UG Mathematics Booster Test 3 - Algebra of Continuous Functions -

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UG Mathematics Booster Test 3 - Algebra of Continuous Functions

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 3 - Algebra of Continuous Functions

1 / 20

QUESTION 1 OF 20

Arrange the logical steps to prove that \(\left(f+g)(x\right)\)is continuous at \(x=c\), given f and g are continuous at c:
1. Evaluate \({lim⁡}_{x\rightarrow c}f(x)\)and \({lim⁡}_{x\rightarrow c}g(x)\)
2. Use limit law: \({lim⁡}_{x\rightarrow c}(f+g)(x)=lim⁡f(x)+lim⁡g(x)\)
3. Start by evaluating \({lim⁡}_{x\rightarrow c}(f(x)+g(x))\)
4. Note that \(=f(c)+g(c)=(f+g)(c)\), hence continuous

Answer Review

1 Arrange the logical steps to prove that \(\left(f+g)(x\right)\)is continuous at \(x=c\), given f and g are continuous at c:
1. Evaluate \({lim⁡}_{x\rightarrow c}f(x)\)and \({lim⁡}_{x\rightarrow c}g(x)\)
2. Use limit law: \({lim⁡}_{x\rightarrow c}(f+g)(x)=lim⁡f(x)+lim⁡g(x)\)
3. Start by evaluating \({lim⁡}_{x\rightarrow c}(f(x)+g(x))\)
4. Note that \(=f(c)+g(c)=(f+g)(c)\), hence continuous

A 1, 2, 3, 4

B 4, 3, 2, 1

C 3, 1, 2, 4

D 2, 3, 4, 1

💡 Short Explanation

Begin with sum limit expression Apply continuity of both functions Conclude limit equals function value

✅ Correct Answer Explanation

To prove continuity of: \(\left(f+g)(x\right)\) first evaluate: \({lim⁡}_{x\rightarrow c}(f(x)+g(x))\) Then evaluate individual limits using continuity of f and g. Apply the sum law of limits, and finally show: \({lim⁡}_{x\rightarrow c}(f+g)(x)=(f+g)(c)\) Hence the correct order is 3,1,2,4. Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Begins with separate limits before introducing total expression.
Option B → Conclusion appears before proof process.
Option D → Limit law should follow identification of required limit.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange proof steps in logical mathematical sequence.

Final Logic:

Expression → individual limits → law → continuity conclusion.

🧠 Memory Trick

"Whole → Parts → Law → Conclude"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Sum Rule of Continuity

2 Let \(f(x)\)and \(g(x)\)be continuous probability density functions.
If a combined function is defined as
\(h(x)=f(x)+g(x)\)
then
\({lim⁡}_{x\rightarrow c}h(x)=\)

A \(f(c)\)

B \(g(c)\)

C 1

D \(f(c)+g(c)\)

💡 Short Explanation

Sum rule applies to limits Both functions are continuous Add function values directly

✅ Correct Answer Explanation

Since: \(h(x)=f(x)+g(x)\) and both functions are continuous at c, \({lim⁡}_{x\rightarrow c}h(x)={lim⁡}_{x\rightarrow c}f(x)+{lim⁡}_{x\rightarrow c}g(x)\) Thus: \(=f(c)+g(c)\) Hence option D correctly applies the limit law for sums.

❓ Why Other Options Are Incorrect

Option A → Ignores contribution of \(g(x)\).
Option B → Ignores contribution of \(f(x)\).
Option C → Sum of densities need not equal 1 at a point.

🎯 Strategy

Used: Substitution

Application:

Replace limits using continuity property.

Final Logic:

Limit of sum equals sum of limits.

