JEE Main 26 June 2022 Shift 1 question paper with solutions

Given $f(x) = \frac{x-1}{x+1}$. Therefore, $f^2(x) = f(f(x)) = \frac{x-1}{x+1} \cdot \frac{-1}{x-1} = \frac{-1}{x}$. Then, $f^3(x) = f(f^2(x)) = f\left(\frac{-1}{x}\right) = \frac{x+1}{1-x}$. Thus, $f^4(x) = f\left(\frac{x+1}{1-x}\right) = \frac{-1}{x}$. Therefore, $f^6(x) = \frac{-1}{x} \Rightarrow f^6(6) = \frac{-1}{8}$. Next, $f^7(x) = \left(\frac{-1}{x}\right) = \frac{x+1}{1-x}$. Thus, $f^7(7) = \frac{8}{-6} = \frac{-4}{3}$. Therefore, $\frac{1}{6} + \frac{4}{3} = \frac{3}{2}$.

Set A implies $\($ $\left$| $\frac{z+1}{z-1}$ $\right$| < 1 $\)$ which implies $\($ |z+1| < |z-1| $\)$. This leads to $\($ (x+1)^2 + y^2 < (x-1)^2 + y^2 $\)$ which implies $\($ x < 0 $\)$. Set B implies $\($ $\arg$ $\left$( $\frac{z-1}{z+1}$ $\right$) = $\frac{2\pi}{3}$ $\)$. This leads to $\($ $\tan$^{-1} $\left$( $\frac{y}{x-1}$ $\right$) - $\tan$^{-1} $\left$( $\frac{y}{x+1}$ $\right$) = $\frac{2\pi}{3}$ $\)$. This results in $\($ x^2 + y^2 + $\frac{2y}{\sqrt{3}}$ - 1 = 0 $\)$. The intersection $\($ A $\cap$ B $\)$ implies the center is $\($ $\left$( 0, -$\frac{1}{\sqrt{3}}$ $\right$) $\)$.

Given $adj(24A)=adj(3(adj(2A)))$ $\Rightarrow |24A|^2=(3\,adj(2A))^2$ $\Rightarrow (24^3|A|)^2=(3^3|adj(2A)|)^2$ $=3^6(|2A|^2)^2$ $\Rightarrow 24^6|A|^2=3^6\times2^{12}|A|^4$ $\Rightarrow |A|^2=\frac{24^6}{3^6\times2^{12}}$ $=64$

Given the determinant: $$\begin{vmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{vmatrix} = 0$$ Expanding, we have: $$3(-8a - 9) + 2(5a - 18) + 1(21) = 0$$ This implies $a = -3$. Also, consider $$\Delta_2 = \begin{vmatrix} 3 & -2 & b \\ 5 & 8 & 3 \\ 2 & 1 & -1 \end{vmatrix}$$ If $b = \frac{1}{3}$, then $\Delta_2 = 0$. So $b$ must be equal to $$-\frac{1}{3}$$

Given $ (2021)^{2023} = (7\lambda - 2)^{2023} $. $$ = \binom{2023}{0} (7A)^{2023} - \ldots - \binom{2023}{2023} 2^{2023} $$ $$ = 7t - 2^{2023} $$ Therefore, $ -2^{2023} = -2 \times 2^{2022} $. $$ = -2 \times (2^3)^{674} $$ $$ = -2 (1 + 7\mu)^{674} $$ $$ = -(7\alpha + 2) $$ Thus, the remainder is $ -2 $ or $ +5 $.

Evaluate the limit: $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin(\cos^{-1} x) - x}{1 - \tan(\cos^{-1} x)}$$ Substitute the trigonometric identities: $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sin\left(\sin^{-1}\sqrt{1-x^2}\right) - x}{1 - \tan\left(\tan^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right)}$$ Simplify the expression: $$\lim_{x \to \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^2} - x}{1 - \frac{\sqrt{1-x^2}}{x}}$$ Evaluate the limit: $$\lim_{x \to \frac{1}{\sqrt{2}}} (-x) = -\frac{1}{\sqrt{2}}$$

Given $$f(x) = \begin{cases} x+3 & ; \ x 0 \end{cases}$$ Check continuity at $x = 0$ $$gof(0) = g\left(f(0^-)\right) = g\left(f(0^+)\right)$$ $$4 + k_2 = 9 - 3k_1 = 4 + k_2$$ $$3k_1 + k_2 = 5 \ldots (a)$$ Differentiate $$\left(g(f(x))\right)' = \begin{cases} 2(x+3) + k_1 & ; \ x < -3 \\ 2(x+3) - k_1 & ; \ -3 \leq x < 0 \\ 4e^x & ; \ x \geq 0 \end{cases}$$ $$6 - k_1 = 4$$ $$k_1 = 2 \ldots (b)$$ Therefore, $k_1 = 2, k_2 = -1$ $$gof(x) = \begin{cases} (x+3)^2 + 2(x+3) & ; \ x < -3 \\ (x+3)^2 - 2(x+3) & ; \ -3 \leq x < 0 \\ 4e^x - 1 & ; \ x \geq 0 \end{cases}$$ $$gof(-4) + gof(4) = 4e^4 - 2$$ $$\Rightarrow 2\left(2e^4 - 1\right)$$

Given $$f(x) = \begin{cases} x^2 - 4x - 2, & \forall x \in \left(-1, \frac{3 - \sqrt{17}}{2}\right) \\ -x^2 + 2x + 2, & \forall x \in \left(\frac{3 - \sqrt{17}}{2}, 2\right) \end{cases}$$ $f'(x)$ when $x \in \left(-1, \frac{3 - \sqrt{17}}{2}\right)$ $$f'(x) = 2x - 4 = 0 \implies x = 2$$ $$f'(x) = 2(x - 2) \implies f'(x) is always \downarrow$$ $$f(2) = 2$$ $$f(-1) = 3$$ $$f\left(\frac{3 - \sqrt{17}}{2}\right) = \frac{\sqrt{17} - 3}{2}$$ $f'(x)$ when $x \in \left(\frac{3 - \sqrt{17}}{2}, 2\right)$ $$f'(x) = -2x + 2$$ $$f'(x) = -2(x - 1)$$ $$f'(x) = 0 when x = 1$$ $$f(1) = 3$$ absolute minimum value $= \frac{\sqrt{17} - 3}{2}$ absolute maximum value $= 3$ Sum $= \frac{\sqrt{17} - 3}{2} + 3 = \frac{\sqrt{17} + 3}{2}$

