$$ \frac{(598+479)^2 - (598-479)^2}{598 \times 479} = ? $$ Join @Apnapdfs logo

Explanation

## Page 41 $$ \frac{(598 + 479)^2 - (598 - 479)^2}{598 \times 479} $$ $$ = \frac{4 \times 598 \times 479}{598 \times 479} \quad \{\because (a+b)^2 - (a-b)^2 = 4ab\} $$ $$ = 4 $$ (c) 3 (d) 2 (e) None of the above/More than one of the above 65th B.P.S.C. (Pre) 2019 Ans. (d) (5)$^{x^2+2x+7}$ = (125)$^{2x+1}$ (5)$^{x^2+2x+7}$ = (5$^3$)$^{2x+1}$ $\Rightarrow$ (5)$^{(x^2+2x+7)}$ = (5)$^{6x+3}$ Bases of both sides are equal, so the powers is equal. $x^2 + 2x + 7 = 6x + 3$ $x^2 + 2x - 6x + 7 - 3 = 0$ $x^2 - 4x + 4 = 0$ $(x-2)^2 = 0$ $x - 2 = 0$ $x = 2$