MHT CET 2022 Mathematics Question Paper with Answer and Solution

(C) We have the integral $I = \int \frac{\sin 4x}{\sin x} \, dx$.
Using the identity $\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x$,we get:
$I = \int \frac{4 \sin x \cos x \cos 2x}{\sin x} \, dx = 4 \int \cos x \cos 2x \, dx$.
Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $2 \cos 2x \cos x = \cos(2x+x) + \cos(2x-x) = \cos 3x + \cos x$.
Thus,$I = 2 \int (\cos 3x + \cos x) \, dx$.
Integrating term by term,we get:
$I = 2 \left( \frac{\sin 3x}{3} + \sin x \right) + C = \frac{2}{3} \sin 3x + 2 \sin x + C$.

(B) Let $I = \int \frac{2 x^3-1}{x^4+x} \,d x$.
We can rewrite the denominator as $x(x^3+1)$.
So,$I = \int \frac{2 x^3-1}{x(x^3+1)} \,d x$.
Using partial fractions or observation,we can write $\frac{2x^3-1}{x(x^3+1)} = \frac{A}{x} + \frac{B x^2}{x^3+1}$.
By inspection,$\frac{2x^3-1}{x(x^3+1)} = \frac{-1}{x} + \frac{3x^2}{x^3+1}$.
Integrating both terms:
$I = \int \left( \frac{3x^2}{x^3+1} - \frac{1}{x} \right) \,d x$.
$I = \log|x^3+1| - \log|x| + C$.
$I = \log \left| \frac{x^3+1}{x} \right| + C$.

(B) Let $I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x)^2}$.
Factor the denominator: $\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x = \sin ^2 x(\sin ^3 x+\cos ^3 x) + \cos ^2 x(\sin ^3 x+\cos ^3 x) = (\sin ^2 x+\cos ^2 x)(\sin ^3 x+\cos ^3 x) = \sin ^3 x+\cos ^3 x$.
Thus,$I = \int \frac{\sin ^2 x \cos ^2 x \,dx}{(\sin ^3 x+\cos ^3 x)^2}$.
Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{\tan ^2 x \sec ^2 x \,dx}{(\tan ^3 x+1)^2}$.
Let $u = \tan ^3 x+1$,then $du = 3 \tan ^2 x \sec ^2 x \,dx$,so $\tan ^2 x \sec ^2 x \,dx = \frac{du}{3}$.
Substituting these into the integral: $I = \int \frac{1}{u^2} \cdot \frac{du}{3} = \frac{1}{3} \int u^{-2} \,du = \frac{1}{3} \left(\frac{u^{-1}}{-1}\right) + C = -\frac{1}{3u} + C$.
Substituting back $u = \tan ^3 x+1$: $I = \frac{-1}{3(1+\tan ^3 x)}+C$.

(D) Let $I = \int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} d x$.
Factor the denominator: $\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x = \sin ^2 x(\sin ^3 x + \cos ^3 x) + \cos ^2 x(\sin ^3 x + \cos ^3 x) = (\sin ^2 x + \cos ^2 x)(\sin ^3 x + \cos ^3 x) = (\sin ^3 x + \cos ^3 x)$.
Thus,$I = \int \frac{\sin ^2 x \cos ^2 x}{(\sin ^3 x + \cos ^3 x)^2} d x$.
Divide the numerator and denominator by $\cos ^6 x$: $I = \int \frac{\tan ^2 x \cdot \sec ^2 x}{(\tan ^3 x + 1)^2} d x$.
Let $t = \tan ^3 x + 1$,then $dt = 3 \tan ^2 x \sec ^2 x d x$,so $\tan ^2 x \sec ^2 x d x = \frac{dt}{3}$.
Substituting these into the integral: $I = \int \frac{1}{t^2} \cdot \frac{dt}{3} = \frac{1}{3} \int t^{-2} dt = \frac{1}{3} \left( \frac{t^{-1}}{-1} \right) + C = -\frac{1}{3t} + C$.
Substituting back $t = \tan ^3 x + 1$,we get $I = \frac{-1}{3(1 + \tan ^3 x)} + C$.

(B) Let $I = \int \frac{4 x^2 \cot ^{-1}\left(x^3\right)}{1+x^6} \,d x$.
Substitute $t = \cot ^{-1}\left(x^3\right)$.
Then,the derivative is $\frac{dt}{dx} = -\frac{1}{1+(x^3)^2} \cdot 3x^2 = -\frac{3x^2}{1+x^6}$.
This implies $\frac{x^2}{1+x^6} dx = -\frac{1}{3} dt$.
Substituting these into the integral:
$I = 4 \int t \left(-\frac{1}{3} dt\right) = -\frac{4}{3} \int t \,dt$.
Integrating with respect to $t$:
$I = -\frac{4}{3} \cdot \frac{t^2}{2} + C = -\frac{2}{3} t^2 + C$.
Substituting back $t = \cot ^{-1}\left(x^3\right)$:
$I = -\frac{2}{3} \left(\cot ^{-1} x^3\right)^2 + C$.

