Question

Answer the following by appropriately matching the lists based on the information in Column I and Column II​​​​​​
$\begin{array}{cc}ColumnI& ColumnII\\ \text{a}.y=f\left(x\right)\text{is given by}& \\ x=t^5-5t^3-20t+7& \\ \text{and}y=4t^3-3t^2-18t+3.& \\ \text{Then}-5\times \frac{dy}{dx}\text{at}t=1& \text{p.}0\\ \text{b}.\text{Let}P\left(x\right)\text{be a polynomial of degree}4,& \\ \text{with}P\left(2\right)=-1,P^{\prime }\left(2\right)=0,& \\ P^{\prime \prime }\left(2\right)=2,P^{\prime \prime \prime }\left(2\right)=-12\text{and}& \\ P^{iv}\left(2\right)=24,\text{then}{P}^{\prime \prime }\left(3\right)\text{is}& \text{q.}-2\\ \text{c}.y=\frac{1}{x},\text{then}\frac{\frac{dy}{\sqrt{1+y^4}}}{\frac{dx}{\sqrt{1+x^4}}}& r.2\\ \text{d}.f\left(\frac{2x+3y}{5}\right)=\frac{2f\left(x\right)+3f\left(y\right)}{5}\text{and}& \\ f^{\prime }\left(0\right)=p\text{and}f\left(0\right)=q.\text{Then},f^{\prime \prime }\left(0\right)\text{is}& \text{s.}-1\end{array}$

• A. $a\to p,b\to r,c\to s,d\to q$
• B. $a\to q,b\to r,c\to s,d\to p$
• C. $a\to q,b\to s,c\to r,d\to p$
• D. $a\to q,b\to r,c\to p,d\to s$

Answer 1

The correct option is B $a\to q,b\to r,c\to s,d\to p$

a.
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{12t^2-6t-18}{5t^4-15t^2-20}$
Putting $t=1$
$\Rightarrow \frac{dy}{dx}=\frac{12-6-18}{5-15-20}=\frac{2}{5}\Rightarrow -5\times \frac{dy}{dx}=-2$
$a\to q$

b.
$P\left(x\right)=a(x-2{)}^4+b(x-2{)}^3+c(x-2{)}^2+d(x-2)+e$
Given
$P\left(2\right)=-1\Rightarrow e=-1$
$P^{\prime }\left(2\right)=0\Rightarrow d=0P^{\prime \prime }\left(2\right)=2\Rightarrow 2c=2\Rightarrow c=1P^{\prime \prime \prime }\left(2\right)=-12\Rightarrow 6b=-12\Rightarrow b=-2P^{iv}\left(2\right)=24\Rightarrow 24a=24\Rightarrow a=1$
So,
$P^{\prime \prime }\left(x\right)=12a(x-2{)}^2+6b(x-2)+2c\Rightarrow P^{\prime \prime }(x)=12(x-2{)}^2-12(x-2)+2\Rightarrow P^{\prime \prime }\left(3\right)=12-12+2=2$
$b\to r$

c.
$\sqrt{1+y^4}=\sqrt{1+\frac{1}{x^4}}[\because y=\frac{1}{x}]$
$\Rightarrow \sqrt{1+y^4}=\frac{\sqrt{1+x^4}}{x^2}\Rightarrow \frac{\sqrt{1+y^4}}{\sqrt{1+x^4}}=\frac{1}{x^2}\cdots \left(1\right)$
Now,
$y=\frac{1}{x}\Rightarrow \frac{dy}{dx}=\frac{-1}{x^2}$
Using equation (1),
$\Rightarrow \frac{dy}{dx}=-\frac{\sqrt{1+y^4}}{\sqrt{1+x^4}}\Rightarrow \frac{\frac{dy}{\sqrt{1+y^4}}}{\frac{dx}{\sqrt{1+x^4}}}=-1$
$c\to s$

d.
Obviously, $f\left(x\right)$ is a linear function,
Also $f^{\prime }\left(0\right)=p,f\left(0\right)=q$
$f\left(x\right)=px+q\Rightarrow f^{\prime \prime }\left(x\right)=0$
$d\to p$