CUET UG Applied Mathematics Booster Test 1

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CUET UG Applied Mathematics Booster Test 1 - 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 1 - Tangents and Normals

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QUESTION 1 OF 20

If the strictly measured angle of inclination of a secant line AB is theta, and A is the initial point (x, f(x)), what does the algebraic form \([f(x+dx)-f(x)]/dx\) geometrically equal?

Answer Review

1 If the strictly measured angle of inclination of a secant line AB is theta, and A is the initial point (x, f(x)), what does the algebraic form \([f(x+dx)-f(x)]/dx\) geometrically equal?

A sin(theta)

B cos(theta)

C tan(theta)

D cot(theta)

💡 Short Explanation

The expression \(\frac{f(x+dx)-f(x)}{dx}\)represents the slope of the secant line. Geometrically, slope equals the tangent of the angle of inclination. Slope = rise/run. Inclination defines slope. Slope equals tan θ.

✅ Correct Answer Explanation

The algebraic expression \(\frac{f(x+dx)-f(x)}{dx}\) represents the slope of the secant line joining two nearby points. The slope of any line making an angle θ with the positive x-axis equals tan θ. Therefore Option C is correct. Options A, B, and D represent other trigonometric ratios unrelated to slope.

❓ Why Other Options Are Incorrect

Option A → sin(theta) represents the sine ratio, not the slope of a line.
Option B → cos(theta) gives the cosine ratio and does not measure inclination.
Option D → cot(theta) is the reciprocal of the slope, not the slope itself.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recognize that "angle of inclination" is directly associated with the slope of a line.

Final Logic: Slope of a line = tan θ, so Option C is correct.

🧠 Memory Trick

"Slope ⇒ Tan θ."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Geometrical Meaning of Derivative

2 Match the algebraic curves in List I with the numerical slope of their tangent at the given point in List II:

List I	List II
1. y = x³ − 3x + 5 at (2,7)	a. 1
2. y = (x³ − 1)(x − 2) at (1,0)	b. 6
3. y = 3x² − 6x at (2,0)	c. -3
4. y = eˣ + x² + 1 at (0,2)	d. 9

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

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

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

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

💡 Short Explanation

Differentiate each function and substitute the given x-coordinate. The resulting slopes are 9, −3, 6, and 1 respectively, matching Option C. Differentiate each curve. Evaluate at given x-values. Match corresponding slopes.

✅ Correct Answer Explanation

Differentiate each function: (1) \(3x^{2}-3=9\) at \(x=2\). (2) Product rule gives \(-3\) at \(x=1\). (3) \(6x-6=6\) at \(x=2\). (4) \(e^{x}+2x=1\) at \(x=0\). Hence the correct matching is 1-b, 2-a, 3-d, 4-c, making Option C correct.

❓ Why Other Options Are Incorrect

Option A assigns incorrect slopes to the functions.
Option B mismatches every derivative value.
Option D does not correspond to the evaluated derivatives obtained through differentiation.

🎯 Strategy

Used: Substitution

Application: Differentiate each expression first and substitute the specified x-value.

Final Logic: Only Option C correctly matches all four calculated slopes.

🧠 Memory Trick

"Differentiate → Substitute → Match."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Derivative as Slope of Tangent

3 Which of the following conditions correctly describe the geometric properties of a normal to a curve y = f(x)?
1) The normal is a straight line.
2) It is absolutely perpendicular to the tangent at the exact point of contact.
3) Its slope is strictly defined as zero when the tangent is parallel to the y-axis.
4) Its slope is universally positive across all coordinate quadrants.

A 1, 2, 3

B 1, 2

C 2, 3, 4

D 1, 2, 3, 4

💡 Short Explanation

A normal is a straight line perpendicular to the tangent. If the tangent is vertical, the normal becomes horizontal, giving slope zero. Statement 4 is incorrect. Normal is straight. Perpendicular to tangent. Horizontal normal has zero slope.

✅ Correct Answer Explanation

A normal is the line perpendicular to the tangent at the point of contact, so 1 and 2 are true. If the tangent is parallel to the y-axis (vertical), the normal is horizontal and its slope is zero, making 3 true. 4 is false because a normal's slope may be positive, negative, zero, or undefined. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option B → 1, 2 ignores the true statement that a vertical tangent gives a horizontal normal with slope zero.
Option C → 2, 3, 4 wrongly includes statement 4 while excluding statement 1.
Option D → 1, 2, 3, 4 is incorrect because statement 4 is false.

