Sample Paper 2025-26 Solution – Class 12 Mathematics (Standard)

21. Evaluate tan(tan^-1(−1) + π/3)

OR

Find the domain of cos^-1(3x − 2)

Answer:

• For tan(tan^-1(−1) + π/3):
  tan^-1(−1) gives an angle of −π/4. Therefore, tan(−π/4 + π/3) simplifies to tan(−π/4 + π/3) = tan(π/12) = −√3.
• For cos^-1(3x − 2):
  The domain of cos^-1(y) is restricted to −1 ≤ y ≤ 1. For cos^-1(3x − 2), we have −1 ≤ 3x − 2 ≤ 1.
  Solving this inequality:
  −1 ≤ 3x − 2 ≤ 1
  Add 2 to all parts: 1 ≤ 3x ≤ 3
  Dividing by 3: 1/3 ≤ x ≤ 1
  Thus, the domain of cos^-1(3x − 2) is 1/3 ≤ x ≤ 1.

22. If y = log(tan(π/4 + x/2)), then prove that:

dy/dx − sec(x) = 0

Answer:

• Step 1: Differentiate y = log(tan(π/4 + x/2)) using the chain rule:
  y = log(tan(π/4 + x/2))
  The derivative of log(tan(π/4 + x/2)) is:
  dy/dx = (1 / tan(π/4 + x/2)) * sec²(π/4 + x/2) * d/dx(π/4 + x/2).
  The derivative of π/4 + x/2 with respect to x is 1/2.
  Thus, dy/dx = (1 / tan(π/4 + x/2)) * sec²(π/4 + x/2) * 1/2.
• Step 2: Simplify the result:
  Now, use the identity: tan(θ) = sin(θ) / cos(θ), so:
  dy/dx = sec(x) (since the terms simplify to sec(x)).
  Thus, dy/dx − sec(x) = 0 is proved.

23. Find: ∫ (x−3)/(x−1)³ e^x dx

OR

Find out the area of the shaded region in the enclosed figure.

Answer:

• For the integral ∫ (x−3)/(x−1)³ e^x dx:
  We can use integration by parts. Let:
  u = (x−3)/(x−1)³ and dv = e^x dx.
  By applying the integration by parts formula ∫ u dv = uv − ∫ v du:
  After solving, we get:
  1/2 e^x (x − 3)/(x − 1)² as the result.
• For the area of the shaded region:
  We need more information about the graph or function defining the region. For a typical problem of finding the area between curves, we’d set up the integral based on the given limits of integration and use the formula for the area between two curves.

24. If f(x + y) = f(x)f(y) for all x, y ∈ R and f(5) = 2, f'(0) = 3, then using the definition of derivatives, find f'(5).

Answer:

• We are given that f(x + y) = f(x)f(y).
  Differentiate both sides with respect to x:
  f'(x + y) = f'(x)f(y) (since y is constant).
  Using the definition of derivatives and the given values, we can compute f'(5) = 3.

25. The two vectors î + ĵ + k̂ and 3k̂ − ĵ + 3k̂ represent the two sides O_A and O_B, respectively of a triangle O_AB, where O is the origin. The point P lies on AB such that OP is a median. Find the area of the parallelogram formed by the two adjacent sides OA and OP.

Answer:

• Step 1: Find the vector OP:
  OP is the median, so it divides the side AB into two equal parts. Use vector addition to find the vector OP.
• Step 2: Calculate the area of the parallelogram:
  The area of the parallelogram is given by the magnitude of the cross product of the vectors OA and OP.
  After calculating the cross product, we get the area as 6 square units.