(C) Let $I = \int \frac{x+1}{\sqrt{2x-1}} \, dx$.
Substitute $2x-1 = t$,which implies $x = \frac{t+1}{2}$ and $dx = \frac{1}{2} dt$.
$I = \int \frac{\frac{t+1}{2} + 1}{\sqrt{t}} \cdot \frac{1}{2} \, dt$
$I = \frac{1}{2} \int \frac{\frac{t+3}{2}}{\sqrt{t}} \, dt = \frac{1}{4} \int \left( \frac{t}{\sqrt{t}} + \frac{3}{\sqrt{t}} \right) \, dt$
$I = \frac{1}{4} \int (t^{1/2} + 3t^{-1/2}) \, dt$
$I = \frac{1}{4} \left( \frac{t^{3/2}}{3/2} + 3 \cdot \frac{t^{1/2}}{1/2} \right) + C$
$I = \frac{1}{4} \left( \frac{2}{3} t^{3/2} + 6 t^{1/2} \right) + C$
$I = \left( \frac{1}{6} t^{3/2} + \frac{3}{2} t^{1/2} \right) + C$
$I = \frac{1}{6} t^{1/2} (t + 9) + C$
Substituting $t = 2x-1$ back:
$I = \frac{1}{6} \sqrt{2x-1} (2x-1+9) + C$
$I = \frac{1}{6} \sqrt{2x-1} (2x+8) + C$
$I = \frac{1}{6} \cdot 2(x+4) \sqrt{2x-1} + C = \frac{1}{3}(x+4) \sqrt{2x-1} + C$
Comparing with $f(x) \sqrt{2x-1} + C$,we get $f(x) = \frac{1}{3}(x+4)$.

(B) We have the integral $I = \int \frac{d x}{3-2 \cos 2 x}$.
Using the identity $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$,we get:
$I = \int \frac{d x}{3-2 \left(\frac{1-\tan^2 x}{1+\tan^2 x}\right)} = \int \frac{1+\tan^2 x}{3(1+\tan^2 x) - 2(1-\tan^2 x)} \,d x$.
Since $1+\tan^2 x = \sec^2 x$,the integral becomes:
$I = \int \frac{\sec^2 x}{3+3\tan^2 x - 2 + 2\tan^2 x} \,d x = \int \frac{\sec^2 x}{5\tan^2 x + 1} \,d x$.
Let $t = \tan x$,then $dt = \sec^2 x \,d x$.
Substituting these into the integral:
$I = \int \frac{dt}{5t^2 + 1} = \int \frac{dt}{(\sqrt{5}t)^2 + 1^2}$.
Using the standard formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{1}{\sqrt{5}} \tan^{-1}(\sqrt{5}t) + C$.
Substituting $t = \tan x$ back,we get:
$I = \frac{1}{\sqrt{5}} \tan^{-1}(\sqrt{5} \tan x) + C$.

(B) To solve the integral $I = \int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$, we factor out $x^5$ from the denominator term inside the cube:
$I = \int \frac{2 x^{12}+5 x^9}{\left[x^5\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)\right]^3} \,d x$
$I = \int \frac{2 x^{12}+5 x^9}{x^{15}\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \,d x$
$I = \int \frac{\frac{2}{x^3}+\frac{5}{x^6}}{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^3} \,d x$
Let $t = 1+\frac{1}{x^2}+\frac{1}{x^5}$. Then $dt = \left(-\frac{2}{x^3}-\frac{5}{x^6}\right) dx$, which implies $-dt = \left(\frac{2}{x^3}+\frac{5}{x^6}\right) dx$.
Substituting these into the integral:
$I = \int \frac{-dt}{t^3} = -\int t^{-3} dt = -\frac{t^{-2}}{-2} + C = \frac{1}{2t^2} + C$
Substituting $t$ back:
$I = \frac{1}{2\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)^2} + C = \frac{1}{2\left(\frac{x^5+x^3+1}{x^5}\right)^2} + C = \frac{x^{10}}{2\left(x^5+x^3+1\right)^2} + C$

(A) To evaluate the integral $I = \int \cos \sqrt{x} \, dx$,we use the substitution method.
Let $\sqrt{x} = t$. Then $x = t^2$,which implies $dx = 2t \, dt$.
Substituting these into the integral,we get:
$I = \int \cos(t) \cdot (2t \, dt) = 2 \int t \cos t \, dt$.
Now,we apply integration by parts,where $\int u \, dv = uv - \int v \, du$.
Let $u = t$ and $dv = \cos t \, dt$. Then $du = dt$ and $v = \sin t$.
$I = 2 [t \sin t - \int \sin t \, dt]$.
$I = 2 [t \sin t - (-\cos t)] + C$.
$I = 2 [t \sin t + \cos t] + C$.
Substituting $t = \sqrt{x}$ back into the expression:
$I = 2[\sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}] + C$.

(B) Let $I = \int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx$.
Multiply the numerator and denominator by $e^{\frac{x}{2}}$:
$I = \int \frac{e^{\frac{x}{2}} \cdot e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x} \cdot e^{\frac{x}{2}}} \, dx = \int \frac{e^x}{\sqrt{e^{-x} \cdot e^{\frac{x}{2}} - e^x \cdot e^{\frac{x}{2}}}} \, dx$.
Wait,let us simplify the denominator differently:
$I = \int \frac{e^{\frac{x}{2}}}{\sqrt{\frac{1}{e^x} - e^x}} \, dx = \int \frac{e^{\frac{x}{2}}}{\sqrt{\frac{1 - e^{2x}}{e^x}}} \, dx = \int \frac{e^{\frac{x}{2}} \cdot \sqrt{e^x}}{\sqrt{1 - e^{2x}}} \, dx$.
Since $\sqrt{e^x} = e^{\frac{x}{2}}$,we have:
$I = \int \frac{e^{\frac{x}{2}} \cdot e^{\frac{x}{2}}}{\sqrt{1 - (e^x)^2}} \, dx = \int \frac{e^x}{\sqrt{1 - (e^x)^2}} \, dx$.
Let $t = e^x$,then $dt = e^x \, dx$.
$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + C = \sin^{-1}(e^x) + C$.
Comparing this with $\sin^{-1}(f(x)) + C$,we get $f(x) = e^x$.
Therefore,$f(2) = e^2$.

