CUET 2026 Mathematics May 29 Shift 2 Question Paper- Download Answer Key and Solution PDF

Concept:


Order = Highest order derivative present in the differential equation.
Degree = Power of the highest order derivative after removing radicals and fractions in derivatives.


Step 1: Identify the highest order derivative

Given equation: \[ \left(\frac{d^3y}{dx^3}\right)^2 + 4\left(\frac{dy}{dx}\right)^4 + y = \sin(x) \]

The derivatives present are: \[ \frac{dy}{dx} \quad and \quad \frac{d^3y}{dx^3} \]

Among these, the highest order derivative is: \[ \frac{d^3y}{dx^3} \]

Hence, \[ \boxed{Order=3} \]

Step 2: Determine the degree

The highest order derivative appears as: \[ \left(\frac{d^3y}{dx^3}\right)^2 \]

Therefore, its power is \(2\).

Hence, \[ \boxed{Degree=2} \]

Final Answer: \[ \boxed{Order 3,\ Degree 2} \] Quick Tip: Degree depends only on the power of the highest order derivative, not on lower order derivatives.

Concept:

For an \(n\times n\) matrix: \[ \det(\operatorname{adj}A)=(\det A)^{n-1} \]

Since \(A\) is a \(3\times3\) matrix: \[ \det(\operatorname{adj}A)=(\det A)^2 \]

Step 1: Find determinant of adjoint matrix
 

Given:

Concept:

Shortest distance between two skew lines is: \[ d= \frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|} {|\vec b_1\times\vec b_2|} \]

where:

\(\vec a_1,\vec a_2\) are points on the lines
\(\vec b_1,\vec b_2\) are direction vectors


Step 1: Identify position and direction vectors

From first line: \[ \vec a_1=(1,2,1) \]
\[ \vec b_1=(1,-1,1) \]

From second line: \[ \vec a_2=(2,-1,-1) \]
\[ \vec b_2=(2,1,2) \]

Step 2: Find cross product of direction vectors
\[ \vec b_1\times\vec b_2= \begin{vmatrix} \hat i&\hat j&\hat k
1&-1&1
2&1&2 \end{vmatrix} \]

Expanding determinant: \[ = \hat i((-1)(2)-(1)(1)) -\hat j((1)(2)-(1)(2)) +\hat k((1)(1)-(-1)(2)) \]
\[ = \hat i(-2-1)-\hat j(2-2)+\hat k(1+2) \]
\[ =-3\hat i+0\hat j+3\hat k \]

Thus: \[ \vec b_1\times\vec b_2=(-3,0,3) \]

Magnitude: \[ |\vec b_1\times\vec b_2| = \sqrt{(-3)^2+0^2+3^2} \]
\[ =\sqrt{9+9} \]
\[ =\sqrt{18} \]

Step 3: Find connecting vector
\[ \vec a_2-\vec a_1 =(2-1,-1-2,-1-1) \]
\[ =(1,-3,-2) \]

Step 4: Apply shortest distance formula
\[ d= \frac{|(1,-3,-2)\cdot(-3,0,3)|} {\sqrt{18}} \]

Compute dot product: \[ =(1)(-3)+(-3)(0)+(-2)(3) \]
\[ =-3+0-6 \]
\[ =-9 \]

Taking modulus: \[ |{-9}|=9 \]

Thus: \[ d=\frac{9}{\sqrt{18}} \]

Rationalized form: \[ =\frac{9}{\sqrt{54}} \]

Final Answer: \[ \boxed{\frac{9}{\sqrt{54}}} \] Quick Tip: Cross product gives a vector perpendicular to both lines and is essential in shortest-distance problems.

Concept:

In Linear Programming Problems (LPP), the maximum or minimum value of the objective function occurs at the corner points of the feasible region.

Step 1: Identify constraints

Given: \[ x+y\le10 \]

and \[ x\ge0,\qquad y\ge0 \]

These inequalities represent the feasible region in the first quadrant.

Step 2: Find corner points

The line: \[ x+y=10 \]

cuts the axes at: \[ (10,0)\quad and\quad (0,10) \]

Including the origin, corner points are: \[ (0,0),\ (10,0),\ (0,10) \]

Step 3: Evaluate objective function

Objective function: \[ Z=3x+4y \]

At \((0,0)\): \[ Z=3(0)+4(0)=0 \]

At \((10,0)\): \[ Z=3(10)+4(0)=30 \]

At \((0,10)\): \[ Z=3(0)+4(10)=40 \]

Step 4: Choose maximum value

Largest value is: \[ 40 \]

obtained at: \[ (0,10) \]

Final Answer: \[ \boxed{(0,10)} \] Quick Tip: In LPP, always test all corner points because optimum values occur only at vertices.

