1. Differentiation Questions

Questions:

How do i do the part $sec^{-1}$

should the angle in $sec^{-1}$ be inverse of what it is because answer is (b) and it is only possible when y is some constant which will be using the identity $sin^{-1}x + cos^{-1}x$

2. Hello, cupid!

There's something wrong with the problem
. . or it is a "trick question".

. . $\sec^{-1}\!\left(\dfrac{2x}{1+x^2}\right) \:\text{ and }\:\sin^{-1}\left(\dfrac{x-1}{x+1}\right)\:\text{ are }constants.$

$\text{Let }\:\theta \;=\;\sec^{-1}\!\left(\dfrac{2x}{1+x^2}\right) \quad\Rightarrow\quad \sec\theta \;=\;\dfrac{2x}{1+x^2} \;=\;\dfrac{hyp}{adj}$

Hence, $\,\theta$ is in a right triangle with: $adj = 1+x^2,\;hyp = 2x$

Since the hypotenuse is the longest side of the right triangle:

. . $adj \:\le\:hyp \quad\Rightarrow\quad 1+x^2 \:\le \:2x$

. . $1 - 2x + x^2 \:\le \:0 \quad\Rightarrow\quad (1-x)^2 \:\le \:0$

The inequality is true only if $x = 1.$

Hence: . $\sec^{-1}\!\left(\dfrac{1+x^2}{2x}\right) \;=\;\sec^{-1}(1) \;=\;0$

Then: . $\sin^{-1}\!\left(\dfrac{x-1}{x+1}\right) \;=\;\sin^{-1}(0) \;=\;0$

3. Hi Soroban
i think you are right but book says that correct answer is (d) None of these

4. Originally Posted by Soroban

. . $adj \:\le\:hyp \quad\Rightarrow\quad 1+x^2 \:\le \:2x$

Couldn't you look at it this way:

. $x^2 - 2x + 1 \:\le \:0 \quad\Rightarrow\quad (x-1)^2 \:\le \:0$

In this case, the inequality would be true if $x \le 1$ and would essentially result in the domain restriction given in the problem.

When I take the derivative I get $\frac{x+1}{x-1}$. I must go to class now, but will post my work this afternoon if nobody else gets to it first - I'll check my work too (I could be wrong at this point).

5. Originally Posted by pflo
Couldn't you look at it this way:

. $x^2 - 2x + 1 \:\le \:0 \quad\Rightarrow\quad (x-1)^2 \:\le \:0$

In this case, the inequality would be true if $x \le 1$ and would essentially result in the domain restriction given in the problem.

When I take the derivative I get $\frac{x+1}{x-1}$. I must go to class now, but will post my work this afternoon if nobody else gets to it first - I'll check my work too (I could be wrong at this point).
$(x-1)^2 \:\le \:0$ is true only for x=1 as well. Something squared can never be less than zero. . .

6. Yes, yes. Of course you're right. Perhaps the authors of this problem made the same mistake I did in overlooking this.

7. No i'm pretty sure that 0 is not the solution.

it may be so for sec-1 term

but sin-1 (acc. to Soroban's post) gives that $x+1 > x-1 \implies 1>-1$ which is true for all x.