🎯 Strategy

Used: Elimination

Application: Check each statement using the geometric definition of the normal.

Final Logic: Only statements 1, 2, and 3 are true, so Option A is correct.

🧠 Memory Trick

"Normal = Perpendicular Partner."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Tangent and Normal

4 Identify the incorrect statement concerning normal slopes and their specific geometric interpretations.

A If dy/dx = 0 at a point, the corresponding normal line is entirely parallel to the y-axis.

B The equation of the normal strictly utilizes the negative reciprocal of the tangent slope.

C The slope of the normal is always numerically greater than the algebraic slope of the tangent.

D The tangent and normal equations share the exact same coordinate contact point (x0, y0).

💡 Short Explanation

The normal slope is the negative reciprocal of the tangent slope whenever defined. Its magnitude is not always greater than the tangent's. Hence statement C is incorrect. Normal uses negative reciprocal. Magnitude is not always greater. Compare reciprocal values carefully.

✅ Correct Answer Explanation

If the tangent slope is \(m\), the normal slope is \(-\frac{1}{m}\)(provided \(m\neq 0\)). There is no rule stating that the numerical value of the normal slope is always greater than the tangent slope. Thus Option C is incorrect. Options A, B, and D correctly describe standard properties of tangent and normal lines.

❓ Why Other Options Are Incorrect

Option A → Correct because when \(dy/dx=0\), the tangent is horizontal and the normal is vertical (parallel to the y-axis).
Option B → Correct since the normal slope is the negative reciprocal of the tangent slope whenever it is defined.
Option D → Correct because both the tangent and normal pass through the same point of contact on the curve.

🎯 Strategy

Used: Extreme Word Filter

Application: Look for absolute words like "always." Such statements are often incorrect unless they hold for every possible case.

Final Logic: The word "always" makes Option C false because the normal's slope is not universally greater than the tangent's.

🧠 Memory Trick

"Normal = −1/m, not always bigger."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Tangent and Normal

5 Evaluate the exact numerical value of the tangent slope at the given coordinate point (2, 0) for the polynomial curve \(y=(x^{3}-1)(x-2)\).

A 7

B -3

C 0

D 1

💡 Short Explanation

Differentiate the function using the product rule. Substitute \(x=2\) into the derivative. The resulting slope of the tangent equals 7. Apply the product rule. Substitute \(x=2\). Evaluate the derivative.

✅ Correct Answer Explanation

Using the product rule, \(\frac{dy}{dx}=3x^{2}(x-2)+(x^{3}-1).\) At \(x=2\), \(\frac{dy}{dx}=3(4)(0)+(8-1)=7.\) Hence the tangent slope is 7, making Option A correct. Options B, C, and D result from incorrect differentiation or substitution.

❓ Why Other Options Are Incorrect

Option B → -3 results from an incorrect application of the product rule or substitution.
Option C → 0 ignores the contribution of the second product-rule term.
Option D → 1 is obtained through incorrect derivative evaluation and does not equal the actual slope.

🎯 Strategy

Used: Substitution

Application: Differentiate the function first, then substitute the given coordinate into the derivative.

Final Logic: Evaluating the derivative at \(x=2\) gives 7, so Option A is correct.

🧠 Memory Trick

"Product Rule → Differentiate both factors."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Slope of Tangent Using Derivatives

6 For the rational algebraic function \(y=\frac{x-1}{x-2}\), evaluating the tangent through differentiation yields a dynamic slope. At which specific constraint region is the tangent's slope strictly evaluated as \(-\frac{1}{64}\)?

A x = 10

B x = 2

C x = -8

D x = 8

💡 Short Explanation

Differentiate using the quotient rule. The derivative becomes \(-1/(x-2)^{2}\). Substituting the given x-values shows only \(x=10\) gives the slope \(-1/64\). Apply quotient rule. Simplify derivative. Substitute options.

