MHT CET 2024 9th May Morning Shift (Mathematics) Question Paper with Solutions

$\begin{array}{cc}& 𝐵={𝐴}^{-1}\\ & ⎡⎣⎢\begin{array}{ccc}-2& 0& 𝑏\\ 7& -1& -2\\ 𝑐& 1& 1\end{array}⎤⎦⎥={𝐴}^{-1}\\ & ⎡⎣⎢\begin{array}{ccc}-2& 0& 𝑏\\ 7& -1& -2\\ 𝑐& 1& 1\end{array}⎤⎦⎥𝐴={𝐴}^{-1}𝐴\\ & ⎡⎣⎢\begin{array}{ccc}-2& 0& 𝑏\\ 7& -1& -2\\ 𝑐& 1& 1\end{array}⎤⎦⎥⎡⎣⎢\begin{array}{ccc}1& 1& 1\\ 1& 𝑎& 3\\ 3& 2& 2\end{array}⎤⎦⎥=𝐼\\ & ⎡⎣⎢\begin{array}{ccc}-2+3𝑏& -2+2𝑏& -2+2𝑏\\ 7-1-6& 7-𝑎-4& 7-3-4\\ 𝑐+1+3& 𝑐+𝑎+2& 𝑐+3+2\end{array}⎤⎦⎥=⎡⎣⎢\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}⎤⎦⎥\end{array}$

On equality of matrix,

$\begin{array}{cc}& -2+3𝑏=1,7-𝑎-4=1,𝑐+3+2=1\\ \Rightarrow 𝑏=1,𝑎& =2,𝑐=-4\\ \therefore 4𝑎+2𝑏-𝑐& =4\left(2\right)+2\left(1\right)+4\\ & =8+2+4=14\end{array}$

$\int f\left(𝑥\right)\mathrm{dx}=𝜓\left(𝑥\right)$ ...[Given]

$\begin{array}{cc}& \text{ Consider }\\ & \begin{array}{cc}& I=\int {𝑥}^{5}f\left({𝑥}^3\right)d𝑥\\ & =\int {𝑥}^3\cdot {𝑥}^{2}f\left({𝑥}^3\right)d𝑥\\ & \text{ Let }{𝑥}^3=t\\ & \Rightarrow 3{𝑥}^{2}d𝑥=\mathrm{dt}\\ & \therefore I=\int \frac{t}{3}f\left(t\right)\mathrm{dt}\\ & =\frac{1}{3}\int tf\left(t\right)\mathrm{dt}\\ & =\frac{1}{3}[t\int f(t)\mathrm{dt}-\int \frac{d}{\mathrm{dt}}(t)\cdot \int f(t\left)\mathrm{dt}\right]\\ & =\frac{1}{3}[t\cdot 𝜓(t)-\int 𝜓(t\left)\mathrm{dt}\right]+c\\ & =\frac{1}{3}\left({𝑥}^3𝜓\left({𝑥}^3\right)\right)-\int 𝜓\left({𝑥}^3\right){𝑥}^{2}d𝑥+c\end{array}\end{array}$

Let p be the population at time $𝑡$ years.

$\begin{array}{cc}& \text{ Then }\frac{𝑑𝑝}{𝑑𝑡}=𝑘𝑝\\ & \Rightarrow \frac{𝑑𝑝}{𝑝}=𝑘𝑑𝑡\end{array}$

Integrating on both sides, we get

$\log p=\mathrm{kt}+c$

when $𝑡=0,𝑝=40000$

$\begin{array}{cc}& \therefore \log 40000=0+𝑐\\ & \Rightarrow 𝑐=\log 40000\end{array}$

$\begin{array}{cc}\therefore & \log p=\mathrm{kt}+\log 40000\\ & \Rightarrow \log \left(\frac{p}{40000}\right)=\mathrm{kt}\text{.... (i)}\end{array}$

When $t=40$ years, $p=80000$

$\begin{array}{cc}& \Rightarrow \log \left(\frac{80000}{40000}\right)=40k\\ & \Rightarrow k=\frac{1}{40}\log 2\\ \therefore & \log \left(\frac{p}{40000}\right)=\frac{1}{40}\log 2\times t\text{.... [From (i)]}\end{array}$

$\therefore$ Population after another 40 years, i.e., $t=80$ years, we have

$\begin{array}{cc}& \log \left(\frac{𝑝}{40000}\right)=\frac{1}{40}\log 2\times 80\\ & \Rightarrow \log \frac{𝑝}{40000}=2\log 2\\ & \Rightarrow \frac{𝑝}{40000}=4\\ & \Rightarrow 𝑝=16,0000\end{array}$

The sum of all the probabilities in a probability distribution is always unity.