Concept:

This is a conditional probability problem.

Since balls are drawn without replacement, the total number of balls changes after the first draw.

Step 1: Write initial number of balls

Initially: \[ 3 red balls \] \[ 5 blue balls \]

Total balls: \[ 3+5=8 \]

Step 2: Apply given condition

It is given that the first ball drawn was blue.

So one blue ball is removed.

Remaining balls: \[ 3 red \] \[ 4 blue \]

Total remaining: \[ 7 \]

Step 3: Find required probability

Probability that second ball is red: \[ P(Red) = \frac{Number of red balls remaining} {Total balls remaining} \]
\[ =\frac37 \]

Final Answer: \[ \boxed{\frac37} \] Quick Tip: Without replacement means the denominator decreases after each draw.

Concept:

A local maximum or minimum occurs at points where: \[ f'(x)=0 \]

These points are called critical points.

To determine whether the point is maximum or minimum, we use the second derivative test.

Step 1: Find first derivative

Given: \[ f(x)=x^3-3x+2 \]

Differentiate with respect to \(x\): \[ f'(x)=3x^2-3 \]

Step 2: Find critical points

Set: \[ f'(x)=0 \]
\[ 3x^2-3=0 \]

Divide by \(3\): \[ x^2-1=0 \]
\[ x^2=1 \]
\[ x=\pm1 \]

Thus the critical points are: \[ x=1,\quad x=-1 \]

Step 3: Find second derivative

Differentiate again: \[ f''(x)=6x \]

Step 4: Apply second derivative test

At: \[ x=-1 \]
\[ f''(-1)=6(-1)=-6 \]

Since: \[ f''(-1)<0 \]

the curve is concave downward.

Hence: \[ x=-1 \]
gives a local maximum.

Final Answer: \[ \boxed{x=-1} \] Quick Tip: If \[ f''(x)<0 \] then the function has a local maximum at that point.

Concept:

For inverse sine function: \[ \sin^{-1}(u) \]
the input value must satisfy: \[ -1\le u\le1 \]

Step 1: Apply inverse sine condition

Given: \[ f(x)=\sin^{-1}(2x-1) \]

Therefore: \[ -1\le2x-1\le1 \]

Step 2: Solve the inequality

Add \(1\) throughout: \[ 0\le2x\le2 \]

Divide throughout by \(2\): \[ 0\le x\le1 \]

Step 3: Write domain

Hence the domain is: \[ [0,1] \]

Final Answer: \[ \boxed{[0,1]} \] Quick Tip: For inverse trigonometric functions, always restrict the inner expression to the valid range.

Concept:

A function is differentiable at a point only if: \[ LHD=RHD \]

Absolute value functions create a sharp corner where the expression inside modulus becomes zero.

Step 1: Find left hand derivative

For: \[ x<2 \]
\[ |x-2|=-(x-2) \]
\[ =-x+2 \]

Differentiate: \[ \frac{d}{dx}(-x+2)=-1 \]

Thus: \[ LHD=-1 \]

Step 2: Find right hand derivative

For: \[ x>2 \]
\[ |x-2|=x-2 \]

Differentiate: \[ \frac{d}{dx}(x-2)=1 \]

Thus: \[ RHD=1 \]

Step 3: Compare derivatives
\[ LHD=-1 \] \[ RHD=1 \]

Since: \[ LHD\ne RHD \]

the function is not differentiable at: \[ x=2 \]

Final Answer: \[ \boxed{No} \] Quick Tip: Functions of the form \(|x-a|\) are non-differentiable at \(x=a\).

Concept:

Use Chain Rule: \[ \frac{d}{dx}(e^u)=e^u\frac{du}{dx} \]

where \(u\) is a function of \(x\).

Step 1: Choose inner function

Let: \[ u=x^2 \]

Differentiate: \[ \frac{du}{dx}=2x \]

Step 2: Apply chain rule
\[ \frac{d}{dx}(e^{x^2}) = e^{x^2}\cdot\frac{d}{dx}(x^2) \]
\[ = e^{x^2}\cdot2x \]
\[ = 2xe^{x^2} \]

Step 3: Write final derivative

Hence: \[ \boxed{2xe^{x^2}} \]

Final Answer: \[ \boxed{2xe^{x^2}} \] Quick Tip: Whenever an exponential contains another function inside it, always apply the chain rule.