CUET UG Mathematics Booster Test 1 - Variable Separable Method -

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CUET UG Mathematics Booster Test 1 - Variable Separable Method

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Variable Separable Method

1 / 20

QUESTION 1 OF 20

Express the differential equation
\(x(x^{2}+1)\frac{dy}{dx}=y\)
in its cleanly separated variable form.

Answer Review

1 Express the differential equation
\(x(x^{2}+1)\frac{dy}{dx}=y\)
in its cleanly separated variable form.

A \(\frac{dy}{x}=\frac{dx}{y}\)

B \(\frac{dy}{y}=\frac{dx}{x(x^{2}+1)}\)

C \(y dy=x(x^{2}+1) dx\)

D \(\frac{dy}{dx}=\frac{y}{x}\)

💡 Short Explanation

Move \(y\) to the left side. Move \(dx\) to the right side. Separate variables completely.

✅ Correct Answer Explanation

Starting from \(x(x^{2}+1)\frac{dy}{dx}=y,\) divide by \(y\) and multiply by \(dx\): \(\frac{dy}{y}=\frac{dx}{x(x^{2}+1)}.\) This places all \(y\)-terms on one side and all \(x\)-terms on the other, satisfying the condition for variable separation. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Does not follow from algebraic rearrangement of the given equation.
Option C → Incorrect separation; \(y\) should appear in the denominator.
Option D → Omits the factor \(\left(x^{2}\ +\ 1\right)\), so it is not equivalent.

🎯 Strategy

Used: Elimination

Application:

Rearrange the equation and eliminate options that do not isolate \(x\) and \(y\).

Final Logic:

Separated form must contain only \(y\)-terms on one side and \(x\)-terms on the other.

🧠 Memory Trick

"Separate → y left, x right."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Variable Separable Form

2 Identify the INCORRECT statement regarding the equation
\(\frac{dy}{dx}=sin⁡(x+y).\)

A It cannot be solved directly using the basic variable separable method.

B The RHS cannot be written directly as the product form \(g(x)h(y)\).

C It is a perfect direct example of the variable separable product form condition.

D We must substitute \(x+y=v\) to make the variables separable.

💡 Short Explanation

RHS contains \(x+y\). Not factorizable into \(g(x)h(y)\). Direct separation fails.

✅ Correct Answer Explanation

A separable equation requires \(\frac{dy}{dx}=g(x)h(y).\) The expression \(sin⁡(x+y)\)depends on the combination \(x+y\), not on separate functions of \(x\) and \(y\). Therefore it is not directly separable. Hence Option C is the incorrect statement.

❓ Why Other Options Are Incorrect

Option A → Correct because direct separation is impossible.
Option B → Correct since \(sin⁡(x+y)\)is not a product \(g(x)h(y)\).
Option D → Correct; substitution \(v=x+y\) is a standard approach.

🎯 Strategy

Used: Odd One Out

Application:

Identify the statement contradicting the definition of separable equations.

Final Logic:

Only Option C incorrectly labels the equation as directly separable.

🧠 Memory Trick

"x+y together ⇒ not separable."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Variable Separation Method

3 Arrange the correct sequence of steps to solve \(\frac{dy}{dx}=(x+1)(2-y).\)

A 4, 2, 3, 1

B 2, 3, 1, 4

C 4, 1, 2, 3

D 4, 3, 2, 1

💡 Short Explanation

Start with given DE. Separate variables. Integrate both sides. Write final solution.

✅ Correct Answer Explanation

The logical sequence is: 1. Write the DE. 2. Separate variables. 3. Integrate both sides. 4. Obtain the final relation. Thus: \(4\rightarrow 3\rightarrow 2\rightarrow 1.\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Integrates before separation.
Option B → Starts from an intermediate step.
Option C → Incorrect ordering of integration and separation.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Follow the standard NCERT solution process.

Final Logic:

DE → Separation → Integration → Solution.

🧠 Memory Trick

"Given → Separate → Integrate → Solve."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Separation of Variables

4 Match the differential equations with their correct isolated terms.

List I	List II
1. \(y^{'}=e^{x}e^{-y}\)	a. \(e^{y} dy=e^{x} dx\)
2. \(y^{'}=\frac{x^{2}}{y^{2}}\)	b. \(y^{2} dy=x^{2} dx\)
3. \(y^{'}=xy\)	c. \(\frac{dy}{y}=x dx\)
4. \(y^{'}=\frac{x}{y}\)	d. \(y dy=x dx\)

A 1–a, 2–b, 3–c, 4–d

B 1–b, 2–a, 3–c, 4–d

C 1–a, 2–c, 3–b, 4–d

D 1–d, 2–c, 3–b, 4–a

💡 Short Explanation

Rearrange each equation. Separate variables. Match corresponding forms.