$\begin{array}{cc}& \therefore k+3k+3k+k=1\\ & \Rightarrow 8k=1\\ & \Rightarrow 𝑘=\frac{1}{8}\\ & E\left(X\right)=\sum {𝑥}_1\cdot P\left({𝑥}_1\right)\\ & =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right)=\frac{3}{2}\\ & \mathrm{Var}\left(X\right)=E\left(X^2\right)-\left[𝐸\right(X){]}^2\\ & =0^2\left(\frac{1}{8}\right)+1^2\left(\frac{3}{8}\right)+2^2\left(\frac{3}{8}\right)+3^2\left(\frac{1}{8}\right)-{\left(\frac{3}{2}\right)}^2\\ & =\frac{3}{4}\end{array}$

$\begin{array}{cc}& \text{ Required probability }=\frac{{}^3{𝐶}_1\times {}^4{𝐶}_1\times {}^2{𝐶}_2}{{}^9{𝐶}_3}\\ & =\frac{3\times 4\times 2}{\left(\frac{9!}{3!6!}\right)}=\frac{24\times 6}{9\times 8\times 7}=\frac{2}{7}\end{array}$

$\begin{array}{cc}\text{ Required area }& =2{\int }_{\frac{1}{2}}^0\left[\right(𝑥-1{)}^2-\frac{1}{4}]d𝑥\\ & =2{\left[\frac{(𝑥-1{)}^3}{3}\right]}_{\frac{1}{2}}^0-\frac{1}{2}[𝑥{]}_{\frac{1}{2}}^0\\ & =\frac{2}{3}(-\frac{1}{8}+1)-\frac{1}{2}(\frac{1}{2}-0)\\ & =\frac{1}{3}\text{ sq. units }\end{array}$

$\begin{array}{cc}\text{ Let }I& =\int \frac{d𝑥}{\text{exception 25:}}\\ & =\int \frac{d𝑥}{{\sin }^{\frac{11}{3}}𝑥\cdot {\cos }^{\frac{1}{3}}𝑥}\\ & =\int \frac{{\sec }^4𝑥}{{\tan }^{\frac{11}{3}}𝑥}\mathrm{dx}\end{array}$

... [Dividing numerator and denominator by ${\cos }^{\frac{11}{3}}𝑥]$

$=\int \frac{(1+{\tan }^2𝑥){\sec }^2𝑥}{{\tan }^{\frac{11}{3}}𝑥}𝑑𝑥$

$\begin{array}{cc}& \text{ Let }\tan 𝑥=t\\ & {\sec }^2𝑥d𝑥=\mathrm{dt}\\ & \begin{array}{cc}I& =\int \frac{(1+t^2)\mathrm{dt}}{t^{\frac{11}{3}}}\\ & =\int t^{\frac{-11}{3}}\mathrm{dt}+\int t^{\frac{-5}{3}}\mathrm{dt}\\ & =\frac{-3}{8}{t}^{\frac{-8}{3}}-\frac{3}{2}{t}^{\frac{-2}{3}}+c\\ & =-[\frac{3}{8}{t}^{\frac{-8}{3}}+\frac{3}{2}{t}^{\frac{-2}{3}}]+c\\ & =-[\frac{3}{8}{\tan }^{\frac{-8}{3}}𝑥+\frac{3}{2}{\tan }^{\frac{-2}{3}}𝑥]+c\\ \therefore f\left(𝑥\right)& ={\tan }^{\frac{-8}{3}}𝑥,g\left(𝑥\right)={\tan }^{\frac{-2}{3}}𝑥\end{array}\end{array}$

Here, $\frac{1+3𝑝}{4},\frac{1-𝑝}{4},\frac{1+2𝑝}{4}$ and $\frac{1-4𝑝}{4}$ are probabilities when X takes values $-1,0,1$ and 2 respectively. Therefore, each is greater than or equal to 0 and less than or equal to 1.