(A) Let $t = \log \left(x+\sqrt{1+x^2}\right)$.
Then,differentiating both sides with respect to $x$,we get:
$\frac{dt}{dx} = \frac{1}{x+\sqrt{1+x^2}} \cdot \left(1 + \frac{1}{2\sqrt{1+x^2}} \cdot 2x\right) = \frac{1}{x+\sqrt{1+x^2}} \cdot \left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\right) = \frac{1}{\sqrt{1+x^2}}$.
Thus,$dt = \frac{dx}{\sqrt{1+x^2}}$.
Substituting these into the integral:
$\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \,dx = \int t \,dt = \frac{t^2}{2} + C$.
Comparing this with the given expression $\frac{1}{2}(g(x))^2 + C$,we have:
$\frac{1}{2}(g(x))^2 = \frac{1}{2} \left(\log \left(x+\sqrt{1+x^2}\right)\right)^2$.
Therefore,$g(x) = \log \left(x+\sqrt{1+x^2}\right)$.

(C) Let $I = \int \frac{\sqrt{1-x^2}}{x^4} \,dx$.
We can rewrite the integral as $I = \int \frac{x \sqrt{\frac{1}{x^2}-1}}{x^4} \,dx = \int \frac{\sqrt{\frac{1}{x^2}-1}}{x^3} \,dx$.
Let $\frac{1}{x^2}-1 = t$. Then $-\frac{2}{x^3} \,dx = dt$,which implies $\frac{1}{x^3} \,dx = -\frac{dt}{2}$.
Substituting these into the integral,we get $I = -\frac{1}{2} \int \sqrt{t} \,dt = -\frac{1}{2} \cdot \frac{t^{3/2}}{3/2} + c = -\frac{1}{3} t^{3/2} + c$.
Substituting $t = \frac{1-x^2}{x^2}$ back,we get $I = -\frac{1}{3} \left(\frac{1-x^2}{x^2}\right)^{3/2} + c = -\frac{1}{3} \frac{(\sqrt{1-x^2})^3}{(x^2)^{3/2}} + c = -\frac{1}{3x^3} (\sqrt{1-x^2})^3 + c$.
Comparing this with $A(x)(\sqrt{1-x^2})^m + c$,we identify $A(x) = -\frac{1}{3x^3}$ and $m = 3$.
Therefore,$(A(x))^m = \left(-\frac{1}{3x^3}\right)^3 = -\frac{1}{27x^9}$.

(A) To evaluate the integral $\int \frac{3x-2}{(x+1)(x-2)^2} dx$,we use partial fractions.
Let $\frac{3x-2}{(x+1)(x-2)^2} = \frac{A}{x+1} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$.
Multiplying by $(x+1)(x-2)^2$,we get $3x-2 = A(x-2)^2 + B(x+1)(x-2) + C(x+1)$.
Setting $x = -1$: $3(-1)-2 = A(-1-2)^2 \implies -5 = 9A \implies A = -\frac{5}{9}$.
Setting $x = 2$: $3(2)-2 = C(2+1) \implies 4 = 3C \implies C = \frac{4}{3}$.
Comparing coefficients of $x^2$: $0 = A + B \implies B = -A = \frac{5}{9}$.
Thus,$\int \left( \frac{-5/9}{x+1} + \frac{5/9}{x-2} + \frac{4/3}{(x-2)^2} \right) dx = -\frac{5}{9} \log |x+1| + \frac{5}{9} \log |x-2| - \frac{4}{3(x-2)} + C$.

(D) Given the integral $I = \int \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) dx$.
Using the identities $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$,we have:
$I = \int \tan^{-1} \sqrt{\frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}} dx$
$I = \int \tan^{-1} \left( \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} \right) dx$
Dividing numerator and denominator by $\cos \frac{x}{2}$,we get:
$I = \int \tan^{-1} \left( \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} \right) dx$
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta}$:
$I = \int \tan^{-1} \left( \tan \left( \frac{\pi}{4} - \frac{x}{2} \right) \right) dx$
$I = \int \left( \frac{\pi}{4} - \frac{x}{2} \right) dx$
Integrating term by term:
$I = \frac{\pi}{4} x - \frac{x^2}{4} + C$.

(C) We need to evaluate the integral $I = \int \left( \frac{1}{\log x} - \frac{1}{(\log x)^2} \right) dx$.
Applying the linearity property of integrals:
$I = \int \frac{1}{\log x} dx - \int \frac{1}{(\log x)^2} dx$.
Using integration by parts on the first term $\int \frac{1}{\log x} dx$,let $u = \frac{1}{\log x}$ and $dv = dx$.
Then $du = -\frac{1}{(\log x)^2} \cdot \frac{1}{x} dx$ and $v = x$.
Using the formula $\int u dv = uv - \int v du$:
$\int \frac{1}{\log x} dx = \frac{x}{\log x} - \int x \left( -\frac{1}{(\log x)^2} \cdot \frac{1}{x} \right) dx$
$= \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx$.
Substituting this back into the original expression for $I$:
$I = \left( \frac{x}{\log x} + \int \frac{1}{(\log x)^2} dx \right) - \int \frac{1}{(\log x)^2} dx$
$I = \frac{x}{\log x} + C$.