Given $\($ $\left$( $\frac{x}{a}$ $\right$)^n + $\left$( $\frac{y}{b}$ $\right$)^n = 2 $\)$. Slope of tangent at $\($(a, b)$\)$ is calculated as follows: $\[$ n $\left$( $\frac{x}{a}$ $\right$)^{n-1} $\cdot$ $\frac{1}{a}$ + n $\left$( $\frac{y}{b}$ $\right$)^{n-1} $\cdot$ $\frac{1}{b}$ $\frac{dy}{dx}$ = 0 $\]$ $\[$ $\left$. $\frac{dy}{dx}$ $\right$|_{(a,b)} = -$\frac{b}{a}$ $\]$ Therefore, the equation of the tangent is: $\[$ y - b = -$\frac{b}{a}$ (x - a) $\]$ $\[$ $\frac{x}{a}$ + $\frac{y}{b}$ = 2 $\forall$ $\ $n $\in$ $\mathbb{N}$ $\]$

Area of shaded region $$= 2 \int_{0}^{3} \left( \sqrt{9+y} - \sqrt{9-y} \right) \, dy + 2 \int_{3}^{9} \left( \sqrt{9-y} \right) \, dy$$ $$= 2 \left[ \int_{0}^{3} (9+y)^{1/2} \, dy - \int_{0}^{3} (9-y)^{1/2} \, dy + \int_{3}^{9} (9-y)^{1/2} \, dy \right]$$ $$= 2 \left[ \frac{2}{3} (9+y)^{3/2} \bigg|_{0}^{3} + \frac{2}{3} (9-y)^{3/2} \bigg|_{3}^{9} - \frac{2}{3} (9-y)^{3/2} \bigg|_{0}^{3} \right]$$ $$= \frac{4}{3} \left[ 12\sqrt{12} - 27 + 6\sqrt{6} - 27 - (0 - 6\sqrt{6}) \right]$$ $$= \frac{4}{3} \left[ 24\sqrt{3} + 12\sqrt{6} - 54 \right]$$ $$= 8 \left( 4\sqrt{3} + 2\sqrt{6} - 9 \right)$$

Given $x + y = 5$ and $\frac{5}{\sqrt{2}}$. We have $\sin 60^\circ = \frac{\frac{5}{\sqrt{2}}}{a}$. Solving for $a$, we get $a = \frac{5\sqrt{2}}{3}$. The area of $\triangle PQR$ is $\frac{\sqrt{3}}{4} a^2 = \frac{25}{2\sqrt{3}}$.

AB = $\sqrt{26}$. The equation for the radius squared is given by $r^2 = CM^2 + AM^2$. Substituting the values, we have: $$r^2 = \left(2 \times \frac{\sqrt{13}}{2}\right)^2 + \left(\frac{\sqrt{13}}{2}\right)^2$$ Simplifying, we find: $$r^2 = \frac{65}{2}$$

Equation of normal: $y = -tx + 2at + at^3$ $\($ $\left$( a = $\frac{3}{2}$ $\right$) $\)$ since passing through $(5, -8)$, we get $t = -2$. Co-ordinate of $Q$: $(6, -6)$. Equation of tangent at $Q$: $x + 2y + 6 = 0$. Put $x = -\frac{3}{2}$ to get $R \left( -\frac{3}{2}, -\frac{9}{4} \right)$.

For $l_1$: $\($ $\frac{x-2}{3}$ = $\frac{y+1}{-2}$ = $\frac{z-2}{0}$ $\)$ For $l_2$: $\($ $\frac{x-1}{1}$ = $\frac{y+3/2}{\alpha/2}$ = $\frac{z+5}{2}$ $\)$ For $l_3$: $\($ $\frac{x-1}{-3}$ = $\frac{y-1/2}{-2}$ = $\frac{z-0}{4}$ $\)$ $l_1 \perp l_2 \Rightarrow \frac{|3-\alpha+0|}{\sqrt{13} \sqrt{1+\frac{\alpha^2}{4}+4}} = 0 \Rightarrow \alpha = 3$ Angle between $l_2$ and $l_3$: $$\cos \theta = \frac{|1 \times (-3) + (-2)(\alpha/2) + 2 \times 4|}{\sqrt{1+4+\frac{\alpha^2}{4}} \sqrt{9+16+4}}$$ $$\cos \theta = \frac{|-3-\alpha+8|}{\sqrt{5+\frac{\alpha^2}{4}} \sqrt{29}}$$ Put $\alpha = 3$ $$\cos \theta = \frac{2}{\sqrt{\frac{29}{4}}} = \frac{4}{29}$$ $$\theta = \cos^{-1} \left( \frac{4}{29} \right) \Rightarrow \theta = \sec^{-1} \left( \frac{29}{4} \right)$$

Let equation of rotated plane be: $$(2x + 3y + z + 20) + \lambda (x - 3y + 5z - 8) = 0$$ $$(2 + \lambda)x + (3 - 3\lambda)y + (1 + 5\lambda)z + 20 - 8\lambda = 0$$ Above plane is perpendicular to $2x + 3y + z + 20 = 0$. So, $(2 + \lambda) \cdot 2 + (3 - 3\lambda) \cdot 3 + (1 + 5\lambda) \cdot 1 = 0 \Rightarrow \lambda = 7$. Therefore, equation of rotated plane is: $$x - 2y + 4z - 4 = 0$$ Mirror image of $A \left(2, \frac{-1}{2}, 2\right)$ in rotated plane is $B(a, b, c)$. Equation of $AB$: $$\frac{x - 2}{1} = \frac{y + 1/2}{-2} = \frac{z - 2}{4} = k$$ Let coordinate of $B$ be $(2 + k, \frac{-1}{2} - 2k, 2 + 4k)$. Midpoint of $AB$ is: $$\left(2 + \frac{k}{2}, \frac{-1}{2} - k, 2 + 2k\right)$$ which will lie on the plane $x - 2y + 4z - 4 = 0$. Hence $k = \frac{-2}{3}$. Therefore $B$ is: $$\left(\frac{4}{3}, \frac{5}{6}, \frac{-2}{3}\right) \equiv \left(\frac{8}{6}, \frac{5}{6}, \frac{-4}{6}\right)$$ So, $\frac{a}{8} = \frac{b}{5} = \frac{c}{-4}$.

