CUET UG Applied Mathematics Booster Test 2

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CUET UG Applied Mathematics Booster Test 2 - Tangents and Normals

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Tangents and Normals

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QUESTION 1 OF 20

If the slope of a non-vertical tangent to the curve y = f(x) at point A is geometrically defined completely by the limiting position of a secant, which limit expression below strictly represents this secant and tangent explicit relation?

Answer Review

1 If the slope of a non-vertical tangent to the curve y = f(x) at point A is geometrically defined completely by the limiting position of a secant, which limit expression below strictly represents this secant and tangent explicit relation?

A Limit dx → 0 [f(x-dx) + f(x)] / dx

B Limit dx → 0 [f(x+dx) - f(x)] / dx

C Limit dx → 0 f(x+dx) / dx

D Limit dx → 0 [f(x) - f(x+dx)] / dx

💡 Short Explanation

Tangent slope is defined by derivative Based on secant slope difference quotient Forward difference form is standard definition

✅ Correct Answer Explanation

The slope of a tangent is defined as the limit of the slope of a secant line: ${lim⁡}_{dx\rightarrow 0}\frac{f(x+dx)-f(x)}{dx}$ This is the fundamental definition of derivative in NCERT calculus.

❓ Why Other Options Are Incorrect

Option A → Incorrect form; addition does not represent slope change
Option C → Missing reference point subtraction, not a secant slope
Option D → Reversed sign gives negative of correct slope

🎯 Strategy

Used: Elimination → Identify standard derivative definition form

Application: By explicitly looking for the standard forward difference formula for a derivative, this elimination strategy prevents confusion with options that have reversed signs, incorrect additions, or limits lacking proper reference points.

Final Logic: Secant slope → limit of difference quotient in forward form

🧠 Memory Trick

"Future minus present over step = derivative"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Derivative Definition

2 Match the complex algebraic curves strictly in List I with the exact analytical slope of their tangent evaluated at the explicitly given point in List II:

List I	List II
1. y = 4x^3 - 2x^5 at (1,2)	a. 3
2. 3x^2 - y^2 = 8 at (2,2)	b. 2
3. x^2 = 4y at (2,1)	c. 1
4. y = (x-1)/(x-2) at (10, 9/8)	d. -1/64

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

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

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

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

💡 Short Explanation

Differentiate each curve Substitute given points Match slopes with list

✅ Correct Answer Explanation

1 → derivative = 12x² − 10x⁴ → at x=1 gives 2 → (b) 2 → 6x − 2y y' = 0 → y' = 3x/y = 3 → (a) 3 → 2x = 4y' → y' = x/2 = 1 → (c) 4 → quotient rule → slope = -1/64 → (d)

❓ Why Other Options Are Incorrect

Option A → incorrect assignment for 1 and 2
Option B → swaps correct 2 and 3 mapping
Option C → incorrect placement of 2 and 3 slopes

🎯 Strategy

Used: Substitution + Matching → Compute slope individually then map

Application: Differentiating each equation separately and substituting the specific points guarantees an accurate slope for each curve. This systematic matching eliminates the risk of mixing up similar numeric values or guessing wrong combinations.

Final Logic: Direct differentiation + point substitution

🧠 Memory Trick

"Differentiate → Plug → Match"

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Tangents and Normals

3 Which of the following strict analytical properties explicitly define the mathematical definition of normal and its absolute characteristics?
1) The normal is perpendicular to tangent
2) If tangent slope is m, normal slope is -1/m
3) Normal always passes through origin
4) If tangent is vertical, normal is horizontal

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 4

💡 Short Explanation

Normal is perpendicular to tangent Slope relation holds Special vertical/horizontal cases apply

✅ Correct Answer Explanation

1 is true (definition of normal). 2 is true (negative reciprocal slope relation). 3 is false (normal does not necessarily pass origin). 4 is true (vertical tangent ⇒ horizontal normal).

❓ Why Other Options Are Incorrect

Option A → includes 3 (incorrect)
Option C → includes 3 (incorrect)
Option D → incomplete without 2

🎯 Strategy

Used: Elimination → Remove false universal statement

Application: By identifying the definitively false claim (that normals must pass through the origin), you can instantly eliminate any option grouping containing that statement, quickly isolating the correct set of perpendicular properties.

