MHT CET 2020 Mathematics Question Paper with Answer and Solution

(A) Let $I = \int \frac{dx}{1+\sqrt{x}}$.
Substitute $\sqrt{x} = t$,which implies $x = t^2$,so $dx = 2t \, dt$.
Substituting these into the integral:
$I = \int \frac{2t}{1+t} \, dt$.
$I = 2 \int \frac{t}{1+t} \, dt$.
$I = 2 \int \frac{(t+1) - 1}{1+t} \, dt$.
$I = 2 \int \left( 1 - \frac{1}{1+t} \right) \, dt$.
$I = 2 \left( \int 1 \, dt - \int \frac{1}{1+t} \, dt \right)$.
$I = 2(t - \log|1+t|) + c$.
Substituting $t = \sqrt{x}$ back:
$I = 2\sqrt{x} - 2\log|1+\sqrt{x}| + c$.

(C) Let $I = \int \frac{\sin x \cdot \cos x}{\sin ^{4} x + \cos ^{4} x} dx$.
Dividing the numerator and the denominator by $\cos ^{4} x$,we get:
$I = \int \frac{\frac{\sin x \cdot \cos x}{\cos ^{4} x}}{\frac{\sin ^{4} x}{\cos ^{4} x} + 1} dx = \int \frac{\tan x \sec ^{2} x}{\tan ^{4} x + 1} dx$.
Let $\tan ^{2} x = t$. Then,differentiating both sides with respect to $x$,we get $2 \tan x \sec ^{2} x dx = dt$,which implies $\tan x \sec ^{2} x dx = \frac{1}{2} dt$.
Substituting these into the integral:
$I = \int \frac{1}{1 + t^{2}} \cdot \frac{1}{2} dt = \frac{1}{2} \int \frac{1}{1 + t^{2}} dt$.
Using the standard integral formula $\int \frac{1}{1 + t^{2}} dt = \tan ^{-1} t + c$,we get:
$I = \frac{1}{2} \tan ^{-1} t + c = \frac{1}{2} \tan ^{-1}(\tan ^{2} x) + c$.

(A) Let $I = \int \frac{(\sin^{-1} x)^{\frac{3}{2}}}{\sqrt{1-x^2}} dx$.
Substitute $\sin^{-1} x = t$.
Then,differentiating both sides with respect to $x$,we get $\frac{1}{\sqrt{1-x^2}} dx = dt$.
Substituting these into the integral,we have:
$I = \int t^{\frac{3}{2}} dt$.
Using the power rule for integration $\int t^n dt = \frac{t^{n+1}}{n+1} + c$:
$I = \frac{t^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} + c = \frac{t^{\frac{5}{2}}}{\frac{5}{2}} + c = \frac{2}{5} t^{\frac{5}{2}} + c$.
Substituting back $t = \sin^{-1} x$,we get:
$I = \frac{2}{5}(\sin^{-1} x)^{\frac{5}{2}} + c$.

(B) Let $I = \int x^{3} e^{x^{2}} dx$.
Substitute $x^{2} = t$,then $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$.
Substituting these into the integral,we get $I = \frac{1}{2} \int t e^{t} dt$.
Using integration by parts,$\int u dv = uv - \int v du$,where $u = t$ and $dv = e^{t} dt$.
Then $du = dt$ and $v = e^{t}$.
So,$I = \frac{1}{2} [t e^{t} - \int e^{t} dt] = \frac{1}{2} [t e^{t} - e^{t}] + c$.
Factoring out $e^{t}$,we get $I = \frac{1}{2} e^{t}(t - 1) + c$.
Substituting back $t = x^{2}$,we obtain $I = \frac{1}{2} e^{x^{2}}(x^{2} - 1) + c$.

(B) Let $I = \int \frac{e^{x}}{\sqrt{x}}(1+2x) dx$.
We can rewrite the integrand as:
$I = \int e^{x} \left( \frac{1}{\sqrt{x}} + \frac{2x}{\sqrt{x}} \right) dx = \int e^{x} \left( \frac{1}{\sqrt{x}} + 2\sqrt{x} \right) dx$.
Rearranging the terms:
$I = \int e^{x} \left( 2\sqrt{x} + \frac{1}{\sqrt{x}} \right) dx$.
Let $f(x) = 2\sqrt{x}$. Then $f'(x) = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$.
Using the standard integral formula $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$,we get:
$I = e^{x} (2\sqrt{x}) + c = 2\sqrt{x} e^{x} + c$.

(B) Let $I = \int e^{\tan ^{-1} x} \left(1 + \frac{x}{1 + x^2} \right) dx$.
Substitute $t = \tan ^{-1} x$,which implies $x = \tan t$ and $dx = (1 + x^2) dt = \sec^2 t dt$.
Then,$I = \int e^t \left(1 + \frac{\tan t}{1 + \tan^2 t} \right) \sec^2 t dt$.
Since $1 + \tan^2 t = \sec^2 t$,the expression becomes $I = \int e^t \left(1 + \frac{\tan t}{\sec^2 t} \right) \sec^2 t dt$.
$I = \int e^t (\sec^2 t + \tan t) dt$.
Using the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = \tan t$ and $f'(t) = \sec^2 t$.
Thus,$I = e^t \tan t + c$.
Substituting back $t = \tan ^{-1} x$,we get $I = x e^{\tan ^{-1} x} + c$.