✅ Correct Answer Explanation

For \(y=\frac{x-1}{x-2},\frac{dy}{dx}=\frac{\left(x-2)-(x-1\right)}{{\left(x-2\right)}^{2}}=-\frac{1}{{\left(x-2\right)}^{2}}.\) At \(x=10\), \(-\frac{1}{{\left(10-2\right)}^{2}}=-\frac{1}{64}.\) Hence Option A is correct. Option B is undefined since the function is not defined at \(x=2\). Options C and D give different derivative values.

❓ Why Other Options Are Incorrect

Option B → x = 2 is invalid because the denominator becomes zero, making the function and derivative undefined.
Option C → x = -8 gives \(-1/100\), not \(-1/64\).
Option D → x = 8 gives \(-1/36\), not the required value.

🎯 Strategy

Used: Substitution

Application: Differentiate once and substitute each option into the derivative.

Final Logic: Only \(x=10\) satisfies the required derivative value.

🧠 Memory Trick

"Quotient Rule → Check denominator square."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Derivatives of Rational Functions

7 Based purely on the numerical tangent slope for the curve \(y=(x^{3}-1)(x-2)\)at \(x=1\), what is the exact normal slope evaluated at that specific given point?

A -3

B 1/3

C -1/3

D 3

💡 Short Explanation

The tangent slope at \(x=1\) is \(-3\). The normal slope is the negative reciprocal of the tangent slope, giving \(1/3\). Find tangent slope. Take negative reciprocal. Simplify.

✅ Correct Answer Explanation

Using the derivative, \(\frac{dy}{dx}=3x^{2}(x-2)+(x^{3}-1).\) At \(x=1\), \(\frac{dy}{dx}=-3.\) The normal slope is \(-\frac{1}{-3}=\frac{1}{3}.\) Thus Option B is correct. Options A, C, and D do not satisfy the negative reciprocal relationship.

❓ Why Other Options Are Incorrect

Option A → -3 is the tangent slope, not the normal slope.
Option C → -1/3 has the wrong sign.
Option D → 3 is neither the tangent slope nor its negative reciprocal.

🎯 Strategy

Used: Substitution

Application: Compute the tangent slope first and then use the negative reciprocal formula.

Final Logic: Normal slope \(=-1/m=1/3\), making Option B correct.

🧠 Memory Trick

"Normal = Negative Reciprocal."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Tangent and Normal

8 An unknown curve has a tangent strictly perpendicular to the given line \(x+14y+3=0\). To properly satisfy the perpendicular slope relation, what must be the derived algebraic slope of the curve's tangent?

A -1/14

B 14

C -14

D 1/14

💡 Short Explanation

The given line has slope \(-1/14\). Perpendicular lines have slopes whose product equals \(-1\). Therefore, the tangent slope must be \(14\). Find given slope. Apply perpendicular condition. Compute reciprocal.

✅ Correct Answer Explanation

The line \(x+14y+3=0\) can be written as \(y=-\frac{1}{14}x-\frac{3}{14}.\) Its slope is \(-1/14\). A perpendicular line has slope \(14.\) Hence Option B is correct. Options A, C, and D do not satisfy the perpendicular slope condition.

❓ Why Other Options Are Incorrect

Option A → -1/14 is the slope of the given line itself.
Option C → -14 has the wrong sign.
Option D → 1/14 is not the negative reciprocal.

🎯 Strategy

Used: Elimination

Application: Determine the given slope first, then eliminate options that are not negative reciprocals.

Final Logic: Only 14 satisfies \(m_{1}m_{2}=-1\).

🧠 Memory Trick

"Perpendicular → Multiply slopes = -1."

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

9 If a differentiable curve passes identically through the origin (0,0) and its tangent possesses a constant slope of 5, what is the point slope form equation summarizing this tangent?

A y = 5

B y = 5x

C x = 5y

D y - 5 = x

💡 Short Explanation

Using the point-slope equation with slope \(5\) through the origin gives \(y-0=5(x-0)\), which simplifies to \(y=5x\). Use point-slope form. Substitute origin. Simplify.

✅ Correct Answer Explanation

The point-slope equation is \(y-y_{1}=m(x-x_{1}).\) Using point \(\left(0\ ,\ 0\right)\)and slope \(5\), \(y=5x.\) Thus Option B is correct. Options A, C, and D do not pass through the origin with slope \(5\).