Given $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = 3 \vec{b} - \vec{c}$. $\vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} = \vec{c} - 2 \vec{a}$. $\vec{c} \times (\vec{b} \times \vec{a}) = (\vec{c} \cdot \vec{a}) \vec{b} - (\vec{c} \cdot \vec{b}) \vec{a} = 3 \vec{b} - 2 \vec{a}$. $[3 \vec{b} - \vec{c}, \vec{c} - 2 \vec{a}, 3 \vec{b} - 2 \vec{a}]$. $(3 \vec{b} - \vec{c}) \cdot [(\vec{c} - 2 \vec{a}) \times (3 \vec{b} - 2 \vec{a})]$. $(3 \vec{b} - \vec{c}) \cdot [3 (\vec{c} \times \vec{b}) - 2 (\vec{c} \times \vec{a}) - 6 (\vec{a} \times \vec{b})]$. $-6 [\vec{b} \ \vec{c} \ \vec{a}] + 6 [\vec{c} \ \vec{a} \ \vec{b}]$

Given $\mathrm{P(H)} = x$, $\mathrm{P(T)} = 1 - x$. $\mathrm{P(4H, 1T)} = \mathrm{P(5H)}$ $$\binom{5}{1} x^4 (1-x)^1 = \binom{5}{5} x^5$$ $$5(1-x) = x$$ $$6x = 5 = 0 x = \frac{5}{6}$$ $\mathrm{P(atmost\ 2H)}$ $= \mathrm{P(0H, 5T)} + \mathrm{P(1H, 4T)} + \mathrm{P(2H, 3T)}$ $$= \binom{5}{0} \left(\frac{1}{6}\right)^5 + \binom{5}{1} \cdot \frac{5}{6} \cdot \left(\frac{1}{6}\right)^4 + \binom{5}{2} \left(\frac{5}{6}\right)^3 \left(\frac{1}{6}\right)^3$$ $$= \frac{1}{6^5} (1 + 25 + 250) = \frac{276}{6^5}$$ $$= \frac{46}{6^4}$$

Given $$\sigma^2 = \frac{\sum_{i=1}^{5} (x_i - \bar{x})^2}{n}$$ Mean = 6 $$\frac{a + b + 8 + 5 + 10}{5} = 6$$ $$a + b = 7$$ $$b = 7 - a$$ $$6.8 = \frac{(a-6)^2 + (b-6)^2 + (8-6)^2 + (5-6)^2 + (10-6)^2}{5}$$ $$34 = (a-6)^2 + (7-a-6)^2 + 4 + 1 + 18$$ $$a^2 - 7a + 12 = 0 \Rightarrow a = 4 or a = 3$$ $$a = 4 a = 3$$ $$b = 3 b = 4$$ $$M = \frac{\sum_{i=1}^{5} |x_i - \bar{x}|}{n}$$ $$M = \frac{|a-6| + |b-6| + |8-6| + |5-6| + |10-6|}{5}$$ When $$a = 3, b = 4$$ $$M = \frac{3 + 2 + 2 + 1 + 4}{5}$$ When $$a = 4, b = 3$$ $$M = \frac{2 + 3 + 2 + 1 + 7}{5}$$ $$M = \frac{12}{5}$$ $$M = \frac{12}{5}$$ $$25M = 25 \times \frac{12}{5} = 60$$

Case-I If $\Delta \equiv \nabla \equiv \wedge$ then $\left( p \land q \right) \to \left( \left( p \land q \right) \land r \right)$ it can be false if $r$ is false, so not a tautology. Case-II If $\Delta \equiv \nabla \equiv \lor$ then $\left( p \lor q \right) \to \left( \left( p \lor q \right) \lor r \right) \equiv$ tautology then $\left( p \lor q \right) \lor r \equiv \left( p \Delta r \right) \lor q$. Case-III if $\Delta = \lor$, $\nabla = \land$ then $\left( p \land q \right) \to \left\{ \left( p \lor q \right) \land r \right\}$ Not a tautology (Check $p \to T$, $q \to T$, $r \to F$). Case-IV if $\Delta = \land$, $\nabla = \lor$ then $\left( p \land q \right) \to \left\{ \left( p \land q \right) \lor r \right\}$ Not a tautology.

Given the equation $x^4 - 3x^3 - 2x^2 + 3x + 1 = 10$. Since $x = 0$ is not the root of this equation, divide it by $x^2$. $$x^2 - 3x - 2 + \frac{3}{x} + \frac{1}{x^2} = 0$$ Rewriting, we have: $$x^2 + \frac{1}{x^2} - 2 + 2 - 3 \left( x - \frac{1}{x} \right) - 2 = 0$$ Simplifying further: $$\left( x - \frac{1}{x} \right)^2 - 3 \left( x - \frac{1}{x} \right) = 0$$ Solving for $x$: $$x - \frac{1}{x} = 0, x - \frac{1}{x} = 3$$ For $x - \frac{1}{x} = 0$: $$x^2 - 1 = 0$$ Thus, $x = \pm 1$. For $x - \frac{1}{x} = 3$: $$x^2 - 3x - 1 = 0$$ Let $\alpha = 1$, $\beta = -1$, $\gamma + \delta = 3$, $\gamma \delta = -1$. Calculating $\alpha^3 + \beta^3 + \gamma^3 + \delta^3$: $$1 - 1 + (\gamma + \delta)((\gamma + \delta)^2 - 3\gamma \delta)$$ $$0 + 3(9 - 3(-1))$$ $$+ 3(12) = 36$$

Given $n(B) = 10$ and $n(a) = 5$. The number of ways of forming a group of 3 girls and 3 boys is given by: $$\binom{10}{3} \times \binom{5}{3}$$ Calculating this, we have: $$= \frac{10 \times 9 \times 8}{3 \times 2} \times \frac{5 \times 4}{2} = 1200$$ The number of ways when two particular boys $B_1$ and $B_2$ are members of the group together is: $$\binom{8}{1} \times \binom{5}{3} = 8 \times 10 = 80$$ The number of ways when boys $B_1$ and $B_2$ are in the same group together is: $$= 1200 \times 80 = 1120$$