(C) Let $I = \int \cot x \cdot \log [\log (\sin x)] d x$.
Substitute $t = \log (\sin x)$.
Then $dt = \frac{1}{\sin x} \cdot \cos x \, dx = \cot x \, dx$.
Thus,the integral becomes $I = \int \log t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log t$ and $dv = dt$:
$I = t \log t - \int t \cdot \frac{1}{t} \, dt$.
$I = t \log t - \int 1 \, dt$.
$I = t \log t - t + c = t(\log t - 1) + c$.
Substituting back $t = \log (\sin x)$:
$I = \log (\sin x) [\log (\log (\sin x)) - 1] + c$.

(B) Let $I = \int \left[ \frac{\log x - 1}{1 + (\log x)^2} \right]^2 dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
The integral becomes $I = \int \left[ \frac{t - 1}{1 + t^2} \right]^2 e^t dt = \int e^t \left[ \frac{t^2 - 2t + 1}{(1 + t^2)^2} \right] dt$.
We can rewrite the numerator as $(t^2 + 1) - 2t$.
So,$I = \int e^t \left[ \frac{t^2 + 1}{(1 + t^2)^2} - \frac{2t}{(1 + t^2)^2} \right] dt = \int e^t \left[ \frac{1}{1 + t^2} - \frac{2t}{(1 + t^2)^2} \right] dt$.
This is in the form $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$,where $f(t) = \frac{1}{1 + t^2}$.
Thus,$I = e^t \left( \frac{1}{1 + t^2} \right) + c$.
Substituting back $t = \log x$,we get $I = \frac{x}{1 + (\log x)^2} + c$.

(A) Let $I = \int \log x \cdot (\log x + 2) dx$.
Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$.
Then,$I = \int t(t + 2) e^t dt = \int (t^2 + 2t) e^t dt$.
Recall the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$.
Here,let $f(t) = t^2$,then $f'(t) = 2t$.
Thus,$I = e^t (t^2) + c$.
Substituting back $t = \log x$,we get $I = x(\log x)^2 + c$.

(A) Let $I = \int \sin^{-1} x \cdot 1 \, dx$.
Using integration by parts,$\int u \cdot v \, dx = u \int v \, dx - \int (u' \cdot \int v \, dx) \, dx$.
Here,$u = \sin^{-1} x$ and $v = 1$.
$I = \sin^{-1} x \cdot x - \int \frac{1}{\sqrt{1 - x^2}} \cdot x \, dx$.
$I = x \sin^{-1} x - \int \frac{x}{\sqrt{1 - x^2}} \, dx$.
To solve $\int \frac{x}{\sqrt{1 - x^2}} \, dx$,let $1 - x^2 = t$,then $-2x \, dx = dt$,so $x \, dx = -\frac{1}{2} dt$.
$\int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot \frac{t^{1/2}}{1/2} = -\sqrt{t} = -\sqrt{1 - x^2}$.
Substituting this back into the expression for $I$:
$I = x \sin^{-1} x - (-\sqrt{1 - x^2}) + c = x \sin^{-1} x + \sqrt{1 - x^2} + c$.

(B) Let $I = \int (1+x) \log x \, dx$.
Using integration by parts,let $u = \log x$ and $dv = (1+x) \, dx$.
Then $du = \frac{1}{x} \, dx$ and $v = x + \frac{x^2}{2}$.
Using the formula $\int u \, dv = uv - \int v \, du$:
$I = \log x \left(x + \frac{x^2}{2}\right) - \int \left(x + \frac{x^2}{2}\right) \cdot \frac{1}{x} \, dx$.
$I = \left(x + \frac{x^2}{2}\right) \log x - \int \left(1 + \frac{x}{2}\right) \, dx$.
$I = \left(x + \frac{x^2}{2}\right) \log x - \left(x + \frac{x^2}{4}\right) + C$.

(B) We know the standard integral form $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + c$.
Given the integral $\int e^{x} \sec x(1+\tan x) dx$.
Distributing $\sec x$,we get $\int e^{x} (\sec x + \sec x \tan x) dx$.
Let $f(x) = \sec x$. Then $f'(x) = \sec x \tan x$.
Substituting this into the standard form,we get $e^{x} \sec x + c$.

(B) Let $I = \int_{0}^{\infty} \frac{dx}{(x^{2}+4)(x^{2}+9)}$.
Using partial fractions,we write $\frac{1}{(x^{2}+4)(x^{2}+9)} = \frac{A}{x^{2}+4} + \frac{B}{x^{2}+9}$.
$1 = A(x^{2}+9) + B(x^{2}+4)$.
Setting $x^{2} = -4$,we get $1 = A(5) \implies A = \frac{1}{5}$.
Setting $x^{2} = -9$,we get $1 = B(-5) \implies B = -\frac{1}{5}$.
So,$I = \frac{1}{5} \int_{0}^{\infty} \left( \frac{1}{x^{2}+4} - \frac{1}{x^{2}+9} \right) dx$.
Using the formula $\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$,we get:
$I = \frac{1}{5} \left[ \frac{1}{2} \tan^{-1}(\frac{x}{2}) - \frac{1}{3} \tan^{-1}(\frac{x}{3}) \right]_{0}^{\infty}$.
Evaluating the limits:
$I = \frac{1}{5} \left[ (\frac{1}{2} \cdot \frac{\pi}{2} - 0) - (\frac{1}{3} \cdot \frac{\pi}{2} - 0) \right]$.
$I = \frac{1}{5} \left( \frac{\pi}{4} - \frac{\pi}{6} \right) = \frac{1}{5} \left( \frac{3\pi - 2\pi}{12} \right) = \frac{1}{5} \cdot \frac{\pi}{12} = \frac{\pi}{60}$.