🧠 Memory Trick

"Sum limit = sum values"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Sum Law of Limits

3 Let \(f(x)\)and \(g(x)\)be continuous functions such that
\(f(3)=10,g(3)=1\)
Evaluate
\({lim⁡}_{x\rightarrow 3}[f(x)-g(x)]\)

A 9

B 15

C 6

D 21

💡 Short Explanation

Difference rule applies directly Use continuity at x=3 \(10-1=9\)

✅ Correct Answer Explanation

Because f and g are continuous: \({lim⁡}_{x\rightarrow 3}[f(x)-g(x)]=f(3)-g(3)\) Substitute the given values: \(10-1=9\) Therefore the required limit equals 9. Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect subtraction arithmetic.
Option C → Does not match function values.
Option D → Incorrect operation performed.

🎯 Strategy

Used: Substitution

Application:

Directly replace limits by function values.

Final Logic:

Continuous difference evaluated pointwise.

🧠 Memory Trick

"Subtract the outputs"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Difference Law of Limits

4 Let \(f(x)\)and \(g(x)\)be continuous functions such that
\(f(2)=8,g(2)=2\)
Find
\({lim⁡}_{x\rightarrow 2}[f(x)-g(x)]\)

A 14

B 6

C 40

D 2.5

💡 Short Explanation

Use continuity of both functions Apply subtraction rule \(8-2=6\)

✅ Correct Answer Explanation

Using continuity: \({lim⁡}_{x\rightarrow 2}[f(x)-g(x)]=f(2)-g(2)\) Substitute values: \(8-2=6\) Thus the limit equals 6. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect arithmetic.
Option C → Product operation incorrectly used.
Option D → Division operation incorrectly used.

🎯 Strategy

Used: Substitution

Application:

Replace limit expressions by known function values.

Final Logic:

Difference law simplifies directly.

🧠 Memory Trick

"Continuous means substitute"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Difference Rule

5 If width \(f(x)\)and length \(g(x)\)are continuous functions, then area
\(A(x)=f(x)g(x)\)
is:

A Discontinuous

B Always constant

C Continuous wherever f and g are continuous

D Undefined

💡 Short Explanation

Area formed by product Product rule preserves continuity Continuous dimensions give continuous area

✅ Correct Answer Explanation

The area function: \(A(x)=f(x)g(x)\) is the product of two continuous functions. By the product continuity theorem, the product remains continuous wherever both factors are continuous. Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Product of continuous functions stays continuous.
Option B → Product need not remain constant.
Option D → Product is properly defined under continuity assumptions.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Interpret geometric quantity through continuity rules.

Final Logic:

Continuous × Continuous = Continuous.

🧠 Memory Trick

"Smooth sides → Smooth area"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Product Rule

6 Let voltage \(V(x)=4\) and current \(I(x)=4\).
Power is given by
\(P(x)=V(x)⋅I(x)\)
Find
\({lim⁡}_{x\rightarrow c}P(x)\)

A 8

B 12

C 16

D 4

💡 Short Explanation

Power equals product Constants remain unchanged under limits \(4\times 4=16\)

✅ Correct Answer Explanation

Given: \(P(x)=V(x)I(x)\) with: \(V(x)=4,I(x)=4\) Thus: \(P(x)=16\) Since constant functions are continuous, \({lim⁡}_{x\rightarrow c}P(x)=16\) Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → Adds instead of multiplying.
Option B → Incorrect arithmetic value.
Option D → Gives only one factor value.

🎯 Strategy

Used: Substitution

Application:

Replace functions with constant values directly.

Final Logic:

Product of constants remains constant.

🧠 Memory Trick

"Power = Voltage × Current"

📚 Class XII – NCERT Physics/Mathematics Link – Chapter: Continuity and Differentiability – Topic: Product Rule Applications

7 Match operations:

List I	List II
1. \(f+g\)	a. Continuous if \(g(c)\neq 0\)
2. \(f-g\)	b. Always continuous
3. \(f⋅g\)	c. Continuous if both functions are continuous
4. \(f/g\)	d. Defined using limit laws

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

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

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

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

💡 Short Explanation

Sum and difference use limit laws Product preserves continuity Quotient needs nonzero denominator

✅ Correct Answer Explanation

For continuous functions: \(f+g\) remains continuous. \(f-g\) follows limit laws. \(f⋅g\) stays continuous. \(f/g\) is continuous when denominator is nonzero. Thus correct matching becomes: \(1-b, 2-d, 3-c, 4-a\) Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Quotient condition incorrectly assigned.
Option B → Product continuity mismatched.
Option C → Quotient continuity incorrectly generalized.