Final Logic: Normal = perpendicular line properties

🧠 Memory Trick

"Normal = Negative reciprocal + 90° rule"

📚 Class XI – NCERT Mathematics – Chapter: Straight Lines – Topic: Perpendicular Lines

4 Identify the explicitly incorrect statement strictly regarding the algebraic behaviour of the slope of normal.

A slope of normal = -dx/dy

B product of tangent and normal slope = 1

C tangent → ∞ implies normal → 0

D normal undefined when tangent horizontal

💡 Short Explanation

Tangent-normal product is -1 Not +1 Other properties are correct

✅ Correct Answer Explanation

Correct relation: $m_{t}⋅m_{n}=-1$ So, statement B is incorrect.

❓ Why Other Options Are Incorrect

A → correct reciprocal derivative form
C → valid limiting behaviour
D → correct due to vertical normal case

🎯 Strategy

Used: Concept Check → Verify slope identity

Application: Checking the fundamental mathematical condition for perpendicular lines

${(m}_{1}\times m_{2}=-1)$ immediately exposes the error in the option claiming a positive product, removing any confusion about basic normal-tangent relations.

Final Logic: Perpendicular slopes multiply to -1

🧠 Memory Trick

"Perpendicular means minus reciprocal"

📚 Class XI – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Normal Slope

5 For (x/a)^n + (y/b)^n = 2, evaluate dy/dx at (a,b).

A -b/a

B b/a

C -a/b

D a/b

💡 Short Explanation

Implicit differentiation Substitute point Simplify slope

✅ Correct Answer Explanation

Differentiating gives: $\frac{dy}{dx}=-\frac{b}{a}$

❓ Why Other Options Are Incorrect

B → wrong sign
C → inverted ratio
D → wrong direction

🎯 Strategy

Used: Direct differentiation

Application: Using implicit differentiation directly on the equation avoids the messy and error-prone process of trying to isolate $y$ algebraically, safely leading straight to the correct slope ratio.

Final Logic: Power rule + substitution

🧠 Memory Trick

"Power curve gives negative ratio"

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

6 Domain of y = (x-1)/(x-2)

A x > 2

B x = 2

C x ≠ 2

D x < 2

💡 Short Explanation

Denominator cannot be zero Exclude x = 2 All other real values allowed

✅ Correct Answer Explanation

Function undefined at x = 2, so domain excludes it.

❓ Why Other Options Are Incorrect

A → unnecessarily restrictive
B → incorrect single value
D → incomplete domain

🎯 Strategy

Used: Elimination

Application: By focusing solely on the restriction that a denominator cannot be zero, you can rapidly eliminate options that represent inequalities or single values, easily identifying the correct domain exclusion.

Final Logic: Denominator restriction

🧠 Memory Trick

"Denominator zero = forbidden value"

📚 Class XI – NCERT Mathematics – Chapter: Relations and Functions – Topic: Domain

7 Normal slope for (x/a)^n + (y/b)^n = 2 at (a,b)

A -b/a

B a/b

C -a/b

D b/a

💡 Short Explanation

Normal = negative reciprocal of tangent Tangent = -b/a So normal = a/b

✅ Correct Answer Explanation

From Q5, tangent slope = -b/a So normal slope = a/b

❓ Why Other Options Are Incorrect

A → tangent value, not normal
C → wrong sign
D → swapped relation

🎯 Strategy

Used: Reciprocal relation

Application: Taking the previously calculated tangent slope and directly applying the negative reciprocal rule bypasses a full recalculation, eliminating the chance of sign errors or swapped variables.

Final Logic: Normal = -1/(tangent slope)

🧠 Memory Trick

"Flip and sign change"

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

8 If normal slope = 3x/y, tangent slope is:

A 3x/y

B -y/(3x)

C y/(3x)

D -3x/y

💡 Short Explanation

Tangent = negative reciprocal of normal Apply inversion Maintain sign rule

✅ Correct Answer Explanation

$m_{t}=-\frac{1}{m_{n}}=-\frac{y}{3x}$

❓ Why Other Options Are Incorrect

A → same as normal
C → missing negative
D → incorrect inversion

🎯 Strategy

Used: Reciprocal relation

Application: Applying the perpendicular slope rule (negative reciprocal) directly to the given algebraic normal slope prevents confusion with inverse formulas that either forget the negative sign or fail to flip the fraction.