(C) We are given the integral $I = \int_{2}^{3} \frac{dx}{x^{2}+x}$.
First,we factor the denominator: $x^{2}+x = x(x+1)$.
Using partial fractions,we can write $\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$.
Solving for $A$ and $B$,we get $1 = A(x+1) + Bx$. Setting $x=0$ gives $A=1$,and setting $x=-1$ gives $B=-1$.
Thus,$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$.
Now,integrate: $I = \int_{2}^{3} \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = [\log|x| - \log|x+1|]_{2}^{3} = [\log|\frac{x}{x+1}|]_{2}^{3}$.
Evaluating at the limits: $I = \log(\frac{3}{4}) - \log(\frac{2}{3}) = \log(\frac{3}{4} \div \frac{2}{3}) = \log(\frac{3}{4} \times \frac{3}{2}) = \log(\frac{9}{8})$.

(A) Let $I = \int \frac{x^{2}}{(x+1)(x+2)^{2}} dx$.
Using partial fractions,we write:
$\frac{x^{2}}{(x+1)(x+2)^{2}} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{(x+2)^{2}}$.
Multiplying both sides by $(x+1)(x+2)^{2}$,we get:
$x^{2} = A(x+2)^{2} + B(x+1)(x+2) + C(x+1)$.
Setting $x = -2$,we get $(-2)^{2} = C(-2+1) \Rightarrow 4 = -C \Rightarrow C = -4$.
Setting $x = -1$,we get $(-1)^{2} = A(-1+2)^{2} \Rightarrow 1 = A(1)^{2} \Rightarrow A = 1$.
Comparing the coefficients of $x^{2}$ on both sides,we get $1 = A + B$.
Since $A = 1$,we have $1 = 1 + B \Rightarrow B = 0$.
Thus,the integral becomes:
$I = \int \left( \frac{1}{x+1} - \frac{4}{(x+2)^{2}} \right) dx$.
$I = \int \frac{1}{x+1} dx - 4 \int (x+2)^{-2} dx$.
$I = \log |x+1| - 4 \left( \frac{(x+2)^{-1}}{-1} \right) + c$.
$I = \log |x+1| + \frac{4}{x+2} + c$.

(A) Given integral: $I = \int \frac{x^{2}+1}{(x-3)(x-2)} d x$
Since the degree of the numerator is equal to the degree of the denominator,we perform division:
$\frac{x^{2}+1}{x^{2}-5x+6} = 1 + \frac{5x-5}{x^{2}-5x+6}$
So,$I = \int (1 + \frac{5x-5}{(x-3)(x-2)}) d x = \int 1 d x + 5 \int \frac{x-1}{(x-3)(x-2)} d x$
Using partial fractions for $\frac{x-1}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$:
$x-1 = A(x-2) + B(x-3)$
For $x=3$: $3-1 = A(3-2) \Rightarrow A=2$
For $x=2$: $2-1 = B(2-3) \Rightarrow B=-1$
Thus,$I = x + 5 \int (\frac{2}{x-3} - \frac{1}{x-2}) d x = x + 10 \log |x-3| - 5 \log |x-2| + c$
Comparing with $Px + Q \log |x-3| + R \log |x-2| + c$,we get $P=1, Q=10, R=-5$.

(A) Let $I = \int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x$.
Using partial fractions,let $x^2 = t$. Then $\frac{2t+3}{(t-1)(t+4)} = \frac{A}{t-1} + \frac{B}{t+4}$.
$2t+3 = A(t+4) + B(t-1)$.
For $t=1$,$5 = 5A \implies A=1$.
For $t=-4$,$-5 = -5B \implies B=1$.
Thus,$\frac{2x^2+3}{(x^2-1)(x^2+4)} = \frac{1}{x^2-1} + \frac{1}{x^2+4}$.
Integrating both sides,$I = \int \frac{1}{x^2-1} dx + \int \frac{1}{x^2+2^2} dx$.
$I = \frac{1}{2(1)} \log \left| \frac{x-1}{x+1} \right| + \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C$.
Comparing this with the given expression $a \log \left| \frac{x-1}{x+1} \right| + b \tan^{-1} \left( \frac{x}{2} \right) + C$,we get $a = \frac{1}{2}$ and $b = \frac{1}{2}$.

(A) Let $I = \int \frac{x+1}{x^{2}+5x+6} dx$.
Factorizing the denominator: $x^{2}+5x+6 = (x+2)(x+3)$.
Using partial fractions: $\frac{x+1}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3}$.
$x+1 = A(x+3) + B(x+2)$.
Setting $x = -2$: $-2+1 = A(-2+3) \implies A = -1$.
Setting $x = -3$: $-3+1 = B(-3+2) \implies -2 = B(-1) \implies B = 2$.
Thus,$I = \int \left( \frac{-1}{x+2} + \frac{2}{x+3} \right) dx$.
$I = -\int \frac{1}{x+2} dx + 2 \int \frac{1}{x+3} dx$.
$I = -\log |x+2| + 2 \log |x+3| + C$.

