CUET 2026 Mathematics May 31 Shift 2 Question Paper- Download Answer Key and Solution PDF

Concept:

First evaluate the determinant and then differentiate the obtained function.

Step 1: Simplify the determinant

Apply the column operation:

Then

Expanding along the third column,

Step 2: Find the derivatives

Step 3: Evaluate \(y''+y\)

Final Answer:

Hence the correct option is

Quick Tip: For determinants containing expressions like

first use column operations to simplify before expansion.

Concept:

The given equation is a first-order linear differential equation of the form
\[ \frac{dy}{dx}+P(x)y=Q(x) \]

which can be solved using the integrating factor method.

Step 1: Find the integrating factor
\[ P(x)=\frac{x}{x^2-1} \]

Hence,
\[ IF=e^{\int \frac{x}{x^2-1}dx} \]

Let
\[ t=x^2-1 \]

Then
\[ dt=2x\,dx \]

Therefore,
\[ IF=e^{\frac12\ln(1-x^2)} \]
\[ =\sqrt{1-x^2} \]

Step 2: Solve the differential equation

Multiplying throughout by the integrating factor,
\[ \frac{d}{dx}\Big(y\sqrt{1-x^2}\Big) = x^6+4x \]

Integrating,
\[ y\sqrt{1-x^2} = \frac{x^7}{7}+2x^2+C \]

Using \(f(0)=0\),
\[ C=0 \]

Thus,
\[ f(x)=\frac{\frac{x^7}{7}+2x^2}{\sqrt{1-x^2}} \]

Step 3: Evaluate the definite integral

After simplification,
\[ \int_{-\frac12}^{\frac12}f(x)\,dx = \frac{\pi}{3}-1 \]

Hence,
\[ 6\int_{-\frac12}^{\frac12}f(x)\,dx = 2\pi-6 \]

Comparing with
\[ 2\pi-\alpha \]

gives
\[ \alpha=6 \]
\[ \alpha^2=36 \]

Final Answer:
\[ \boxed{36} \] Quick Tip: For linear differential equations, always compute the integrating factor first and then convert the equation into an exact derivative.

Concept:

For infinitely many solutions,
\[ \operatorname{Rank}(A)=\operatorname{Rank}(A|B)<3 \]

Therefore one equation must be a linear combination of the other two.

Step 1: Assume the third equation is a linear combination of the first two

Let
\[ a(2x+\lambda y+3z)+b(3x+2y-z)=4x+5y+\mu z \]

Comparing coefficients,
\[ 2a+3b=4 \]
\[ 5a+7b=9 \]

Step 2: Find \(a\) and \(b\)

Solving,
\[ a=-1,\qquad b=2 \]

Step 3: Find \(\lambda\) and \(\mu\)

Using
\[ a\lambda+2b=5 \]
\[ -\lambda+4=5 \]
\[ \lambda=-1 \]

Also,
\[ \mu=3a-b \]
\[ =-3-2 \]
\[ =-5 \]

Step 4: Calculate required value
\[ \lambda^2+\mu^2 = (-1)^2+(-5)^2 \]
\[ =1+25 \]
\[ =26 \]

Final Answer:
\[ \boxed{26} \] Quick Tip: For infinitely many solutions, one row must be expressible as a linear combination of the remaining rows.

Concept:

The general solution of
\[ f''=f \]

is
\[ f(x)=Ae^x+Be^{-x} \]

Step 1: Use the odd function property

Since \(f\) is odd,
\[ f(x)=A(e^x-e^{-x}) \]

Step 2: Use the condition \(f'(0)=3\)

Differentiating,
\[ f'(x)=A(e^x+e^{-x}) \]

At \(x=0\),
\[ 2A=3 \]
\[ A=\frac32 \]

Hence,
\[ f(x)=\frac32(e^x-e^{-x}) \]

Step 3: Evaluate \(f(\ln3)\)
\[ f(\ln3) = \frac32\left(3-\frac13\right) \]
\[ = \frac32\cdot\frac83 \]
\[ =4 \]

Therefore,
\[ 9f(\ln3)=9\times4=36 \]

Final Answer:
\[ \boxed{36} \] Quick Tip: An odd solution of \(f''=f\) is always proportional to \(\sinh x\).

Concept:

Convert the equation into a linear differential equation and use the integrating factor method.

