CBSE 12th Maths Question Paper 2025 (Set 1)

A is the identity matrix of order 3. The inverse of the identity matrix is itself, so A^{-1} = I_3 = [[1,0,0],[0,1,0],[0,0,1]]. Hence option (a) is correct. Since all options shown are the same matrix, any one of them is equivalent, but the correct mathematical answer is the identity matrix.

Let a = (3, 2, -1) and b = (1, -1, 1). Their dot product is a · b = 3(1) + 2(-1) + (-1)(1) = 3 - 2 - 1 = 0. Hence the vectors are perpendicular, so option (b) is correct. Also, they are not parallel, and their magnitudes are |a| = sqrt(14) and |b| = sqrt(3), so the other options are false.

For x in (0,1], |x|/x = 1, and for x in [-1,0), |x|/x = -1. Therefore, ∫[-1 to 0] |x|/x dx = ∫[-1 to 0] (-1) dx = -1 and ∫[0 to 1] |x|/x dx = ∫[0 to 1] 1 dx = 1. Adding both parts gives -1 + 1 = 0. Hence option (b) is correct.

A homogeneous function has all its terms of the same total degree, or is a function that satisfies f(tx, ty) = t^n f(x, y). Here, y^2 - xy is homogeneous of degree 2, and x - 3y is homogeneous of degree 1. Also, (y/x) sin(y/x) depends only on y/x, so it is homogeneous of degree 0. But tan x - sec y is not homogeneous because it is not of the form f(tx, ty) = t^n f(x, y). Therefore option (d) is not homogeneous.

Given f(x) = |x| + |x - 1|. Absolute value functions are continuous everywhere, so f(x) is continuous at x = 0 and x = 1. However, differentiability fails at points where the expression inside an absolute value changes sign: at x = 0 for |x| and at x = 1 for |x - 1|. Hence f(x) is continuous but not differentiable at x = 0 and x = 1. So option (c) is correct.