(A) Let $I = \int \frac{x^{2}+1}{x^{4}+x^{2}+1} dx$.
Divide the numerator and denominator by $x^{2}$:
$I = \int \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} dx$.
Rewrite the denominator as $(x^{2} + \frac{1}{x^{2}} - 2) + 3 = (x - \frac{1}{x})^{2} + (\sqrt{3})^{2}$.
So,$I = \int \frac{1 + \frac{1}{x^{2}}}{(x - \frac{1}{x})^{2} + (\sqrt{3})^{2}} dx$.
Let $t = x - \frac{1}{x}$,then $dt = (1 + \frac{1}{x^{2}}) dx$.
Substituting these into the integral,we get:
$I = \int \frac{dt}{t^{2} + (\sqrt{3})^{2}}$.
Using the standard integral formula $\int \frac{dx}{x^{2} + a^{2}} = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + c$:
$I = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{t}{\sqrt{3}}) + c$.
Replacing $t$ with $x - \frac{1}{x}$,we get:
$I = \frac{1}{\sqrt{3}} \tan^{-1}(\frac{x - \frac{1}{x}}{\sqrt{3}}) + c$.

(A) Let $I = \int \frac{\sin 2x}{\sin^2 x \cos^2 x} dx$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int \frac{2 \sin x \cos x}{\sin^2 x \cos^2 x} dx$
$I = 2 \int \frac{1}{\sin x \cos x} dx$
Multiply numerator and denominator by $2$:
$I = 2 \int \frac{2}{2 \sin x \cos x} dx = 4 \int \frac{1}{\sin 2x} dx$
$I = 4 \int \operatorname{cosec} 2x dx$
Using the formula $\int \operatorname{cosec} u du = \log |\tan(u/2)| + c$,we get:
$I = 4 \times \frac{1}{2} \log |\tan x| + c = 2 \log |\tan x| + c$
Using the property $n \log a = \log a^n$,we get:
$I = \log |\tan^2 x| + c$.

(B) Let $I = \int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx$.
Put $\log x = t$,then $x = e^{t}$,so $dx = e^{t} dt$.
Substituting these into the integral,we get $I = \int e^{t} \left[\frac{1-t}{1+t^{2}}\right]^{2} dt$.
Expanding the square,$I = \int e^{t} \frac{1+t^{2}-2t}{(1+t^{2})^{2}} dt = \int e^{t} \left[\frac{1}{1+t^{2}} - \frac{2t}{(1+t^{2})^{2}}\right] dt$.
Using the standard integral formula $\int e^{t} [f(t) + f'(t)] dt = e^{t} f(t) + c$,where $f(t) = \frac{1}{1+t^{2}}$ and $f'(t) = -\frac{2t}{(1+t^{2})^{2}}$.
Thus,$I = e^{t} \left(\frac{1}{1+t^{2}}\right) + c$.
Substituting back $t = \log x$,we get $I = \frac{x}{1+(\log x)^{2}} + c$.

(A) Let $I = \int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x$.
Substitute $x = (x - \frac{\pi}{4}) + \frac{\pi}{4}$,so $I = \int \frac{\sin \left((x-\frac{\pi}{4}) + \frac{\pi}{4}\right)}{\sin \left(x-\frac{\pi}{4}\right)} d x$.
Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $I = \int \frac{\sin (x-\frac{\pi}{4}) \cos \frac{\pi}{4} + \cos (x-\frac{\pi}{4}) \sin \frac{\pi}{4}}{\sin (x-\frac{\pi}{4})} d x$.
Since $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,we have $I = \int \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \cot (x-\frac{\pi}{4}) \right) d x$.
Integrating term by term,$I = \frac{1}{\sqrt{2}} \int 1 d x + \frac{1}{\sqrt{2}} \int \cot (x-\frac{\pi}{4}) d x$.
The integral of $\cot u$ is $\log |\sin u|$,so $I = \frac{1}{\sqrt{2}} [x + \log |\sin (x-\frac{\pi}{4})|] + c$.

(D) Let $I = \int \left[ \log (1+\cos x) - x \tan \left( \frac{x}{2} \right) \right] dx$.
Using integration by parts on the first term,$\int u \cdot v dx = u \int v dx - \int (u' \int v dx) dx$,where $u = \log(1+\cos x)$ and $v = 1$.
$I = x \log(1+\cos x) - \int x \cdot \frac{d}{dx} [\log(1+\cos x)] dx - \int x \tan \left( \frac{x}{2} \right) dx$.
Since $\frac{d}{dx} [\log(1+\cos x)] = \frac{-\sin x}{1+\cos x} = \frac{-2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)} = -\tan(x/2)$.
$I = x \log(1+\cos x) - \int x \left( -\tan \frac{x}{2} \right) dx - \int x \tan \frac{x}{2} dx$.
$I = x \log(1+\cos x) + \int x \tan \frac{x}{2} dx - \int x \tan \frac{x}{2} dx$.
$I = x \log(1+\cos x) + c$.

(A) We know that $\int e^{x}[f(x) + f'(x)] dx = e^{x}f(x) + C$.
Given integral is $I = \int e^{x} \left( \frac{1-x}{1+x^{2}} \right)^{2} dx$.
Expanding the numerator: $I = \int e^{x} \frac{1 + x^{2} - 2x}{(1+x^{2})^{2}} dx$.
Splitting the fraction: $I = \int e^{x} \left[ \frac{1+x^{2}}{(1+x^{2})^{2}} - \frac{2x}{(1+x^{2})^{2}} \right] dx$.
$I = \int e^{x} \left[ \frac{1}{1+x^{2}} - \frac{2x}{(1+x^{2})^{2}} \right] dx$.
Let $f(x) = \frac{1}{1+x^{2}}$. Then $f'(x) = -\frac{1}{(1+x^{2})^{2}} \cdot (2x) = -\frac{2x}{(1+x^{2})^{2}}$.
Thus,$I = \int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$.
Therefore,$I = \frac{e^{x}}{1+x^{2}} + C$.