🎯 Strategy

Used: Option Grouping

Application:

Match continuity theorem with operation type.

Final Logic:

Quotients uniquely require denominator nonzero.

🧠 Memory Trick

"Division needs nonzero bottom"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Algebra of Continuous Functions

8 A unit vector is defined as
\(\hat{v}(x)=\frac{v(x)}{∣v(x)∣}\)
This is continuous at \(x=c\) provided:

A \(v(c)=0\)

B \(∣v(c)∣=\infty\)

C \(v(c)\)is zero vector

D \(∣v(c)∣\neq 0\)

💡 Short Explanation

Denominator must stay nonzero Zero magnitude makes quotient undefined Quotient continuity rule applies

✅ Correct Answer Explanation

The unit vector: \(\hat{v}(x)=\frac{v(x)}{∣v(x)∣}\) contains a denominator: \(∣v(x)∣\) For continuity, denominator cannot become zero. Therefore: \(∣v(c)∣\neq 0\) is required. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Zero vector causes undefined denominator.
Option B → Infinite magnitude is unnecessary.
Option C → Zero vector cannot define unit vector.

🎯 Strategy

Used: Elimination

Application:

Identify denominator restriction in quotient continuity.

Final Logic:

Nonzero denominator ensures continuity.

🧠 Memory Trick

"Unit vector needs nonzero length"

📚 Class XII – NCERT Mathematics – Chapter: Vector Algebra/Continuity – Topic: Quotient Continuity

9 Let \(f(x)\)be continuous and
\(g(x)=\lambda f(x)\)
Then
\({lim⁡}_{x\rightarrow c}g(x)=\)

A \(\lambda f(c)\)

B \(f(c)\)

C \(\lambda\)

D 0

💡 Short Explanation

Constant factor comes outside limit Continuity allows substitution Multiply by scalar afterward

✅ Correct Answer Explanation

Since: \(g(x)=\lambda f(x)\) and f is continuous, \({lim⁡}_{x\rightarrow c}g(x)=\lambda {lim⁡}_{x\rightarrow c}f(x)\) Thus: \(=\lambda f(c)\) Hence option A is correct.

❓ Why Other Options Are Incorrect

Option B → Omits scalar multiplier.
Option C → Ignores function value.
Option D → No reason limit becomes zero.

🎯 Strategy

Used: Substitution

Application:

Apply scalar multiple law of limits.

Final Logic:

Constants factor outside limits.

🧠 Memory Trick

"Scalar comes outside"

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Constant Multiple Law

10 Let \(f(x)\)be a continuous function and \(\lambda \in R\).
Which of the following statements about \(\lambda f(x)\)is incorrect?

A The function becomes discontinuous because it is inverted graphically

B \({lim⁡}_{x\rightarrow c}\lambda f(x)=\lambda f(c)\)

C Scalar multiplication preserves continuity

D \(\left(\lambda f)(x)=\lambda f(x\right)\)

💡 Short Explanation

Scaling does not destroy continuity Reflection/stretching still preserves smoothness Scalar multiplication obeys limit laws

✅ Correct Answer Explanation

Multiplying by a scalar changes graph size or orientation but does not create discontinuities. Thus: \(\lambda f(x)\) remains continuous whenever: \(f(x)\) is continuous. Therefore option A is incorrect, while the remaining statements correctly describe scalar multiplication.

❓ Why Other Options Are Incorrect

Option B → Correct scalar limit property.
Option C → Continuity is preserved under scalar multiplication.
Option D → Standard definition of scalar multiplication.

