3 Marks Question

Question 13 Marks

Differentiate $f(x)=x^5 e^x+x^3 \log x-2^x$ with respect to $x$.

Answer

Given, $\quad f(x)=x^5 e^x+x^3 \log x-2^x$
$\therefore \quad f^{\prime}(x)=\frac{d}{d x}\left(x^5 e^x+x^3 \log x-2^x\right)$
$=\frac{d}{d x}\left(x^5 e^x\right)+\frac{d}{d x}\left(x^3 \log x\right)-\frac{d}{d x} 2^x$
$=e^x \frac{d}{d x} x^5+x^5 \frac{d}{d x} e^x+\log x \frac{d}{d x} x^3$$+x^3 \frac{d}{d x} \log x-2^x \log _e^2$
$\begin{array}{l}=e^x \cdot 5 x^4+x^5 \cdot e^x+\log x \cdot 3 x^2+x^3 \cdot \frac{1}{x}-2^x \log _e^2 \\ =5 x^4 e^x+e^x x^5+3 \log x \cdot x^2+x^2-2^x \log _e^2 \\ =\left(5 x^4+x^5\right) e^x+(3 \log x+1) x^2-2^x \log _e^2\end{array}$

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Question 23 Marks

Find $\lim _{x \rightarrow 1} f(x)$, where $f(x)=\left\{\begin{array}{cc}x^2-1, & x \leq 1 \\ -x^2-1, & x>1\end{array}\right.$

Answer

$\begin{aligned} \text { L.H.L } & =\lim _{x \rightarrow 1-} f(x)=\lim _{x \rightarrow 1-}\left(x^2-1\right) \\ & =(1)^2-1 \\ & =1-1 \\ & =0\end{aligned}$
and,
$ \begin{aligned} \text { R.H.L } & =\lim _{x \rightarrow 1+} f(x)=\lim _{x \rightarrow 1+}\left(-x^2-1\right) \\ & =-1^2-1 \\ & =-2\end{aligned}$
Since $\quad \lim _{x \rightarrow 1-} f(x) \neq \lim _{x \rightarrow 1+} f(x)$,
So $\lim _{x \rightarrow 1} f(x)$ does not exist.

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Question 33 Marks

If $\left(x^2+y^2\right)^2=x y$, find $\frac{d y}{d x}$.

Answer

Given, $\left(x^2+y^2\right)^2=x y$
Differentiating with respect to $x$.
$\Rightarrow \quad \frac{d}{d x}\left(x^2+y^2\right)^2=\frac{d}{d x}(x y)$
$\Rightarrow \quad 2\left(x^2+y^2\right) \frac{d}{d x}\left(x^2+y^2\right)=x \frac{d}{d x}(y)+y \frac{d}{d x}(x)$
$\Rightarrow 2\left(x^2+y^2\right)\left[\frac{d}{d x}\left(x^2\right)+\frac{d}{d x}\left(y^2\right)\right]=x \frac{d y}{d x}+y$
$\Rightarrow \quad 2\left(x^2+y^2\right)\left[2 x+2 y \cdot \frac{d y}{d x}\right]=x \cdot \frac{d y}{d x}+y$
$\Rightarrow 4 x\left(x^2+y^2\right)+4 y\left(x^2+y^2\right) \frac{d y}{d x}-x \frac{d y}{d x}=y$
$\Rightarrow \quad\left[4 y\left(x^2+y^2\right)-x\right] \frac{d y}{d x}=y-4 x\left(x^2+y^2\right)$
$\therefore \quad \frac{d y}{d x}=\frac{\left[y-4 x\left(x^2+y^2\right)\right]}{\left[4 y\left(x^2+y^2\right)-x\right]}$

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Question 43 Marks

Differentiate $e^x \sin x+x^n \cos x$ with respect to $x$.

Answer

Let, $\quad f(x)=e^x \sin x+x^n \cos x$
$\therefore \quad f^{\prime}(x)=\frac{d}{d x}\left\{e^x \sin x+x^n \cos x\right\}$
$=\frac{d}{d x}\left(e^x \sin x\right)+\frac{d}{d x}\left(x^n \cos x\right)$
$\begin{array}{l}=\sin x \frac{d}{d x} e^x+e^x \frac{d}{d x} \sin x+\cos x \frac{d}{d x} x^n+x^n \frac{d}{d x} \cos x \\ =\sin x \cdot e^x+e^x \cdot \cos x+\cos x \cdot n x^{n-1}+x^n \cdot(-\sin x) \\ =e^x(\sin x+\cos x)+x^{n-1}[n \cos x-x \sin x] .\end{array}$

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Question 53 Marks

$\lim _{x \rightarrow 3}\left(\frac{x^4-81}{2 x^2-5 x-3}\right)$

Answer

$\begin{aligned} \lim _{x \rightarrow 3}\left(\frac{x^4-81}{2 x^2-5 x-3}\right) & =\lim _{x \rightarrow 3} \frac{\left(x^2+9\right)\left(x^2-9\right)}{(2 x+1)(x-3)} \\ & =\lim _{x \rightarrow 3} \frac{\left(x^2+9\right)(x+3)(x-3)}{(2 x+1)(x-3)} \\ & =\lim _{x \rightarrow 3} \frac{\left(x^2+9\right)(x+3)}{(2 x+1)} \quad[x \neq 3] \\ & =\frac{\lim _{x \rightarrow 3}\left(x^2+9\right)(x+3)}{\lim _{x \rightarrow 3}(2 x+1)} \\ & =\frac{\left(3^2+9\right)(3+3)}{2 \times 3+1} \\ & =\frac{18 \times 6}{7} \\ & =\frac{108}{7}\end{aligned}$

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Applied Maths 3 Marks Question

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