$\begin{array}{cc}& \text{ i.e., }0\le \frac{1+3𝑝}{4}\le 1,0\le \frac{1-𝑝}{4}\le 1\\ & 0\le \frac{1+2𝑝}{4}\le 1\text{ and }0\le \frac{1-4𝑝}{4}\le 1\\ & \Rightarrow -\frac{1}{3}\le 𝑝\le \frac{1}{4}\end{array}$

$\begin{array}{cc}\mathrm{Mean}\left(𝑋\right)=-1\times \frac{1+3𝑝}{4}+0\times \frac{1-𝑝}{4}& +1\times \frac{1+2𝑝}{4}\\ & +2\times \frac{1-4𝑝}{4}\end{array}$

$\begin{array}{cc}& =\frac{2-9𝑝}{4}\\ & \text{ Now, }-\frac{1}{3}\le p\le \frac{1}{4}\\ & \Rightarrow 3\ge -9𝑝\ge -\frac{9}{4}\\ & \Rightarrow -\frac{1}{4}\le 2-9p\le 5\\ & \Rightarrow -\frac{1}{16}\le \frac{2-9𝑝}{4}\le \frac{5}{4}\end{array}$

$\begin{array}{cc}& 𝑦\left(𝑥\right)=1+\sqrt{\surd 4𝑥-3}\\ \therefore & \frac{d𝑦}{d𝑥}=\frac{4}{2\sqrt{\surd 4𝑥-3}}=\frac{2}{\sqrt{\surd 4𝑥-3}}\\ & \Rightarrow \frac{2}{\sqrt{\surd 4𝑥-3}}=\frac{2}{3}\\ & \Rightarrow 𝑥=3\\ \therefore & 𝑦=4\end{array}$

$\therefore$ Equation of normal is

$\begin{array}{cc}& 𝑦-4=\frac{-3}{2}(𝑥-3)\\ & \Rightarrow 2𝑦-8=-3𝑥+9\\ & \Rightarrow 3𝑥+2𝑦-17=0\end{array}$

$\therefore$ Option (A) i.e., $(1,7)$ satisfies above equation.

Center of circle ${𝑥}^2+{𝑦}^2+6𝑥-14𝑦+5=0$ is $(-3,7)$ and centre of circle ${𝑥}^2+{𝑦}^2-4𝑥+10𝑦-4=0$ is $(2,-5)$

$\therefore$ Equation of the required circle is

$\begin{array}{cc}& (𝑥+3)(𝑥-2)+(𝑦-7)(𝑦+5)=0\\ & \Rightarrow {𝑥}^2+{𝑦}^2+𝑥-2𝑦-41=0\end{array}$

The equation of plane passing through $(1,1,1)$ is

$𝑎(𝑥-1)+𝑏(𝑦-1)+𝑐(𝑧-1)=0\text{... (i)}$

Since plane (i) is perpendicular to the planes

$2𝑥+𝑦-2𝑧=5\text{ and }3𝑥-6𝑦-2𝑧=7$

$\begin{array}{cc}\therefore & 2𝑎+𝑏-2𝑐=5\text{.... (ii)}\\ & 3𝑎-6𝑏-2𝑐=7\text{.... (iii)}\end{array}$

On solving (i), (ii) and (iii), we get

$𝑎=14,𝑏=2,𝑐=15$

Substituting the values of $𝑎,𝑏,𝑐$ in (i), we get

$\begin{array}{cc}& 14(𝑥-1)+2(𝑦-1)+15(𝑧-1)=0\\ & \Rightarrow 14𝑥+2𝑦+15𝑧=31\end{array}$

Let $𝑦={\tan }^{-1}\left(\frac{6𝑥\sqrt{\surd 𝑥}}{1-9{𝑥}^3}\right)$

$\begin{array}{cc}& ={\tan }^{-1}\left(\frac{3𝑥\sqrt{\surd 𝑥}+3𝑥\sqrt{\surd 𝑥}}{1-3𝑥\sqrt{\surd 𝑥}\cdot 3𝑥\sqrt{\surd 𝑥}}\right)\\ & ={\tan }^{-1}\left(3𝑥\sqrt{\surd 𝑥}\right)+{\tan }^{-1}3𝑥\sqrt{\surd 𝑥}\end{array}$

