Logarithms JEE Main Mathematics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (d)

Explanation: Half life period of bismath is 5 days. Initial mass = 1000 mg. $\therefore$ Mass left after 5 days = 500 mg Mass left after 10 days = 250 mg Mass left after 15 days = 125 mg Mass left after 20 days = 62.5 mg Mass left after 25 days = 31.25 mg Mass left after 30 days = 15.625 mg

Correct Answer: (d)

Explanation: $\begin{aligned} f\left(\frac{2 x}{1+x^2}\right) & =\log _e\left(\frac{1-\frac{2 x}{1+x^2}}{1+\frac{2 x}{1+x^2}}\right) \\ & =\log _e\left(\frac{(1-x)^2}{(1+x)^2}\right) \\ & =2 \log _e\left(\frac{1-x}{1+x}\right) \\ & =2 \mathrm{f}(x) \end{aligned}$

Correct Answer: (d)

Explanation: ${\log _{225}}70 = {{{{\log }_e}70} \over {{{\log }_e}225}} = {{{{\log }_e}7 + {{\log }_e}5 + {{\log }_e}2} \over {{{\log }_e}25 + {{\log }_e}9}}$ $ = {{1 + {{{{\log }_e}5} \over {{{\log }_e}7}} + {{{{\log }_e}2} \over {{{\log }_e}7}}} \over {2{{{{\log }_e}5} \over {{{\log }_e}7}} + 2{{{{\log }_e}3} \over {{{\log }_e}7}}}}$ $ = {{1 + {{\log }_7}5 + {{\log }_3}2 \times {{\log }_5}3 \times {{\log }_7}5} \over {2[{{\log }_7}5 + {{\log }_5}3 \times {{\log }_7}5]}}$ $ = {{1 + a + abc} \over {2[a + ab]}} = {{1 + a + abc} \over {2a(1 + b)}}$

Correct Answer: (a)

Explanation: To solve this problem, let's use the relationship for the inverse hyperbolic cosine function: $ \cosh^{-1} x = \log \left(x + \sqrt{x^2 - 1}\right) $ Applying this to $\cosh^{-1} 2$, we get: $ \cosh^{-1} 2 = \log \left(2 + \sqrt{2^2 - 1}\right) $ Simplifying under the square root: $ \sqrt{2^2 - 1} = \sqrt{4 - 1} = \sqrt{3} $ So, the expression becomes: $ \cosh^{-1} 2 = \log (2 + \sqrt{3}) $

Correct Answer: (a)

Explanation: The inequality is $\log _{0.2} \frac{x+2}{x} \leq 1$. The LHS is valid if $\frac{x+2}{x}>0$ $ \begin{array}{ll} \Rightarrow & x(x+2)>0 \\ \Rightarrow & x0 \end{array} $ Solving the inequality, we get (note that base $0$ $ \Rightarrow \quad x \in\left(-\infty,-\frac{5}{2}\right] \cup(0, \infty) $