(A) We have the integral $I = \int \frac{\sin \theta}{\sin 3 \theta} d \theta$.
Using the identity $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$,we get:
$I = \int \frac{\sin \theta}{\sin \theta(3 - 4 \sin^2 \theta)} d \theta = \int \frac{1}{3 - 4 \sin^2 \theta} d \theta$.
Dividing the numerator and denominator by $\cos^2 \theta$:
$I = \int \frac{\sec^2 \theta}{3 \sec^2 \theta - 4 \tan^2 \theta} d \theta = \int \frac{\sec^2 \theta}{3(1 + \tan^2 \theta) - 4 \tan^2 \theta} d \theta = \int \frac{\sec^2 \theta}{3 - \tan^2 \theta} d \theta$.
Let $\tan \theta = t$,then $\sec^2 \theta d \theta = dt$.
$I = \int \frac{1}{(\sqrt{3})^2 - t^2} dt$.
Using the formula $\int \frac{1}{a^2 - x^2} dx = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + c$:
$I = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + t}{\sqrt{3} - t} \right| + c = \frac{1}{2 \sqrt{3}} \log \left| \frac{\sqrt{3} + \tan \theta}{\sqrt{3} - \tan \theta} \right| + c$.
Comparing this with the given expression $\frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan \theta}\right|+c$,we find $k = \sqrt{3}$.

(B) We know that $\sin ^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$ (or $30^{\circ}$).
We know that $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ (or $30^{\circ}$).
We know that $\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \pi - \cot ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ (or $120^{\circ}$).
Adding these values together:
$\frac{\pi}{6} + \frac{\pi}{6} + \frac{2\pi}{3} = \frac{2\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{3} + \frac{2\pi}{3} = \frac{3\pi}{3} = \pi$.

(D) Let $y = \sin ^{-1}\left(-\frac{1}{2}\right)$.
Then,$\sin y = -\frac{1}{2}$.
We know that the range of the principal value branch of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$,we have $\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}$.
Thus,the principal value is $-\frac{\pi}{6}$.

(C) Let $\alpha = \sin ^{-1}\left(-\frac{1}{2}\right)$. Since the range of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,we have $\sin \alpha = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right)$,so $\alpha = -\frac{\pi}{6}$.
Let $\beta = \cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$. Since the range of $\cos ^{-1} x$ is $[0, \pi]$,we have $\cos \beta = -\frac{\sqrt{3}}{2} = -\cos\left(\frac{\pi}{6}\right) = \cos\left(\pi - \frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right)$,so $\beta = \frac{5\pi}{6}$.
Adding these values,we get $\alpha + \beta = -\frac{\pi}{6} + \frac{5\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$.
Now,checking the options:
Option $C$ is $\cos ^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.
Thus,the correct option is $C$.

(A) Given $y = \tan^{-1}(\sec x + \tan x)$.
We can rewrite the expression inside the inverse tangent function as:
$y = \tan^{-1}\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right)$.
Using the half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$,and $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$,we get:
$y = \tan^{-1}\left[\frac{\cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}}\right]$
$y = \tan^{-1}\left[\frac{(\cos\frac{x}{2} + \sin\frac{x}{2})^2}{(\cos\frac{x}{2} - \sin\frac{x}{2})(\cos\frac{x}{2} + \sin\frac{x}{2})}\right]$
$y = \tan^{-1}\left[\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}\right]$
Dividing numerator and denominator by $\cos\frac{x}{2}$:
$y = \tan^{-1}\left(\frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\right) = \tan^{-1}\left(\tan(\frac{\pi}{4} + \frac{x}{2})\right)$
$y = \frac{\pi}{4} + \frac{x}{2}$.
Now,differentiating with respect to $x$:
$\frac{dy}{dx} = 0 + \frac{1}{2} = \frac{1}{2}$.

(B) Given $y = \tan^{-1} \left[ \sqrt{\frac{1 + \cos(x/2)}{1 - \cos(x/2)}} \right]$.
Using the trigonometric identities $1 + \cos \theta = 2 \cos^2(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$,we have:
$y = \tan^{-1} \sqrt{\frac{2 \cos^2(x/4)}{2 \sin^2(x/4)}}$
$y = \tan^{-1} \sqrt{\cot^2(x/4)}$
$y = \tan^{-1} (\cot(x/4))$
Since $\cot \theta = \tan(\pi/2 - \theta)$,we get:
$y = \tan^{-1} \left[ \tan \left( \frac{\pi}{2} - \frac{x}{4} \right) \right]$
$y = \frac{\pi}{2} - \frac{x}{4}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{2} - \frac{x}{4} \right) = 0 - \frac{1}{4} = -\frac{1}{4}$.

(A) Given $u=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$.
Let $x=\tan \theta$,then $\theta=\tan ^{-1} x$.
$u=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) = \tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) = \tan ^{-1}\left(\frac{2 \sin ^{2}(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)}\right) = \tan ^{-1}(\tan(\theta/2)) = \frac{\theta}{2} = \frac{1}{2} \tan ^{-1} x$.
Thus,$\frac{d u}{d x} = \frac{1}{2(1+x^{2})}$.
Given $v=\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)$.
Let $x=\sin \theta$,then $\theta=\sin ^{-1} x$.
$v=\tan ^{-1}\left(\frac{2 \sin \theta \cos \theta}{1-2 \sin ^{2} \theta}\right) = \tan ^{-1}\left(\frac{\sin 2 \theta}{\cos 2 \theta}\right) = \tan ^{-1}(\tan 2 \theta) = 2 \theta = 2 \sin ^{-1} x$.
Thus,$\frac{d v}{d x} = \frac{2}{\sqrt{1-x^{2}}}$.
Then $\frac{d u}{d v} = \frac{d u/d x}{d v/d x} = \frac{1}{2(1+x^{2})} \times \frac{\sqrt{1-x^{2}}}{2} = \frac{\sqrt{1-x^{2}}}{4(1+x^{2})}$.
At $x=0$,$\frac{d u}{d v} = \frac{\sqrt{1-0}}{4(1+0)} = \frac{1}{4}$.

