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NIMCET Previous Year Questions (PYQs)
NIMCET Exponential Series PYQ
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ
The value of the series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+\cdots$ is
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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ
Solution
Given the infinite series:
\[ S = \frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!} \]
This is a known convergent series, and its sum is:
\[ \boxed{e^{-1}} \]
✅ Final Answer: \(\boxed{e^{-1}}\)
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ
Writing in terms of $2^x$:
$(2^x)^2 - 3 \cdot 2^x \cdot 2^3 + 128 = 0$
$(2^x)^2 - 24(2^x) + 128 = 0$
Let $t = 2^x$
$\Rightarrow t^2 - 24t + 128 = 0$
$\Rightarrow (t - 8)(t - 16) = 0$
$\Rightarrow t = 8$ or $t = 16$
When $t = 8$:
$2^x = 8 = 2^3 \Rightarrow x = 3$
When $t = 16$:
$2^x = 16 = 2^4 \Rightarrow x = 4$
Sum of roots $= 3 + 4$
$\therefore \boxed{\text{Sum of roots} = 7}$
Sum of the roots of the equation $4^x - 3(2^{x+3}) + 128 = 0$ is
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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ
Solution
Given: $4^x - 3(2^{x+3}) + 128 = 0$Writing in terms of $2^x$:
$(2^x)^2 - 3 \cdot 2^x \cdot 2^3 + 128 = 0$
$(2^x)^2 - 24(2^x) + 128 = 0$
Let $t = 2^x$
$\Rightarrow t^2 - 24t + 128 = 0$
$\Rightarrow (t - 8)(t - 16) = 0$
$\Rightarrow t = 8$ or $t = 16$
When $t = 8$:
$2^x = 8 = 2^3 \Rightarrow x = 3$
When $t = 16$:
$2^x = 16 = 2^4 \Rightarrow x = 4$
Sum of roots $= 3 + 4$
$\therefore \boxed{\text{Sum of roots} = 7}$
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