Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 71 Mark
MCQ
The integral part of $(\sqrt{2}+1)^6$ is
A
98
B
96
C
99
D
100
✓
Answer
(c) 99 Explanation: We have $(1+ x )^{ n }=1+{ }^{ n } C_1(x)+{ }^{ n } C_2(x)^2+\ldots .+( x )^{ n }$ Hence $(\sqrt{2}+1)^6=1+{ }^6 C_1(\sqrt{2})+{ }^6 C_2(\sqrt{2})^2+{ }^6 C_3(\sqrt{2})^3+{ }^6 C_4(\sqrt{2})^4+{ }^6 C_5(\sqrt{2})^5+(\sqrt{2})^6$ $\begin{array}{l}\Rightarrow(\sqrt{2}+1)^6=1+6(\sqrt{2})+15 \times 2+20 \times 2(\sqrt{2})+15 \times 4+6 \times 4(\sqrt{2})+8 \\ =99+70 \sqrt{2}\end{array}$ Hence integral part of $(\sqrt{2}+1)^6=99$`
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