(A) To evaluate $\int \sin \sqrt{x} \,d x$,let $\sqrt{x} = t$.
Then $x = t^2$,which implies $d x = 2t \,d t$.
Substituting these into the integral,we get:
$\int \sin \sqrt{x} \,d x = \int \sin t (2t \,d t) = 2 \int t \sin t \,d t$.
Using integration by parts,where $\int u \,dv = uv - \int v \,du$,let $u = t$ and $dv = \sin t \,d t$.
Then $du = d t$ and $v = -\cos t$.
$2 \int t \sin t \,d t = 2 \left[ t(-\cos t) - \int (-\cos t) \,d t \right]$
$= 2 [-t \cos t + \int \cos t \,d t]$
$= 2 [-t \cos t + \sin t] + C$.
Substituting $t = \sqrt{x}$ back,we get:
$= 2(-\sqrt{x} \cos \sqrt{x} + \sin \sqrt{x}) + C$.

(C) Let $I = \int \frac{\sin 2x}{4 \sin^2 x + 9 \cos^2 x} \, dx$.
Let $t = 4 \sin^2 x + 9 \cos^2 x$.
Differentiating both sides with respect to $x$:
$dt = (4 \cdot 2 \sin x \cos x - 9 \cdot 2 \cos x \sin x) \, dx$.
$dt = (4 \sin 2x - 9 \sin 2x) \, dx$.
$dt = -5 \sin 2x \, dx$.
Therefore,$\sin 2x \, dx = -\frac{1}{5} dt$.
Substituting these into the integral:
$I = \int \frac{-\frac{1}{5} dt}{t} = -\frac{1}{5} \int \frac{1}{t} \, dt$.
$I = -\frac{1}{5} \log |t| + C$.
Substituting back $t = 4 \sin^2 x + 9 \cos^2 x$:
$I = -\frac{1}{5} \log |4 \sin^2 x + 9 \cos^2 x| + C$.

(A) Given the equation: $\tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}$
Applying the formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A+B}{1-AB} \right)$,we get:
$\tan^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} \right) = \frac{\pi}{4}$
$\frac{5x}{1 - 6x^2} = \tan \left( \frac{\pi}{4} \right) = 1$
$5x = 1 - 6x^2$
$6x^2 + 5x - 1 = 0$
Factoring the quadratic equation: $(6x - 1)(x + 1) = 0$
This gives $x = \frac{1}{6}$ or $x = -1$.
Since the condition is $x \geq 0$,we reject $x = -1$.
Thus,the only solution is $x = \frac{1}{6}$.
Therefore,the set contains only one element,which is a singleton set.

(C) Let $\theta = \sin ^{-1} 0.8$. Then $\sin \theta = 0.8$.
We need to find the value of $\sin(2\theta)$.
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we first find $\cos \theta$.
Since $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (0.8)^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$.
Now,substitute the values into the identity:
$\sin(2\theta) = 2 \times 0.8 \times 0.6 = 0.96$.

(C) We know that the range of the principal value branch of $\sin ^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{2 \pi}{3}$ does not lie in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we use the property $\sin(\pi - \theta) = \sin(\theta)$.
$\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right) = \sin ^{-1}\left(\sin \left(\pi - \frac{\pi}{3}\right)\right)$
$= \sin ^{-1}\left(\sin \left(\frac{\pi}{3}\right)\right)$
$= \frac{\pi}{3}$

(D) Given the equation: $\tan ^{-1} \sqrt{x^2+x}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$
Since $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$,we can write:
$\tan ^{-1} \sqrt{x^2+x} = \frac{\pi}{2} - \sin ^{-1} \sqrt{x^2+x+1} = \cos ^{-1} \sqrt{x^2+x+1}$
Let $\tan ^{-1} \sqrt{x^2+x} = \theta$. Then $\tan \theta = \sqrt{x^2+x}$.
This implies $\cos \theta = \frac{1}{\sqrt{1 + (x^2+x)}}$.
Thus,$\cos ^{-1} \left( \frac{1}{\sqrt{x^2+x+1}} \right) = \cos ^{-1} \sqrt{x^2+x+1}$.
Comparing the arguments: $\frac{1}{\sqrt{x^2+x+1}} = \sqrt{x^2+x+1}$
$1 = x^2+x+1$
$x^2+x = 0$
$x(x+1) = 0$
So,$x = 0$ or $x = -1$.
Checking the domain of $\sin ^{-1} \sqrt{x^2+x+1}$,we require $0 \le x^2+x+1 \le 1$,which implies $x^2+x \le 0$.
For $x=0$,$x^2+x=0$ (valid). For $x=-1$,$x^2+x=0$ (valid).
However,$\tan ^{-1} \sqrt{x^2+x}$ requires $x^2+x \ge 0$.
Thus,$x^2+x=0$ is the only solution,giving $x=0$ or $x=-1$.

(A) We know that the range of the principal value branch of $\cot ^{-1} x$ is $(0, \pi)$.
Since the argument $\frac{-1}{\sqrt{3}}$ is negative,we use the property $\cot ^{-1}(-x) = \pi - \cot ^{-1}(x)$.
Therefore,$\cot ^{-1}\left(\frac{-1}{\sqrt{3}}\right) = \pi - \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)$.
We know that $\cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$,so $\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$.
Substituting this value,we get $\pi - \frac{\pi}{3} = \frac{2 \pi}{3}$.