🎯 Strategy

Used: Extreme Word Filter

Application:

Detect exaggerated discontinuity claim.

Final Logic:

Graph inversion does not break continuity.

🧠 Memory Trick

"Stretch ≠ Break"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Scalar Multiplication Continuity

11 Assertion (A): If \(f(c)\neq 0\), then the function
\(g(x)=\frac{1}{f(x)}\)
is continuous at \(x=c\).
Reason (R): The function
\(\frac{1}{f(x)}\)
is continuous only at points where \(f(x)\neq 0\).

A Both true, R explains A

B A false, R true

C Both false

D A true, R false

💡 Short Explanation

Reciprocal continuity needs nonzero denominator \(f(c)\neq 0\) ensures definition Reason directly explains assertion

✅ Correct Answer Explanation

If: \(f(c)\neq 0\) and f is continuous at c, then: \(\frac{1}{f(x)}\) is also continuous at c. Reciprocal functions remain continuous wherever the denominator is nonzero. Thus both Assertion and Reason are true, and the Reason correctly explains the Assertion. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion is actually true.
Option C → Reciprocal continuity theorem is valid.
Option D → Reason correctly supports Assertion.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Check whether theorem directly justifies statement.

Final Logic:

Nonzero denominator guarantees reciprocal continuity.

🧠 Memory Trick

"No zero below fraction"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Reciprocal Function Continuity

12 Which of the following reciprocal functions are continuous everywhere in their domains?
I. \(f(x)=1/x, x\neq 0\)
II. \(f(x)=1/(x^{2}+1)\), for all real x
III. \(f(x)=1/sin⁡x,\)for all real x

A I only

B I and II

C II and III

D I, II, and III

💡 Short Explanation

Reciprocal continuity needs nonzero denominator \(x^{2}+1\) never becomes zero \(sin⁡x=0\) at many points

✅ Correct Answer Explanation

Function I is continuous on its domain: \(x\neq 0\) Function II is continuous for all real numbers because: \(x^{2}+1>0\) always. Function III is not continuous for all real numbers since: \(sin⁡x=0\) at integer multiples of \(\pi\). Hence only I and II satisfy the condition. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Function II is also continuous everywhere.
Option C → Function III fails where \(sin⁡x=0\).
Option D → Reciprocal of sine is undefined at multiples of \(\pi\).

🎯 Strategy

Used: Elimination

Application:

Remove functions with zero denominators.

Final Logic:

Reciprocal continuity survives only where denominator never vanishes.

🧠 Memory Trick

"Check denominator first"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Reciprocal Functions

13 A high-degree polynomial function such as
\(f(x)=x^{100}+x^{2}+1\)
is continuous for all real numbers because:

A It contains only even powers

B The degree is 100

C Every polynomial function is continuous everywhere

D The constant term is 1

💡 Short Explanation

Polynomial operations preserve continuity Degree does not affect continuity All polynomials are continuous on \(R\)

✅ Correct Answer Explanation

Polynomial functions are formed using addition, multiplication, and powers of x. These operations preserve continuity. Hence every polynomial function is continuous for all real numbers regardless of degree or power parity. Therefore option C is correct.

❓ Why Other Options Are Incorrect

Option A → Odd-power polynomials are also continuous.
Option B → Degree size does not determine continuity.
Option D → Constant term value is irrelevant to continuity.

🎯 Strategy

Used: Extreme Word Filter

Application:

Avoid unnecessary restrictions on polynomial continuity.

Final Logic:

Every polynomial is continuous over all real numbers.

🧠 Memory Trick

"Polynomial = Always smooth"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Polynomial Continuity

14 If \(f(x)\)is a polynomial function, what is the geometric consequence of its continuity over \(R\)?

A Its graph can be drawn without lifting the pen

B It always passes through the origin

C It has sharp discontinuities

D It has vertical asymptotes

💡 Short Explanation

Continuous graphs have no breaks Polynomials remain smooth everywhere Pen-lifting heuristic indicates discontinuity

✅ Correct Answer Explanation

Since polynomial functions are continuous everywhere on: \(R\) their graphs contain no jumps or breaks. Hence the graph can be drawn in one continuous stroke without lifting the pen. Therefore option A is correct.