Step 1: Rewrite in standard form

Dividing by
\[ \cos x(\ln(\cos x))^2 \]

gives
\[ \frac{dy}{dx} -\frac{3\tan x}{\ln(\cos x)}y = -\frac{\tan x}{(\ln(\cos x))^2} \]

Step 2: Find the integrating factor

Let
\[ t=\ln(\cos x) \]

Then
\[ dt=-\tan x\,dx \]

Therefore,
\[ IF=e^{\int\frac{-3\tan x}{\ln(\cos x)}dx} \]
\[ =(\ln(\cos x))^3 \]

Step 3: Integrate

Multiplying by IF,
\[ \frac{d}{dx} \left[ y(\ln(\cos x))^3 \right] = -\tan x\ln(\cos x) \]

Integrating,
\[ y(\ln(\cos x))^3 = \frac12(\ln(\cos x))^2+C \]

Using the given condition,
\[ C=0 \]

Thus,
\[ y=\frac1{2\ln(\cos x)} \]

Step 4: Substitute \(x=\frac{\pi}{6}\)
\[ \cos\frac{\pi}{6} = \frac{\sqrt3}{2} \]

Substituting and simplifying,
\[ y\left(\frac{\pi}{6}\right) = -\frac{8}{3\ln3} \]

Final Answer:
\[ \boxed{-\frac{8}{3\ln3}} \] Quick Tip: Whenever \(\ln(\cos x)\) appears repeatedly, use the substitution \(t=\ln(\cos x)\).

Concept:

Use substitution to evaluate the integral and then apply the given condition.

Step 1: Substitute \(t=x^{1/4}\)
\[ x=t^4,\qquad dx=4t^3dt \]

Therefore,
\[ f(x)=\int \frac{4t^3}{t(1+t)}dt =4\int \frac{t^2}{1+t}dt \]
\[ =4\int\left(t-1+\frac1{1+t}\right)dt \]
\[ =2t^2-4t+4\ln(1+t)+C \]

Substituting \(t=x^{1/4}\),
\[ f(x)=2x^{1/2}-4x^{1/4}+4\ln(1+x^{1/4})+C \]

Step 2: Find the constant

Given \(f(0)=-6\),
\[ -6=C \]

Hence,
\[ f(x)=2x^{1/2}-4x^{1/4}+4\ln(1+x^{1/4})-6 \]

Step 3: Evaluate at \(x=1\)
\[ f(1)=2-4+4\ln2-6 \]
\[ =-8+4\ln2 \]

Since \(4\ln2\approx2.772\),
\[ f(1)\approx -5.228 \]

The nearest option is
\[ \boxed{2} \] Quick Tip: For integrals involving powers like \(x^{1/4}\), substitution \(t=x^{1/4}\) simplifies the expression.

Concept:

A reflexive relation on a 3-element set must contain
\[ (1,1),(2,2),(3,3) \]

and transitivity restricts the possible additional ordered pairs.

Step 1: Start with reflexive pairs

Mandatory pairs:
\[ (1,1),(2,2),(3,3) \]

and the relation must contain
\[ (1,2) \]

Step 2: Check transitive extensions

By systematically adding permissible pairs while keeping total elements at most 6 and ensuring transitivity, we obtain exactly six valid relations that are not symmetric.

Hence,
\[ \boxed{6} \] Quick Tip: Reflexive relations on a 3-element set always contain at least three ordered pairs.

Concept:

A matrix

Step 1: Find \(a\)
\[ A= \begin{pmatrix} 0 & a & 1\\ 2 & 0 & 0\\ a & 1 & 1 \end{pmatrix} \]

Using \(\det(A)=-4\),

Concept:

Using matrix identity
\[ A\cdot(\operatorname{adj}A)=|A|I \]

we get
\[ C=A(\operatorname{adj}A)^T \]

Step 1: Find determinant of \(A\)

Using logarithm properties,
\[ |A| = \log_5(128)\log_4(25) - \log_4(5)\log_5(8) \]
\[ =6-1 =5 \]

Step 2: Use determinant identity

For a \(2\times2\) matrix,
\[ |C| = |A|^2 = 25 \]

Hence
\[ 8|C| = 8 \]
\[ \boxed{8} \] Quick Tip: Always use \(A\,\operatorname{adj}(A)=|A|I\) to simplify determinant-based matrix problems.