NEB Class 12 Science Mathematics (NEB Released Items) Question Paper 2081 (Set A) Nepal

Given data ($n=6$):

a) Correlation coefficient

$$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{n\sum X^2-(\sum X)^2}\;\sqrt{n\sum Y^2-(\sum Y)^2}}$$

Numerator $= 6(1896)-(128)(92)=11376-11776=-400$.

$n\sum X^2-(\sum X)^2 = 6(2874)-128^2 = 17244-16384 = 860$.

$n\sum Y^2-(\sum Y)^2 = 6(1542)-92^2 = 9252-8464 = 788$.

$$r=\frac{-400}{\sqrt{860}\,\sqrt{788}}=\frac{-400}{\sqrt{677680}}\approx\frac{-400}{823.2}\approx -0.49$$

(The official marking scheme records the working values $\sum X=128,\ \sum Y=92,\ \sum XY=1896,\ \sum X^2=2874,\ \sum Y^2=1542$.) [3]

b) Regression coefficient of $Y$ on $X$

$$b_{YX}=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\frac{-400}{860}\approx -0.465$$

Thus the regression coefficient of $Y$ on $X$ is approximately $-0.47$. [2]

a) Binomial coefficient identity

Consider $(1+x)^n(1+x)^n=(1+x)^{2n}$.

Write $(1+x)^n=C_0+C_1x+C_2x^2+\cdots+C_nx^n$ and also $(x+1)^n=C_0x^n+C_1x^{n-1}+\cdots+C_n$.

Multiplying these two expansions, the coefficient of $x^n$ on the left is

$$C_0C_n+C_1C_{n-1}+C_2C_{n-2}+\cdots+C_nC_0.$$

On the right, $(1+x)^{2n}=\sum_{r=0}^{2n}\binom{2n}{r}x^r$, so the coefficient of $x^n$ is $\binom{2n}{n}=\dfrac{(2n)!}{n!\,n!}$.

Equating the coefficients of $x^n$:

$$C_0C_n+C_1C_{n-1}+\cdots+C_nC_0=\binom{2n}{n}=\frac{(2n)!}{n!\,n!}.\qquad\blacksquare$$

[3]

b) Mathematical induction

Let $P(n):\ 1+2+3+\cdots+n=\dfrac{n(n+1)}{2}$.

Base case $P(1)$: LHS $=1$, RHS $=\dfrac{1(1+1)}{2}=1$. So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true, i.e. $1+2+\cdots+k=\dfrac{k(k+1)}{2}$.

Then for $P(k+1)$:

$$1+2+\cdots+k+(k+1)=\frac{k(k+1)}{2}+(k+1)=(k+1)\left(\frac{k}{2}+1\right)=\frac{(k+1)(k+2)}{2}.$$

This is exactly $P(k+1)$. Hence by the principle of mathematical induction, $P(n)$ is true for every natural number $n$. $\blacksquare$ [2]

c) Row-equivalent (Gaussian elimination) method

The augmented matrix is

$$\left[\begin{array}{ccc|c}1&1&1&6\\1&-1&1&2\\1&1&-1&0\end{array}\right].$$

$R_2\to R_2-R_1,\ R_3\to R_3-R_1$:

$$\left[\begin{array}{ccc|c}1&1&1&6\\0&-2&0&-4\\0&0&-2&-6\end{array}\right].$$

From $R_2:\ -2y=-4\Rightarrow y=2$. From $R_3:\ -2z=-6\Rightarrow z=3$.

From $R_1:\ x+y+z=6\Rightarrow x+2+3=6\Rightarrow x=1$.

$$\boxed{x=1,\quad y=2,\quad z=3}$$

[3]