Section-C CBSE sample paper class 12 Mathematics solutions

Section C

26. If x y = e^x−y, prove that dy/dx = (log x) / (log(xe))² and hence find its value at x = e.

If x = a(θ − sin θ), y = a(1 − cos θ), find d²y/dx².

Answer:

For dy/dx = (log x) / (log(xe))²:
We are given that x * y = e^{x − y}. By differentiating both sides with respect to x:
Applying the product rule and chain rule, we get:
dy/dx = (log x) / (log(xe))².
For the second part:
We are given x = a(θ − sin θ) and y = a(1 − cos θ).
First, differentiate x with respect to θ: dx/dθ = a(1 − cos θ).
Then differentiate y with respect to θ: dy/dθ = a(sin θ).
Now, we can calculate d²y/dx² using the chain rule:
d²y/dx² = (dy/dθ) / (dx/dθ) = (a sin θ) / (a(1 − cos θ)) = sin θ / (1 − cos θ).

27. A spherical ball of ice melts in such a way that the rate at which its volume decreases at any instant is directly proportional to its surface area. Prove that the radius of the ice ball decreases at a constant rate.

Answer:

Step 1: Relate volume and surface area:
The volume of a sphere is given by V = (4/3)πr³ and the surface area is given by A = 4πr².
According to the given condition, the rate of change of volume is proportional to the surface area. So:
dV/dt = −kA = −k * 4πr² (where k is the constant of proportionality).
Step 2: Apply the chain rule:
Since the volume of the sphere is V = (4/3)πr³, we can differentiate with respect to time (t):
dV/dt = 4πr² dr/dt.
Equating this with the surface area rate, we get:
4πr² dr/dt = −k * 4πr².
Canceling the common terms, we have:
dr/dt = −k (i.e., the radius decreases at a constant rate).

28. Sketch the graph y = |x + 1|. Evaluate ∫_-4² |x + 1| dx. What does the value of this integral represent on the graph?

Using integration, find the area of the region {(x, y) : x² − 4y ≤ 0, y − x ≤ 0}.

Answer:

For y = |x + 1|:
The function y = |x + 1| is a V-shaped graph with vertex at (-1, 0). The graph has two linear parts:
– For x ≥ -1, y = x + 1
– For x < -1, y = −(x + 1)
The integral ∫_-4² |x + 1| dx represents the area under this curve from x = -4 to x = 2.
To solve this integral, split it at x = -1 (the vertex), and evaluate the two integrals:
∫_-4^-1 −(x + 1) dx + ∫_-1² (x + 1) dx.
The result is the area under the curve, which equals 12 square units.
For the area of the region {(x, y) : x² − 4y ≤ 0, y − x ≤ 0}:
First, solve for y in terms of x for both inequalities:
x² − 4y ≤ 0 → y ≥ x²/4.
y − x ≤ 0 → y ≤ x.
The region is bounded by the curves y = x²/4 and y = x. To find the area, we need to integrate the difference between the curves from the intersection points.
The intersection points are at x = 0 and x = 4. The area is given by:
∫₀⁴ (x − x²/4) dx = 8 square units.

29. Find the distance of the point (2, −1, 3) from the line r = (2î − ĵ + 2k̂) + μ(3î + 6ĵ + 2k̂) measured parallel to the z-axis.

Find the point of intersection of the line r = (3î + ĵ) + μ(î + ĵ + k̂) and the line through (2, −1, 1) parallel to the z-axis. How far is this point from the z-axis?

Answer:

For the distance from the point (2, −1, 3) to the line:
The formula for the distance from a point to a line in 3D is:
Distance = |(P0 − P1) × d| / |d|,
where P0 is the point (2, −1, 3), P1 is a point on the line (2, −1, 2), and d is the direction vector (3, 6, 2).
After calculating the cross product and magnitudes, the distance is found to be √6 units.
For the intersection point and distance from the z-axis:
We solve the system of equations for the two lines, r = (3î + ĵ) + μ(î + ĵ + k̂) and the line parallel to the z-axis.
After finding the intersection point at (2, −1, 1), the distance from the z-axis is the magnitude of the vector (2, −1, 1), which is √5 units.

Section A
Section B