Given $x^2 + y^2 = \frac{9}{4}$ and $y = 4x$. The equation of the line is $y = mx + \frac{1}{m}$. Compare equations (1) and (2): $$\pm \frac{3}{2} \sqrt{(1 + m^2)} = \frac{1}{m^2}$$ $$9m^2(1 + m^2) = 4$$ $$9m^4 + 9m^2 - 4 = 0$$ $$9m^4 + 12m^2 - 3m^2 - 4 = 0$$ $$3m^2(3m^2 + 4) - (3m^2 + 4) = 0$$ $$m^2 = -\frac{4}{3} (Rejected)$$ $$m^2 = \frac{1}{3} \implies m = \pm \frac{1}{\sqrt{3}}$$ Equation of common tangent: $$y = \frac{1}{\sqrt{3}} x + \sqrt{3}$$ On the X-axis, $y = 0$. $OQ = -3$ $b = |OQ| = 3$ $a = 6$ $$b^2 = a^2(1 - e^2) \implies e^2 = 1 - \frac{9}{36} = \frac{3}{4}$$ $$e = \frac{2b^2}{a} = \frac{2 \times 9}{6} = 3$$ $$\frac{e}{e^2} = \frac{3}{3/4} = 4$$

Given $f(x) = \max\{|x+1|, |x+2|, |x+3|, |x+4|, |x+5|\}$. The integral is evaluated as follows: $$\int_{-6}^{0} f(x) \, dx = \int_{-6}^{-3} |x+1| \, dx + \int_{-3}^{0} |x+5| \, dx$$ This simplifies to: $$= -\int_{-6}^{-3} (x+1) \, dx + \int_{-3}^{0} (x+5) \, dx$$ Evaluating the integrals: $$= -\left[ \frac{x^2}{2} + x \right]_{-6}^{-3} + \left[ \frac{x^2}{2} + 5x \right]_{-3}^{0}$$ Substituting the limits: $$= -\left( \frac{9}{2} - 3 \right) - (18 - 6) + \left( 0 - \left( \frac{9}{2} - 15 \right) \right)$$ Simplifying further: $$= -\left[ \frac{3}{2} - 12 \right] + \frac{21}{2} = \frac{21}{2} + \frac{21}{2} = 21$$

Given the differential equation $(4 + x^2) dy - 2x(x^2 + 3y + 4) dx$. Rewriting, we have: $$(x^2 + 4) \frac{dy}{dx} = 2x^3 + 6xy + 8x$$ $$(x^2 + 4) \frac{dy}{dx} - 6xy = 2x^3 + 8x$$ This can be written as: $$\frac{dy}{dx} - \frac{6x}{x^2 + 4} y = \frac{2x^3 + 8x}{x^2 + 4}$$ This is a linear differential equation of the form $\frac{dy}{dx} + py = \phi$. The integrating factor is given by: $$e^{\log_e(x^2 + 4)^{-3}} = \frac{1}{(x^2 + 4)^3}$$ Solution: Multiply through by the integrating factor: $$\frac{1}{(x^2 + 4)^3} y = \int \frac{2x^3 + 8x}{(x^2 + 4)^3} (x^2 + 4) \, dx$$ Simplifying, we have: $$\frac{y}{(x^2 + 4)^3} = \int \frac{2x(x^2 + 4)}{(x^2 + 4)^3} \, dx$$ Let $x^2 + 4 = t$, then $2x \, dx = dt$. Thus, $$\frac{y}{(x^2 + 4)^3} = \int \frac{dt}{t^3}$$ Integrating, we get: $$\frac{y}{(x^2 + 4)^3} = \frac{-1}{2(x^2 + 4)^2} + C$$ Since the curve passes through the origin $(0, 0)$, $$0 = \frac{-1}{2 \times 16} + C$$ Solving for $C$, $$C = \frac{1}{32}$$ Thus, $$\frac{y}{(x^2 + 4)^3} = \frac{-1}{2(x^2 + 4)^2} + \frac{1}{32}$$ Simplifying, $$y = \frac{-(x^2 + 4)}{2} + \frac{(x^2 + 4)^3}{32}$$ Evaluating at $x = 2$, $$y(2) = -8 + \frac{8 \times 8 \times 8}{32}$$ $$y(2) = -8 + 16 = 8$$

Given $\sin 10^\circ \left( \frac{1}{2} \cdot 2 \sin 20^\circ \sin 40^\circ \right) \cdot \sin 10^\circ \sin(60^\circ - 10^\circ) \sin(60^\circ + 10^\circ)$. $$\sin 10^\circ \frac{1}{2} \left( \cos 20^\circ - \cos 60^\circ \right) \cdot \frac{1}{4} \sin 30^\circ$$ $$\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \sin 10^\circ \left( \cos 20^\circ - \frac{1}{2} \right)$$ $$= \frac{1}{32} \left( 2 \sin 10^\circ \cos 20^\circ - \sin 10^\circ \right)$$ $$= \frac{1}{32} \left( \sin 30^\circ - \sin 10^\circ - \sin 10^\circ \right)$$ $$= \frac{1}{32} \left( \frac{1}{2} - 2 \sin 10^\circ \right)$$ $$= \frac{1}{64} \left( 1 - 4 \sin 10^\circ \right)$$ $$= \frac{1}{64} - \frac{1}{16} \sin 10^\circ$$ Hence $\alpha = \frac{1}{64}$

Sum of elements in $A \cap B$ $$= (2 + 4 + 6 + \ldots + 200) - (6 + 12 + \ldots + 198)$$ Multiple of 2, Multiple of 2 $\&$ 3 i.e. 6 $$- (10 + 20 + \ldots + 200) + (30 + 60 + \ldots + 180)$$ Multiple of 5 $\&$ 2 i.e. 10, Multiple of 2, 5 $\&$ 3 i.e. 30 $$= 5264$$