Final Logic: Perpendicular slope rule

🧠 Memory Trick

"Normal flips to tangent with minus sign"

📚 Class XI – NCERT Mathematics – Chapter: Straight Lines – Topic: Slope Relations

9 Equation of tangent at (x0, y0):

A f'(x0)(x-x0)

B f(x)(x-x0)

C (x-x0)/f'(x0)

D -1/f'(x0)(x-x0)

💡 Short Explanation

Point-slope form Uses derivative as slope Standard NCERT formula

✅ Correct Answer Explanation

$y-y_{0}=f^{'}(x_{0})(x-x_{0})$

❓ Why Other Options Are Incorrect

B → uses function instead of derivative
C → inverted form
D → incorrect slope inversion

🎯 Strategy

Used: Formula recall

Application: Recalling the standard point-slope equation utilizing the derivative $f'(x_0)$ confirms the precise algebraic structure, preventing selection of inverted slopes or non-differentiated function forms.

Final Logic: Tangent = point-slope equation

🧠 Memory Trick

"Derivative × displacement = tangent"

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

10 Tangent to (x/a)^n + (y/b)^n = 2 at (a,b):

A x/a - y/b = 2

B x/a + y/b = 2

C x/a + y/b = 1

D ax + by = 2

💡 Short Explanation

Standard superellipse tangent form Substitute point structure Linear form emerges

✅ Correct Answer Explanation

At (a,b), tangent simplifies to: $\frac{x}{a}+\frac{y}{b}=2$

❓ Why Other Options Are Incorrect

A → wrong sign
C → incorrect constant
D → incorrect scaling form

🎯 Strategy

Used: Standard result recognition

Application: Recognizing the established linear tangent format for superellipses evaluated at identical constants allows you to immediately pinpoint the simplified equation, bypassing lengthy algebraic derivation and potential arithmetic mistakes.

Final Logic: Known tangent form of superellipse

🧠 Memory Trick

"Same structure, same sum form"

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

11 For the implicit parabolic curve x² = 4y passing identically through (1,2), find the proper algebraic equation of normal line strictly intersecting that precise point.

A x + y - 3 = 0

B x - y + 1 = 0

C 2x + y - 4 = 0

D x - 2y + 3 = 0

💡 Short Explanation

Differentiate implicitly Find tangent slope Use perpendicular relation for normal

✅ Correct Answer Explanation

Given x² = 4y Differentiate: 2x = 4 dy/dx → dy/dx = x/2 At (1,2): slope = 1/2 Normal slope = -2 Equation: y - 2 = -2(x - 1) → x + y - 3 = 0

❓ Why Other Options Are Incorrect

B → incorrect slope structure
C → not passing through point correctly
D → wrong slope ratio

🎯 Strategy

Used: Substitution + slope conversion

Application: Finding the tangent slope first and consciously converting it to the perpendicular normal slope ensures the correct gradient is plugged into the linear equation, avoiding the common mistake of building the line with the tangent's slope.

Final Logic: normal = perpendicular to tangent

🧠 Memory Trick

"Differentiate → flip → line form"

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Tangent and Normal

12 The unified explicit tangent and normal pair for a point on a differentiable curve will always intersect at:

A 0°

B 45°

C 90°

D 180°

💡 Short Explanation

Tangent and normal are perpendicular Angle always right angle Definition-based result

✅ Correct Answer Explanation

By definition, normal is perpendicular to tangent, so angle between them is 90°.

❓ Why Other Options Are Incorrect

A → parallel lines case (not valid)
B → incorrect geometric relation
D → straight line, not perpendicular

🎯 Strategy

Used: Concept definition

Application: Relying strictly on the geometric definition—that a normal line is always perpendicular to its tangent—instantly confirms a 90-degree intersection, completely eliminating the need for complex analytical coordinate geometry.

Final Logic: perpendicular lines form right angle

🧠 Memory Trick

"Normal means right angle"

📚 Class XI – NCERT Mathematics – Chapter: Straight Lines – Topic: Angle Between Lines

📄 Passage

Implicit curve tangents and structurally formulated normal to implicit curves completely require differentiating the composite algebraic equation directly on both sides, as the dependent variable y is not strictly isolated. For example, considering two separate implicit curves xy = a^2 and x^2 + y^2 = 2a^2, finding their respective tangents logically requires the product rule and standard chain rule respectively. For the first inverse curve xy = a^2, implicit differentiation analytically yields x(dy/dx) + y = 0, properly giving a tangent slope of -y/x. For the circular second curve x^2 + y^2 = 2a^2, the resultant slope is strictly isolated as -x/y. When these two varying curves share a common coordinate point entirely, we can safely evaluate if they mathematically touch each other by strictly comparing their derived respective tangent slopes at the exact intersection.

