Q 12 a) Define binomial expansion.
Solution: Binomial expansion is defined as a way for expansion of an algebraic expression containing two terms raised to its nth power. For example, binomial expansion of $$\rm (a + x)^n = ^nC_0 a^nx^0 + ^nC_1 a^{n-1}x^1 + ….. + ^nC_r a^{n – r} x^r + ….. + ^nC_n a^0 x^n$$
Q 12 b) Find the middle term of $(2 + 3x)^{30}$.
Solution: The index of the expression is an even number so, the middle term is at r = $\rm \frac{30}{2} = 15$, i.e., it is the 16th term.
$$\rm t_{15 + 1} = \ ^{30}C_{15} \cdot (2)^{30 – 15} \cdot (3x)^{15}$$$$\rm t_{16} = \ ^{30} C_{15} \cdot (6x)^{15}$$Hence, the required middle term is found.
Q 12 c) Find the sum of n binomial coefficients in the expansion of (1 +x)^n when x = 1.
Solution: Let us consider $^nC_0 = C_0, ^nC_1 = C_1$, and so on.
The expansion of $$\rm (1 + x)^n = C_0 + C_1 x + C_2 x^2 + ….. + C_n x^n$$
When x = 1
$$\rm (1 + 1)^n = C_0 + C_1 + C_2 + …. + C_n$$$$\Rightarrow \rm C_0 + C_1 + C_2 + … + C_n = 2^n$$
Hence, the required sum of n binomial coefficients is $2^n$.
Q 12 d) What do you mean by argument of complex number?
Solution: Argument of a complex number is defined as the angle ‘$\rm \theta $’ present in its polar form.
Q 13 a) Prove that $\rm 1^2 + 2^2 + 3^2 + … + n^2 = \frac{n(n+1)(2n+1)}{6}$ by mathematical induction.
Solution: To prove: $\rm 1^2 + 2^2 + 3^2 + … + n^2 = \frac{n(n+1)(2n+1)}{6}$.
BASIC STEP: Let us consider n = 1.
$$\rm 1^2 = \frac{1(1+1)(2 \times 1 + 1)}{6}$$$$\rm 1 = \frac{6}{6} = 1$$Hence, the basic step is verified.
INDUCTIVE STEP: Let us consider it is true for a real number, $\rm n = k, k \in R$.
$$\rm 1^2 + 2^2 + 3^2 + … + k^2 = \frac{k(k+1)(2k + 1)}{6}$$
PROOF STEP: Using the inductive step, we need to prove that: $\rm 1^2 + 2^2 + 3^2 + … + k^2 + (k + 1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
We know,
$$\rm 1^2 + 2^2 + 3^2 + … + k^2 + (k + 1)^2$$$$\rm = \frac{k(k+1)(2k + 1)}{6} + (k+1)^2$$$$\rm = \frac{k(k+1)(2k +1) + 6(k+1)^2}{6}$$$$\rm = \frac{(k+1) \left ( k(2k + 1) + 6(k+1) \right ) }{6}$$$$\rm = \frac{(k+1)(2k^2 + k + 6k + 6)}{6}$$$$\rm = \frac{(k+1)(2k^2 + 7k + 6)}{6}$$$$\rm = \frac{(k+1)(2k^2 + 4k + 3k + 6)}{6}$$$$\rm = \frac{(k+1)(2k + 3)(k + 2)}{6}$$$$\rm = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$Hence, by the principle of mathematical induction, it the given statement is verified.
Q 13 b) Find the sum of the squares of the first 20 natural numbers.
Solution: We have, sum of squares of the first n natural numbers = $\rm \frac{n(n +1)(2n + 1)}{6}$
Put n = 20$$\rm S_n = \frac{20(20 + 1)(2 \times 20 + 1)}{6}$$$$\rm S_n = 2870$$Hence, the required sum of the squares of the first 20 natural numbers is 2870.
Q 14 a) Find the equation of plane through (-3,2,-1) and parallel to the plane 2x + 5y – 2 = 0.
Solution: A plane parallel to the plane $\rm 2x + 5y – 2 = 0$ will have an equation $\rm 2x + 5y – 2 + k = 0$ where k is a constant.
Put $(x,y) = (-3,2)$ then$$\rm 2(-3) + 5(2) – 2 + k = 0$$$$\rm k = -2$$
Put value of k in equation of plane to get $\rm 2x + 5y – 2 – 2 = 0$.
Hence, the required equation of the plane is $\rm 2x + 5y – 4 = 0$.
Q 14 b) Using vector method, prove that $\rm \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ in any $\triangle$ ABC.
Solution:
Q 15 a) Find the derivative of $(\tan x)^{\log x}$.
Solution:
Q 15 b) Solve: $(1 + x)dy = dx(1 + xy – x)$.
Solution:
Q 16) The regression equations related to age (x) and blood pressure (y) are given as below:
0.22 x – y + 81.1175 = 0 and x – 1.623 y + 105.695 = 0
a) Find the regression coefficients.
Solution:
b) Find the correlation coefficient.
Solution:
c) Find the mean of x and mean of y.
Solution:
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