Question bank/JEE Main/Mathematics

JEE MainMathematicsModerateBinomial Theorem

If the third term in the binomial expansion of
(1+xlog⁡2x)5\left( 1 + x^{\log_{2}x} \right)^{5} equals 2560, then a possible
value of x is

Show answer and solution
In the expansion of (1+xlog⁡2x)5\left( 1 + x^{\log_{2}x} \right)^{5}
Third term say T3=(xlog⁡2x)2=2560T_{3} = \left( x^{\log_{2}x} \right)^{2} = 2560
⇒(xlog⁡2x)2=256\Rightarrow \left( x^{\log_{2}x} \right)^{2} = 256
Taking logarithm to the base 2 on both sides
⇒2(log⁡2x)2=8⇒(log⁡2x)=±2\Rightarrow 2\left( \log_{2}x \right)^{2} = 8 \Rightarrow \left( \log_{2}x \right) = \pm 2
⇒x=4,14\Rightarrow x = 4,\frac{1}{4}
Here, x=14x = \frac{1}{4}

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx

More JEE Main questionsChapter practice

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.

Scan a doubtView plans