❓ Why Other Options Are Incorrect

Option B → Not every polynomial passes through origin.
Option C → Polynomials do not have discontinuities.
Option D → Vertical asymptotes occur in rational functions, not polynomials.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate continuity to graphical interpretation.

Final Logic:

Continuous graph means no pen lifting.

🧠 Memory Trick

"No breaks → No lifts"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Graphical Meaning of Continuity

📄 Passage

Theorem: If a function \(g\) is continuous at \(x=c\) and a function \(f\) is continuous at \(g(c)\), then the composite function
\(\left(f\circ g)(x)=f(g(x)\right)\)
is continuous at \(x=c\).
For example, if
\(f(x)=sin⁡x,g(x)=x^{2}\)
then
\(\left(f\circ g)(x)=sin⁡(x^{2}\right)\)
is continuous for all real x.

If instead of composition we form a function
\(h(x)=\frac{p(x)}{q(x)}\)
where \(p(x)\)and \(q(x)\)are polynomials, this function is:

A A trigonometric composite function

B A constant function

C Discontinuous everywhere

D A rational function

💡 Short Explanation

Ratio of polynomials defines rational function Numerator and denominator are polynomials Standard algebraic definition applies

✅ Correct Answer Explanation

A function of the form: \(h(x)=\frac{p(x)}{q(x)}\) where both p and q are polynomials, is called a rational function. It is continuous wherever: \(q(x)\neq 0\) Hence option D correctly identifies the function type.

❓ Why Other Options Are Incorrect

Option A → No trigonometric operation appears here.
Option B → Quotient need not be constant.
Option C → Rational functions are continuous on their domains.

🎯 Strategy

Used: Odd One Out

Application:

Identify definition matching quotient of polynomials.

Final Logic:

Polynomial ratio = Rational function.

🧠 Memory Trick

"Poly over poly = Rational"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Rational Functions

📄 Passage

A rational function
\(h(x)=\frac{p(x)}{q(x)}\)
is discontinuous at:

A All real numbers

B Points where \(q(x)=0\)

C Points where \(p(x)=0\)

D Only integers

💡 Short Explanation

Denominator zero causes undefined value Undefined points break continuity Numerator zero is allowed

✅ Correct Answer Explanation

A rational function: \(\frac{p(x)}{q(x)}\) is continuous wherever: \(q(x)\neq 0\) If: \(q(x)=0\) the function becomes undefined and discontinuous there. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Rational functions are continuous on valid domains.
Option C → Zero numerator does not create discontinuity.
Option D → Discontinuity depends on denominator, not integers.

🎯 Strategy

Used: Elimination

Application:

Identify the condition making denominator undefined.

Final Logic:

Zero denominator creates discontinuity.

🧠 Memory Trick

"Bottom zero → Break"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Rational Function Discontinuity

17 Evaluate
\({lim⁡}_{x\rightarrow \pi /6}sin⁡x\)

A 1

B 0

C \(1/2\)

D \(\sqrt{3}/2\)

💡 Short Explanation

Sine function is continuous Direct substitution allowed \(sin⁡(\pi /6)=1/2\)

✅ Correct Answer Explanation

Since: \(\sin\,x\) is continuous for all real x, \({lim⁡}_{x\rightarrow \pi /6}sin⁡x=sin⁡\left(\frac{\pi }{6}\right)\) Now: \(sin⁡\left(\frac{\pi }{6}\right)=\frac{1}{2}\) Hence option C is correct.

❓ Why Other Options Are Incorrect

Option A → \(sin⁡(\pi /2)=1\), not \(sin⁡(\pi /6)\).
Option B → Sine at \(\pi /6\) is nonzero.
Option D → \(\sqrt{3}/2\) equals \(sin⁡(\pi /3)\).

