1

### JEE Main 2017 (Online) 9th April Morning Slot

The function f defined by

f(x) = x3 $-$ 3x2 + 5x + 7 , is :
A
increasing in R.
B
decreasing in R.
C
decreasing in (0, $\infty$) and increasing in ($-$ $\infty$, 0)
D
increasing in (0, $\infty$) and decreasing in ($-$ $\infty$, 0)
2

### JEE Main 2018 (Online) 15th April Morning Slot

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm2) of this cone is :
A
$6\sqrt 2 \pi$
B
$6\sqrt 3 \pi$
C
$8\sqrt 2 \pi$
D
$8\sqrt 3 \pi$

## Explanation

Sphere of radius r = 3 cm

Let b, h be base radius and height of cone respectively.

So, volume of cone = ${1 \over 2}$ $\pi$b2h

In right $\Delta$ABC by Pythagoras theorem

(h $-$ r)2 + b2 = r2

$\Rightarrow$ b2 = r2 $-$ (h $-$ r)2 = r2 $-$ (h2 $-$ 2hr + r2) = 2hr $-$ h2

$\therefore\,\,\,$ Volume (v) = ${1 \over 3}$ $\pi$h[2hr $-$ h2] = ${1 \over 3}$ [ 2h2r $-$ h3]

${{dv} \over {dh}}$ = ${1 \over 3}$ [4hr $-$ 3h2] = 0

$\Rightarrow$ h (4r $-$ 3h) = 0

${{{d^2}v} \over {d{h^2}}}$ = ${1 \over 3}$ [4r $-$ 6h]

At h = ${{4r} \over 3}$, ${{{d^2}y} \over {d{h^2}}}$ = ${1 \over 3}$ $\left[ {4r - {{4r} \over 3} \times 6} \right] = {1 \over 3}\left[ {4r - 8r} \right] < 0$

$\Rightarrow$ maximum volume cours at h = ${{4r} \over 3}$ = ${4 \over 3}$ $\times$ 3 = 4 cm

As from (1),

(h $-$ r)2 + b2 = r2

$\Rightarrow$ b2 = 2hr $-$ h2 = 2.${{4r} \over 3}$ r $-$ ${{16{r^2}} \over 9}$ = ${{8{r^2}} \over 3}$ $-$ ${{16{r^2}} \over 9}$

= ${{\left( {24 - 16} \right){r^2}} \over 9}$ = ${{8{r^2}} \over 9}$

$\Rightarrow$ b = ${{2\sqrt 2 } \over 3}$ r = 2 $\sqrt 2 \,\,m$

Therefore curved surface area = $\pi bl$

= $\pi$b$\sqrt {{h^2} + {r^2}}$ = $\pi$2$\sqrt 2$ $\sqrt {{4^2} + 8}$ = 8$\sqrt 3$$\pi$ cm2
3

### JEE Main 2018 (Online) 15th April Morning Slot

If $\beta$ is one of the angles between the normals to the ellipse, x2 + 3y2 = 9 at the points (3 cos $\theta$, $\sqrt 3 \sin \theta$) and ($-$ 3 sin $\theta$, $\sqrt 3 \,\cos \theta$); $\theta \in \left( {0,{\pi \over 2}} \right);$ then ${{2\,\cot \beta } \over {\sin 2\theta }}$ is equal to :
A
${2 \over {\sqrt 3 }}$
B
${1 \over {\sqrt 3 }}$
C
$\sqrt 2$
D
${{\sqrt 3 } \over 4}$

## Explanation

Since, x2 + 3y2 = 9

$\Rightarrow$ 2x + 6y ${{dy} \over {dx}}$ = 0

$\Rightarrow$ ${{dy} \over {dx}}$ = ${{ - x} \over {3y}}$

Slope of normal is $-$ ${{dx} \over {dy}}$ = ${{3y} \over x}$

$\Rightarrow$ ${\left( { - {{dx} \over {dy}}} \right)_{\left( {3\cos \theta ,\sqrt 3 \sin \theta } \right)}}$

= ${{3\sqrt 3 \sin \theta } \over {3\cos \theta }}$ = $\sqrt 3 \tan \theta$ = m1

&   ${\left( { - {{dx} \over {dy}}} \right)_{\left( { - 3\sin \theta ,\sqrt 3 \cos \theta } \right)}}$

= ${{3\sqrt 3 \cos \theta } \over { - 3\sin \theta }}$ = $- \sqrt 3 \cot \theta$ = m2

As, $\beta$ is the angle between the normals to the given ellipse then

tan$\beta$ = $\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$

= $\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3\tan \theta \cot \theta }}} \right|$ = $\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3}}} \right|$

So, tan $\beta$ = ${{\sqrt 3 } \over 2}$ $\left| {\tan \theta + \cot \theta } \right|$

$\Rightarrow$ ${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}\left| {{{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}} \right|$

$\Rightarrow$ ${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}$ $\left| {{1 \over {\sin \theta \cos \theta }}} \right|$

$\Rightarrow$ ${1 \over {\cot \beta }} = {{\sqrt 3 } \over {\sin 2\theta }}$

$\Rightarrow$ ${{2\cot \beta } \over {\sin 2\theta }}$ = ${2 \over {\sqrt 3 }}$
4

### JEE Main 2018 (Online) 16th April Morning Slot

Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x3 $-$ 9x2 + 12x + 5 in the interval [0, 3]. Then M $-$m is equal to :
A
5
B
9
C
4
D
1

## Explanation

Here, f(x) = 2x3 $-$ 9x2 + 12x + 5

$\Rightarrow$ f'(x) = 6x2 $-$ 18x + 12 = 0

For maxima or minima put f'(x) = 0

$\Rightarrow$ x2 $-$ 3x + 2 = 0

$\Rightarrow$ x = 1 or x = 2

Now, f''(x) = 12x $-$ 18

$\Rightarrow$ f''(1) = 12(1) $-$ 18 = $-$ 6 < 0

Hence, f(x) has maxima at x = 1

$\therefore$ maximum value = M = f(1) = 2 $-$ 9 + 12 + 5 = 10

And, f''(2) = 12(2) $-$ 18 = 6 > 0.

Hence, f(x) has minima at x = 2.

$\therefore$ minimum value = m = f(2) = 2(8) $-$ 9(4) + 12(2) + 5 = 9

$\therefore$ M $-$ m = 10 $-$ 9 = 1