If S- x - N-2 - log2-x - 1 - 1 - log1-2-4 - x2- then

The correct option is C S=N 2+log _2√(x+1) gt;1 log _1/2√(4 x^2) 1+log_2√(x+1) gt;log _2√(4 x^2) log _22+log _2√(x+1) gt;log _2√(4 x^2) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of Set Operation

If S= x ∈ N...

Question

If S= ${x \in N : 2 + {log}_{2} \sqrt{x + 1} > 1 - {log}_{1 / 2} \sqrt{4 - x^{2}}}$ , then

S = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

S = Z

No worries! We‘ve got your back. Try BYJU‘S free classes today!

S = N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

S = {x \in N : 2 + {log}_{2} \sqrt{x + 1} > 1 - {log}_{1 / 2} \sqrt{4 - x^{2}}},

S = {x \in N : 2 + {log}_{2} \sqrt{x + 1} > 1 - {log}_{1 / 2} \sqrt{4 - x^{2}}}

S

S = {x \in N : 2 + {log}_{2} \sqrt{x + 1} > 1 - {log}_{\frac{1}{2}} \sqrt{4 - x^{2}}}

{x \in N : 2 + {log}_{2} \sqrt{x + 1} > 1 - {log}_{1 / 2} \sqrt{4 - x^{2}}}

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

n \sum r = 0 n \sum s = 0 (r + s) C_{r} C_{s}

Join BYJU'S Learning Program