✅ Correct Answer Explanation

Separating: 1 → \(e^{y}dy=e^{x}dx\)→ a 2 → \(y^{2}dy=x^{2}dx\)→ b 3 → \(\frac{dy}{y}=xdx\)→ c 4 → \(ydy=xdx\)→ d Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly swaps equations 1 and 2.
Option C → Misplaces equations 2 and 3.
Option D → Several mismatches occur.

🎯 Strategy

Used: Option Grouping

Application:

Solve each correspondence independently.

Final Logic:

Each equation has a unique separated form.

🧠 Memory Trick

"Separate first, match later."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Variable Separation

5 Evaluate the general solution resulting from \(\int y dy=\int x^{2} dx.\)

A \(y=x^{3}/3+C\)

B \(y^{2}/2=x^{3}/3+C\)

C \(y^{2}=x^{3}+C\)

D \(y=x^{2}+C\)

💡 Short Explanation

Integrate both sides directly. Use standard formulas. Add integration constant.

✅ Correct Answer Explanation

\(\int y dy=\frac{y^{2}}{2}\) and \(\int x^{2}dx=\frac{x^{3}}{3}+C.\) Therefore, \(\frac{y^{2}}{2}=\frac{x^{3}}{3}+C.\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect integration of \(y dy\).
Option C → Missing factor \(1/2\).
Option D → Not obtained from the given integrals.

🎯 Strategy

Used: Substitution

Application:

Apply standard integration formulas.

Final Logic:

\(\int ydy=y^{2}/2\).

🧠 Memory Trick

"Power rule on both sides."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Integration Method

6 Consider \(\frac{dy}{dx}=\frac{1}{x}.\)
Which expressions correctly incorporate the arbitrary constant?

A 1 only

B 3 only

C 1 and 2

D 1, 2, and 3

💡 Short Explanation

Integrate \(1/x\). Constants can be absorbed. Compare equivalent forms.

✅ Correct Answer Explanation

Integration gives \(y=ln⁡∣x∣+C.\) Also, \(y=ln⁡∣Cx∣\) is equivalent because \(ln⁡∣Cx∣=ln⁡∣x∣+ln⁡∣C∣.\) Statement 3 is not equivalent. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores another valid form.
Option B → Statement 3 is incorrect.
Option D → Includes an invalid expression.

🎯 Strategy

Used: Option Grouping

Application:

Check equivalence of logarithmic forms.

Final Logic:

Statements 1 and 2 represent the same solution family.

🧠 Memory Trick

"Log constant can move inside."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: General Solution

7 Assertion (A): The solution of \(\frac{dy}{dx}=-\frac{x}{y}\)is \(x^{2}+y^{2}=C\).
Reason (R): \(y dy=-x dx\) integrates to \(x^{2}+y^{2}=C_{1}\), forming concentric circles.

A Both false

B A true, R false

C A false, R true

D Both true, R explains A

💡 Short Explanation

Separate variables. Integrate. Obtain circle family.

✅ Correct Answer Explanation

\(y dy=-x dx.\) Integrating: \(\frac{y^{2}}{2}=-\frac{x^{2}}{2}+C.\) Rearranging: \(x^{2}+y^{2}=C_{1}.\) The solution represents concentric circles. Thus both Assertion and Reason are true, and the Reason explains the Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are correct.
Option B → Reason is not false.
Option C → Assertion is also true.

🎯 Strategy

Used: Substitution

Application:

Directly solve the differential equation.

Final Logic:

Integration produces \(x^{2}+y^{2}=C\).

🧠 Memory Trick

"−x/y ⇒ circles."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Variable Separation

8 The general solution \(y=Cx\)
corresponding to \(\frac{dy}{dx}=\frac{y}{x}\)
represents which region?

A Lines passing through the origin (excluding the vertical axis for finite slopes)

B Parallel horizontal lines

C Circles around the origin

D Parabolas passing through the origin

💡 Short Explanation

Solution is \(y=Cx\). Slope varies with \(C\). All lines pass through origin.

✅ Correct Answer Explanation

The solution family \(y=Cx\) represents straight lines with different slopes. Every member passes through the origin. Hence Option A correctly describes the geometric family.