Given $$I = \frac{48}{\pi^4} \int_0^\pi x^2 \left( \frac{3\pi}{2} - x \right) \frac{\sin x}{1 + \cos^2 x} \, dx \ldots (1)$$ Apply king property $$I = \frac{48}{\pi^4} \int_0^\pi (\pi - x)^2 \left( \frac{\pi}{2} + x \right) \frac{\sin x}{1 + \cos^2 x} \, dx \ldots (2)$$ (1) + (2) $$I = \frac{12}{\pi^3} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \left[ \pi^2 + (\pi - 2).x(\pi - 2x) \right] dx \ldots (3)$$ Apply king again $$I = \frac{12}{\pi^3} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \left[ \pi^2 + (\pi - 2)(\pi - x)(2x - \pi) \right] dx \ldots (4)$$ (3) + (4) $$I = \frac{6}{\pi^2} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \left[ 2\pi + (\pi - 2)(\pi - 2x) \right] dx \ldots (5)$$ Apply king $$I = \frac{6}{\pi^2} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \left[ 2\pi + (\pi - 2)(2x - \pi) \right] dx \ldots (6)$$ (5) + (6) $$I = \frac{12}{\pi} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx$$ Let $\($ $\cos$ x = t $\Rightarrow$ $\sin$ x dx = -dt $\)$ $$I = \frac{12}{\pi} \int_1^{-1} \frac{-dt}{1 + t^2} = 6$$

Given $$A = \sum_{i=1}^{10} \sum_{j=1}^{10} \min \{i, j\}$$ $$B = \sum_{i=1}^{10} \sum_{j=1}^{10} \max \{i, j\}$$ For A: $$A = \sum_{j=1}^{10} \min(i, 1) + \min(j, 2) + \ldots + \min(i, 10)$$ $$= (1 + 1 + 1 + \ldots + 1)_{19 times} + (2 + 2 + \ldots + 2)_{17 times} + (3 + 3 + \ldots + 3)_{15 times}$$ $$+ \ldots + (1)_{1 time}$$ For B: $$B = \sum_{j=1}^{10} \max(i, 1) + \max(j, 2) + \ldots + \max(i, 10)$$ $$= (10 + 10 + \ldots + 10)_{19 times} + (9 + 9 + \ldots + 9)_{17 times} + \ldots + 1_{1 time}$$ Combining A and B: $$A + B = 20(1 + 2 + 3 + \ldots + 10)$$ $$= 20 \times \frac{10 \times 11}{2} = 10 \times 110 = 1100$$

Given $\($ $\frac{dy}{dx}$ = $\frac{1}{1 + \sin 2x}$ $\)$. Integrating both sides, we have: $$ y(x) = -\frac{1}{1 + \tan x} + C $$ Given $\($ y$\left$( $\frac{\pi}{4}$ $\right$) = $\frac{1}{2}$ = -$\frac{1}{2}$ + C $\)$, we find $\($ C = 1 $\)$. Thus, $$ y(x) = \frac{-1}{1 + \tan x} + 1 $$ Simplifying, $$ y(x) = \frac{-1 + 1 + \tan x}{1 + \tan x} $$ $$ y(x) = \frac{\tan x}{1 + \tan x} $$ Solving with $\($ y = $\sqrt{2}$ $\sin$ x $\)$, $$ \frac{\tan x}{1 + \tan x} = \sqrt{2} \sin x $$ For $\($ $\sin$ x = 0 $\)$, $\($ $\frac{1}{\sqrt{2}}$ = $\sin$ x + $\cos$ x $\)$. Thus, $\($ x = $\pi$ $\)$ and $\($ $\frac{1}{2}$ = $\sin$ $\left$( x + $\frac{\pi}{4}$ $\right$) $\)$. Also, $\($ $\sin$ $\frac{\pi}{6}$ = $\sin$ $\left$( x + $\frac{\pi}{4}$ $\right$) $\)$. Therefore, $$ x + \frac{\pi}{4} = \pi - \frac{\pi}{6}, 2\pi + \frac{\pi}{6} $$ The sum of solutions is: $$ x = \frac{5\pi}{12}, \frac{13\pi}{12} - \frac{\pi}{12} $$ Thus, $$ = \pi + \frac{7\pi}{12} + \frac{23\pi}{12} $$ $$ = \frac{12\pi + 7\pi + 23}{12} = \frac{42\pi}{12} = \frac{k\pi}{12} $$ Therefore, $\($ k = 42 $\)$.

Given $P = \frac{\alpha}{\beta} \log_e \left( \frac{kt}{\beta x} \right)$. $$\frac{kt}{\beta x} = 1 \implies \beta = \frac{kt}{x} = \frac{\mathrm{ML}^2 \mathrm{T}^{-2}}{\mathrm{L}}$$ Therefore, $E = \frac{1}{2} kt$. As $P$ is dimensionless, $$\Rightarrow [\alpha] = [\beta] = [\mathrm{MLT}^{-2}]$$

Given the forces acting on the person in the lift, we have: $$mg - N = ma$$ Solving for $N$, we get: $$N = m(g - a)$$ Therefore, the person experiences weight loss when the acceleration of the lift is downward.

At maximum height, $V = 0$. Therefore, the momentum of the object is zero.

Given $V = \sqrt{2gR \sin \alpha}$. $N - mg \sin \alpha = \frac{mv^2}{R} = 2mg \sin \alpha$ $$\frac{N}{2mg \sin \alpha} = \frac{1}{2} + 1 = \frac{3}{2}$$ Therefore, $A = constant$.

Applying conservation of angular momentum $$MR^2 \omega = (MR^2 + 2mR^2) \omega'$$ $$\omega' = \frac{2M}{M + 2m}$$

The gravitational field strength $g$ is given by the piecewise function: $$g = \begin{cases} \frac{GMr}{R^3}, & r \leq R \\ \frac{GM}{r^2}, & r \geq R \end{cases}$$ The graph shows $g$ as a function of $r$, with a linear increase up to $R$ and a decrease following an inverse square law beyond $R$.

The period is given by the formula $$T = 2\pi \sqrt{\frac{\ell}{g_{eff}}}$$. (a) When $a = 0$, the period is $$T = 2\pi \sqrt{\frac{\ell}{g}}$$. (b) When $a = \frac{g}{6}$, the period is $$T' = 2\pi \sqrt{\frac{\ell}{g + \frac{g}{6}}}$$. Therefore, $$T' = \sqrt{\frac{6}{7}} T$$.

Charge on capacitor $C_2$ $$= \frac{C_2 \times Q_{total}}{C_{total}} = \frac{C_2 [C_1 V]}{C_1 + C_2} = \frac{C_1 C_2 V}{C_1 + C_2}$$

S1: In nonpolar molecules, centre of positive charge coincides with centre of negative charge, hence net dipole moment comes to zero. S2: When nonpolar material is placed in external field, centre of charges does not coincide, hence gives non-zero moment in field.