$\begin{array}{cc}& 𝑦=2{\tan }^{-1}(3𝑥\cdot \sqrt{\surd 𝑥})\\ & \text{ Differentiating w.r.t. }𝑥\text{, we get }\\ & \begin{array}{cc}\frac{𝑑𝑦}{𝑑𝑥}& =\frac{2}{1+(3𝑥\sqrt{\surd 𝑥}{)}^2}\cdot \frac{𝑑}{𝑑𝑥}(3𝑥\cdot \sqrt{\surd 𝑥})\\ & =\frac{2}{1+9{𝑥}^3}\cdot \frac{9}{2}\sqrt{\surd 𝑥}\\ \frac{𝑑𝑦}{𝑑𝑥}& =\frac{9\sqrt{\surd 𝑥}}{1+9{𝑥}^3}\end{array}\end{array}$

$\begin{array}{cc}& \therefore \text{ Comparing with }\frac{d𝑦}{d𝑥}=\sqrt{\surd 𝑥}\cdot g\left(𝑥\right)\\ & \therefore 𝑔\left(𝑥\right)=\frac{9}{1+9{𝑥}^3}\end{array}$

Since $\mid \begin{array}{ccc}𝑎& 1& 1\\ 1& 𝑏& 1\\ 1& 1& 𝑐\end{array}\mid =0$ Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we get

$\begin{array}{cc}& \mid \begin{array}{ccc}𝑎& 1& 1\\ 1-𝑎& 𝑏-1& 0\\ 1-𝑎& 0& 𝑐-1\end{array}\mid =0\\ & \Rightarrow 𝑎(𝑏-1)(𝑐-1)-(1-𝑎)(𝑐-1)-(1-𝑎)(𝑏-1)=0\end{array}$

Dividing by $(1-𝑎)(1-𝑏)(1-𝑐)$,

we get $\frac{𝑎}{1-𝑎}+\frac{1}{1-𝑏}+\frac{1}{1-𝑐}=0\text{.... (i)}$

Consider, $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$

$\begin{array}{cc}& =\frac{1}{1-𝑎}-\frac{𝑎}{1-𝑎}\text{.... [From (i)]}\\ & =1\end{array}$

$a=\sin 𝜃,b=\cos 𝜃\text{ and }c=\sqrt{\surd 1+\sin 𝜃\cos 𝜃}$

Since $\sqrt{\surd 1+\sin 𝜃\cos 𝜃}$ is greater than $\sin 𝜃$ and $\cos 𝜃$.

$\therefore C$ is the greatest angle,

$\begin{array}{cc}\therefore \cos 𝐶& =\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}\\ & =\frac{{\sin }^2𝜃+{\cos }^2𝜃-(1+\sin 𝜃\cos 𝜃)}{2\sin 𝜃\cos 𝜃}\\ & =-\frac{1}{2}=\cos {120}^{\circ }\\ \therefore 𝐶=& {120}^{\circ }=\frac{2𝜋}{3}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \underset{𝑥\to \infty }{lim}(\frac{{𝑥}^2+𝑥+1}{𝑥+1}-\mathrm{ax}-b)=4\\ & \Rightarrow \underset{𝑥\to \infty }{lim}\left(\frac{{𝑥}^2+𝑥+1-\mathrm{ax}(𝑥+1)-b(𝑥+1)}{𝑥+1}\right)=4\\ & \Rightarrow \underset{𝑥\to \infty }{lim}\left[\frac{(1-a){𝑥}^2+(1-a-b)𝑥+(1-b)}{𝑥+1}\right]=4\end{array}\\ & \text{ Limit is finite, it exists when }(1-𝑎)=0\\ & \begin{array}{cc}\therefore & 1-𝑎=0\\ & \Rightarrow 𝑎=1\\ \therefore & \underset{𝑥\to \infty }{lim}\left[\frac{1-𝑎-𝑏+\frac{(1-𝑏)}{𝑥}}{1+\frac{1}{𝑥}}\right]=4\\ & \Rightarrow 1-𝑎-𝑏=4\\ & \Rightarrow 𝑏=-4\\ \therefore & 𝑎=1,𝑏=-4\end{array}\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& 𝑥\frac{𝑑𝑦}{𝑑𝑥}+2𝑦={𝑥}^2\\ & \frac{𝑑𝑦}{𝑑𝑥}+\left(\frac{2}{𝑥}\right)𝑦=𝑥\\ & \text{ I.F. }=e^{\int \frac{2}{𝑥}\mathrm{dx}}=e^{2\log 𝑥}={𝑥}^2\end{array}\\ & \therefore \text{ Solution of differential equation is }\end{array}$