❓ Why Other Options Are Incorrect

Option B → Horizontal lines have form \(y=C\).
Option C → Circles require quadratic terms.
Option D → Parabolas require degree-two expressions.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recognize the graph of \(y=Cx\).

Final Logic:

Straight-line family through origin.

🧠 Memory Trick

"y = Cx ⇒ line family."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Family of Curves

9 If the velocity components satisfy \(\frac{dy}{dx}=-\frac{y}{x},\)
what is the equation of the trajectory?

A \(y^{2}-x^{2}=C\)

B \(xy=C\)

C \(y=Cx\)

D \(y^{2}+x^{2}=C\)

💡 Short Explanation

Separate variables. Integrate logarithms. Simplify solution.

✅ Correct Answer Explanation

\(\frac{dy}{y}=-\frac{dx}{x}.\) Integrating: \(ln⁡∣y∣=-ln⁡∣x∣+C.\) Therefore \(ln⁡∣xy∣=C\) or \(xy=C_{1}.\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Not obtained from integration.
Option C → Corresponds to \(\frac{dy}{dx}=\frac{y}{x}\).
Option D → Represents circle family.

🎯 Strategy

Used: Substitution

Application:

Use separation of variables.

Final Logic:

\(ln⁡∣xy∣=C\Rightarrow xy=C\).

🧠 Memory Trick

"dy/y = −dx/x ⇒ xy = constant."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Variable Separation

10 Find the geometric area enclosed by the coordinate axes and the curve satisfying \(\frac{dy}{dx}=-1\) which passes through \(\left(4\ ,\ 0\right)\).

A 4

B 16

C 8

D 12

💡 Short Explanation

Integrate differential equation. Determine line equation. Compute triangular area.

✅ Correct Answer Explanation

\(\frac{dy}{dx}=-1\) gives \(y=-x+C.\) Using \(\left(4\ ,\ 0\right)\), \(C=4.\) Thus \(y=-x+4.\) The intercepts are 4 and 4. Area enclosed with axes: \(\frac{1}{2}(4)(4)=8.\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Area calculation is too small.
Option B → Equals rectangle area, not triangle area.
Option D → Incorrect geometric computation.

🎯 Strategy

Used: Substitution

Application:

Find the particular line, then apply area formula.

Final Logic:

Area \(=\frac{1}{2}\times 4\times 4=8\).

🧠 Memory Trick

"Intercept triangle ⇒ ½ × base × height."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Particular Solution and Applications

11 Solve the equation \(\frac{dy}{dx}=\frac{1}{1+x^{2}}\)
given the initial condition \(y(0)=\pi /4\).

A \(y={tan⁡}^{-1}x\)

B \(y={tan⁡}^{-1}x-\pi /4\)

C \(y={sin⁡}^{-1}x+\pi /4\)

D \(y={tan⁡}^{-1}x+\pi /4\)

💡 Short Explanation

Integrate the differential equation. Apply the initial condition. Determine the constant.

✅ Correct Answer Explanation

Integrating, \(y=\int \frac{dx}{1+x^{2}}={tan⁡}^{-1}x+C.\) Using \(y(0)=\pi /4\), \(\pi /4={tan⁡}^{-1}(0)+C=C.\) Thus \(y={tan⁡}^{-1}x+\frac{\pi }{4}.\) Hence Option D is correct. Options A and B use incorrect constants, while C uses the wrong antiderivative.

❓ Why Other Options Are Incorrect

Option A → Does not satisfy \(y(0)=\pi /4\).
Option B → Gives \(y(0)=-\pi /4\).
Option C → Integral of \(\frac{1}{1+x^{2}}\)is not \({sin⁡}^{-1}x\).

🎯 Strategy

Used: Substitution

Application:

Integrate and use the initial value immediately.

Final Logic:

\(C=\pi /4\).

🧠 Memory Trick

"1+x² → tan⁻¹x."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Particular Solution

12 A probabilistic survival variable follows the path \(\frac{dp}{dt}=pcos⁡t.\)
Find \(p(t)\)exactly if \(p(0)=1\).

A \(p(t)=e^{\sin\,t}\)

B \(p(t)=e^{\cos\,t}\)

C \(p(t)=sin⁡t+1\)

D \(p(t)=cos⁡t\)

💡 Short Explanation

Separate variables. Integrate both sides. Apply initial condition.