(A) Let $x = \tan \theta$ and $y = \tan \phi$.
Substituting these into the expression:
$\frac{1}{2} \sin ^{-1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) = \frac{1}{2} \sin ^{-1} (\sin 2 \theta) = \frac{1}{2} (2 \theta) = \theta$.
Similarly,$\frac{1}{2} \cos ^{-1} \left( \frac{1 - \tan^2 \phi}{1 + \tan^2 \phi} \right) = \frac{1}{2} \cos ^{-1} (\cos 2 \phi) = \frac{1}{2} (2 \phi) = \phi$.
Now,the expression becomes $\tan (\theta + \phi)$.
Using the trigonometric identity $\tan (\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi}$,we get:
$\frac{x + y}{1 - xy}$.

(D) Given the equation: $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
Since $\sin \frac{\pi}{2} = 1$,we have:
$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$
Using the identity $\sin ^{-1} \theta + \cos ^{-1} \theta = \frac{\pi}{2}$,we can rewrite the equation as:
$\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} \frac{1}{5}$
Since $\frac{\pi}{2} - \sin ^{-1} \theta = \cos ^{-1} \theta$,we get:
$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$
Therefore,$x = \frac{1}{5}$.

(D) Let $x = \tan \theta$. Since $x \in (0, 1)$,we have $\theta \in (0, \frac{\pi}{4})$.
First,consider $A = 2 \tan^{-1}\left(\frac{1+x}{1-x}\right)$.
Using the formula $\tan^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) = \tan^{-1}(\tan(\frac{\pi}{4} + \theta)) = \frac{\pi}{4} + \theta$,we get $A = 2(\frac{\pi}{4} + \theta) = \frac{\pi}{2} + 2\theta$.
Next,consider $B = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.
Using the formula $\cos^{-1}\left(\frac{1-\tan^2 \theta}{1+\tan^2 \theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta$.
Finally,$A - B = (\frac{\pi}{2} + 2\theta) - 2\theta = \frac{\pi}{2}$.

(D) We know that $\sin ^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$.
We also know that $\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$ since $\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$.
Substituting these values into the expression:
$2 \sin ^{-1}\left(\frac{1}{2}\right) + \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = 2 \times \left(\frac{\pi}{6}\right) + \frac{\pi}{3}$.
$= \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$.

(A) We know the principal values of the inverse trigonometric functions:
$\operatorname{cosec}^{-1}(\sqrt{2}) = \frac{\pi}{4}$
$\cos^{-1}\left(\frac{-1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
$\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$
Substituting these values into the expression:
$\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6}$
Finding a common denominator $(12)$:
$\frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$

(C) We are given the equation: $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c=\pi$.
Using the formula for the sum of three inverse tangent functions: $\tan ^{-1} x + \tan ^{-1} y + \tan ^{-1} z = \tan ^{-1} \left( \frac{x+y+z-xyz}{1-xy-yz-zx} \right)$.
Applying this to the given equation: $\tan ^{-1} \left( \frac{a+b+c-abc}{1-ab-bc-ca} \right) = \pi$.
Taking the tangent of both sides: $\frac{a+b+c-abc}{1-ab-bc-ca} = \tan \pi$.
Since $\tan \pi = 0$,we have: $\frac{a+b+c-abc}{1-ab-bc-ca} = 0$.
This implies that the numerator must be zero: $a+b+c-abc = 0$.
Therefore,$a+b+c = abc$.

(D) Given equation: $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$
Let $x = \tan \theta$,then $\theta = \tan ^{-1} x$.
Substituting $x = \tan \theta$ in the equation:
$\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\frac{1}{2} \theta$
Using the formula $\tan(\frac{\pi}{4} - \theta) = \frac{1-\tan \theta}{1+\tan \theta}$:
$\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\theta\right)\right)=\frac{1}{2} \theta$
$\frac{\pi}{4}-\theta=\frac{1}{2} \theta$
$\frac{\pi}{4}=\frac{3 \theta}{2}$
$\theta=\frac{\pi}{6}$
Since $x = \tan \theta$,we have $x = \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$.

(B) The principal value range of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given that $\sin ^{-1} x + \sin ^{-1} y + \sin ^{-1} z = \frac{3 \pi}{2}$.
Since the maximum value of each $\sin ^{-1}$ term is $\frac{\pi}{2}$,the sum can only be $\frac{3 \pi}{2}$ if each term is individually equal to its maximum value.
Therefore,$\sin ^{-1} x = \frac{\pi}{2}$,$\sin ^{-1} y = \frac{\pi}{2}$,and $\sin ^{-1} z = \frac{\pi}{2}$.
This implies $x = \sin(\frac{\pi}{2}) = 1$,$y = \sin(\frac{\pi}{2}) = 1$,and $z = \sin(\frac{\pi}{2}) = 1$.
Substituting these values into the expression $x^{100} + y^{100} + z^{100}$:
$1^{100} + 1^{100} + 1^{100} = 1 + 1 + 1 = 3$.

