



L-Glucose is the enantiomer of D-Glucose, which means it has the same structure as D-Glucose but with opposite configurations at each chiral center.
In L-Glucose, the -OH groups are positioned in such a way that it is a mirror image of D-Glucose.
For L-Glucose, the configuration of the hydroxyl (-OH) groups should be as follows from top to bottom:
Based on the given configurations, Option (1) matches the structure of L-Glucose.
The correct structure of L-Glucose is represented by Option (1).
Fat soluble vitamins are :
A. Vitamin B\( _1 \)
B. Vitamin C
C. Vitamin E
D. Vitamin B\( _{12} \)
E. Vitamin K
Choose the correct answer from the options given below :

Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: