JEE Main 2020 Mathematics Question Paper with Answer and Solution

(C) Let $z = \sin \frac{2 \pi}{9} + i \cos \frac{2 \pi}{9}$.
Note that $|z|^2 = \sin^2 \frac{2 \pi}{9} + \cos^2 \frac{2 \pi}{9} = 1$,so $\bar{z} = \frac{1}{z}$.
The expression becomes $\left(\frac{1+z}{1+\bar{z}}\right)^3 = \left(\frac{1+z}{1+\frac{1}{z}}\right)^3 = \left(\frac{1+z}{\frac{z+1}{z}}\right)^3 = z^3$.
Now,$z = i(\cos \frac{2 \pi}{9} - i \sin \frac{2 \pi}{9}) = i e^{-i \frac{2 \pi}{9}}$.
Then $z^3 = (i e^{-i \frac{2 \pi}{9}})^3 = i^3 e^{-i \frac{6 \pi}{9}} = -i e^{-i \frac{2 \pi}{3}}$.
$z^3 = -i (\cos \frac{2 \pi}{3} - i \sin \frac{2 \pi}{3}) = -i (-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = i \frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1}{2}(\sqrt{3}-i)$.

(D) The implication $(p \wedge q) \rightarrow (\sim q \vee r)$ is false $(F)$ only when the antecedent $(p \wedge q)$ is true $(T)$ and the consequent $(\sim q \vee r)$ is false $(F)$.
For $(p \wedge q)$ to be true,both $p$ and $q$ must be true $(T)$.
For $(\sim q \vee r)$ to be false,both $\sim q$ and $r$ must be false $(F)$.
Since $q$ is true,$\sim q$ is false. For $r$ to be false,$r$ must be $F$.
Thus,the truth values are $p = T, q = T, r = F$.

(A) Given: Original mean $\bar{x} = 20$,original standard deviation $\sigma = 2$.
If each observation $x_i$ is transformed to $y_i = p x_i - q$,then the new mean $\bar{y} = p \bar{x} - q$ and the new standard deviation $\sigma_y = |p| \sigma$.
Given that the new mean is half of the original mean: $\bar{y} = \frac{20}{2} = 10$.
So,$p(20) - q = 10 \implies 20p - q = 10$ $(i)$.
Given that the new standard deviation is half of the original standard deviation: $\sigma_y = \frac{2}{2} = 1$.
So,$|p| \times 2 = 1 \implies |p| = \frac{1}{2} \implies p = \pm \frac{1}{2}$.
Case $1$: If $p = \frac{1}{2}$,then $20(\frac{1}{2}) - q = 10 \implies 10 - q = 10 \implies q = 0$. This contradicts the condition $q \neq 0$.
Case $2$: If $p = -\frac{1}{2}$,then $20(-\frac{1}{2}) - q = 10 \implies -10 - q = 10 \implies q = -20$.
Thus,$q = -20$.

(B) Let the unknown numbers be $x$ and $y$.
Given mean $\bar{x} = 8$.
$\frac{2+4+10+12+14+x+y}{7} = 8$
$42 + x + y = 56$
$x + y = 14$ $... (i)$
Given variance $\sigma^2 = 16$.
We know $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$.
$16 = \frac{2^2+4^2+10^2+12^2+14^2+x^2+y^2}{7} - 8^2$
$16 + 64 = \frac{4+16+100+144+196+x^2+y^2}{7}$
$80 \times 7 = 460 + x^2 + y^2$
$560 = 460 + x^2 + y^2$
$x^2 + y^2 = 100$ $... (ii)$
From $(x+y)^2 = x^2 + y^2 + 2xy$,we have $14^2 = 100 + 2xy$.
$196 - 100 = 2xy$ $\Rightarrow 2xy = 96$ $\Rightarrow xy = 48$.
Now,$(x-y)^2 = (x+y)^2 - 4xy = 14^2 - 4(48) = 196 - 192 = 4$.
$|x-y| = \sqrt{4} = 2$.

(D) Given the equation $\cos^4 \theta + \sin^4 \theta + \lambda = 0$.
We know that $\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2 \sin^2 \theta \cos^2 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta$.
Substituting this into the equation: $1 - 2 \sin^2 \theta \cos^2 \theta + \lambda = 0$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we have $\sin^2 \theta \cos^2 \theta = \frac{\sin^2 2\theta}{4}$.
Thus,$1 - 2 \left(\frac{\sin^2 2\theta}{4}\right) + \lambda = 0$,which simplifies to $1 - \frac{\sin^2 2\theta}{2} + \lambda = 0$.
Therefore,$\lambda = \frac{\sin^2 2\theta}{2} - 1$.
Since $0 \leq \sin^2 2\theta \leq 1$,we have $0 \leq \frac{\sin^2 2\theta}{2} \leq \frac{1}{2}$.
Subtracting $1$ from all parts,we get $-1 \leq \frac{\sin^2 2\theta}{2} - 1 \leq -\frac{1}{2}$.
Hence,$\lambda \in \left[-1, -\frac{1}{2}\right]$.