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#### Let 'a' be a real number such that the function is increasing in and decreasing in . Then the function has aOption: 1Option: 2Option: 3Option: 4

\begin{aligned} &f(x)=a x^{2}+6 x-15 . \\ &f^{\prime}(x)=2 a x+6 . \end{aligned}

$f(x)$ increasing in $\left(-\infty, \frac{3}{4}\right)$ & decreasing in $\left(\frac{3}{4}, \infty\right)$

$f^{\prime}(x)>0 \text { in }\left(-\infty, \frac{3}{4}\right) \& f^{\prime}(x)<0 \text { in }\left(\frac{3}{4}, \infty\right)$

As $f'(x)$ is linear, so $f^{\prime}(x)=0 \text { at } x=\frac{3}{4}$

\begin{aligned} &\Rightarrow \quad 2 \cdot a \cdot\left(\frac{3}{4}\right)+6=0 \\ &\Rightarrow \quad a=-4 . \end{aligned}

Now, $g(x)=-4 x^{2}-6 x+15$

It is a downward parabola, so it will have maximum at $x=\frac{-b}{2 a} \text { (vertex) }$

So maximum at $x=\frac{-(-6)}{2 \cdot(-4)}=-\frac{3}{4}$

Hence, the correct answer is option (1)

y=2 has minima

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#### Find the slope of tangent to the curve at

Hence the slope of the tangent is zero.

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#### Show that the function $f(x) = \frac{x}{3} + \frac{3}{x}$ decreases in the intervals

$f(x) = \frac{x}{3} + \frac{3}{x}$

$f'(x) = \frac{1}{3} - \frac{3}{x^2}$

For f(x) decreases

$\frac{1}{3} - \frac{3}{x^2} <0$

$\frac{1}{3} < \frac{3}{x^2}$

$x^2<9$

$x\epsilon (-3, 0)\cup (0, 3).$

#### Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its side. Also, find the maximum volume.

Let the sides of the rectangle are x cm and y cm.

Perimeter = 36 cm

2x + 2y = 36

y = 18 -x

$\\ \text {Volume of ractnagle when it rotating around one axis }( V)=\pi x^{2} y$

$V=\pi x^{2}(18-x)$

$V=\pi\left(18 x^{2}-x^{3}\right)$

Diffentiate w.r.t. x

$\frac{d V}{d x}=\pi\left(36 x-3 x^{2}\right)$

For critical point

$\frac{d V}{d x}=\pi\left(36 x-3 x^{2}\right) =0$

$x(12- x) =0$

$x =0 \ cm , 1 2 \ cm$

For minima and maxima

$\frac{d^2 V}{d x^2}=\pi\left(36 -6 x\right)$

$\frac{d^2 V}{d x^2}_{at x =12}=\pi\left(36 -72\right) <0$

It has maximum volume at x = 12 cm.

Hence the dimension of the rectangle is 12cm x 6 cm.

$\text { Maximum volume }=\pi x^{2} (18-x)$

$V_{max}= \pi 12^{2} \times 6 = 864 \pi \mathrm{cm}^{3}$

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#### Find the intervals on which the function is (a) strictly increasing (b) strictly decreasing.

$f(x) = (x - 1)^{3} (x - 2)^2$

$f'(x) = 3(x - 1)^{2} (x - 2)^2 + 2(x - 1)^{3} (x - 2)$

$f'(x) = (x - 1)^{2} (x - 2)(3x-6 + 2x - 2)$

$f'(x) = (x - 1)^{2} (x - 2)(5x- 8)$

(a) strictly increasing

$f'(x) = (x - 1)^{2} (x - 2)(5x- 8)>0$

$x \in(-\infty, 1) \cup\left(1, \frac{8}{5}\right) \cup(2, \infty)$

(b) strictly decreasing.

$f'(x) = (x - 1)^{2} (x - 2)(5x- 8)<0$

$x \in\left(\frac{8}{5}, 2\right)$

#### Show that the function f defined by is an increasing function for all x > 0.

$f^{\prime}(x)=x e^{x}$

$f^{\prime}(x)=x e^{x} >0 for all x>0$

$f is an increasing function for x>0$