Given $\phi = 5t^3 + 4t^2 + 2t - 5$. The magnitude of $e$ is given by $\left| e \right| = \left| \frac{d\phi}{dt} \right| = 15t^2 + 8t + 2$. At $t = 2$, $\left| e \right| = 15 \times 2^2 + 8 \times 2 + 2$. Therefore, $e = 78 \, \mathrm{V}$ implies $I = \frac{e}{R} = \frac{78}{5} = 15.60$.

Given the formula for resistance, $$R = \frac{\rho \ell}{A}$$ we have the relative change in resistance as $$\frac{\Delta R}{R} = \frac{\Delta \ell}{\ell} - \frac{\Delta A}{A}$$ Given that $$\ell A = k$$ we have $$\frac{\Delta \ell}{\ell} + \frac{\Delta A}{A} = 0$$ Therefore, $$\frac{\Delta R}{R} = \frac{2 \Delta \ell}{\ell}$$ Substituting the given values, $$\frac{\Delta R}{R} = 2 \times 0.4 = 0.8\%$$

Given $\($ $\frac{R_{\alpha}}{R_{P}}$ = $\frac{M_{\alpha}}{M_{P}}$ $\times$ $\frac{q_{P}}{q_{\alpha}}$ $\)$. $\($ $\frac{R_{\alpha}}{R_{P}}$ = $\frac{4}{1}$ $\times$ $\frac{1}{2}$ = 2 $\)$

Given $\($ $\mathbf{E}$ = 301.6 $\sin$ (kz - $\omega$ t)(-$\hat{a}$_x) + 452.4 $\sin$ (kz - $\omega$ t) $\hat{a}$_y $\)$. $\($ $\mathbf{B}$ = $\frac{301.6}{C}$ $\sin$ (kz - $\omega$ t)(-$\hat{a}$_y) + $\frac{452.4}{C}$ $\sin$ (kz - $\omega$ t)(-$\hat{a}$_x) $\)$. $\($ $\mathbf{H}$ = $\frac{\mathbf{B}}{\mu_0}$ = $\frac{301.6}{\mu C}$ $\sin$ (kz - $\omega$ t)(-$\hat{a}$_y) + $\frac{452.4}{\mu C}$ $\sin$ (kz - $\omega$ t)(-$\hat{a}$_x) $\)$. $\($ $\mathbf{H}$ = -0.8 $\sin$ (kz - $\omega$ t) $\hat{a}$_y - 1.2 $\sin$ (kz - $\omega$ t) $\hat{a}$_x $\)$. For direction $\($ $\mathbf{E}$ $\times$ $\mathbf{B}$ $\)$ is direction of $\($ $\mathbf{C}$ $\)$. For first part $\($ $\hat{E}$ = -$\hat{i}$, $\hat{B}$ = ? $\)$ $\($ $\hat{E}$ $\times$ $\hat{B}$ = $\hat{k}$ $\Rightarrow$ $\hat{B}$ = -$\hat{j}$ $\)$ Similarly for second $\($ $\hat{E}$ = $\hat{j}$, $\hat{B}$ = ? $\)$ $\($ $\hat{E}$ $\times$ $\hat{B}$ = $\hat{k}$ $\Rightarrow$ $\hat{B}$ = -$\hat{i}$ $\)$

For reflection, size of obstacle must be much larger than wavelength, for diffraction size should be order of wavelength. Since the object is of size $\frac{\lambda}{100}$, much smaller than wavelength, so scattering will occur.

Given $\lambda_e = \lambda_{photon}$. $$\frac{h}{mv} = \frac{h}{P_{photon}} \implies P_{photon} = mv$$ $$\frac{E_e}{E_{ph}} = \frac{\frac{1}{2}mv^2}{\frac{hc}{\lambda}} = \frac{1}{2} \frac{mv}{P_{ph} C} \times v = \frac{v}{2C}$$

The reaction is given by: $$^{238}_{92} \mathrm{U} \rightarrow \boxed{8} \; ^{4}_{2} \mathrm{He} + \boxed{6} \; ^{0}_{-1} \mathrm{e} + ^{206}_{82} \mathrm{Pb}$$ 8 $\alpha$ particles and 6 $\beta$ particles are emitted.

Given the formula for resistance, $R = \frac{\Delta V}{\Delta i}$. We have: $$\frac{R_1}{R_2} = \frac{\Delta v_1}{\Delta v_2} \frac{\Delta i_2}{\Delta i_1} = \frac{0.1}{0.2} \times \frac{50}{5} = 5$$

In amplitude modulation, the amplitude of high frequency carrier wave is varied in accordance with message signal.

Both should have same horizontal component of velocity. $$200 = 400 \cos \theta$$ $$\theta = 60^\circ$$

Given $v^2 = u^2 + 2as$. $$100 = 0 + 2(10)s$$ Therefore, $S = 5 \, \mathrm{m}$. Height from ground $= 10 - 5 = 5 \, \mathrm{m}$.

Given $\($ y = $\frac{stress}{strain}$ = 2.0 $\times$ 10^{10} $\)$. Energy density = $\($ $\frac{1}{2}$ stress $\times$ strain $\)$ $\[$ = $\frac{1}{2}$ (strain)^2 $\cdot$ y = $\frac{1}{2}$ (5 $\times$ 10^{-4})^2 $\times$ 20 $\times$ 10^{10} $\]$ $\[$ = 25 $\times$ 10^2 $\times$ 10 = 25 $\frac{kJ}{m^3}$ $\]$ Ans. 25

Given $\Delta \ell \propto g$. $$\frac{\Delta \ell_{earth}}{\Delta \ell_{planet}} = \frac{g_{earth}}{g_{planet}} = \frac{10^{-4}}{6 \times 10^{-5}}$$ Therefore, $g_{planet} = 6 \, \mathrm{m/s^2}$.

Given $\delta_{total} = 360^\circ - 2\theta$. $$= 360^\circ - 2 \times 75^\circ$$ $$\delta_{total} = 210^\circ$$ From the diagram, $\theta_1 = 45^\circ$. In the second diagram, $\delta = 120^\circ + 90^\circ = 210^\circ$.