❓ Why Other Options Are Incorrect

Option A → y = 5 is a horizontal line with slope zero.
Option C → x = 5y has slope \(1/5\), not \(5\).
Option D → y - 5 = x has slope \(1\) and does not pass through the origin.

🎯 Strategy

Used: Substitution

Application: Apply the point-slope formula using the given point and slope.

Final Logic: The equation simplifies directly to \(y=5x\).

🧠 Memory Trick

"Origin + slope m ⇒ y = mx."

📚 Class XI – NCERT Mathematics – Chapter: Straight Lines – Topic: Point-Slope Form

10 Determine the precise equation of tangent isolated for the polynomial curve \(y=3x^{2}-6x\) evaluated at the explicit coordinate \(x=2\).

A 6x - y - 12 = 0

B x - 6y + 12 = 0

C 6x + y - 12 = 0

D y = 6x + 12

💡 Short Explanation

Differentiate the curve to obtain slope \(6\) at \(x=2\). The point is \(\left(2\ ,\ 0\right)\). Applying point-slope form gives the tangent equation. Differentiate. Find point. Apply point-slope equation.

✅ Correct Answer Explanation

For \(y=3x^{2}-6x,\frac{dy}{dx}=6x-6.\) At \(x=2\), the slope is \(6\). Since \(y(2)=0,\) the tangent is \(y=6(x-2),\) or \(6x-y-12=0.\) Hence Option A is correct. Options B, C, and D have incorrect slope or intercept.

❓ Why Other Options Are Incorrect

Option B → x - 6y + 12 = 0 has slope \(1/6\), not \(6\).
Option C → 6x + y - 12 = 0 has slope \(-6\), opposite to the required slope.
Option D → y = 6x + 12 has the correct slope but incorrect intercept, so it does not pass through \(\left(2\ ,\ 0\right)\).

🎯 Strategy

Used: Substitution

Application: Differentiate, substitute the given coordinate, and use the point-slope formula.

Final Logic: Slope \(=6\) through \(\left(2\ ,\ 0\right)\)gives \(6x-y-12=0\).

🧠 Memory Trick

"Derivative → Slope → Point-Slope."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Equation of Tangent

11 For the curve \(y=3x^{2}-6x\) resting at the coordinate \(x=2\), find the strict explicit equation of normal.

A x + 6y - 2 = 0

B 6x + y + 2 = 0

C x - 6y + 2 = 0

D 6x - y - 2 = 0

💡 Short Explanation

Differentiate the curve to obtain the tangent slope. The normal slope is the negative reciprocal of the tangent slope. Using point-slope form gives the required normal equation. Differentiate the curve. Find the normal slope. Apply point-slope form.

✅ Correct Answer Explanation

For \(y=3x^{2}-6x,\frac{dy}{dx}=6x-6.\) At \(x=2\), the tangent slope is \(6\), so the normal slope is \(-\frac{1}{6}\). The point is \(\left(2\ ,\ 0\right)\). Using \(y-0=-\frac{1}{6}(x-2),\) we obtain \(x+6y-2=0.\) Hence Option A is correct. Options B, C, and D have incorrect slopes or do not pass through the given point.

❓ Why Other Options Are Incorrect

Option B → 6x + y + 2 = 0 has slope \(-6\), not \(-1/6\).
Option C → x - 6y + 2 = 0 has slope \(1/6\), giving the wrong sign.
Option D → 6x - y - 2 = 0 has slope \(6\), which is the tangent slope, not the normal slope.

🎯 Strategy

Used: Substitution

Application: Differentiate to find the tangent slope, determine the normal slope, and substitute the point into the point-slope equation.

Final Logic: Normal slope \(=-1/6\) through \(\left(2\ ,\ 0\right)\)gives \(x+6y-2=0\).

🧠 Memory Trick

"Normal = Negative Reciprocal + Point."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Equation of Normal

12 For the polynomial curve \(y=x^{3}-3x+5\) at the geometric point \(\left(2\ ,\ 7\right)\), the tangent and normal pair equations are respectively \(y=9x-11\) and \(x+9y-65=0\). What is the explicit value of the tangent's slope integrated into forming these formulas?

