We have: \(y={ \left( { sin }^{ -1 }x \right) }^{ 2 }\)
\({ y }_{ 1 }=2\left( { sin }^{ -1 }x \right) .\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } \)
\(\sqrt { 1-{ x }^{ 2 } } { y }_{ 1 }=2\left( { sin }^{ -1 }x \right) \)
Squaring \(\left( 1-{ x }^{ 2 } \right) { { y }_{ 1 } }^{ 2 }=4{ \left( { sin }^{ -1 }x \right) }^{ 2 }\)
\( \left( 1-{ x }^{ 2 } \right) { { y }_{ 1 } }^{ 2 }=4y\)
Diff. w.r.t.x, \(\left( 1-{ x }^{ 2 } \right) 2{ y }_{ 1 }{ y }_{ 2 }+(-2x){ { y }_{ 1 } }^{ 2 }=4{ y }_{ 1 }\)
Hence, \(\left( 1-{ x }^{ 2 } \right) \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -x\frac { dy }{ dx } =2\)

We have: \({ x }^{ 16 }{ y }^{ 9 }={ \left( { x }^{ 2 }+y \right) }^{ 17 },\)
Taking logs., \(log({ x }^{ 16 }{ y }^{ 9 })=log{ \left( { x }^{ 2 }+y \right) }^{ 17 }\)
\(16logx+9logy=17log\left( { x }^{ 2 }+y \right) \)
\( \frac { 16 }{ x } +\frac { 9 }{ y } .\frac { dy }{ dx } =17\frac { 1 }{ { x }^{ 2 }+y } \left[ 2x+\frac { dy }{ dx } \right] \)
\(=\left[ \frac { 9 }{ y } -\frac { 17 }{ { x }^{ 2 }+y } \right] \frac { dy }{ dx } =\frac { 34x }{ { x }^{ 2 }+y } -\frac { 16 }{ x } \)
\(=\frac { { 9x }^{ 2 }+9y-17y }{ y\left( { x }^{ 2 }+y \right) } \frac { dy }{ dx } =\frac { { 34x }^{ 2 }-{ 16x }^{ 2 }-16y }{ x\left( { x }^{ 2 }+y \right) }\)
\(=\frac { 1 }{ y } \left( { 9x }^{ 2 }-8y \right) \frac { dy }{ dx } =\frac { 1 }{ x } \quad \left( { 18x }^{ 2 }-16y \right) \)
\(\frac { 1 }{ y } \quad \frac { dy }{ dx } =\frac { 2 }{ x } \)
\(\frac { dy }{ dx } =\quad \frac { 2y }{ x } \)

A = {1, 2, 3, 4, 5}
is even R = {ab |a-b∣ is even}
It is clear that for any element a ∈A, we have|a-a| = 0 (which is even).
∴ R is reflexive.
Let (a, b) ∈ R.
⇒ |a-b| is even
⇒ |-(a-b)| = |b-a| is also even
⇒ (b,a) ∈ R
∴ R is symmetric.
Now, let (a, b) ∈ R and (b, c) ∈ R.
⇒ |a-b| is even and |b -c| is even
⇒ (a - b) is even and (b - c) is even
⇒ (a - c) = (a - b) + (b - c) is even     [Sum of two even integer is even]
⇒ |a - c | is even
∴ R is transitive.
Hence, R is an equivalence relation.
Now, all elements of the set {1, 3, 5} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even.
Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even.
Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

One-one function from {1, 2, 3} to itself is simply a permutation on three symbols 1, 2, 3. Therefore, total number of one-one maps from {1, 2, 3} to itself is same as total number of permutations on three symbols 1, 2, 3 which is 3! = 6.

Let y be an arbitrary element of range f. Then y = 4x² + 12x + 15, for some
x in N, which implies that y = (2x + 3)² + 6. This gives \(x=\frac{((\sqrt{y-6})-3)}{2}, \text { as } y \geq 6\)
\(\text { Let us define } g: \mathrm{S} \rightarrow \mathbf{N} \text { by } g(y)=\frac{((\sqrt{y-6})-3)}{2}\)
Now gof (x) = g(f (x)) = g(4x² + 12x + 15) = g((2x + 3)² + 6)
\(=\frac{\left(\left(\sqrt{(2 x+3)^{2}+6-6}\right)-3\right)}{2}=\frac{(2 x+3-3)}{2}=x\)
\(\text { and }\operatorname{fog}(y)=f\left(\frac{((\sqrt{y-6})-3)}{2}\right)=\left(\frac{2((\sqrt{y-6})-3)}{2}+3\right)^{2}+6\)
\(=((\sqrt{y-6})-3+3))^{2}+6=(\sqrt{y-6})^{2}+6=y-6+6=y\)
Hence, gof = I_N and fog =I_S. This implies that f is invertible with f ^–1 = g.

We have gof (x) = g(f (x)) = g(cos x) = 3 (cos x)² = 3 cos² x. Similarly, fog(x) = f (g(x)) = f (3x²) = cos (3x²). Note that 3cos² x \(\ne\) cos 3x², for x = 0. Hence, gof \(\ne\) fog.