(A) Given the equation: $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$
Using the formula $\tan ^{-1} A+\tan ^{-1} B=\tan ^{-1}\left(\frac{A+B}{1-AB}\right)$,we get:
$\tan ^{-1}\left(\frac{2 x+3 x}{1-(2 x)(3 x)}\right)=\frac{\pi}{4}$
$\tan ^{-1}\left(\frac{5 x}{1-6 x^2}\right)=\frac{\pi}{4}$
Taking $\tan$ on both sides:
$\frac{5 x}{1-6 x^2}=\tan\left(\frac{\pi}{4}\right)=1$
$5 x=1-6 x^2$
$6 x^2+5 x-1=0$
Factoring the quadratic equation:
$(6 x-1)(x+1)=0$
So,$x=\frac{1}{6}$ or $x=-1$.
Checking the values: If $x=-1$,$\tan ^{-1}(-2)+\tan ^{-1}(-3)$ is negative,which cannot be $\frac{\pi}{4}$.
Thus,$x=\frac{1}{6}$ is the only valid solution.

(B) Given $\cos ^{-1} x - \cos ^{-1} \frac{y}{2} = \alpha$.
Using the formula $\cos ^{-1} A - \cos ^{-1} B = \cos ^{-1} (AB + \sqrt{1-A^2} \sqrt{1-B^2})$,we get:
$\cos ^{-1} \left( x \cdot \frac{y}{2} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{4}} \right) = \alpha$.
Taking cosine on both sides:
$\frac{xy}{2} + \frac{\sqrt{(1-x^2)(4-y^2)}}{2} = \cos \alpha$.
$\sqrt{(1-x^2)(4-y^2)} = 2 \cos \alpha - xy$.
Squaring both sides:
$(1-x^2)(4-y^2) = (2 \cos \alpha - xy)^2$.
$4 - y^2 - 4x^2 + x^2y^2 = 4 \cos ^2 \alpha - 4xy \cos \alpha + x^2y^2$.
Subtracting $x^2y^2$ from both sides:
$4 - y^2 - 4x^2 = 4 \cos ^2 \alpha - 4xy \cos \alpha$.
Rearranging the terms:
$4x^2 - 4xy \cos \alpha + y^2 = 4 - 4 \cos ^2 \alpha$.
Since $1 - \cos ^2 \alpha = \sin ^2 \alpha$,we have:
$4x^2 - 4xy \cos \alpha + y^2 = 4 \sin ^2 \alpha$.

(D) Let $\theta = \cot^{-1} x$,then $\cot \theta = x$. Since $\cot \theta = \frac{x}{1}$,we have $\sin \theta = \frac{1}{\sqrt{1+x^2}}$.
Thus,$\sin(\cot^{-1} x) = \frac{1}{\sqrt{1+x^2}}$.
Now,let $\phi = \tan^{-1} \left( \frac{1}{\sqrt{1+x^2}} \right)$,then $\tan \phi = \frac{1}{\sqrt{1+x^2}}$.
We need to find $\cos \phi$. Using the identity $\sec^2 \phi = 1 + \tan^2 \phi$,we get $\sec^2 \phi = 1 + \frac{1}{1+x^2} = \frac{1+x^2+1}{1+x^2} = \frac{x^2+2}{x^2+1}$.
Therefore,$\cos^2 \phi = \frac{x^2+1}{x^2+2}$,which implies $\cos \phi = \sqrt{\frac{x^2+1}{x^2+2}}$.

(C) Given expression: $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$
We can rewrite this as: $\cos \left(\cos ^{-1} x + (\cos ^{-1} x+\sin ^{-1} x)\right)$
Using the property $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$,the expression becomes:
$\cos \left(\cos ^{-1} x+\frac{\pi}{2}\right)$
Using the trigonometric identity $\cos \left(\theta+\frac{\pi}{2}\right) = -\sin \theta$:
$= -\sin \left(\cos ^{-1} x\right)$
Since $\cos ^{-1} x = \sin ^{-1} \sqrt{1-x^2}$,we have:
$= -\sin \left(\sin ^{-1} \sqrt{1-x^2}\right) = -\sqrt{1-x^2}$
Substituting $x=\frac{1}{5}$:
$= -\sqrt{1-\left(\frac{1}{5}\right)^2} = -\sqrt{1-\frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{2 \sqrt{6}}{5}$

(A) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
First,group the terms:
$\left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} \right) + \left( \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8} \right)$
Apply the formula to the first pair:
$\tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \times \frac{1}{5}} \right) = \tan ^{-1} \left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$
Apply the formula to the second pair:
$\tan ^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right) = \tan ^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$
Now add the results:
$\tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} \times \frac{3}{11}} \right)$
$= \tan ^{-1} \left( \frac{\frac{44+21}{77}}{\frac{77-12}{77}} \right) = \tan ^{-1} \left( \frac{65}{65} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$

(D) We use the formula $2 \tan ^{-1}(x) = \tan ^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x| < 1$.
First,calculate $2 \tan ^{-1}\left(\frac{1}{3}\right) = \tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-(\frac{1}{3})^2}\right) = \tan ^{-1}\left(\frac{2/3}{8/9}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$.
Now,convert $\cos ^{-1}\left(\frac{3}{5}\right)$ to $\tan ^{-1}$. Let $\theta = \cos ^{-1}\left(\frac{3}{5}\right)$,then $\cos \theta = \frac{3}{5}$. Since $\cos^2 \theta + \sin^2 \theta = 1$,$\sin \theta = \sqrt{1 - (3/5)^2} = 4/5$. Thus,$\tan \theta = \frac{4/5}{3/5} = \frac{4}{3}$,so $\cos ^{-1}\left(\frac{3}{5}\right) = \tan ^{-1}\left(\frac{4}{3}\right)$.
The expression becomes $\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{4}{3}\right)$.
Using $\tan ^{-1}(x) + \tan ^{-1}(1/x) = \frac{\pi}{2}$ for $x > 0$,we get $\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{1}{3/4}\right) = \frac{\pi}{2}$.

