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(i) we have, f(x) = x².

f(x) = 2x.

andf(x) = 2.

AR extreme point, f(x) = 0

2x = 0

x = 0.

When x = 0, f(0) = 2 > 0.

∴x = 0 is a point of local minima and value of local minimum is given by f(0) = 0² = 0.

(ii) g(x) = x³ 3x

A(ii)

we have, g(x) = x³- 3x

$\Rightarrow$ g'(x) = 3x²- 3

$\Rightarrow$ g''(x) = 6x.

At extreme point,

g'(x) = 0

3x²- 3 = 0.

3(x²- 1) = 0 ⇒ 3(x - 1)(x + 1) = 0.

x = 1 or x = -1.

At x = 1, g"(1) = 6.1 = 6 > 0.

So, x = 1 is a point of local minima and value of local minimum is given by g(1) = 1³- 3.1 = 1 - 3 = - 2.

And at x = -1, g"( -1) = 6 ( -1) = 6 < 0.

So, x = -1 is a point of local minima and value of local minimum is given by

g(- 1) = (- 1)³- 3(- 1) = 1 + 3 = 2.

(iii) $h\left(x\right)=sinx+cosx.0<x<\frac{\pi }{2}$

A(iii)

we have, h(x) = sin x + cos x.

$\Rightarrow$ h'(x) = cos x - sin x

$\Rightarrow$ h'(x) = -sin x - cos x.

At extreme points,

h'(x) = 0

cosx - sin x = 0

cosx = sin x

$\Rightarrow \frac{cosx}{cosx}=\frac{sinx}{cosx}$

$\Rightarrow$ 1 = tan x

$\Rightarrow tanx=tan\frac{\pi }{4}$

$\Rightarrow x=\frac{\pi }{4}.$

(iv) $f\left(x\right)=sinx-cosx,0<x<2\pi$

A(iv)

we have, f(x) = sin x cos x.

$\Rightarrow$ f(x) = cos x + sin x

$\Rightarrow$ f(x) = sin x + cos x.

At extreme points,

f(x) = 0

$\Rightarrow$ cosx + sin x = 0.

$\Rightarrow \frac{sinx}{cosx}=\frac{-cosx}{cosx}$

$\Rightarrow tanx=-1=-tan\frac{\pi }{4}=tan\left(\pi -\frac{\pi }{4}\right)=tan\left(2\pi -\frac{\pi }{4}\right)$

$\therefore x=\left(\pi -\frac{\pi }{4}\right)$ IInd quadrate or $\left(2\pi -\frac{\pi }{4}\right)$ IIIth quadrate

$\Rightarrow x=\frac{4\pi -\pi }{4}\frac{8x-\pi }{4}$

$\Rightarrow x=\frac{3\pi }{4}orx=\frac{f\pi }{4}.$

$Atx=\frac{3\pi }{4},f^{\prime \prime }\left(\frac{3\pi }{4}\right)=-sin\frac{3\pi }{4}+cos\frac{3\pi }{4}$

$=-sin\left(\pi -\frac{\pi }{4}\right)+cos\left(\pi -\frac{\pi }{4}\right)$

$=-sin\frac{\pi }{4}-cot\frac{\pi }{4}$

(v) f(x) = x³- 6x² + 9x + 15.

A(v)

we have, f(x) = x³- 6x² + 9x + 15.

$\Rightarrow$ f(x) = 3x²- 12x + 9.

$\Rightarrow$ f'(x) = 6x - 12.

At extreme point, f'(x) = 0.

$\Rightarrow$ 3x²- 12x + 9 = 0.

$\Rightarrow$ x²- 4x + 3 = 0

$\Rightarrow$ x²-x - 3x + 3 = 0

$\Rightarrow$ x(x - 1) -3(x - 1) = 0

$\Rightarrow$ (x - 1)(x - 3) = 0

$\Rightarrow$ x = 1 or x = 3.

At, x = 1,f"(1) = 6 × 1 - 12 = 6 - 12 = - 6 < 0

∴x = 1 is a point of local maxima and the value of local

Maximum is given by f(1) = 1³- 6(1)² + 9(1) + 15

= 1 - 6 + 9 + 15

= 19.

And at x = 3,f"(3) = 6 × 3 - 12 = 18 - 12 = 6 > 0

∴x = 3 is a point of local minima and the value of

local minimum is given by f (3) = 3³- 6(3)² + 9(3) + 15.

