the perimeter of top of rectangular table is 28m.where as its area 48m².what length of diagonal?1) p=50m,area=200, length of diagonal? D=?
Answers
Answer:
Solution−
Given function is
\rm :\longmapsto\:f(x) = {sin}^{ - 1}\bigg[\dfrac{1 + {x}^{2} }{2x} \bigg]:⟼f(x)=sin
−1
[
2x
1+x
2
]
We know,
\boxed{\tt{ Domain \: of \: {sin}^{ - 1}x \: is \: x \: \in \: [ - 1,1] \: }}
Domainofsin
−1
xisx∈[−1,1]
So, using this
\rm :\longmapsto\: - 1 \leqslant \dfrac{1 + {x}^{2} }{2x} \leqslant 1:⟼−1⩽
2x
1+x
2
⩽1
We know,
\boxed{\tt{ - y \leqslant x \leqslant y\rm \implies\: |x| \leqslant y \: }}
−y⩽x⩽y⟹∣x∣⩽y
So, using this
\rm \implies\:\bigg |\dfrac{1 + {x}^{2} }{2x} \bigg| \leqslant 1⟹
∣
∣
∣
∣
∣
2x
1+x
2
∣
∣
∣
∣
∣
⩽1
\rm \implies\:\dfrac{1 + {x}^{2} }{2 |x| } \leqslant 1⟹
2∣x∣
1+x
2
⩽1
\rm \implies\:\dfrac{2 |x| }{1 + {x}^{2} } \geqslant 1⟹
1+x
2
2∣x∣
⩾1
\rm \implies\:1 + {x}^{2} \leqslant 2 |x|⟹1+x
2
⩽2∣x∣
\rm \implies\:1 + {x}^{2} - 2 |x| \leqslant 0⟹1+x
2
−2∣x∣⩽0
\rm \implies\: { |x| }^{2} - 2 |x| + 1\leqslant 0⟹∣x∣
2
−2∣x∣+1⩽0
\rm \implies\: {( |x| - 1) }^{2} \leqslant 0⟹(∣x∣−1)
2
⩽0
As square of number can never be negative.
\rm \implies\: {( |x| - 1) }^{2} = 0⟹(∣x∣−1)
2
=0
\rm \implies\: {|x| - 1 } = 0⟹∣x∣−1=0
\rm \implies\: {|x|} = 1⟹∣x∣=1
\bf\implies \:x = 1 \: \: or \: \: - 1⟹x=1or−1
Hence,
\boxed{\tt{ Domain \: of \: f(x) = {sin}^{ - 1}\bigg[\dfrac{1 + {x}^{2} }{2x} \bigg] \: is \: \{ - 1, \: 1 \} \: }}
Domainoff(x)=sin
−1
[
2x
1+x
2
]is{−1,1}
Answer:
length of diagonal =10m