Question

√-x^2+2x+24 >= √8x-x^2 iit question Pls don't spam it

Answer 1

I assume the question reads:

Find the set of values of x which satisfies the inequality

$\sqrt{-x^{2} + 2x + 24} \geq \sqrt{8x - x^{2}}.$

Solution:

Square both sides and simplify:

$-x^{2} + 2x + 24 \geq 8x - x^{2}$

$\implies 2x + 24 \geq 8x$

$\implies x \geq 4.$

Now this is all good, but we must be careful here. We note that the original inequality has square roots and hence there is the additional requirement that nothing inside the square is negative. To deal with this, we must also solve:

1)

$-x^{2} + 2x + 24 \geq 0$

$\iff x^{2} - 2x - 24 \leq 0$

Critical values:

$x^{2} - 2x - 24 = 0 \iff (x + 4)(x - 6) = 0 \iff x = -4 \ or\ x = 6.$

Now sketch the graph of y = x^{2} - 2x - 24 and note that the solution to y <= 0 is:

$-4 \leq x \leq 6.$

2)

$8x - x^{2} \geq 0$

$\iff x^{2} - 8x \leq 0$

Critical values:

$x^{2} - 8x = 0 \iff x(x - 8) = 0 \iff x = 0 \ or\ x = 8.$

This time sketch a graph of y = x^{2} - 8x to see that the solution to y <= 0 is:

$0 \leq x \leq 8.$

So, our final answer is the set of values which solves the inequality while making sure the things inside the square root are kept positive. This is the set of values which satisfies the three inequalities we have found:

$x \geq 4 \ and\ -4 \leq x \leq 6 \ and\ 0 \leq x \leq 8.$

A number line shows that the set of values satisfying all three inequalities and therefore the solution to this problem, is:

$\underline{\underline{4 \leq x \leq 6}}.$

Answer 2

Answer:

Let f: R -> R be given by f(x) = (x - 1) (x - 2) (x - 5). Define F(x) = \int_0^x f(t) dt, x > 0∫

0

x

f(t)dt,x>0. Then which of the following options is/are correct?

a) F has a local minimum at x = 1

b) F has a local maximum at x = 2

c) F(x) ≠ 0 for all x ϵ (0, 5)

d) F has two local maxima and one local minimum in (0, ∞)

Solution:

F'(x) = (x - 1) (x - 2) (x - 5)

JEE Advanced Question Paper 2019 Maths Paper 2

At x = 1 and x = 5, F'(x) changes from - to +

Therefore, F(x) has two local minima points at x = 1 and x = 5.

F(x) has one local maxima point at x = 2.

Answer: f(x) = (x - 1) (x - 2) (x - 5)