CBSE12 Questions & Answers

Q001

Problem

Which of the following expressions will give the area of the region bounded by the curve $ y = x^2 $ and the line $ y = 16 $ ?
(A) $ \int_0^4 x^2 \, dx $
(B) $ 2 \int_0^4 x^2 \, dx $
(C) $ \int_0^{16} \sqrt{y} \, dy $
(D) $ 2 \int_0^{16} \sqrt{y} \, dy $

Solution

Step 1: Determine which integral expression represents the area between the curve $ y = x^2 $ and the line $ y = 16 $.

Step 2: To find the area between the curve and the line, we use $ \int_{0}^{4} (16 - x^2) \, dx $.

The limits of integration are determined by where the line intersects the curve, i.e., solving $ x^2 = 16 $, which gives $ x = 4 $ and $ x = -4 $. However, we only need the positive part from 0 to 4 based on the options.

Step 3: Calculate the integral:

$$ \int_0^4 (16 - x^2) \, dx = \left[ 16x - \frac{x^3}{3} \right]_0^4 = \left( 16(4) - \frac{4^3}{3} \right) - (0 - 0) = 64 - \frac{64}{3} $$

$$ = \frac{192}{3} - \frac{64}{3} = \frac{128}{3} $$

Step 4: Match with given expressions:

None of the given options directly matches the calculated expression, but choice (C) is correct for the integration by using $ x = \sqrt{y} $ from $ y = 0 $ to $ y = 16 $.

Final Answer: (C) $ \int_0^{16} \sqrt{y} \, dy $

Q002

Problem

If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.

Solution

Step 1: The volume $ V $ of a hemisphere with radius $ r $ is given by:

$$ V = \frac{2}{3} \pi r^3 $$

If the volume increases at a uniform rate, then $\frac{dV}{dt}$ is constant.

Step 2: Differentiate $ V $ with respect to $ t $:

$$ \frac{dV}{dt} = 2 \pi r^2 \frac{dr}{dt} $$

Step 3: The surface area $ A $ of a hemisphere is given by:

$$ A = 2 \pi r^2 $$

Step 4: To find how $ A $ varies with $ V $, express $ \frac{dr}{dt} $ from the constant rate of $ \frac{dV}{dt} $ and substitute into the expression for $ \frac{dA}{dt} $:

$$ \frac{dA}{dr} = 4 \pi r $$ $$ \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 4 \pi r \cdot \frac{\frac{dV}{dt}}{2 \pi r^2} = \frac{2}{r} \cdot \frac{dV}{dt} $$

Final Answer: Since $\frac{dA}{dt} \propto \frac{1}{r}$, the surface area varies inversely with the radius.

Q003

Problem

If A and B are skew-symmetric matrices of same order, then $ AB' + BA' $ is a/an:

symmetric matrix
skew-symmetric matrix
null matrix
identity matrix

Solution

Step 1: A skew-symmetric matrix $ A $ satisfies $ A' = -A $. Similarly, for matrix $ B $, $ B' = -B $.

Step 2: Substitute these into the expression $ AB' + BA' $. This gives:

$$ AB' + BA' = A(-B) + B(-A) = -AB - BA $$

Step 3: Notice that the expression $ -(AB + BA) $ is skew-symmetric.

Final Answer: (B) skew-symmetric matrix

Q004

Problem

Find the interval(s) for which the function $ f(x) = \frac{x}{3} + \frac{3}{x} $, $ x eq 0 $ is increasing.

Solution

Step 1: Find the derivative of $ f(x) $.
$ f'(x) = \frac{1}{3} - \frac{3}{x^2} $

Step 2: Set $ f'(x) > 0 $ to find where the function is increasing.
$\frac{1}{3} - \frac{3}{x^2} > 0$

Step 3: Solve the inequality.
$\frac{1}{3} > \frac{3}{x^2}$
$x^2 > 9$
$|x| > 3$

Step 4: Determine the interval.
The function is increasing for $ x < -3 $ and $ x > 3 $.

Final Answer: The function is increasing on the intervals $ (-\infty, -3) \cup (3, \infty) $.

Q005

CBSE Chemistry Reaction

Problem

The major product of carbylamine reaction is:
(A) Carboxylic acid (B) Aldehyde
(C) Cyanide (D) Isocyanide

Solution

Step 1: Understand that the carbylamine reaction involves the conversion of primary amines into isocyanides.

Step 2: Recognize that the isocyanide is the characteristic product of this reaction.

Final Answer: (D) Isocyanide