✅ Correct Answer Explanation

\(\frac{dp}{p}=cos⁡t dt.\) Integrating, \(ln⁡p=sin⁡t+C.\) Thus \(p=Ce^{\sin\,t}.\) Using \(p(0)=1\), \(1=Ce^{0},\) so \(C=1\). Hence \(p=e^{\sin\,t}.\) Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Differentiation does not satisfy the equation.
Option C → Not obtained from exponential growth form.
Option D → Does not satisfy \(p^{'}=pcos⁡t\).

🎯 Strategy

Used: Substitution

Application:

Convert to separable form and integrate.

Final Logic:

\(ln⁡p=sin⁡t+C\).

🧠 Memory Trick

"p′=pf(t) ⇒ p=e^{\int f(t)dt}."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Variable Separation

13 Solve \(\frac{dy}{dx}=e^{3x}\)
subjected to the initial parameter \(y(0)=1/3\). Compute the value of \(y(1)\).

A \(e^{3}\)

B \(e^{3}/3\)

C \(e^{3}/3+1\)

D \(3e^{3}\)

💡 Short Explanation

Integrate exponential function. Use initial condition. Evaluate at x=1.

✅ Correct Answer Explanation

\(y=\int e^{3x}dx=\frac{e^{3x}}{3}+C.\) Using \(y(0)=1/3\), \(\frac{1}{3}=\frac{1}{3}+C,\) so \(C=0\). Therefore \(y(1)=\frac{e^{3}}{3}.\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Missing factor \(1/3\).
Option C → Adds an unnecessary constant.
Option D → Multiplies instead of divides.

🎯 Strategy

Used: Substitution

Application:

Integrate and apply the given condition.

Final Logic:

\(C=0\Rightarrow y(1)=e^{3}/3\).

🧠 Memory Trick

"∫e^(ax)dx = e^(ax)/a."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Particular Solutions

14 A continuous moving average dataset exhibits a precise rate of change modeled by \(\frac{dy}{dx}=2x.\)
If the data point \(\left(1\ ,\ 4\right)\)is recorded, what is the evaluated integration constant \(C\)?

A 1

B 2

C 3

D 4

💡 Short Explanation

Integrate derivative. Substitute given point. Solve for C.

✅ Correct Answer Explanation

Integrating, \(y=x^{2}+C.\) Using the point \(\left(1\ ,\ 4\right)\), \(4=1+C.\) Hence \(C=3.\) Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Does not satisfy the point.
Option B → Gives y=3 at x=1.
Option D → Gives y=5 at x=1.

🎯 Strategy

Used: Substitution

Application:

Insert the known point into the general solution.

Final Logic:

\(4=1+C\).

🧠 Memory Trick

"Point determines C."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Particular Solution

15 A curve passes through \(\left(-2,3\right)\)and possesses the slope \(\frac{2x}{y^{2}}\)
at any point \(\left(x\ ,\ y\right)\). Calculate the exact value of \(y\) when \(x=2\).

A 1

B -3

C 2

D 3

💡 Short Explanation

Separate variables. Form particular curve. Evaluate at x=2.

✅ Correct Answer Explanation

\(y^{2}dy=2x dx.\) Integrating, \(\frac{y^{3}}{3}=x^{2}+C.\) Using \(\left(-2,3\right)\), \(9=4+C,\) so \(C=5\). Thus \(y^{3}=3x^{2}+15.\) At \(x=2\), \(y^{3}=27,\) giving \(y=3\).

❓ Why Other Options Are Incorrect

Option A → Does not satisfy the equation.
Option B → Gives \(y^{3}=-27\), inconsistent with the curve.
Option C → Gives \(y^{3}=8\), incorrect.

🎯 Strategy

Used: Substitution

Application:

Determine the particular curve first.

Final Logic:

\(y^{3}=27\Rightarrow y=3\).

🧠 Memory Trick

"Find C before evaluating."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Particular Curves

16 The variable separable differential equation \(\frac{dy}{dx}=-\frac{x}{y}\)
defines which orthogonal family related to the linear equations \(y=kx\)?

A Concentric circles

B Hyperbolas

C Parabolas

D Ellipses

💡 Short Explanation

Separate variables. Integrate. Identify geometric family.

✅ Correct Answer Explanation

Integrating, \(ydy=-xdx,\) gives \(x^{2}+y^{2}=C.\) This is the family of circles centered at the origin. These circles are orthogonal to radial lines \(y=kx\). Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Hyperbolas arise from differences of squares.
Option C → Parabolas require quadratic-variable relations.
Option D → Ellipses have unequal coefficients.

🎯 Strategy

Used: Substitution

Application:

Solve the DE and recognize the curve family.

Final Logic:

\(x^{2}+y^{2}=C\).