A 11

B 9

C -9

D 1/9

💡 Short Explanation

Differentiate the polynomial to obtain the tangent slope. Evaluating the derivative at the given point gives \(9\), matching the coefficient in the tangent equation. Differentiate the function. Substitute \(x=2\). Match with tangent equation.

✅ Correct Answer Explanation

For \(y=x^{3}-3x+5,\frac{dy}{dx}=3x^{2}-3.\) At \(x=2\), \(3(2)^{2}-3=12-3=9.\) The tangent equation \(y=9x-11\) also confirms slope \(9\). Thus Option B is correct. Options A, C, and D do not equal the derivative at the specified point.

❓ Why Other Options Are Incorrect

Option A → 11 is the intercept-related value, not the tangent slope.
Option C → -9 has the wrong sign and contradicts the derivative.
Option D → 1/9 is the reciprocal, corresponding to neither the tangent slope nor the given equation.

🎯 Strategy

Used: Substitution

Application: Differentiate the polynomial and substitute the given x-coordinate.

Final Logic: The derivative at \(x=2\) equals 9, making Option B correct.

🧠 Memory Trick

"Differentiate First, Then Substitute."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Tangent and Normal

📄 Passage

When a curve is defined by a generalized implicit equation where y cannot be easily isolated, such as x² = 4y or x^(2/3) + y^(2/3) = 2, we must differentiate both sides uniformly with respect to x, structurally treating y as a composite function of x. This allows us to systematically extract dy/dx directly from the expression. Once dy/dx is successfully obtained, we substitute the specific coordinate values (x0, y0) to find the strict numerical slope of the tangent. From there, forming the respective tangent to implicit curves and the corresponding normal to implicit curves follows the standard point-slope algebraic formulation without exception.

Based strictly on the passage, what is the resultant evaluated slope of the tangent to the implicit curve x² = 4y explicitly at the specific point (2, 1)?

A 1

B 2

C 1/2

D -1

💡 Short Explanation

Differentiate the implicit equation with respect to x. Evaluate the derivative at the given point to obtain the tangent slope directly using implicit differentiation. Differentiate both sides. Find \(dy/dx\). Substitute the point.

✅ Correct Answer Explanation

Given \(x^{2}=4y,\) differentiate both sides: \(2x=4\frac{dy}{dx}.\) Thus, \(\frac{dy}{dx}=\frac{x}{2}.\) At \(x=2\), \(\frac{dy}{dx}=1.\) Therefore Option A is correct. Options B, C, and D do not satisfy the derivative evaluated at the given point.

❓ Why Other Options Are Incorrect

Option B → 2 results from incorrect simplification of the derivative.
Option C → 1/2 ignores substituting the correct x-coordinate.
Option D → -1 has the wrong sign since the derivative is positive.

🎯 Strategy

Used: Substitution

Application: Differentiate implicitly first, then substitute the coordinates into the derivative.

Final Logic: \(\frac{dy}{dx}=x/2\), and at \(x=2\), the slope is 1.

🧠 Memory Trick

"Implicit → Differentiate Both Sides."

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

📄 Passage

Based structurally on the passage, what is the resultant explicit equation of the normal to the implicit curves x² = 4y directly passing through the contact point (2, 1)?

A x + y - 3 = 0

B x - y - 1 = 0

C 2x + y - 5 = 0

D x + 2y - 4 = 0

💡 Short Explanation

The tangent slope at the point is 1. Therefore, the normal slope is −1. Using point-slope form through (2,1) gives the required equation. Find tangent slope. Compute normal slope. Use point-slope form.

✅ Correct Answer Explanation

From Question 13, the tangent slope is \(m=1.\) Hence the normal slope is \(-\frac{1}{1}=-1.\) Using point-slope form, \(y-1=-1(x-2).\) Simplifying, \(x+y-3=0.\) Thus Option A is correct. Options B, C, and D either have incorrect slopes or fail to pass through the given point.

❓ Why Other Options Are Incorrect

Option B → x − y − 1 = 0 has slope \(1\), which is the tangent slope.
Option C → 2x + y − 5 = 0 has slope \(-2\), not the required normal slope.
Option D → x + 2y − 4 = 0 has slope \(-1/2\), which is incorrect.