$\begin{array}{cc}& 𝑦\cdot {𝑥}^2=\int 𝑥\cdot {𝑥}^{2}d𝑥+c\\ & 𝑦{𝑥}^2=\frac{{𝑥}^4}{4}+c\text{... (i)}\\ \therefore & 𝑦=\frac{{𝑥}^2}{4}+\frac{c}{{𝑥}^2}\text{... (ii)}\end{array}$

Given, $𝑦\left(1\right)=1$

$\Rightarrow 1=\frac{1}{4}+c$

$\Rightarrow c=\frac{3}{4}$

$\therefore$ equation (i) becomes,

$\begin{array}{cc}& 𝑦=\frac{{𝑥}^2}{4}+\frac{3}{4{𝑥}^2}\\ \therefore & 𝑦\left(\frac{1}{2}\right)=\frac{{\left(\frac{1}{2}\right)}^2}{4}+\frac{3}{4\times {\left(\frac{1}{2}\right)}^2}=\frac{1}{16}+3=\frac{49}{16}\end{array}$

$\begin{array}{cc}& {\cos }^{-1}\left\{\frac{1}{\sqrt{\surd 2}}[\cos \left(\frac{9𝜋}{10}\right)-\sin \left(\frac{9𝜋}{10}\right)]\right\}\\ & ={\cos }^{-1}\left\{\frac{1}{\sqrt{\surd 2}}(\cos (𝜋-\frac{𝜋}{10})-\sin (𝜋-\frac{𝜋}{10}))\right\}\\ & ={\cos }^{-1}\left\{\frac{1}{\sqrt{\surd 2}}[-\cos \left(\frac{𝜋}{10}\right)-\sin \left(\frac{𝜋}{10}\right)]\right\}\end{array}$

$\begin{array}{cc}& ={\cos }^{-1}\left\{\right(-1\left)[\frac{1}{\sqrt{\surd 2}}\cos \left(\frac{𝜋}{10}\right)+\frac{1}{\sqrt{\surd 2}}\sin \left(\frac{𝜋}{10}\right)]\right\}\\ & ={\cos }^{-1}\left\{\right(-1\left)[\cos \frac{𝜋}{4}\cos \frac{𝜋}{10}+\sin \frac{𝜋}{4}\sin \frac{𝜋}{10}]\right\}\\ & ={\cos }^{-1}\left\{\right(-1\left)[\cos (\frac{𝜋}{4}-\frac{𝜋}{10})]\right\}\\ & ={\cos }^{-1}\left\{\right(-1\left)[\cos \left(\frac{3𝜋}{20}\right)]\right\}\\ & =𝜋-{\cos }^{-1}(\cos \frac{3𝜋}{20})\\ & =𝜋-\frac{3𝜋}{20}\\ & =\frac{17𝜋}{20}\end{array}$

$\begin{array}{c}\text{(A) Let }𝑝:3+3=7\\ 𝑞:4+3=8\end{array}$

Truth value of $𝑝$ and $𝑞$ are $𝐹$ and $𝐹$.

$\therefore p\to q\equiv T$

(B) Let $p:5+3=8$,

q ; earth is flat

Truth value of $𝑝$ and $𝑞$ are $𝑇$ and $𝐹$.

$\therefore p\to q\equiv F$

(C) Let p : Both (A) and (B) are true

$q:5+6=17$

Truth values of $𝑝,𝑞$ are false and false.

$\therefore p\to q\equiv T$

$\therefore \left(A\right)$ and $\left(C\right)$ are true, while $\left(B\right)$ is false.

$\begin{array}{cc}& \text{ Given, }𝛼=𝛽+𝛾\\ & \begin{array}{cc}& \therefore 𝛾=𝛼-𝛽\\ & \tan 𝛾=\tan (𝛼-𝛽)\\ & =\frac{\tan 𝛼-\tan 𝛽}{1+\tan 𝛼\cdot \tan 𝛽}\end{array}\end{array}$

$=\frac{\tan 𝛼-\tan 𝛽}{1+\tan 𝛼\cdot \tan (\frac{𝜋}{2}-𝛼)}\cdots \left[\begin{array}{c}𝛼+𝛽=\frac{𝜋}{2}\\ \therefore 𝛽=\frac{𝜋}{2}-𝛼\end{array}\right]$

