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STD 11 - basic of algoritham — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - basic of algoritham4 Marks
MCQ
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$
  • A
    $A = 1$
  • B
    $B = -3$
  • C
    $C = 2$
  • ✓
    All of these
✓

Answer

Correct option: D.
All of these
d
(d) ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}} = {{7\,.\,({2^{1/4}} - 1)} \over {({2^{1/4}} - 1)\,[{{({2^{1/4}})}^2} + {2^{1/4}}.1 + {1^2}]}}$

$= {{7\,.\,({2^{1/4}} - 1)} \over {{2^{3/4}} - 1}} = A + B\,.\,{2^{1/4}} + C.\,{2^{1/2}} + D{.2^{3/4}}$

==> $7\,.\,{2^{1/4}} - 7 = (A - D)\,{2^{3/4}} + (2B - A) + (2C - B){.2^{1/4}}$$ + (2D - C){2^{1/2}}$

==> $(2B - A + 7) + (A - D){2^{3/4}} + (2C - B - 7){2^{1/4}}$ $+ (2D - C){2^{1/2}} = 0$

==> $2B - A + 7 = A - D = 2C - B - 7 = 2D - C = 0$

==> $A = D = 1,\,B = - 3,\,C = 2$.

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