= 27 - 54 + 27 + 15.

= 15.

(vi) $g(x)=\frac{x}{2}+\frac{2}{x},x>0$

A(vi)

We have, $g(x)=\frac{x}{2}+\frac{2}{x},x>0$

$\Rightarrow g^{\prime }(x)=\frac{1}{2}-\frac{2}{x^2}$

$\Rightarrow g{g}^{\prime z}(x)=\frac{x^2-4}{2x^2}=\frac{(x-2)(x+2)}{2x^2}$

$g^{\prime \prime }(x)=\frac{4}{x^3}.$

At, extreme point, g'(x) = 0

$\Rightarrow \frac{(x+2)(x-2)}{x^2}=0$

$\Rightarrow$ x = -2 or x = 2

Given, that x > 0, hence we have x = 2.

At, x = 2, ${\stackrel{?}{g}}^{\prime }(x)=\frac{4}{2^3}=\frac{4}{8}=\frac{1}{2}>0$

∴x = 2 is a point of local minima and value of local

Minimum is given by g(2) = $\frac{2}{2}+\frac{2}{2}=1+1=2.$

(vii) $f(x)=\frac{1}{x^2+2}$

A(vii)

Given, $f(x)=\frac{1}{x^2+2}$

$\Rightarrow g^{\prime }(x)=-\frac{1}{{\left(x^2+2\right)}^2}\frac{d}{dx}\left(x^2+2\right)=\frac{-2x}{{\left(x^2+2\right)}^2}.$

$\Rightarrow g^{\prime \prime }(x)=-\left[\frac{{\left(x^2+2\right)}^2\frac{d}{dx}2x-2x\frac{d}{dx}{\left(x^2+2\right)}^2}{{\left(x^2+2\right)}^4}\right]$

$=\frac{-\left[2{\left(x^2+2\right)}^2-2x\cdot 2\left(x^2+2\right)\cdot 2x\right]}{{\left(x^2+2\right)}^4}.$

$=\frac{-\left[2{\left(x^2+2\right)}^2\left[x^2+2-4x^2\right]\right]}{{\left(x^2+2\right)}^4}$

$=\frac{-2\left(x^2+2\right)\left(2-3x^2\right)}{{\left(x^2+2\right)}^4}=-\frac{2\left(2-3x^2\right)}{{\left(x^2+2\right)}^3}.$

At extreme points, g'(x) = 0.

$\Rightarrow \frac{-2x}{{\left(x^2+2\right)}^2}=0.$

$\Rightarrow$ x = 0.

At, x = 0, $g^{\prime \prime }(x)=\frac{-2\left(2-2\times 0^2\right)}{{\left(0^2+2\right)}^3}=\frac{-2(2-0)}{2^3}=-\frac{1}{2}<0$

∴x = 0 is a point of local maxima and value of local maximum is given by $g(0)=\frac{1}{0^2+2}=\frac{1}{2}.$

(i) we have, f(x) = |x + 2| - 1

We know that, for all $x\in ?,|x+2|\ge 0$

$\Rightarrow |x+2|-1\ge -1.$

$\Rightarrow$ f(x)$\ge$- 1.

∴ Minimum value of f(x) = -1 when x + 2 = 0 $\Rightarrow$ x = - 2.

And maximum value of f(x) does not exist.

(ii) $g(x)=-|x+1|+3$

A(ii)

We have, $g(x)=-|x+1|+3$

For all $x\in ,|x+1|\ge 0.$

$\Rightarrow -|x+1|\le 0$

$\Rightarrow -|x+1|+3\le 0+3$

$\Rightarrow$ g(x) $\le \mathrm{}$3.

∴ Maximum value of g(x) = 3 when $|x+1|=0\Rightarrow x=-1.$

And minimum value does not exist.

(iii) h(x) = sin (2x) + 5.

A(iii)

we have, h(x) = sin (2x) + 5.

For all $x\in ?,-1\le sin2x\le 1.$ {range of sine function is [-1, 1]}

$\Rightarrow$ -1 + 5 $\le$sin 2x + 5 $\le$1 + 5.

$\Rightarrow$ 4 $\le$h(x) $\le$6.

∴ Maximum value of h(x) = 6.

Minimum value of h(x) = 4.