🧠 Memory Trick

"−x/y ⇒ circle family."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Geometric Interpretation

17 If a restricted population grows according to \(\frac{dP}{dt}=0.1P\)
and \(P(0)=1000\), calculate the approximate population at \(t=10\)(use \(e\approx 2.718\)).

A 1000

B 2718

C 2000

D 2500

💡 Short Explanation

Use exponential growth model. Substitute t=10. Approximate using e.

✅ Correct Answer Explanation

The solution is \(P=1000e^{0.1t}.\) At \(t=10\), \(P=1000e.\) Using \(e\approx 2.718\), \(P\approx 2718.\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores growth.
Option C → Underestimates the exponential increase.
Option D → Not the calculated value.

🎯 Strategy

Used: Substitution

Application:

Apply the growth formula directly.

Final Logic:

\(P(10)=1000e\).

🧠 Memory Trick

"Growth ⇒ multiply by e^(kt)."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Exponential Growth

18 If ₹100 doubles in 10 years under continuous compounding, which expression solves for the rate \(r\%\)?

A \(100ln⁡10\)

B \(10ln⁡10\)

C \(10ln⁡2\)

D \(100ln⁡2\)

💡 Short Explanation

Use continuous compounding. Apply doubling condition. Solve for r.

✅ Correct Answer Explanation

\(200=100e^{(r/100)⋅10}.\) Thus \(2=e^{r/10}.\) Taking logarithms, \(ln⁡2=\frac{r}{10}\) and \(r=10ln⁡2.\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Uses wrong logarithmic quantity.
Option B → Uses \(\ln\,10\) instead of \(\ln\,2\).
Option D → Incorrect scaling factor.

🎯 Strategy

Used: Substitution

Application:

Apply the doubling condition in the compounding model.

Final Logic:

\(r=10ln⁡2\).

🧠 Memory Trick

"Double ⇒ ln2."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Continuous Compounding

📄 Passage

Differential equations with variables separable vividly represent physical phenomena where spatial and temporal variations successfully decouple. For example, when inflating a spherical balloon, the volume \(V=\frac{4}{3}\pi r^{3}\) changes at a strictly constant rate \(\frac{dV}{dt}=k.\) By differentiating, we obtain \(4\pi r^{2}\frac{dr}{dt}=k,\) revealing a variable separable equation: \(4\pi r^{2}dr=kdt.\) Integrating both sides provides the robust functional relationship \(\frac{4\pi r^{3}}{3}=kt+C,\) cleanly linking geometry to dynamic rates.

In the equation
\(4\pi r^{2}dr=kdt,\)
separating variables and integrating yields a relationship primarily between:

A Surface area and volume

B Radius and circumference

C Time and temperature

D Radius cubed and time

💡 Short Explanation

Integrate both sides. Observe resulting expression. Identify linked variables.

✅ Correct Answer Explanation

The passage shows \(\frac{4\pi r^{3}}{3}=kt+C.\) The final equation directly relates \(r^{3}\)and \(t\). Therefore the primary relationship is between radius cubed and time. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Volume and surface area are not the final integrated variables.
Option B → Circumference does not appear.
Option C → Temperature is unrelated.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Read the integrated result directly from the passage.

Final Logic:

Integrated equation contains \(r^{3}\)and \(t\).

🧠 Memory Trick

"Sphere volume ∝ r³."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Applications of Variable Separation

📄 Passage

If initially \(r(0)=0\) and at \(t=3\) seconds, \(r=3\), deduce the constant \(k\).

A \(12\pi\)

B \(36\pi\)

C \(9\pi\)

D \(4\pi\)

💡 Short Explanation

Use integrated equation. Apply initial condition. Substitute known values.

✅ Correct Answer Explanation

From \(\frac{4\pi r^{3}}{3}=kt+C.\) Since \(r(0)=0\), \(C=0.\) Using \(r=3\) at \(t=3\), \(\frac{4\pi (27)}{3}=3k.\) Thus \(36\pi =3k\) and \(k=12\pi .\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Equals the left side before dividing by 3.
Option C → Arithmetic error.
Option D → Underestimates the constant.

🎯 Strategy

Used: Substitution

Application:

Insert the given data into the integrated relation.

Final Logic:

\(36\pi =3k\Rightarrow k=12\pi\).

🧠 Memory Trick

"Apply conditions after integration."

📚 Class XII – NCERT Mathematics – Chapter: Differential Equations – Topic: Applications of Variable Separation

CUET UG Mathematics Booster Test 1 - Variable Separable Method

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