🎯 Strategy

Used: Substitution

Application: Use the tangent slope from implicit differentiation and substitute the contact point into the point-slope equation.

Final Logic: Normal slope = −1 through (2,1), giving \(x+y-3=0\).

🧠 Memory Trick

"Normal = Negative Reciprocal."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Tangent and Normal to Implicit Curves

15 Extract the generalized tangent from parametric equations \(x=at^{2}, y=2at\) at the theoretical generic point 't'. What is the simplified linear equation of the tangent?

A ty = x + at²

B y = tx + at²

C ty = x - at²

D x - ty + at² = 0

💡 Short Explanation

Differentiate the parametric equations to find the tangent slope. Apply the point-slope equation at the parameter \(t\) and simplify to obtain the tangent equation. Differentiate parametrically. Find slope. Simplify the equation.

✅ Correct Answer Explanation

Given \(x=at^{2},y=2at,\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac{1}{t}.\) Using point-slope form at \(\left(at^{2}\ ,\ 2at\right)\), \(y-2at=\frac{1}{t}(x-at^{2}),\) which simplifies to \(ty=x+at^{2}.\) Hence Option A is correct. Option D is an equivalent rearrangement of Option A, but the standard simplified form given in NCERT is Option A.

❓ Why Other Options Are Incorrect

Option B → y = tx + at² uses an incorrect slope and does not satisfy the tangent equation.
Option C → ty = x − at² has the wrong constant term.
Option D → x − ty + at² = 0 is algebraically equivalent to Option A, but the question asks for the simplified linear equation, making Option A the preferred answer.

🎯 Strategy

Used: Substitution

Application: Differentiate the parametric equations and substitute the parametric point into the point-slope form.

Final Logic: The tangent simplifies to \(ty=x+at^{2}\), so Option A is selected.

🧠 Memory Trick

"Parametric → dy/dx first, then Point-Slope."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Tangents to Parametric Curves

16 Considering the normal from parametric equations for \(x=at^{2}, y=2at\) at the exact generic point 't', the explicit slope of the normal evaluated algebraically is:

A 1/t

B t

C -t

D -1/t

💡 Short Explanation

For the parametric curve, the tangent slope is \(1/t\). The normal slope is the negative reciprocal of the tangent slope, giving \(-t\). Find \(dy/dx\). Take the negative reciprocal. Simplify.

✅ Correct Answer Explanation

For \(x=at^{2}, y=2at\), \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2a}{2at}=\frac{1}{t}.\) Hence the normal slope is \(-\frac{1}{1/t}=-t.\) Therefore Option C is correct. Options A, B, and D do not represent the negative reciprocal of the tangent slope.

❓ Why Other Options Are Incorrect

Option A → 1/t is the tangent slope, not the normal slope.
Option B → t has the wrong sign.
Option D → -1/t is not the reciprocal of the tangent slope.

🎯 Strategy

Used: Substitution

Application: Find the tangent slope first, then apply the negative reciprocal rule.

Final Logic: Normal slope \(=-t\), so Option C is correct.

🧠 Memory Trick

"Normal = Negative Reciprocal."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Parametric Curves

17 Locate the exact x-coordinates of the points on the polynomial curve \(y=2x^{3}-15x^{2}+36x-21\) where the tangents parallel to axis (x-axis) dynamically naturally occur.

A 2, 3

B 1, 4

C 0, 5

D -2, -3

💡 Short Explanation

Tangents parallel to the x-axis occur where the derivative equals zero. Differentiate the polynomial and solve the resulting quadratic equation. Differentiate. Set derivative to zero. Solve for x.

✅ Correct Answer Explanation

\(y^{'}=6x^{2}-30x+36=6(x-2)(x-3).\) Setting \(y^{'}=0\) gives \(x=2, 3.\) Hence Option A is correct. Options B, C, and D do not satisfy the derivative equation. 3Why Other Options Are Incorrect Option B → 1, 4 do not make the derivative zero. Option C → 0, 5 are not stationary points. Option D → -2, -3 also fail the derivative condition.

🎯 Strategy

Used: Substitution

Application: Differentiate and solve the stationary-point equation.

Final Logic: Only \(x=2\) and \(x=3\) satisfy \(y^{'}=0\).

