C-11-Q094mediumsingle_mcq

The solutions of Example 5, $\dfrac{1 - \sqrt{1+bx}}{1 + \sqrt{1-bx}} \cdot \ldots = 1$ where $0 < b < 2a < 2b$ , include—

a $x = 0$ only
b $x = 0,\ \pm \dfrac{1}{a}\sqrt{\dfrac{2a}{b} - 1}$
c $x = 1$ only
dno real root

ব্যাখ্যা

Solving the equation leads to $x=0$ as one root and, after applying componendo–dividendo and simplifying the quadratic, the symmetric pair $x=\pm\frac{1}{a}\sqrt{\frac{2a}{b}-1}$ . All three satisfy the original relation.

এই অধ্যায়ের আরও প্রশ্ন

C-11-Q095The condition in Example 5 ensures—
C-11-Q096By componendo-dividendo on , the next simplified form is—
C-11-Q097If , then equals—
C-11-Q098If , with positive, equals—
C-11-Q099If , then equals—
C-11-Q100The compound ratio of and is—
C-11-Q101Two ratios and can be combined into the continued ratio—
C-11-Q102To express and in the form , the subsequent of the first is made equal to the antecedent of the second using t
C-11-Q103Therefore and in continued ratio form is—
C-11-Q104If and , then

Algebraic Ratio and Proportion — সব প্রশ্ন দেখুন (180টি) →

The solutions of Example 5, $\dfrac{1 - \sqrt{1+bx}}{1 + \sqrt{1-bx}} \cdot \ldots = 1$ where $0 < b < 2a < 2b$ , include—

x = 0

only

x = 0,\ \pm \dfrac{1}{a}\sqrt{\dfrac{2a}{b} - 1}

x = 1

only

dno real root

The solutions of Example 5, 1−1+bx1+1−bx⋅…=1\dfrac{1 - \sqrt{1+bx}}{1 + \sqrt{1-bx}} \cdot \ldots = 11+1−bx​1−1+bx​​⋅…=1 where 0<b<2a<2b0 < b < 2a < 2b0<b<2a<2b, include—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The solutions of Example 5, 1−1+bx1+1−bx⋅…=1\dfrac{1 - \sqrt{1+bx}}{1 + \sqrt{1-bx}} \cdot \ldots = 11+1−bx​1−1+bx​​⋅…=1 where 0<b<2a<2b0 < b < 2a < 2b0<b<2a<2b, include—

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The solutions of Example 5, $\dfrac{1 - \sqrt{1+bx}}{1 + \sqrt{1-bx}} \cdot \ldots = 1$ where $0 < b < 2a < 2b$ , include—

The solutions of Example 5, $\dfrac{1 - \sqrt{1+bx}}{1 + \sqrt{1-bx}} \cdot \ldots = 1$ where $0 < b < 2a < 2b$ , include—