(A) We know that the range of the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$.
Given expression is $\cos ^{-1}\left(\cos \left(\frac{7 \pi}{6}\right)\right)$.
Since $\frac{7 \pi}{6} > \pi$,we cannot directly write $\cos ^{-1}(\cos \theta) = \theta$.
We use the property $\cos(2 \pi - \theta) = \cos \theta$.
$\cos \left(\frac{7 \pi}{6}\right) = \cos \left(2 \pi - \frac{5 \pi}{6}\right) = \cos \left(\frac{5 \pi}{6}\right)$.
Now,$\cos ^{-1}\left(\cos \left(\frac{5 \pi}{6}\right)\right) = \frac{5 \pi}{6}$,which lies in the interval $[0, \pi]$.
Thus,the correct option is $A$.

(C) We know that the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$.
Given expression is $\cos ^{-1}\left(\cos \frac{8 \pi}{3}\right)$.
Since $\frac{8 \pi}{3} = 2 \pi + \frac{2 \pi}{3}$,we have $\cos \frac{8 \pi}{3} = \cos \left(2 \pi + \frac{2 \pi}{3}\right) = \cos \frac{2 \pi}{3}$.
Since $\frac{2 \pi}{3} \in [0, \pi]$,we have $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right) = \frac{2 \pi}{3}$.
Therefore,the value is $\frac{2 \pi}{3}$.

(A) Given that in $\Delta ABC$,$C=90^{\circ}$,therefore by Pythagoras theorem,$a^{2}+b^{2}=c^{2}$.
We need to evaluate $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$:
$= \tan ^{-1} \left[ \frac{\frac{a}{b+c} + \frac{b}{c+a}}{1 - \left( \frac{a}{b+c} \right) \left( \frac{b}{c+a} \right)} \right]$
$= \tan ^{-1} \left[ \frac{a(c+a) + b(b+c)}{(b+c)(c+a) - ab} \right]$
$= \tan ^{-1} \left[ \frac{ac + a^{2} + b^{2} + bc}{bc + ab + c^{2} + ac - ab} \right]$
Since $a^{2}+b^{2}=c^{2}$,the numerator becomes $ac + c^{2} + bc = c(a+b+c)$.
The denominator becomes $bc + ac + c^{2} = c(b+a+c)$.
$= \tan ^{-1} \left( \frac{c(a+b+c)}{c(a+b+c)} \right)$
$= \tan ^{-1}(1) = \frac{\pi}{4}$.

(B) Let $\tan ^{-1} \frac{3}{4} = \theta$. Then $\tan \theta = \frac{3}{4}$.
Using the identity $\sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}}$,we get $\sin \theta = \frac{3/4}{\sqrt{1 + (3/4)^2}} = \frac{3/4}{\sqrt{25/16}} = \frac{3/4}{5/4} = \frac{3}{5}$.
Thus,$\left[\sin \left(\tan ^{-1} \frac{3}{4}\right)\right]^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$.
Let $\tan ^{-1} \frac{4}{3} = \phi$. Then $\tan \phi = \frac{4}{3}$.
Using the identity $\sin \phi = \frac{\tan \phi}{\sqrt{1 + \tan^2 \phi}}$,we get $\sin \phi = \frac{4/3}{\sqrt{1 + (4/3)^2}} = \frac{4/3}{\sqrt{25/9}} = \frac{4/3}{5/3} = \frac{4}{5}$.
Thus,$\left[\sin \left(\tan ^{-1} \frac{4}{3}\right)\right]^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$.
Adding the two values: $\frac{9}{25} + \frac{16}{25} = \frac{25}{25} = 1$.

(A) We know that for any $\theta \in [-1, 1]$,$\sin^{-1} \theta + \cos^{-1} \theta = \frac{\pi}{2}$.
Given the equation $x^{2} y^{2} = \sin^{-1} \sqrt{x^{2} + y^{2}} + \cos^{-1} \sqrt{x^{2} + y^{2}}$,we can simplify the right side to get $x^{2} y^{2} = \frac{\pi}{2}$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}(x^{2} y^{2}) = \frac{d}{dx}(\frac{\pi}{2})$
$x^{2} \cdot (2y \frac{dy}{dx}) + y^{2} \cdot (2x) = 0$
$2x^{2}y \frac{dy}{dx} = -2xy^{2}$
Dividing both sides by $2xy$ (assuming $x, y \neq 0$):
$\frac{dy}{dx} = \frac{-2xy^{2}}{2x^{2}y} = -\frac{y}{x}$.

(B) We know that $\tan^{-1}(u) + \cot^{-1}(u) = \frac{\pi}{2}$ for any $u > 0$.
Given the integral $I = \int_{-1}^{3} \left[ \tan^{-1} \left( \frac{x}{x^{2}+1} \right) + \tan^{-1} \left( \frac{x^{2}+1}{x} \right) \right] dx$.
Since $\tan^{-1} \left( \frac{x^{2}+1}{x} \right) = \cot^{-1} \left( \frac{x}{x^{2}+1} \right)$ for $x > 0$,the expression inside the integral becomes $\tan^{-1} \left( \frac{x}{x^{2}+1} \right) + \cot^{-1} \left( \frac{x}{x^{2}+1} \right) = \frac{\pi}{2}$.
Thus,$I = \int_{-1}^{3} \frac{\pi}{2} dx = \frac{\pi}{2} [x]_{-1}^{3} = \frac{\pi}{2} (3 - (-1)) = \frac{\pi}{2} (4) = 2 \pi$.