Based implicitly on the passage, what is the proper resultant implicit tangent slope (dy/dx) explicitly yielded for the curve x^2 + y^2 = 2a^2 evaluated at a generic geometric point (x, y)?

A y/x

B -x/y

C -y/x

D x/y

💡 Short Explanation

Implicit differentiation Apply chain rule Solve for dy/dx

✅ Correct Answer Explanation

Differentiate: 2x + 2y dy/dx = 0 dy/dx = -x/y

❓ Why Other Options Are Incorrect

A → wrong sign
C → inverted ratio
D → incorrect direction

🎯 Strategy

Used: Implicit differentiation

Application: Applying the chain rule to the non-isolated variables safely processes the geometry of the circle, ensuring the derived $dy/dx$ ratio carries the correct negative sign and orientation.

Final Logic: circle slope formula

🧠 Memory Trick

"Circle slope = -x/y"

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

📄 Passage

Based solely on the passage's structural formulation, what is the exact normal to implicit curves algebraic slope explicitly for the inverse curve xy = a^2 evaluated at a generic mathematical point (x, y)?

A -y/x

B -x/y

C x/y

D y/x

💡 Short Explanation

First find tangent slope Then take negative reciprocal Simplify

✅ Correct Answer Explanation

For xy = a²: x dy/dx + y = 0 → dy/dx = -y/x Normal slope = x/y

❓ Why Other Options Are Incorrect

A → tangent slope, not normal
B → wrong inversion
D → incorrect sign

🎯 Strategy

Used: Reciprocal relation

Application: Taking the algebraic tangent slope derived from the product rule and simply flipping it with a sign change yields the exact normal slope, eliminating redundant derivative steps and avoiding algebraic inversion errors.

Final Logic: normal = -1/tangent slope

🧠 Memory Trick

"Swap and flip sign"

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

15 If a complex parametric curve is strictly algebraically defined by x = 1/(1+t) and y = t/(1+t), the fully extracted tangent from parametric equations slope dy/dx explicitly evaluates numerically to:

A -1

B 1

C t

D 1/t

💡 Short Explanation

Use parametric differentiation dy/dt ÷ dx/dt Simplify ratio

✅ Correct Answer Explanation

dx/dt = -1/(1+t)² dy/dt = 1/(1+t)² dy/dx = -1

❓ Why Other Options Are Incorrect

B → wrong sign
C → unrelated variable
D → incorrect ratio

🎯 Strategy

Used: Parametric differentiation

Application: Structuring the slope as $dy/dt$ divided by $dx/dt$ correctly merges the independent rates of change. This strategy highlights how identical algebraic magnitude terms cancel out, leaving the exact constant numerical slope without confusion.

Final Logic: identical magnitudes cancel leaving -1

🧠 Memory Trick

"Same structure → minus one"

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

16 Considering strictly the proper mathematical normal from parametric equations isolated for x = at^2 and y = 2at, what is the exact algebraic normal equation explicitly passing through generic moving parameter 't'?

A y + tx = 2at + at³

B y - tx = 2at - at³

C ty = x + at²

D x + ty = at³

💡 Short Explanation

Find dy/dx from parametric form Use normal slope Apply point-slope form

✅ Correct Answer Explanation

dx/dt = 2at, dy/dt = 2a dy/dx = 1/t Normal slope = -t Equation: y - 2at = -t(x - at²) → y + tx = 2at + at³

❓ Why Other Options Are Incorrect

B → wrong sign arrangement
C → not a line equation form
D → incorrect structure

🎯 Strategy

Used: Parametric + slope method

Application: Finding the normal slope from parametric derivatives first, then systematically substituting it along with the given $x$ and $y$ parametric coordinates into the point-slope form structurally secures the correct polynomial expansion.

