NIMCET 2015 — Mathematics PYQ
NIMCET | Mathematics | 2015If P=sin20θ+cos48θ, then the inequality that holds for all values of θ is:
Choose the correct answer:
- A.
P≥1
- B.
0 < P \le 1
(Correct Answer) - C.
1 < P < 3
- D.
0≤P≤1
0 < P \le 1
Explanation
1. Establish the basic range
For any real value of θ, we know:
-
0≤sin2θ≤1
-
0≤cos2θ≤1
Since P is a sum of even powers, P must be non-negative. If sinθ=0 and cosθ=1 (or vice versa), then P=1. If both sinθ and cosθ are non-zero, P will be positive. Thus, P > 0.
2. Compare with the fundamental identity
We know the identity:
Now, consider the individual terms of P:
-
Since 0≤sin2θ≤1, raising it to a higher power (like 10) makes it smaller or equal:
sin20θ=(sin2θ)10≤sin2θ -
Similarly, since 0≤cos2θ≤1:
cos48θ=(cos2θ)24≤cos2θ
3. Combine the inequalities
Adding the two inequalities together:
4. Determine the lower bound
The only way P could be 0 is if sinθ=0 and cosθ=0 simultaneously, which is impossible because sin2θ+cos2θ=1. Therefore, P is always strictly greater than 0.
Combining these findings:
Correct Option: (b)
Explanation
1. Establish the basic range
For any real value of θ, we know:
-
0≤sin2θ≤1
-
0≤cos2θ≤1
Since P is a sum of even powers, P must be non-negative. If sinθ=0 and cosθ=1 (or vice versa), then P=1. If both sinθ and cosθ are non-zero, P will be positive. Thus, P > 0.
2. Compare with the fundamental identity
We know the identity:
Now, consider the individual terms of P:
-
Since 0≤sin2θ≤1, raising it to a higher power (like 10) makes it smaller or equal:
sin20θ=(sin2θ)10≤sin2θ -
Similarly, since 0≤cos2θ≤1:
cos48θ=(cos2θ)24≤cos2θ
3. Combine the inequalities
Adding the two inequalities together:
4. Determine the lower bound
The only way P could be 0 is if sinθ=0 and cosθ=0 simultaneously, which is impossible because sin2θ+cos2θ=1. Therefore, P is always strictly greater than 0.
Combining these findings:
Correct Option: (b)