NIMCET 2017 Previous Year Questions (PYQs) – Page 9 of 12

$\dfrac{d}{dx}(x^3) = 3x^2$

$\dfrac{d}{dx}(e^x) = e^x$

$\dfrac{d}{dx}(3x) = 3$

$\dfrac{d}{dx}(\cot x) = -cosec^2 x$

Therefore, total derivative $= 3x^2 + e^x + 3 - cosec^2 x$

Answer: $\boxed{3x^2 + e^x + 3 - cosec^2 x}$ ✅

$\dfrac{dy}{dx}=e^y(e^x+x^2)\ \Rightarrow\ e^{-y}dy=(e^x+x^2)dx$

$\int e^{-y}dy=\int (e^x+x^2)dx$ $\Rightarrow\ -e^{-y}=e^x+\dfrac{x^3}{3}+C$

Hence, $e^{-y}+e^x+\dfrac{x^3}{3}=C$ (or $y=-\ln\!\big(C-e^x-\dfrac{x^3}{3}\big)$).

Answer: $\boxed{e^{-y}+e^x+\dfrac{x^3}{3}=C}$ ✅

Let $y=-\log(\log x)$

$\dfrac{dy}{dx}=-\dfrac{d}{dx}[\log(\log x)]$

$\dfrac{d}{dx}[\log(\log x)]=\dfrac{1}{\log x}\cdot\dfrac{1}{x}$

Therefore, $\dfrac{dy}{dx}=-\dfrac{1}{x\log x}$

Answer: $\boxed{-\dfrac{1}{x\log x}},\; x>1$ ✅

Given limit is indeterminate of type $\frac{0}{0}$.

(DL Rule) Differentiate numerator and denominator:

$\displaystyle \lim_{x\to0}\frac{x\tan x}{1-\cos x}=\lim_{x\to0}\frac{\tan x+x\sec^2 x}{\sin x}$

As $x\to0$, $\tan x\approx x$, $\sin x\approx x$, $\sec^2 x\approx1$

$\Rightarrow \dfrac{x+x}{x}=2$

Answer: $\boxed{2}$ ✅

Possible cases:

• (5M,0F): $^7C_5 \times ^6C_0$
• (4M,1F): $^7C_4 \times ^6C_1$
• (3M,2F): $^7C_3 \times ^6C_2$

Total = $^7C_5\times^6C_0 + ^7C_4\times^6C_1 + ^7C_3\times^6C_2 $ $= 21 + 210 + 525 = 756$

Answer: $\boxed{756}$ ✅