(C) Given: $\tan ^{-1} x + \tan ^{-1} y + \tan ^{-1} z = \frac{\pi}{2}$
$\tan ^{-1} x + \tan ^{-1} y = \frac{\pi}{2} - \tan ^{-1} z$
Using the identity $\frac{\pi}{2} - \tan ^{-1} z = \cot ^{-1} z = \tan ^{-1} (\frac{1}{z})$:
$\tan ^{-1} (\frac{x+y}{1-xy}) = \tan ^{-1} (\frac{1}{z})$
Comparing both sides:
$\frac{x+y}{1-xy} = \frac{1}{z}$
$z(x+y) = 1 - xy$
$zx + zy = 1 - xy$
$xy + yz + zx = 1$

(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 $\tan ^{-1}(x) + \tan ^{-1}(y) = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$:
$= \tan \left(\tan ^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} \times \frac{2}{3}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{\frac{9+8}{12}}{1 - \frac{6}{12}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right)\right)$
$= \tan \left(\tan ^{-1}\left(\frac{17}{6}\right)\right) = \frac{17}{6}$

(C) Given equation is $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0$.
Rearranging the terms,we get $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x$.
Using the property $\tan ^{-1} a - \tan ^{-1} b = \tan ^{-1}\left(\frac{a-b}{1+ab}\right)$,we can write $\tan ^{-1}(1) - \tan ^{-1}(x) = \frac{1}{2} \tan ^{-1} x$.
This simplifies to $\frac{\pi}{4} = \frac{1}{2} \tan ^{-1} x + \tan ^{-1} x$.
$\frac{\pi}{4} = \frac{3}{2} \tan ^{-1} x$.
$\tan ^{-1} x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6}$.
Therefore,$x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$.

(A) We use the formula $2 \tan ^{-1} x = \tan ^{-1} \left( \frac{2x}{1-x^2} \right)$.
For $x = \frac{1}{3}$,we have:
$2 \tan ^{-1} \left( \frac{1}{3} \right) = \tan ^{-1} \left( \frac{2 \times \frac{1}{3}}{1 - (\frac{1}{3})^2} \right)$
$= \tan ^{-1} \left( \frac{\frac{2}{3}}{1 - \frac{1}{9}} \right)$
$= \tan ^{-1} \left( \frac{\frac{2}{3}}{\frac{8}{9}} \right)$
$= \tan ^{-1} \left( \frac{2}{3} \times \frac{9}{8} \right) = \tan ^{-1} \left( \frac{3}{4} \right)$
Now,substituting this back into the original expression:
$2 \tan ^{-1} \left( \frac{1}{3} \right) - \tan ^{-1} \left( \frac{3}{4} \right) = \tan ^{-1} \left( \frac{3}{4} \right) - \tan ^{-1} \left( \frac{3}{4} \right) = 0$.

(C) We use the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
First,combine the first two terms:
$\tan ^{-1} \left( \frac{1}{3} \right) + \tan ^{-1} \left( \frac{1}{5} \right) = \tan ^{-1} \left( \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \times \frac{1}{5}} \right) = \tan ^{-1} \left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$.
Now,combine the last two terms:
$\tan ^{-1} \left( \frac{1}{7} \right) + \tan ^{-1} \left( \frac{1}{8} \right) = \tan ^{-1} \left( \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \times \frac{1}{8}} \right) = \tan ^{-1} \left( \frac{\frac{15}{56}}{\frac{55}{56}} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$.
Finally,add the results:
$\tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} \times \frac{3}{11}} \right) = \tan ^{-1} \left( \frac{\frac{44+21}{77}}{\frac{77-12}{77}} \right) = \tan ^{-1} \left( \frac{65}{65} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$.
Thus,the correct option is $C$.

(A) We know that the range of $\sin ^{-1} \theta$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since the sum of three such values is $\frac{3 \pi}{2}$,each term must be equal to its maximum value,$\frac{\pi}{2}$.
Thus,$\sin ^{-1} x = \frac{\pi}{2}$,$\sin ^{-1} y = \frac{\pi}{2}$,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,we get $x^{100} + y^{100} + z^{100} = 1^{100} + 1^{100} + 1^{100} = 1 + 1 + 1 = 3$.

(B) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x \to 0} f(x)$.
$f(0) = \lim_{x \to 0} \frac{2^{x}-1}{1-3^{x}}$
$f(0) = \lim_{x \to 0} \frac{2^{x}-1}{-(3^{x}-1)}$
$f(0) = -\frac{\lim_{x \to 0} \frac{2^{x}-1}{x}}{\lim_{x \to 0} \frac{3^{x}-1}{x}}$
Using the standard limit $\lim_{x \to 0} \frac{a^{x}-1}{x} = \log a$,we get:
$f(0) = -\frac{\log 2}{\log 3}$