🎯 Strategy

Used: Substitution

Application:

Use continuity for direct evaluation.

Final Logic:

Continuous trig functions allow substitution.

🧠 Memory Trick

"\(\pi /6\rightarrow 1/2\)"

📚 Class XI – NCERT Mathematics – Chapter: Trigonometric Functions – Topic: Standard Trigonometric Values

18 The function
\(f(x)=tan⁡x\)
is undefined at:

A \(x=0\)

B \(x=\pi /4\)

C \(x=\pi\)

D \(x=\pi /2\)

💡 Short Explanation

Tangent denominator is cosine Undefined when cosine becomes zero \(cos⁡(\pi /2)=0\)

✅ Correct Answer Explanation

Since: \(tan⁡x=\frac{\sin\,x}{\cos\,x}\) the function becomes undefined wherever: \(cos⁡x=0\) At: \(x=\frac{\pi }{2}\) cosine equals zero. Hence tangent is undefined there. Therefore option D is correct.

❓ Why Other Options Are Incorrect

Option A → \(tan⁡0=0\) exists.
Option B → \(tan⁡(\pi /4)=1\) exists.
Option C → \(tan⁡\pi =0\) exists.

🎯 Strategy

Used: Substitution

Application:

Check cosine denominator at given angles.

Final Logic:

Tangent undefined when cosine vanishes.

🧠 Memory Trick

"tan breaks at \(\pi /2\)"

📚 Class XI – NCERT Mathematics – Chapter: Trigonometric Functions – Topic: Tangent Function Domain

19 For continuity of \(f(g(x))\)at \(x=c\), identify the incorrect statement:

A Only g needs to be continuous at c

B g must be continuous at c

C f must be continuous at \(g(c)\)

D If both are continuous everywhere, composite is continuous everywhere

💡 Short Explanation

Composite continuity needs two conditions Inner and outer continuity both required Outer checked at image point

✅ Correct Answer Explanation

For: \(f(g(x))\) to be continuous at c: \(g(x)\)must be continuous at c. \(f(x)\)must be continuous at \(g(c)\). Thus continuity of g alone is insufficient. Therefore option A is the incorrect statement.

❓ Why Other Options Are Incorrect

Option B → Inner function continuity is required.
Option C → Outer function continuity at image point is essential.
Option D → Composite of globally continuous functions stays continuous.

🎯 Strategy

Used: Extreme Word Filter

Application:

Detect incomplete continuity condition.

Final Logic:

Composition needs continuity of both functions.

🧠 Memory Trick

"Inner smooth, outer smooth"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Composite Functions

20 Let
\(f(x)=x+3,g(x)=x^{2}\)
Then
\((f\circ g)(x)=x^{2}+3\)
Find
\({lim⁡}_{x\rightarrow 2}(f\circ g)(x)\)

A 7

B 10

C 5

D 12

💡 Short Explanation

Composite simplifies to polynomial Use direct substitution \(2^{2}+3=7\)

✅ Correct Answer Explanation

Given: \((f\circ g)(x)=x^{2}+3\) Since polynomials are continuous, \({lim⁡}_{x\rightarrow 2}(f\circ g)(x)=2^{2}+3\) Thus: \(=4+3=7\) Hence option A is correct. (The provided answer B was incorrect.)

❓ Why Other Options Are Incorrect

Option B → Arithmetic error; \(4+3\neq 10\).
Option C → Omits square contribution.
Option D → Incorrect substitution calculation.

🎯 Strategy

Used: Substitution

Application:

Evaluate polynomial directly at x=2.

Final Logic:

Continuous polynomials allow direct substitution.

🧠 Memory Trick

"Square then add three"

📚 Class XII – NCERT Mathematics – Chapter: Continuity and Differentiability – Topic: Composite Polynomial Functions

UG Mathematics Booster Test 3 - Algebra of Continuous Functions

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