$\begin{array}{cc}& \frac{\tan 𝛼-\tan 𝛽}{1+\tan 𝛼\cot 𝛼}\\ & =\frac{\tan 𝛼-\tan 𝛽}{2}\\ \Rightarrow 2& \tan 𝛾=\tan 𝛼-\tan 𝛽\\ \Rightarrow \tan 𝛼& =2\tan 𝛾+\tan 𝛽\end{array}$

To determine where the function $𝑓\left(𝑥\right)=\frac{\log 𝑥}{𝑥}$ is increasing, we need to examine its derivative and determine the intervals where the derivative is positive.

First, let's find the first derivative of the function using the quotient rule. The quotient rule states:

${\left(\frac{𝑢}{𝑣}\right)}^{\prime }=\frac{{𝑢}^{\prime }𝑣-𝑢{𝑣}^{\prime }}{{𝑣}^2}$

where $𝑢=\log 𝑥$ and $𝑣=𝑥$.

Calculating the derivatives of $𝑢$ and $𝑣$:

${𝑢}^{\prime }=\frac{1}{𝑥}$

${𝑣}^{\prime }=1$

Applying the quotient rule:

${𝑓}^{\prime }\left(𝑥\right)=\frac{\frac{1}{𝑥}\cdot 𝑥-\log 𝑥\cdot 1}{{𝑥}^2}=\frac{1-\log 𝑥}{{𝑥}^2}$

To find where $𝑓\left(𝑥\right)$ is increasing, we set ${𝑓}^{\prime }\left(𝑥\right)>0$:

$\frac{1-\log 𝑥}{{𝑥}^2}>0$

Since ${𝑥}^2>0$ for $𝑥>0$, the inequality $1-\log 𝑥>0$ simplifies to:

$1>\log 𝑥$

Equivalently,

$\log 𝑥<1⟹𝑥<𝑒$

Thus, the function $𝑓\left(𝑥\right)$ is increasing in the interval $(0,𝑒)$.

Therefore, the correct option is:

Option A: $(0,𝑒)$

$\begin{array}{cc}& \text{ (D) Std.12|Part-1|Ch-5| Exercise-5.5 }\\ & [\bar{a}+\bar{b}+\bar{c}]\cdot \left[\right(\bar{a}+\bar{b})\times (\bar{a}+\bar{c}\left)\right]\\ & =(\bar{a}+\bar{b})\cdot \left[\right(\bar{a}+\bar{b})\times (\bar{a}+\bar{c}\left)\right]+\bar{c}\cdot \left[\right(\bar{a}+\bar{b})\times (\bar{a}+\bar{c}\left)\right]\\ & =0+[\bar{c}\bar{a}+\bar{b}\bar{a}+\bar{c}]\\ & =[\bar{c}\bar{a}\bar{a}+\bar{c}]+[\bar{c}\bar{b}\bar{a}+\bar{c}]\\ & =\left[\bar{c}\bar{a}\bar{a}\right]+\left[\bar{c}\bar{c}\bar{c}\bar{c}\right]+\left[\bar{c}\bar{b}\bar{a}\right]+\left[\bar{c}\bar{b}\bar{c}\right]\\ & =0+0+\left[\bar{c}\bar{b}\bar{a}\right]+0\\ & =-\left[\bar{a}\bar{b}\bar{c}\right]\end{array}$

To find the number of newspapers, we can use the relationship between the total number of readings, the number of students, the number of newspapers, and the conditions given:

Each of the 300 students reads 5 newspapers, which results in:

$\text{Total Readings}=300\times 5=1500$

Each newspaper is read by 60 students. Let $𝑛$ be the number of newspapers, thus:

$\text{Total Readings}=𝑛\times 60$

Setting these two expressions for total readings equal, we have:

$300\times 5=𝑛\times 60$

Simplifying the equation:

$1500=60𝑛$

Dividing both sides by 60:

$𝑛=\frac{1500}{60}=25$

Thus, the number of newspapers is exactly 25.

Option C: exactly 25 is the correct answer.

Let $f\left(𝑥\right)=\frac{\log 𝑥}{𝑥}\Rightarrow f^{\prime }\left(𝑥\right)=\frac{1}{{𝑥}^2}-\frac{\log 𝑥}{{𝑥}^2}$