ISC Class 12 Maths Paper 2026 Solution

Solution:

Using the property |kA| = kⁿ|A|, where n is the order of the matrix:

|-4A| = (-4)³ |A|
|-4A| = -64 × (-3)
|-4A| = 192

Correct Option: (a)

Solution:

The function f(x) = log x is continuous for all x ∈ (0, ∞).

To check differentiability, we find the derivative:

f'(x) = 1/x

For x > 0, the derivative exists. Specifically at x = 1:

• f(1) = log(1) = 0 (Finite and Defined)
• f'(1) = 1/1 = 1 (Finite and Defined)

Since both the function and its derivative are defined at x = 1, the function is both continuous and differentiable.

Correct Option: (a)

Solution:

For mutually exclusive events, the probability of both occurring at the same time is zero:

P(A ∩ B) = 0

The addition rule for the union of two mutually exclusive events is:

P(A ∪ B) = P(A) + P(B)

P(A ∪ B) = 1/5 + 2/3
P(A ∪ B) = (3 + 10) / 15
P(A ∪ B) = 13/15

Correct Option: (b)

Solution:

First, we calculate the derivatives from both sides at x = 2:

Since LHD (1) ≠ RHD (-1), the function is not differentiable at x = 2. The Assertion is true.

The Reason accurately states the mathematical requirement for differentiability (LHD = RHD), which is exactly what we used to prove the Assertion.

Correct Option: (a)

Solution:

We analyze the behavior of the absolute value function |x| within the limits of integration [0, 3/2].

Since the interval is entirely non-negative (x ≥ 0), we can simplify the function:

|x| = x for x ≥ 0

The integral becomes:

Correct Option: (b)

Solution:

Regarding Statement 1:

Using the co-function identity cot x = tan(π/2 – x):

In the interval 0 < x < π, the value of (π/2 - x) falls within the principal value branch of tan^-1, which is (-π/2, π/2). Therefore, the statement is True.

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Regarding Statement 2:

The property tan^-1(tan x) = x is only valid when x is within the principal value branch:

x ∈ (-π/2, π/2)

Since the statement claims this is true for all real numbers (e.g., it fails for x = π), the statement is False.

Correct Option: (a)

Solution:

To find the total number of matrices, we consider the following:

• A 3 × 3 matrix has a total of 3 × 3 = 9 positions (elements).
• Each of these 9 positions can be filled in 2 ways (either with a 0 or a 1).

According to the Fundamental Principle of Counting, the total number of ways is:

Correct Option: (c)

Solution:

We trace the function from left to right to determine where it is rising (increasing) and falling (decreasing):

Now evaluate the statements:

• Statement 1 claims the function increases in (-∞, -1). We determined it is decreasing in this interval. → Statement 1 is False.
• Statement 2 claims the function decreases in (1, ∞). We determined it is increasing in this interval. → Statement 2 is False.

Since both statements are false, the correct option is (d).

Correct Option: (d)

Solution:

For a function to be onto (surjective), every element in the codomain (Set B) must have at least one pre-image in the domain (Set A).

Since 4 is less than 5 (n(A) < n(B)), it is mathematically impossible to cover all 5 elements of Set B using only 4 elements from Set A in a one-to-one manner.

Therefore, the number of such mappings is 0.

Correct Option: (b)

Solution:

This is a linear differential equation of the form:
dy/dx + Py = Q, where P = -1 and Q = 1.

Step 1: Find the Integrating Factor (I.F.)

Step 2: Apply the general solution formula

y × (I.F.) = ∫ (Q × I.F.) dx + C
y(e^-x) = ∫ (1 × e^-x) dx
y(e^-x) = -e^-x + C

Multiply through by e^x to solve for y:

y = -1 + Ce^x

Step 3: Use initial condition y(0) = 1

Substitute x = 0 and y = 1:

Final equation: y = 2e^x – 1

Correct Option: (d)

Solution:

For a square matrix A, being non-singular means that its determinant is non-zero (|A| ≠ 0).

The condition that A is non-singular is precisely what allows us to define the unique solution X = A^-1B. Therefore, both statements are true, and the Reason explains why the Assertion is possible.

Correct Option: (a)

Solution:

The given expression is an infinite nested exponent. We can observe that the part of the exponent starting from the second e is identical to the original definition of x.

Step 1: Simplify the expression

Step 2: Take the natural logarithm (ln) on both sides

ln(x) = ln(e^{y + x})
ln(x) = y + x

Step 3: Differentiate with respect to x

Step 4: Solve for dy/dx

dy/dx = 1/x – 1
dy/dx = (1 – x) / x

Answer: dy/dx = (1 – x) / x

Solution:

We expand the determinant along the first column (C₁):

Solving the Quadratic Equation:

Setting each factor to zero:

• x + 7 = 0 ⇒ x = -7
• x – 3 = 0 ⇒ x = 3

Answer: x = 3 or x = -7

Solution:

Let sec^-1(-√2) = θ. This implies:

sec θ = -√2

Step 1: Identify the Principal Value Branch

The range (principal value branch) of sec^-1 is [0, π] – {π/2}.

Step 2: Solve for θ

We know that sec(π/4) = √2. Since secant is negative in the second quadrant, we use:

Since 3π/4 lies within the interval [0, π], it is the principal value.

Answer: 3π/4

Solution:

Yes, the relation R is symmetric.

Justification:

A relation R on a set A is said to be symmetric if for every element (x, y) ∈ R, the reverse pair (y, x) must also be in R.

Since the condition for symmetry is satisfied for all pairs in the relation, R is symmetric.

Answer: Yes, R is symmetric.