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# Trigonometry Tricks – 4 [Maximum and Minimum Values] NABARD Grade A & B Online Course Get Course Now 150+ RRB NTPC Previous Solved Papers PDF Free Download Now RBI Grade B Complete Online Course Register Today RRB JE Free Study Material PDF Books Free Download Now RRB NTPC Free Study Material PDF Free Download Now All Exam Free Study Group (Free Videos,Pdf,Mock etc) Join Right Now

Few points to remember:

1. sinθ and cosθ both have “1” as their maximum value and “-1” as their minimum value. Hence the values of sinx, sin2x, cosx, cos3x, etc. lie between -1 and 1.
2. For sin2x and cos2x
• Minimum value = 0
• Maximum value = 1
3. For sinxcosx
• Minimum value = -1/2
• Maximum value = 1/2
4. Minimum value of (sin θ cos θ)n = (-1/2)n
5. Some SSC CGL questions :

1. 1) What is the least value of 2sin2θ + 3cos2θ

(A) 1            (B) 2             (C) 3            (D) 0

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Since 2 is less than 3
Minimum value = 2

Answer: (B)  Hence the least value of 4sec2θ + 9cosec2θ = 13 + 12 = 25

Answer: (C)

1. 4) The maximum of 3sinx – 4cosx is

(A) -1               (B) 5                (C) 7               (D) 9

Maximum value = √(3^2 + 4^2) = √25 = 5
Answer: (B)

Sometimes they ask the minimum/maximum value of a function. In CGL, that function would always be quadratic (ax2 + bx + c). Please note that a quadratic function can’t have both maximum and minimum values.

• If “a” is positive, then the quadratic function will only have a minimum value. The maximum value would be infinite.
• If “a” is negative, then the quadratic function will only have a maximum value. The minimum value would be infinite.

But the process to find both minimum and maximum values is same. Hence you shouldn’t be worried about the words “maximum” or “minimum”. When finding the minimum/maximum value of a function, we use the concept of “differentiation”. Although differentiation is a wide topic in itself, but for CGL purpose, we only have to learn the basics.

Just remember following things:

1. Differentiation of axn = a*n*xn-1. Hence differentiation of 4x3 = 12x2 and differentiation of 3x2 = 6x
2. Differentiation of ax = a. Hence differentiation of 4x is 4.
3. Differentiation of any constant is zero. Hence differentiation of 5 is 0.
4. If you have to differentiate ax2 + bx + c, just differentiate each of its term separately and add the result.
5. Process of finding the minimum value of a function :
1. Differentiate the function
2. What ever result you get, equate it with zero
3. Find the value of x
4. Put this value of x in the original function to get the minimum value.

Let us solve a CGL question:
Q. 5) Find the minimum value of (x – 2)(x – 9)
(A) -11/4             (B) 49/4             (C) 0            (D) -49/4

(x – 2)(x – 9) = x^2 – 11x + 18

Now we will differentiate (x^2 – 11x + 18)

Note that in this question, the value of a is positive, i.e., +1 and hence the question has asked you the “minimum” value. Had the value of “a” been negative (like -1), they would have asked you the “maximum” value. Hence, differentiation of (x^2 – 11x + 18) = 2x – 11

Now 2x – 11 = 0

x = 11/2

Put x = 11/2 in (x^2 – 11x + 18) to get the minimum value

Minimum value of x^2 – 11x + 18 = (11/2)^2 – 11*(11/2) + 18 = -49/4

Answer: (D)

If you have any doubt in this article, please drop a comment…
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