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 _____________.
Your Input ________

Answer

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 _____________.
Your Input ________

Answer

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 _________.
Your Input ________

Answer

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 ___________.
Your Input ________

Answer

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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