જો $x$ ની નાની કિંમતો માટે $\frac{(1 - 3x)^{1/2} + (1 - x)^{5/3}}{\sqrt{4 - x}}$ એ $a + bx$ ની આશરે સમાન હોય,તો $(a,b) = $

  • A
    $\left( 1, \frac{35}{24} \right)$
  • B
    $\left( 1, -\frac{35}{24} \right)$
  • C
    $\left( 2, \frac{35}{12} \right)$
  • D
    $\left( 2, -\frac{35}{12} \right)$

Explore More

Similar Questions

જો $x = \frac{3}{10} + \frac{3 \cdot 7}{10 \cdot 15} + \frac{3 \cdot 7 \cdot 9}{10 \cdot 15 \cdot 20} + \ldots$ હોય,તો $5x + 8 = $

જો $x=1+\frac{3}{1!} \times \frac{1}{6}+\frac{3 \times 7}{2!}\left(\frac{1}{6}\right)^2+\frac{3 \times 7 \times 11}{3!}\left(\frac{1}{6}\right)^3+\ldots$ હોય,તો $x^4$ ની કિંમત શોધો.

જો $y = \frac{3}{4} + \frac{3 \times 5}{4 \times 8} + \frac{3 \times 5 \times 7}{4 \times 8 \times 12} + \ldots$ અનંત સુધી હોય,તો

જો $x = \frac{2 \cdot 5}{2! \cdot 3} + \frac{2 \cdot 5 \cdot 7}{3! \cdot 3^2} + \frac{2 \cdot 5 \cdot 7 \cdot 9}{4! \cdot 3^3} + \ldots$ હોય,તો $x^2 + 8x + 8 = $

$1 + \frac{1}{4} + \frac{1 \times 3}{4 \times 8} + \frac{1 \times 3 \times 5}{4 \times 8 \times 12} + \dots = $

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo