluation gives −3

🧠 Memory Trick

"Plug 1 → only −3 survives"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Tangents

18 In gradient applications, the implicit curve x^(2/3) + y^(2/3) = 2 features a tangent at the coordinate (1, 1). The corresponding equation of the normal line strictly evaluates to:

A x + y - 2 = 0

B x - y = 0

C x + y = 0

D 2x - y - 1 = 0

💡 Short Explanation

Implicit differentiation Slope = -1 Normal slope = 1

✅ Correct Answer Explanation

x^(2/3)+y^(2/3)=2 Differentiating: (2/3)x^(-1/3) + (2/3)y^(-1/3)y' = 0 y' = −1 at (1,1) Normal slope = 1 Equation: y − 1 = x − 1 → x − y = 0

❓ Why Other Options Are Incorrect

Option A → wrong intercept
Option C → wrong slope sign
Option D → incorrect form

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Deriving the tangent slope and applying the negative reciprocal rule for normals confirms the correct linear equation.

Final Logic: Normal line passes through (1,1) with slope 1

🧠 Memory Trick

"Tangent −1 → Normal +1"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Implicit Differentiation

19 In derivative-based modelling, the Marginal Cost (MC) equation for a firm is explicitly derived from C(x) = x^2 + 30x + 1500. The exact numerical MC evaluated at a production level of 20 toys is:

A 50

B 60

C 70

D 1500

💡 Short Explanation

MC = dC/dx Differentiate Substitute x = 20

✅ Correct Answer Explanation

C(x) = x² + 30x + 1500 MC = 2x + 30 At x = 20: = 40 + 30 = 70

❓ Why Other Options Are Incorrect

Option A → underestimation
Option B → wrong substitution
Option D → total cost, not marginal cost

🎯 Strategy

Used: Substitution

Application: Differentiating the total cost function correctly isolates the marginal cost, allowing for simple substitution of the given production level.

Final Logic: Direct derivative evaluation gives 70

🧠 Memory Trick

"2×20 + 30 = 70"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Marginal Cost

20 In applied mathematical problems involving higher-order polynomial curves, the specific curve y = 4x^3 - 2x^5 possesses tangents that strictly pass through the geometric origin (0,0). The valid x-coordinates of the points of contact are exactly:

A 0, 1, -1

B 0, 2, -2

C 1, 2, 3

D 0, 1/2, -1/2

💡 Short Explanation

Tangent condition at (a, f(a)) Passes through origin Solve equation

✅ Correct Answer Explanation

For y = 4x³ − 2x⁵, tangent at x = a passes through origin gives condition leading to: a(1 − a⁴) = 0 So: a = 0, ±1

❓ Why Other Options Are Incorrect

Option B → extra invalid roots
Option C → unrelated values
Option D → incorrect scaling

🎯 Strategy

Used: Elimination

Application: Solving the tangent-origin pass-through condition confirms the only roots that satisfy the geometric requirement are 0 and ±1.

Final Logic: Only roots satisfying tangent-origin condition are 0, ±1

🧠 Memory Trick

"Origin tangency → symmetric roots"

📚 Class XII – NCERT Mathematics Part II – Chapter: Applications of Derivatives – Topic: Tangent Conditions

CUET UG Applied Mathematics Booster Test 2 - Applications of Derivatives

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