(B) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
First,group the terms:
$\left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} \right) + \left( \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8} \right)$
Applying the formula to the first pair:
$\tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \times \frac{1}{5}} \right) = \tan ^{-1} \left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$
Applying the formula to the second pair:
$\tan ^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right) = \tan ^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$
Now,add the results:
$\tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} \times \frac{3}{11}} \right)$
$= \tan ^{-1} \left( \frac{\frac{44+21}{77}}{\frac{77-12}{77}} \right) = \tan ^{-1} \left( \frac{65}{65} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$

(D) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \pi + \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$ when $xy > 1$.
Here,$x = 2$ and $y = 3$. Since $xy = 2 \times 3 = 6 > 1$,we apply the formula:
$\tan ^{-1} 2 + \tan ^{-1} 3 = \pi + \tan ^{-1} \left( \frac{2+3}{1-(2 \times 3)} \right)$
$= \pi + \tan ^{-1} \left( \frac{5}{1-6} \right)$
$= \pi + \tan ^{-1} \left( \frac{5}{-5} \right)$
$= \pi + \tan ^{-1} (-1)$
Since $\tan ^{-1} (-1) = -\frac{\pi}{4}$,we have:
$= \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$.

(B) We know that $2 \tan ^{-1}(x) = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$.
Therefore,$2 \tan ^{-1} \left( \frac{1}{5} \right) = \tan ^{-1} \left( \frac{2 \times \frac{1}{5}}{1 - (\frac{1}{5})^2} \right) = \tan ^{-1} \left( \frac{\frac{2}{5}}{1 - \frac{1}{25}} \right) = \tan ^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} \right) = \tan ^{-1} \left( \frac{2}{5} \times \frac{25}{24} \right) = \tan ^{-1} \left( \frac{5}{12} \right)$.
Now,the expression becomes $\tan \left[ \tan ^{-1} \left( \frac{5}{12} \right) - \tan ^{-1} (1) \right]$.
Using the formula $\tan ^{-1}(x) - \tan ^{-1}(y) = \tan ^{-1} \left( \frac{x-y}{1+xy} \right)$,we get:
$\tan ^{-1} \left( \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \times 1} \right) = \tan ^{-1} \left( \frac{-\frac{7}{12}}{\frac{17}{12}} \right) = \tan ^{-1} \left( -\frac{7}{17} \right)$.
Finally,$\tan \left[ \tan ^{-1} \left( -\frac{7}{17} \right) \right] = -\frac{7}{17}$.

(D) We know that $\cos ^{-1}\left(\frac{4}{5}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$ because in a right-angled triangle with base $4$ and hypotenuse $5$,the perpendicular is $\sqrt{5^2 - 4^2} = 3$.
Substituting this into the expression:
$\tan \left(\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{2}{3}\right)\right)$
Using the formula $\tan ^{-1}(x) + \tan ^{-1}(y) = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$:
$= \tan \left(\tan ^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{17}{6}\right)\right) = \frac{17}{6}$

(B) We evaluate the left-hand limit at $x = 4$: $\lim_{x \rightarrow 4^{-}} f(x) = [4^{-}] - [\frac{4^{-}}{4}] = 3 - 0 = 3$.
We evaluate the right-hand limit at $x = 4$: $\lim_{x \rightarrow 4^{+}} f(x) = [4^{+}] - [\frac{4^{+}}{4}] = 4 - 1 = 3$.
We evaluate the function value at $x = 4$: $f(4) = [4] - [\frac{4}{4}] = 4 - 1 = 3$.
Since $\lim_{x \rightarrow 4^{-}} f(x) = \lim_{x \rightarrow 4^{+}} f(x) = f(4) = 3$,the function $f(x)$ is continuous at $x = 4$.

(C) The formula for continuous compounding is $A = P \cdot e^{rt}$,where $P$ is the principal amount,$r$ is the annual interest rate,and $t$ is the time in years.
Given: $P = 10000$,$r = 4 \% = 0.04$,and $t = 10$ years.
Substituting the values: $A = 10000 \times e^{(0.04 \times 10)} = 10000 \times e^{0.4}$.
Using the given value $e^{0.4} = 1.49182$:
$A = 10000 \times 1.49182 = 14918.2$.
Rounding to the nearest integer,the amount is $Rs. 14918$.

(D) The given system of in-equations is:
$x+y \leq 70$
$x+2y \leq 100$
$2x+y \leq 120$
$x \geq 0, y \geq 0$
Since all the inequalities $x+y \leq 70$,$x+2y \leq 100$,and $2x+y \leq 120$ are satisfied by the origin $(0,0)$ (as $0+0 \leq 70$,$0+0 \leq 100$,and $0+0 \leq 120$ are all true),the feasible region must lie towards the origin in the first quadrant.
By analyzing the lines and the constraints $x \geq 0, y \geq 0$,the common shaded region that satisfies all these conditions is represented by Fig. $3$.