20 MSD = 1 cm 1 MSD = $\frac{1}{20}$ cm 10 VSD = 9 MSD 1 VSD = $\frac{9}{10}$ MSD = $\frac{9}{10}$ $\times$ $\frac{1}{20}$ cm 1 VSD = $\frac{9}{200}$ cm VC = 1 MSD - 1 VSD = $\frac{1}{20}$ cm - $\frac{9}{200}$ cm = $\frac{1}{200}$ $\times$ 10 mm VC = 5 $\times$ 10^{-2} mm Ans. 5

Given $V = IR_{net}$. $$10 = I \times 10$$ Therefore, $I = 1 \, \mathrm{A}$. Ans. 1

At resonance $I_L = I_C$. Alternatively, $$\frac{1}{Z} = \sqrt{\left(\frac{1}{X_L} - \frac{1}{X_C}\right)^2}$$ At resonance, $X_L = X_C$ and $Z \to \infty$. Therefore, $Z_{total circuit} \to \infty$, i.e., $I = 0$. Ans. 0

From continuity equation $$a v_1 = \frac{a}{2} v_2$$ $$v_2 = 2 v_1$$ From Bernoulli's theorem, $$P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$$ $$P_1 - P_2 = \rho \left[ \frac{v_2^2 - v_1^2}{2} + g(h_2 - h_1) \right]$$ $$4100 = 800 \left[ \frac{4 v_1^2 - v_1^2}{2} \right] + 10 \times (0 - 1)$$ $$\frac{41}{8} + 10 = \frac{3 v_1^2}{2}$$ $$\frac{121}{8} \times \frac{2}{3} = v_1^2$$ $$v_1 = \sqrt{\frac{121}{4 \times 3} \times \frac{3}{3}}$$ $$v_1 = \frac{\sqrt{363}}{6} \, \mathrm{m/s}$$ $$X = 363.$$

Let total volume = 1000 $\mathrm{mL}$ = 1 $\mathrm{L}$. Total mass of solution = 1460 $\mathrm{g}$. Mass of $\mathrm{HCl}$ = $\frac{35}{100}$ $\times$ 1460. Moles of $\mathrm{HCl}$ = $\frac{35 \times 1460}{100 \times 36.5}$. So molarity = $\frac{35 \times 1460}{100 \times 36.5}$ = 14 $\mathrm{M}$.

Mass of liquid $=135-40=95\,\mathrm{g}$ Volume of liquid $=\frac{\text{mass}}{\text{density}}=\frac{95}{0.95}\,\mathrm{mL}=100\,\mathrm{mL}=0.1\,\mathrm{L}$ Mass of ideal gas $=40.5-40\,\mathrm{g}=0.5\,\mathrm{g}$ $PV=nRT$ $0.82\times0.1=\left(\frac{0.5}{M}\right)\times0.082\times250$ $M=125$

Given the formula for the radius: $$r = 0.529 \times \frac{n^2}{z} \, Å$$ Calculate $r_3$: $$r_3 = 0.529 \times \frac{3^2}{1}$$ Calculate $r_4$: $$r_4 = 0.529 \times \frac{4^2}{1}$$ The ratio $\frac{r_4}{r_3}$ is: $$\frac{r_4}{r_3} = \frac{4^2}{3^2} = \frac{16}{9}$$ Thus, $$r_4 = \frac{16r_3}{9}$$

\begin{tabular}{|l|c|c|c|} \hline \textbf{ion/molecule} & \textbf{Number of $e^-$ in BMO} & \textbf{Number of $e^-$ in ABMO} & \textbf{Bond order} \\ \hline O$_2^+$ & 10 & 5 & 2.5 \\ \hline O$_2$ & 10 & 6 & 2 \\ \hline O$_2^-$ & 10 & 7 & 1.5 \\ \hline O$_2^{2-}$ & 10 & 8 & 1 \\ \hline \end{tabular} Bond order $O_2^{2-}<O_2^-<O_2<O_2^+$

A cell with less variation in EMF with temperature is preferred as a reference electrode because it can be used over a wider temperature range without much deviation from its standard value. Therefore, a cell with a smaller value of $\left(\frac{\partial E}{\partial T}\right)_P$ is preferred.

Moving down the group, stability of lower oxidation state increases. $$\mathrm{Al} < \mathrm{Ga} < \mathrm{In} < \mathrm{Tl}$$

Metal oxide with lower $\Delta G^\circ$ is more stable. Statement II is correct.

The reaction involves electrolysis and hydrolysis of $2\mathrm{HSO_4^- (aq.)}$ to form $2\mathrm{HSO_4^-} + 2\mathrm{H^+} + \mathrm{H_2O_2}$. The structure shown is a solid phase with an angle of $90.2^\circ$.

The melting points (M.P) of the elements are given as follows: Beryllium (Be): $1560 \, \mathrm{K}$ Magnesium (Mg): $924 \, \mathrm{K}$ Calcium (Ca): $1124 \, \mathrm{K}$ Strontium (Sr): $1062 \, \mathrm{K}$

\begin{tabular}{|l|c|} \hline \textbf{Hydride} & \textbf{M.P.} \\ \hline H$_2$O & 273 K \\ \hline H$_2$S & 188 K \\ \hline H$_2$Se & 208 K \\ \hline H$_2$Te & 222 K \\ \hline \end{tabular}

White $\mathrm{P_4}$ + alkali $\rightarrow \mathrm{H_3PO_2}$ Red $\mathrm{P_4}$ + alkali $\rightarrow \mathrm{H_4P_2O_6}$

Polar stratospheric clouds provide surface on which hydrolysis of $\mathrm{ClONO_2}$ takes place to form $\mathrm{HOCl}$ (Hypochlorous acid). $$\mathrm{ClONO_2(g) + H_2O(g) \rightarrow HOCl(g) + HNO_3(g)}$$

Both statement I and statement II are correct.

The reaction involves the photochemical decomposition of a peroxide compound. Upon exposure to light $h\nu$, the peroxide bond breaks, leading to the formation of a phenyl radical and a benzoate radical. The benzoate radical further decomposes to form a phenyl radical and carbon dioxide $\mathrm{CO_2}$. The overall reaction results in the formation of two phenyl radicals and the release of $\mathrm{CO_2}$.

Although Acetyl Acetone predominantly gives Acid base reaction with G.R due to Active methylene group but according to given option ans should be based on nucleophilic addition reaction (NAR).

The compound on the left is a tautomer of the compound on the right, which is aromatic in nature.

All these enamines are interconvertible through their resonating structures. So most stable form is 'C' due to steric factor.