Final Logic: normal via perpendicular slope

🧠 Memory Trick

"Parametric → differentiate → substitute"

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Parametric Tangent/Normal

17 Find the strict explicit x-coordinate locations where mathematical tangents parallel to axis (x-axis) explicitly occur for the cubic algebraic curve y = x³/3 - 5x²/2 + 6x + 4

A 2, 3

B 1, 6

C 0, 5

D -2, -3

💡 Short Explanation

Set dy/dx = 0 Solve quadratic Find x-values

✅ Correct Answer Explanation

dy/dx = x² - 5x + 6 = 0 (x-2)(x-3)=0 → x=2,3

❓ Why Other Options Are Incorrect

B → incorrect roots
C → unrelated values
D → wrong sign roots

🎯 Strategy

Used: Factorization

Application: Setting the first derivative to zero (the condition for a horizontal line) and factoring the resulting quadratic mathematically pinpoints the exact x-coordinates, removing guesswork from finding where the curve flattens out.

Final Logic: slope zero condition

🧠 Memory Trick

"Horizontal tangent → derivative zero"

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

18 A parabolic curve is algebraically strictly defined by y = x² - 4x + 5. Find the strict geometric coordinate point where the unique tangents perpendicular to lines precisely possessing analytical slope -1/2 are strictly formed.

A (3,2)

B (2,1)

C (1,2)

D (0,5)

💡 Short Explanation

Differentiate function Set slope = -1/2 Solve x

✅ Correct Answer Explanation

dy/dx = 2x - 4 2x - 4 = -1/2 → x = 3/2? wait check carefully: Correct solving: 2x - 4 = -1/2 2x = 7/2 → x = 7/4 Now check point: y = (7/4)² - 4(7/4) + 5 = 49/16 - 7 + 5 = 49/16 - 32/16 = 17/16 So correct point is (7/4, 17/16) which is NOT in options.

❓ Why Other Options Are Incorrect

A → slope mismatch
B → slope mismatch
C → incorrect derivative condition
D → not satisfying slope condition

🎯 Strategy

Used: Direct solving

Application: Setting the algebraic derivative directly equal to the required numerical slope explicitly creates an equation to solve for $x$. This provides a concrete mathematical coordinate to check against the options, rather than testing options blindly.

Final Logic: derivative condition mismatch with options

🧠 Memory Trick

"Set slope → solve x → plug back"

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Tangent Slope Condition

19 The generalized parabolic curve y = x² + 3x + 4 possesses specific geometric tangents through external points, specifically originating strictly through the exact origin (0,0). What are the exact strict algebraic equations of these intersecting tangent lines?

A y = 7x and y = -x

B y = 5x and y = -2x

C y = 4x and y = x

D y = 3x and y = -3x

💡 Short Explanation

Use tangent condition through origin Solve quadratic condition Find slopes

✅ Correct Answer Explanation

Let tangent slope m satisfy: y = mx passes through parabola condition → m = 7 and m = -1 Thus tangents: y = 7x, y = -x

❓ Why Other Options Are Incorrect

B → wrong roots
C → incorrect slopes
D → not satisfying tangent condition

🎯 Strategy

Used: Condition of tangency

Application: Assuming the general form of a line through the origin ($y=mx$) and forcing its intersection with the parabola to yield a discriminant of zero completely isolates the specific slopes, cutting out non-tangent secant lines.

Final Logic: line y=mx intersects parabola with discriminant zero

🧠 Memory Trick

"Through origin → assume y = mx"

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Tangent from external point

20 For the higher-order polynomial y = 4x³ - 2x⁵, geometric tangents through origin (0,0) explicitly contact the continuous curve at multiple distinct points. What is the exact derived numerical value of the unique tangent's slope strictly evaluated at the explicit non-zero coordinate (1, 2)?

A 2

B 1

C 0

D 4

💡 Short Explanation

Differentiate function Substitute x = 1 Evaluate slope

✅ Correct Answer Explanation

dy/dx = 12x² - 10x⁴ At x=1 → 12 - 10 = 2

❓ Why Other Options Are Incorrect

B → incorrect evaluation
C → slope not zero
D → incorrect derivative result

🎯 Strategy

Used: Direct differentiation

Application: Computing the derivative and strictly substituting the provided specific coordinate (x=1) cuts through the distracting information about multiple origin tangents, yielding the exact instantaneous slope at that single point.

Final Logic: plug x into derivative

🧠 Memory Trick

"Differentiate then substitute"

📚 Class XII – NCERT Mathematics – Chapter: Applications of Derivatives – Topic: Tangent Slope Evaluation

CUET UG Applied Mathematics Booster Test 2 - Tangents and Normals

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