(iv) $f(x)=|sin4x+3|.$

A(iv)

we have, $f(x)=|sin4x+3|.$

As for all $x\in ?,-1\le sin4x\le 1$

$\Rightarrow$ -1 + 3 $\le$sin 4x + 3$\le$ 1 + 3

$\Rightarrow \left|2\right|\le |sin4x+3|\le \left|4\right|.$

$\Rightarrow$ 2 $\le$f(x) $\le$4.

∴ Maximum value of f(x) = 4.

Minimum value of f(x) = 2.

(v) h(x) = x + 1.,x∈ ( -1, 1).

A(v)

we have, h(x) = x + 1.,x∈ ( -1, 1).

Given, -1

- 1 + 1 <x + 1 < 1 + 1

0 <h (x) < 2.

∴ Maximum value and minimum value of h(x) does not exist.

(i) We have, f(x) = (2x - 1)² + 3.

For all $x\in ?,(2x-1){}^2\ge 0$

(2x - 1)² + 3 ≥ 3.

f(x) ≥ 3.

∴The minimum value of f(x) = 3. When 2x - 1 = 0--> x = $\frac{1}{2}$

Again as $x\to \infty ,f(x)\to \infty$ as there is vouppa bound to ‘x’ value hence, f(x) has no maximum values.

(ii) $f\left(x\right)=9x^2+12x+2$

A(ii)

We have, f(x) = 9² + 12x + 2.

$f(x)=9\left[x^2+\frac{12x}{9}+\frac{2}{9}\right]$ (Taking 9 common from each team).

$f(x)=9\left[x^2+\frac{4x}{3}+\frac{2}{9}\right]$

$f(x)=9\left[x^2+2\times \frac{2x}{3}+{\left(\frac{2}{3}\right)}^2-{\left(\frac{2}{3}\right)}^2+\frac{2}{9}\right]$

$f(x)=9\left[{\left(x+\frac{2}{3}\right)}^2-\frac{4}{9}+\frac{2}{9}\right]$

$f(x)=9\left[\cdot {\left(x+\frac{2}{3}\right)}^2-\frac{2}{9}\right]=9{\left(x+\frac{2}{3}\right)}^2-2$

For all $x\in ?,{\left(x+\frac{2}{3}\right)}^2\ge 0$

$\Rightarrow {\left(x+\frac{2}{3}\right)}^2-2\ge \cdot 0-2$

$\Rightarrow$ f(x)≥ - 2.

∴The minimum value of f(x) = -2 when $x+\frac{2}{3}=0$

And as $x\to \infty ,f(x)\to \infty$ so f(x) has $\Rightarrow x=-\frac{2}{3}.$

no maximum values.

(iii) f(x) = (x - 1)² + 10

A(iii)

we have, f(x) = - (x - 1)² + 10

For all $x\in ?,(x-1){}^2\ge 0.$

$\Rightarrow$ (x - 1)² ≤ 0

$\Rightarrow$ -(x- 1)² + 10 ≤ 10.

$\Rightarrow$ f(x) ≤ 10.

∴maximum value of f(x) = 10 when x - 1 = 0 $\Rightarrow$ x = 1.

And minimum value of f(x) does not exist.

(iv) g(x) = x³ + 1

A(iv)

we have, g(x) = x³ + 1

For the given fxⁿ, $x\in ?,$

$g(x)\to \infty asx\to \infty$

and $g(x)-\infty asx\to -\infty$

Q Maximum and minimum value does not exist.

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Electrical Engineering (EE)
Having this rank makes it possible for you to enter into EE courses at many IITs. As an example, the number of students in EE branch at IITs in 2024 were:

There are 6,505 students at IIT Bhubaneswar.

IIT Jodhpur has 5,852 students.

IIT Ropar has 5,209 students.

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You should be able to get into EE in new IITs and might be accepted in a few mid-tier IITs if the competition is not too tough that year.

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You can study ECE at a few IITs and the rankings are generally quite high. In 2024:

IIT Roorkee has 2,000 students.

At IIT, Hyderabad there are around 2,080 students.

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IIT Bhubaneswar has nearly 6,505 students.

IIT Jodhpur has around 5,852 students

Number of students at IIT Ropar is 5,209.

Admission to IITs might be unlikely, but you could consider applying to newer IITs or considering interdisciplinary programs.

 Recommendations
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Check the main JoSAA website from time to time to find the newest cutoff and the seats that are still available.

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