🧠 Memory Trick

"Horizontal Tangent ⇒ Derivative = 0."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Stationary Points

18 What exactly are the simultaneous equations of the tangents perpendicular to lines possessing the structure \(x+14y+3=0\) directly applied for the curve \(y=x^{3}+2x-4\)?

A \(14x-y-20=0\) and \(14x-y+12=0\)

B \(x+14y-20=0\) and \(x+14y+12=0\)

C \(14x-y=0\)

D \(x-14y+20=0\)

💡 Short Explanation

Tangents perpendicular to the given line must have slope \(14\). Differentiate the curve, locate points with slope \(14\), then form tangent equations. Find required slope. Differentiate. Form tangent equations.

✅ Correct Answer Explanation

The line \(x+14y+3=0\) has slope \(-1/14\). Therefore, perpendicular tangents have slope \(14\). Solving \(3x^{2}+2=14\) gives \(x=\pm 2\). The tangent equations obtained are \(14x-y-20=0\) and \(14x-y+12=0.\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → represents lines with slope \(-1/14\), not \(14\).
Option C → gives only one tangent instead of two.
Option D → has an incorrect slope and intercept.

🎯 Strategy

Used: Elimination

Application: Determine the required slope before comparing equations.

Final Logic: Only Option A has slope \(14\) and both tangents.

🧠 Memory Trick

"Perpendicular → Negative Reciprocal."

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

19 The mathematical curve \(3x^{2}-y^{2}=8\) consistently yields tangents through external points, specifically through \(\left(4/3\ ,\ 0\right)\). Using the contact point algebraically defined as \(\left(h\ ,\ k\right)\), the absolute value of \(h\) structurally resolves to:

A 3

B 4/3

C ±√(44/3)

D ±√44/3

💡 Short Explanation

Use the tangent equation to the hyperbola together with the external point condition. Solving the resulting equations gives the contact-point x-coordinate magnitude. Write tangent equation. Apply external point. Solve for \(h\).

✅ Correct Answer Explanation

Using the tangent equation to \(3x^{2}-y^{2}=8\) at \(\left(h\ ,\ k\right)\)and substituting the external point \(\left(4/3\ ,\ 0\right)\), the resulting equation simplifies to \(h=\pm \frac{\sqrt{44}}{3}.\) Thus Option D is correct. The remaining options do not satisfy the tangent condition.

❓ Why Other Options Are Incorrect

Option A → 3 does not satisfy the derived equation.
Option B → 4/3 is merely the x-coordinate of the external point.
Option C → ±√(44/3) has an incorrect simplification.

🎯 Strategy

Used: Elimination

Application: Use the tangent condition and simplify carefully.

Final Logic: Only Option D matches the algebraic solution.

🧠 Memory Trick

"External Point → Tangent Condition."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Tangents to Curves

20 For the fundamental quadratic curve \(y=x^{2}+3x+4\), identify all unique x-coordinates of the points at which the tangents through origin explicitly touch the curve.

A 0

B ±1

C ±2

D ±3

💡 Short Explanation

Let the tangent touch the curve at \(x=a\). Requiring the tangent to pass through the origin gives \(a^{2}=4\), so \(a=\pm 2\), corresponding to x-coordinates ±2. Form tangent equation. Pass through origin. Solve for contact point.

✅ Correct Answer Explanation

For \(y=x^{2}+3x+4\), the tangent at \(x=a\) is \(y=(2a+3)x-a^{2}+4.\) Passing through the origin gives \(0=-a^{2}+4,\) so \(a=\pm 2.\) Therefore the contact points have x-coordinates ±2. Hence, the correct option is C.

❓ Why Other Options Are Incorrect

Option A → 0 does not satisfy the tangent-through-origin condition.
Option B → ±1 gives tangents that do not pass through the origin.
Option D → ±3 also fails the required condition.

🎯 Strategy

Used: Substitution

Application: Use the tangent equation and substitute the origin.

Final Logic: The tangent passes through the origin only when \(a=\pm 2\), so Option C is correct.

🧠 Memory Trick

"Origin on Tangent ⇒ Constant = 0."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Tangents Through a Given Point

CUET UG Applied Mathematics Booster Test 1 - Tangents and Normals

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