Exercise 11.4-Additional Questions - Chapter 11 Integral Calculus 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
If $f^{\prime}(x)=2 x-7$ and $f(1)=0$ find $f(x)$
Solution:
Given $\mathrm{f}^{\prime}(x)=2 \mathrm{x}-7$
$\Rightarrow \mathrm{f}(\mathrm{x})=\int(2 x-7) d x$
$=\mathrm{x}^2-7 \mathrm{x}+\mathrm{c}$
Given $f(1)=0 \Rightarrow 1-7+c=0$
$
\Rightarrow \mathrm{c}=6
$
So $f(x)=x^2-7 x+6$
Question 2.
Given $f^{\prime}(x)=6 x+6, f^{\prime}(0)=-5$ and $f(1)=6$ find $f(x)$
Solution:
$
\begin{aligned}
& \mathrm{f}^{\prime \prime}(\mathrm{x})=6 \mathrm{x}+6 \\
& \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=(6 \mathrm{x}+6) \mathrm{dx} \\
& =\frac{6 x^2}{2}+6 \mathrm{x}+\mathrm{c} \\
& =3 \mathrm{x}^2+6 \mathrm{x}+\mathrm{c}
\end{aligned}
$
Given $f^{\prime}(0)=-5 \Rightarrow 0+c=-5$
$
\Rightarrow \mathrm{c}=-5
$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^2+6 \mathrm{x}-5$
So $\mathrm{f}(\mathrm{x})=\int\left(3 x^2+6 x-5\right) d x$
$
=\frac{3 x^3}{3}+\frac{6 x^2}{2}-5 \mathrm{x}+\mathrm{c}
$
(i.e.,) $f(x)=x^3+3 x^2-5 x+c$
Given $\mathrm{f}(1)=6$
$
\Rightarrow 1+3+5+\mathrm{c}=6
$
$
\Rightarrow \mathrm{c}-1=6 \Rightarrow \mathrm{c}=7
$
So $f(x)=x^3+3 x^2-5 x+7$