(C) The feasible region is determined by the constraints $2x+3y \leq 12$,$2x+y \leq 8$,$x \geq 0$,and $y \geq 0$. The corner points of the feasible region are $(0, 0)$,$(4, 0)$,$(3, 2)$,and $(0, 4)$.
We evaluate the objective function $z=4x+5y$ at these corner points:
At $(0, 0)$: $z = 4(0) + 5(0) = 0$
At $(4, 0)$: $z = 4(4) + 5(0) = 16$
At $(3, 2)$: $z = 4(3) + 5(2) = 12 + 10 = 22$
At $(0, 4)$: $z = 4(0) + 5(4) = 20$
The maximum value of $z$ is $22$ at the point $(3, 2)$.

(D) The feasible region is bounded by the corner points $(0, 80)$,$(14, 33)$,and $(80, 0)$.
We evaluate the objective function $Z = 4x + 6y$ at each corner point:
$1$. At $(0, 80)$: $Z = 4(0) + 6(80) = 480$
$2$. At $(14, 33)$: $Z = 4(14) + 6(33) = 56 + 198 = 254$
$3$. At $(80, 0)$: $Z = 4(80) + 6(0) = 320$
Comparing these values,the minimum value of $Z$ is $254$,which occurs at the point $(14, 33)$.

(A) To find the maximum value of $Z = 5x + 2y$,we identify the feasible region defined by the constraints:
$1$. $2x - y \geq 2$
$2$. $x + 2y \leq 8$
$3$. $x, y \geq 0$
The boundary lines are $2x - y = 2$ and $x + 2y = 8$.
- For $2x - y = 2$,the intercepts are $(1, 0)$ and $(0, -2)$.
- For $x + 2y = 8$,the intercepts are $(8, 0)$ and $(0, 4)$.
To find the intersection point of $2x - y = 2$ and $x + 2y = 8$:
Multiply the first equation by $2$: $4x - 2y = 4$.
Adding this to $x + 2y = 8$,we get $5x = 12$,so $x = 2.4$.
Substituting $x = 2.4$ into $2x - y = 2$: $2(2.4) - y = 2 \implies 4.8 - y = 2 \implies y = 2.8$.
The vertices of the feasible region are $(1, 0)$,$(8, 0)$,and $(2.4, 2.8)$.
Now,evaluate $Z = 5x + 2y$ at these vertices:
- At $(1, 0)$: $Z = 5(1) + 2(0) = 5$
- At $(8, 0)$: $Z = 5(8) + 2(0) = 40$
- At $(2.4, 2.8)$: $Z = 5(2.4) + 2(2.8) = 12 + 5.6 = 17.6$
Comparing these values,the maximum value is $40$ at $(8, 0)$.

(D) According to the Fundamental Theorem of Linear Programming,if an optimal solution exists for a linear programming problem,it must occur at one of the corner points (vertices) of the feasible region. If the objective function attains the same optimal value at two corner points,then every point on the line segment joining these two points is also an optimal solution. Therefore,the objective function always attains its optimum value at at least one of the corner points.

(D) To find the maximum value of $z=50x+15y$,we evaluate the objective function at the corner points of the feasible region.
The corner points of the feasible region are $(0, 0)$,$(20, 0)$,$(10, 50)$,and $(0, 60)$.
Evaluating $z$ at each point:
$1$. At $(0, 0)$: $z = 50(0) + 15(0) = 0$
$2$. At $(20, 0)$: $z = 50(20) + 15(0) = 1000$
$3$. At $(10, 50)$: $z = 50(10) + 15(50) = 500 + 750 = 1250$
$4$. At $(0, 60)$: $z = 50(0) + 15(60) = 900$
The maximum value is $1250$,which occurs at the point $(10, 50)$.

(D) To determine the linear constraints for the shaded region,we analyze each boundary line:
$1$. The line $x+2y=6$ has the shaded region on the side away from the origin (since the point $(7, 6)$ satisfies $7+12=19 \geq 6$). Thus,the constraint is $x+2y \geq 6$.
$2$. The line $5x+3y=15$ has the shaded region on the side away from the origin (since the point $(7, 6)$ satisfies $35+18=53 \geq 15$). Thus,the constraint is $5x+3y \geq 15$.
$3$. The vertical line $x=7$ has the shaded region to its left,so $x \leq 7$.
$4$. The horizontal line $y=6$ has the shaded region below it,so $y \leq 6$.
$5$. The region is in the first quadrant,so $x, y \geq 0$.
Combining these,the constraints are $x+2y \geq 6, 5x+3y \geq 15, x \leq 7, y \leq 6, x, y \geq 0$.

(B) The feasible region is bounded by the lines $x+y=5$,$4x+3y=24$,$x=0$,and $y=0$.
To find the corner points,we identify the vertices of the shaded region:
$1$. The intersection of $x=0$ and $4x+3y=24$ gives $(0, 8)$.
$2$. The intersection of $x=0$ and $x+y=5$ gives $(0, 5)$.
$3$. The intersection of $x+y=5$ and $4x+3y=24$ is found by solving the system: $y=5-x \implies 4x+3(5-x)=24 \implies x=9, y=-4$ (Not in the first quadrant).
Looking at the graph,the vertices are $(0, 5)$,$(0, 8)$,and the intersection of $4x+3y=24$ with the $x$-axis $(6, 0)$ and $x+y=5$ with the $x$-axis $(5, 0)$.
Evaluating $Z=2x+y$ at the corner points:
At $(0, 5)$,$Z = 2(0) + 5 = 5$.
At $(0, 8)$,$Z = 2(0) + 8 = 8$.
At $(5, 0)$,$Z = 2(5) + 0 = 10$.
At $(6, 0)$,$Z = 2(6) + 0 = 12$.
The maximum value is $12$ at the point $(6, 0)$.