Let frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B cd | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $x = 2^{1/4}$. Then the expression becomes $\frac{7}{x^2 + x + 1}$.Multiplying numerator and denominator by $(x - 1)$,we get $\frac{7(x - 1)}{

Vedclass · Accepted Answer

All of these

Vedclass · Answer

(D) Let $x = 2^{1/4}$. Then the expression becomes $\frac{7}{x^2 + x + 1}$.Multiplying numerator and denominator by $(x - 1)$,we get $\frac{7(x - 1)}{x^3 - 1}$.Since $x = 2^{1/4}$,$x^3 = 2^{3/4}$.So,$\frac{7(2^{1/4} - 1)}{2^{3/4} - 1} = A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4}$.Multiplying by $(2^{3/4} + 1)$,we get $7(2^{1/4} - 1)(2^{3/4} + 1) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2^{3/4} - 1)(2^{3/4} + 1)$.$7(2^{1} + 2^{1/4} - 2^{3/4} - 1) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2^{3/2} - 1)$.$7(1 + 2^{1/4} - 2^{3/4}) = (A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4})(2 \cdot 2^{1/2} - 1)$.Expanding and equating coefficients,we find $A = 1, B = -3, C = 2, D = 1$.Thus,$A + B + C + D = 1 - 3 + 2 + 1 = 1$.

Let $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B \cdot 2^{1/4} + C \cdot 2^{1/2} + D \cdot 2^{3/4}$,then find the value of $A + B + C + D$.

Similar Questions

${\frac{{[4 + \sqrt{15}]}^{3/2} + {[4 - \sqrt{15}]}^{3/2}}{{[6 + \sqrt{35}]}^{3/2} - {[6 - \sqrt{35}]}^{3/2}}} = $

The least number among $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}$ and $\sqrt[3]{8}$ is:

The number of digits in the decimal expansion of $16^5 \times 5^{16}$ is

$\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{14-6 \sqrt{5}}}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module