Maths MCQ

Hint: $y=\alpha \log x+\beta x^2+x$

$\therefore \frac{d y}{d x}=\frac{\alpha}{x}+\beta \times 2 x+1=\frac{\alpha}{x}+2 \beta x+1$

f(x) has extreme values at x = -1 and x = 2 ∴ f'(-1) = 0 and f(2) = 0 α + 2β = 1

and $\frac{\alpha}{2}+4 \beta=-1$

By solving these two equations, we get

$\alpha=2, \beta=\frac{-1}{2}$

$
\begin{aligned}
& \text {(a) : } \frac{d v}{d t}=-72 \pi m ^3 / min , v_0=4500 \pi \\
& v=\frac{4}{3} \pi r^3 \therefore \frac{d v}{d t}=\frac{4}{3} \pi \times 3 r^2 \times \frac{d r}{d t}
\end{aligned}
$
After $49 \min , v=v_0+49 \cdot \frac{d v}{d t}=4500 \pi-49 \times 72 \pi$ $=4500 \pi-3528 \pi=972 \pi$
$\Rightarrow \quad 972 \pi=\frac{4}{3} \pi r^3 \Rightarrow r^3=243 \times 3=3^6 \Rightarrow r=9$
$\therefore-72 \pi=4 \pi \times 81 \times \frac{d r}{d t} \Rightarrow \frac{d r}{d t}=-\frac{18}{81}=-\frac{2}{9}$
Thus, radius decreases at a rate of $\frac{2}{9} m / min$

(a) : According to lagrange's mean value theorem for any function $f(x), a<x<b$, there exist $c \in(a, b)$ such that
$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
$
Here, $a=1, b=5$ and $f(x)=\sqrt{25-x^2}$
$
\therefore f(5)=\sqrt{25-25=0}
$
And $f(1)=\sqrt{25-1}=\sqrt{24}=2 \sqrt{6}$
$
\begin{aligned}
& \therefore f^{\prime}(c)=\frac{0-2 \sqrt{6}}{5-1}, c \in(1,5)=\frac{-2 \sqrt{6}}{4} \\
& \Rightarrow f^{\prime}(c)=-\frac{\sqrt{6}}{2} ...(i)
\end{aligned}
$
Now, $f(x)=\sqrt{25-x^2}$
$
\Rightarrow f^{\prime}(x)=\frac{1}{2 \sqrt{25-x^2}}(-2 x)=\frac{-x}{\sqrt{25-x^2}}
$
$
\Rightarrow f^{\prime}(c)=\frac{-c}{\sqrt{25-c^2}} ...(ii)
$
On equating (i) and (ii), we get
$
\begin{aligned}
& \frac{-c}{\sqrt{25-c^2}}=\frac{-\sqrt{6}}{2} \\
& \Rightarrow 2 c=\sqrt{150-6 c^2} \\
& \Rightarrow 4 c^2=150-6 c^2 \Rightarrow 10 c^2=150 \\
& \Rightarrow c^2=15 \Rightarrow c= \pm \sqrt{15}
\end{aligned}
$
Since, $c \in(1,5)$, So, $c=\sqrt{15}$

 (d) : We have, $y=\log _e x$ and point $P(1,0)$.
$
\therefore \frac{d y}{d x}=\frac{1}{x} \Rightarrow\left[\frac{d y}{d x}\right]_{(1,0)}=1
$
$\therefore$ Equation of normal is $y-0=-1(x-1)$
$\Rightarrow y=1-x \Rightarrow x+y=1$

(a) : $f(x)=x^3+5 x^2-7 x+9$
Differentiate w.r.t. $x$, we get $f^{\prime}(x)=3 x^2+10 x-7$
$
\begin{aligned}
& a+h=1.1=1+0.1 \Rightarrow a=1, h=0.1 \\
\because \quad & f(a+h) \approx f(a)+h f^{\prime}(a) \Rightarrow f(1.1) \approx f(1)+(0.1) f^{\prime}(1) \\
= & {\left[(1)^3+5(1)^2-7(1)+9\right]+(0.1)\left[3(1)^2+10(1)-7\right] } \\
= & (1+5-7+9)+(0.1)(3+10-7)=8+0.6=8.6
\end{aligned}
$