Which of the following set are correct regarding polymer. Bona - 5 is copolymer of butadiene + styrene Nylon 6.6 is condensation polymer of adipic Acid and hexanediamine. Nylon 6.6 is fiber Terylene is fiber not thermosetting polymer Buna-N is copolymer not Homopolymer

Histamine (It is used for secretion of pepsin & HCl in stomach)

Ring is formed due to formation of nitrosoferrous sulphate.

Given $\Delta U = -726 \, \mathrm{KJ/mol}$. $\Delta ng = 1 - 3/2 = -\frac{1}{2}$. $\Delta H = \Delta U + \Delta ng RT$. $$= -726 - \frac{1}{2} \times \frac{8.3 \times 300}{1000}$$ $$= -727.245$$

For a 0.5% solution of KCl, we have: $$\Delta T_f = i \cdot m_f$$ $$0.24 = i \cdot \frac{0.5}{74.6} \times \frac{1.80}{0.1}$$ $$i = \frac{0.24 \times 74.6}{0.5 \times 1.80} \times 0.1$$ $$= 1.989$$ Now, using the equation: $$1.989 = 1 + \alpha (n-1)$$ $$1.989 = 1 + \alpha$$ $$\alpha = 0.989$$ The percentage of $\alpha$ is $98.9\%$. Answer is $99\%$. If the mass of $\mathrm{H_2O} = 99.5$: $$m = \frac{0.5}{74.5} \times \frac{1}{0.0995}$$ $$i = \frac{0.24 \times 74.6 \times 0.0995}{0.5 \times 1.80}$$ $$= 1.979$$ Again, using the equation: $$1.979 = 1 + \alpha (n-1)$$ The percentage of $\alpha$ is $97.9\%$.

Moles of $\mathrm{CH_3COOH} = 5$ m mole. Moles of $\mathrm{NaOH} = 2.5$ m mole. $$\mathrm{NaOH} + \mathrm{CH_3COOH} \longrightarrow \mathrm{CH_3COONa} + \mathrm{H_2O}$$ $2.5$ m mole $2.5$ m mole $0 2.5$ m mole $ 2.5$ m mole So buffer is formed. $$\mathrm{pH} = \mathrm{pKa} + \log \left( \frac{2.5/75}{2.5/75} \right) = \mathrm{pKa}$$ $$\mathrm{pH} = 4.76$$ $$= 476 \times 10^{-2}$$

Given $k_A = \frac{\ln 2}{100}$ and $k_B = \frac{\ln 2}{50}$. $A_t = A_0 \times e^{-k_A t}$ $A_t = A_0 \times e^{\left(-\frac{\ln 2}{100} \times t\right)}$ $B_t = B_0 \times e^{\left(-\frac{\ln 2}{50} \times t\right)}$ $A_0 = B_0$ And $A_t = 4B_t$ $e^{-\frac{\ln 2}{100} \times t} = 4 \times e^{-\frac{\ln 2}{50} \times t}$ $e^{\frac{\ln 2}{100} \times t} = 4$ $e^{\frac{\ln 2}{100} \times t} = 4$ $\frac{\ln 2}{100} \times t = \ln 4 = 2 \ln 2$ $t = 200 \, sec$

Volume of $\mathrm{H_2} = \frac{nRT}{p} = \frac{2}{2} \times \frac{0.083 \times 300}{1}$ $$= 24.92$$ $$= 24900 \, \mathrm{mL}$$ So 'g platinum adsorb $= \frac{24900}{2.5} \, \mathrm{mL} \mathrm{H_2}$ $$= 9960$$

Most basic oxide is $\mathrm{V_2O_3}$. $\mathrm{V^{+3}} \rightarrow [\mathrm{Ar}] \, 3d^2$ $$\mu = \sqrt{2(2+2)} = 2.84 \, \mathrm{BM} \approx 3$$

$\mathrm{CoCl_3 \cdot 4NH_3 \rightarrow [Co(NH_3)_4Cl_2]Cl}$ $\mathrm{NiCl_2 \cdot 6H_2O \rightarrow [Ni(H_2O)_6]Cl_2}$ $\mathrm{PtCl_4 \cdot 2HCl \rightarrow H_2[PtCl_6]}$ $\mathrm{[Ni(H_2O)_6]Cl_2 \xrightarrow{2AgNO_3} 2AgCl \downarrow + [Ni(H_2O)_6](NO_3)_2}$ $\mu = \sqrt{2(2+2)}\,\mathrm{B.M.}$ $= 2.84 \approx 3$

The reaction is given by: $$\mathrm{C_xH_yO_z} + \left( x + \frac{y}{4} - \frac{z}{2} \right) \mathrm{O_2} \rightarrow x\mathrm{CO_2} + \frac{y}{2} \mathrm{H_2O}$$ Given 0.3 g of $\mathrm{C_xH_yO_z}$ and 0.2 g of $\mathrm{O_2}$ produce 1 g of products. The moles of $\mathrm{CO_2}$ and $\mathrm{H_2O}$ are: $$n_{\mathrm{CO_2}} = \frac{x}{44} = \frac{0.2}{44}$$ $$n_{\mathrm{H_2O}} = \frac{y}{2} = \frac{1}{18}$$ From the equations: $$\frac{2x}{y} = \frac{36}{44} = \frac{9}{11}$$ Solving for $x$: $$x = \frac{9y}{22}$$ The ratio of moles is: $$\frac{n_{\mathrm{C_xH_yO_z}}}{n_{\mathrm{CO_2}}} = \frac{1}{x}$$ Calculating: $$\frac{0.3}{12x + y + 16z} \times \frac{44}{0.2} = \frac{1}{x}$$ Simplifying: $$66x = 12x + y + 16z$$ $$54x = y + 16z$$ Substituting $x$: $$54 \times \frac{9y}{22} = y = 16z$$ Solving for $y$: $$\frac{464y}{22} = 16z$$ Solving for $z$: $$z = \frac{29y}{22}$$ The empirical formula is: $$\mathrm{C_xH_yO_z} = \mathrm{C_xH_yO_z}$$ The molecular formula is: $$\mathrm{C_9H_{22}O_{29}}$$ Calculating the percentage of carbon: $$\% of C = \frac{12 \times 9}{(12 \times 9 + 22 \times 1 + 16 \times 29)} \times 100 = \frac{108}{594} \times 100$$ The percentage of carbon is 18.18%.

Uracil is the base which only present is RNA. Structure of nucleotides number of 0-9.