1
Numerical

### JEE Main 2021 (Online) 27th August Morning Shift

The number of distinct real roots of the equation 3x4 + 4x3 $-$ 12x2 + 4 = 0 is _____________.

Correct Answer is 4

## Explanation

3x4 + 4x3 $-$ 12x2 + 4 = 0

So, let f(x) = 3x4 + 4x3 $-$ 12x2 + 4

$\therefore$ f'(x) = 12x(x2 + x $-$ 2)

= 12x (x + 2) (x $-$ 1)

$\therefore$ f'(x) = 12x3 + 12x2 – 24x = 12x(x + 2) (x – 1)

Points of extrema are at x = 0, –2, 1

f(0) = 4

f(–2) = –28

f(1) = –1

So, 4 Real Roots
2
Numerical

### JEE Main 2021 (Online) 26th August Morning Shift

A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then $\left( {{4 \over \pi } + 1} \right)k$ is equal to _____________.

Correct Answer is 36

## Explanation

Let x + y = 36

x is perimeter of square and y is perimeter of circle side of square = x/4

radius of circle = ${y \over {2\pi }}$

Sum Areas = ${\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}$

$= {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}$

For min Area :

$x = {{144} \over {\pi + 4}}$

$\Rightarrow$ Radius = y = 36 $-$ ${{144} \over {\pi + 4}}$

$\Rightarrow$ k = ${{36\pi } \over {\pi + 4}}$

$\left( {{4 \over \pi } + 1} \right)k$ = 36
3
Numerical

### JEE Main 2021 (Online) 17th March Evening Shift

Let f : [$-$1, 1] $\to$ R be defined as f(x) = ax2 + bx + c for all x$\in$[$-$1, 1], where a, b, c$\in$R such that f($-$1) = 2, f'($-$1) = 1 for x$\in$($-$1, 1) the maximum value of f ''(x) is ${{1 \over 2}}$. If f(x) $\le$ $\alpha$, x$\in$[$-$1, 1], then the least value of $\alpha$ is equal to _________.

Correct Answer is 5

## Explanation

$f(x) = a{x^2} + bx + c$

$f'(x) = 2ax + b,$

$f''(x) = 2a$

Given, $f''( - 1) = {1 \over 2}$

$\Rightarrow a = {1 \over 4}$

$f'( - 1) = 1 \Rightarrow b - 2a = 1$

$\Rightarrow b = {3 \over 2}$

$f( - 1) = a - b + c = 2$

$\Rightarrow c = {{13} \over 4}$

Now, $f(x) = {1 \over 4}({x^2} + 6x + 13),x \in [ - 1,1]$

$f'(x) = {1 \over 4}(2x + 6) = 0$

$\Rightarrow x = - 3 \notin [ - 1,1]$

$f(1) = 5,f( - 1) = 2$

$f(x) \le 5$

So, $\alpha$minimum = 5
4
Numerical

### JEE Main 2021 (Online) 26th February Evening Shift

Let a be an integer such that all the real roots of the polynomial
2x5 + 5x4 + 10x3 + 10x2 + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to ___________.

Correct Answer is 2

## Explanation

Let, $f(x) = 2{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 10x + 10$

$\Rightarrow f'(x) = 10({x^4} + 2{x^3} + 3{x^2} + 2x + 1)$

$= 10\left( {{x^2} + {1 \over {{x^2}}} + 2\left( {x + {1 \over x}} \right) + 3} \right)$

$= 10\left( {{{\left( {x + {1 \over x}} \right)}^2} + 2\left( {x + {1 \over x}} \right) + 1} \right)$

$= 10{\left( {\left( {x + {1 \over x}} \right) + 1} \right)^2} > 0;\forall x \in R$

$\therefore$ f(x) is strictly increasing function. Since, it is an odd degree polynomial it will have exactly one real root.

Now, by observation.

$f( - 1) = 3 > 0$

$f( - 2) = - 64 + 80 - 80 + 40 - 20 + 10$

$= - 34 < 0$

$\Rightarrow f(x)$ has at least one root in $( - 2, - 1) \equiv (a,a + 1)$

$\Rightarrow a = - 2$

$\Rightarrow$ |a| = - 2

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