(B) The feasible region is the unbounded region defined by the constraints $5x + y \geq 5$,$x + y \geq 3$,$x \geq 0$,and $y \geq 0$.
To find the corner points,we solve the equations of the lines:
$1$) $5x + y = 5$ and $x + y = 3$. Subtracting the second from the first gives $4x = 2$,so $x = 0.5$. Substituting into $x + y = 3$ gives $y = 2.5$. Thus,point $P = (0.5, 2.5)$.
$2$) The intersection of $x + y = 3$ with the $x$-axis $(y=0)$ is $C = (3, 0)$.
$3$) The intersection of $5x + y = 5$ with the $y$-axis $(x=0)$ is $B = (0, 5)$.
We evaluate $Z = 7x + y$ at these corner points:
At $C(3, 0)$: $Z = 7(3) + 0 = 21$.
At $P(0.5, 2.5)$: $Z = 7(0.5) + 2.5 = 3.5 + 2.5 = 6$.
At $B(0, 5)$: $Z = 7(0) + 5 = 5$.
Since the feasible region is unbounded,we check if the minimum value $5$ is attained. The value $5$ is the minimum among the corner points. Testing a point in the region,e.g.,$(1, 3)$,$Z = 7(1) + 3 = 10 > 5$. Thus,the minimum value is $5$.

(D) To find the feasible region,we first determine the intercepts of the boundary lines:

$\text{Line}$	$\text{Intercepts}$
$8x + 5y = 60$	$A(7.5, 0), B(0, 12)$
$4x + 5y = 40$	$C(10, 0), D(0, 8)$

Solving the equations $8x + 5y = 60$ and $4x + 5y = 40$ simultaneously:
Subtracting the second from the first: $(8x - 4x) = 60 - 40 \Rightarrow 4x = 20 \Rightarrow x = 5$.
Substituting $x = 5$ into $4x + 5y = 40$: $4(5) + 5y = 40 \Rightarrow 20 + 5y = 40 \Rightarrow 5y = 20 \Rightarrow y = 4$.
So,the intersection point is $E(5, 4)$.
The constraints $x \geq 0, y \geq 0$ restrict the region to the first quadrant.
The feasible region is bounded by the vertices $O(0, 0)$,$A(7.5, 0)$,$E(5, 4)$,and $D(0, 8)$.
Since there are four vertices,the feasible region is a quadrilateral.

(B) To find the solution,we first identify the feasible region defined by the constraints:
$1$. $x+y \leq 30$
$2$. $x \leq 15$
$3$. $y \leq 20$
$4$. $x+y \geq 15$
$5$. $x, y \geq 0$
The vertices of the feasible region are determined by the intersection of these lines:
- Intersection of $x=15$ and $y=20$ is $E(15, 20)$.
- Intersection of $x=0$ and $y=20$ is $D(0, 20)$.
- Intersection of $x=0$ and $x+y=15$ is $F(0, 15)$.
- Intersection of $x=15$ and $x+y=15$ is $C(15, 0)$.
The feasible region is the quadrilateral $CDEF$.
We evaluate the objective function $z=x+y$ at these vertices:
- At $C(15, 0): z = 15+0 = 15$
- At $D(0, 20): z = 0+20 = 20$
- At $E(15, 20): z = 15+20 = 35$
- At $F(0, 15): z = 0+15 = 15$
The maximum value of $z$ is $35$,which occurs at the unique vertex $E(15, 20)$. Therefore,the $L$.$P$.$P$. has a unique solution.

(D) $1$. The line $5x + 9y = 90$ passes through $(18, 0)$ and $(0, 10)$. The shaded region is towards the origin,so the constraint is $5x + 9y \leq 90$.
$2$. The line $x + y = 4$ passes through $(4, 0)$ and $(0, 4)$. The shaded region is away from the origin,so the constraint is $x + y \geq 4$.
$3$. The line $y = 8$ is a horizontal line. The shaded region is below this line,so the constraint is $y \leq 8$.
$4$. Since the region is in the first quadrant,$x \geq 0$ and $y \geq 0$.
Combining these,the constraints are $5x + 9y \leq 90, x + y \geq 4, y \leq 8, x, y \geq 0$.

(C) To find the maximum value of $Z=3x+5y$,we first identify the feasible region defined by the constraints $x+4y \leq 24$,$y \leq 4$,$x \geq 0$,and $y \geq 0$.
The vertices of the feasible region are $O(0,0)$,$A(24,0)$,$D(8,4)$,and $C(0,4)$.
We evaluate the objective function $Z=3x+5y$ at each vertex:
$1$. At $O(0,0)$: $Z = 3(0) + 5(0) = 0$
$2$. At $A(24,0)$: $Z = 3(24) + 5(0) = 72$
$3$. At $D(8,4)$: $Z = 3(8) + 5(4) = 24 + 20 = 44$
$4$. At $C(0,4)$: $Z = 3(0) + 5(4) = 20$
Comparing these values,the maximum value of $Z$ is $72$ at point $A(24,0)$.

(D) Given the objective function $Z=10 x+25 y$ subject to the constraints $0 \leq x \leq 3, 0 \leq y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$.
The feasible region is bounded by the vertices $O(0,0), C(3,0), F(3,2), G(2,3), D(0,3)$.
We evaluate $Z$ at each corner point:
$1$. At $O(0,0): Z = 10(0) + 25(0) = 0$
$2$. At $C(3,0): Z = 10(3) + 25(0) = 30$
$3$. At $F(3,2): Z = 10(3) + 25(2) = 30 + 50 = 80$
$4$. At $G(2,3): Z = 10(2) + 25(3) = 20 + 75 = 95$
$5$. At $D(0,3): Z = 10(0) + 25(3) = 75$
The maximum value of $Z$ is $95$ at the point $(2,3)$.