Maximum or Minimum value MCQ Quiz - Objective Question with Answer for Maximum or Minimum value - Download Free PDF

Calculation:

Given,

The expression for p  is:

\( p = |\sin \alpha - \sin(\alpha - 90^\circ)| \)

Using the identity for \(\sin(\alpha - 90^\circ) \)

\( \sin(\alpha - 90^\circ) = \cos \alpha \)

Substituting this into the expression for p :

\( p = |\sin \alpha - \cos \alpha| \)

The expression  \(|\sin \alpha - \cos \alpha|\)  achieves its maximum value when the difference between \(\sin \alpha \) and \(\cos \alpha \) is largest.

We can rewrite \(\sin \alpha - \cos \alpha \) as:

\( \sin \alpha - \cos \alpha = \sqrt{2} \left( \sin \left( \alpha - 45^\circ \right) \right) \)

Since \(\sin(\alpha - 45^\circ) \) has a maximum value of 1, the maximum value of \(\sqrt{2} \left( \sin(\alpha - 45^\circ) \right) \) is:

\( \sqrt{2} \)

Hence, the correct answer is Option 2

Calculation:

Given,

We are given the expression:

\( p = | \sin \alpha - \sin(\alpha - 90^\circ) | \)

Using the identity for sine of a difference:

\( \sin(\alpha - 90^\circ) = \cos \alpha \)

Substituting this identity into the expression for p :

\( p = | \sin \alpha - \cos \alpha | \)

The minimum value of \( | \sin \alpha - \cos \alpha | \) occurs when \(\sin \alpha = \cos \alpha \), which happens when\(\alpha = 45^\circ \). At this point:

\( \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \)

Therefore, at \(\alpha = 45^\circ \), the difference between \(\sin \alpha \) and \(\cos \alpha \) is 0, so:

\( p = 0 \)

∴ The minimum value of p  is 0.

Hence, the correct answer is option (a) 0.

\(\mathrm { y } = 3 \cos \theta + 5 \left( \sin \theta \dfrac { \sqrt { 3 } } { 2 } - \cos \theta \dfrac { 1 } { 2 } \right)\)

\(\dfrac { 5 \sqrt { 3 } } { 2 } \sin \theta + \dfrac { 1 } { 2 } \cos \theta\)

\(y _ { \max } = \sqrt { \dfrac { 75 } { 4 } + \dfrac { 1 } { 4 } } = \sqrt { 19 }\)

Concept:

• 1 + 2 + 3 + ... + n = \(n(n+1) \over 2\)
• 12 + 22 + 32 + ...+ n² = \(n(n+1)(2n+1) \over 6\)
• f(x) is minimum at x = a if f '(a) = 0 and f "(a) > 0

Calculation:

Given, f(x) = \(\displaystyle\sum_{j=1}^7\)(x − j)2 

⇒ f(x) = (x − 1)2 + (x − 2)2 + (x − 3)2 + (x − 4)2 + (x − 5)2 + (x − 6)2 + (x − 7)2 

⇒ f(x) = (x² − 2x + 1²) + (x2 − 4x + 2²) + (x2 − 6x + 3²) + (x2 − 8x + 4²) + (x2 − 10x + 5²) + (x2 − 12x + 6²) +(x2 − 14x + 7²)

⇒ f(x) = 7x² - 2(1 + 2 + 3 + 4 + 5 + 6 + 7)x + (1² + 22 + 32 + 42 + 52 + 62 + 72)

⇒ f(x) = 7x2 - 2(\(7(7+1) \over 2\))x + (\(7(7+1)(2(7)+1) \over 6\))

⇒ f(x) = 7x2 - 2(7(4))x + (\(7(8)(15) \over 6\))

⇒ f(x) = 7x2 - 56x + 140

⇒ f '(x) = 14x - 56 

⇒ f "(x) = 14

So f '(x) = 0 at x = 4 and f "(4) > 0

⇒ f(x) is minimum at x = 4

∴ The correct option is (2).

Concept:

• 1 + tan²x = sec²
• 1 + cot²x = cosec²x
• f(x) is minimum at x = a, if f '(a) = 0 and f "(a) > 0
• Chain Rule: (f(g(x))' = f '(g(x)).g'(x)
• (f(x)g(x))' = f '(x).g(x) + f(x).g'(x)

Calculation:

Given: f(x) = (p sec x)2 + (q cosec x)2 

⇒ f(x) = p² sec²x + q² cosec²x

⇒ f(x) = p2 + p2tan2x + q2 + q2cot2x

⇒ f '(x) = 0 + p2(2tan x sec²x) + q2(2cot x (-cosec²x)

⇒ f '(x) = 2p2tan x sec2x - 2q2cot x cosec2​x __(i)

Again differentiating (i),

⇒ f ''(x) = 2p2((tan x)' sec2x + tan x (sec²x)') - 2q2((cot x)' cosec2​x + cot x (cosec²x)')

⇒ f ''(x) = 2p2(sec4x + tan x ( 2sec x tan x sec x)) - 2q2(-cosec4​x + cot x (-2cosec x cosec x cot x))

⇒ f ''(x) = 2p2(sec4x + 2tan²x sec²x) + 2q2(cosec4​x + 2cot²x cosec²x )

⇒ f :"(x) > 0 for every x {∵ every term is of even power}

And from (i),

f ' (x) = 0 when 2p2tan x sec2x - 2q2cot x cosec2​x = 0

⇒ p2tan x sec2x = q2cot x cosec2​x

⇒ \({\tan x \sec^2 x \over \cot x\ \text {cosec}^2 x} = {q^2 \over p^2}\)

⇒ \({\tan^2 x \sin^2 x \over \text {cos}^2 x} = {q^2 \over p^2}\)

⇒ \({\tan^4 x } = {q^2 \over p^2}\)

⇒ tan2 x = \(\frac{q}{p}\)

so when tan2 x = \(\frac{q}{p}\), f "(x) = 0 and f "(x) > 0

⇒ f(x) is minimum when tan2 x = \(\frac{q}{p}\)

∴ The correct option is (1).

Calculation:

Given: f(x) = 4 sin² x + 1

To find: Maximum value of f(x)

We know that, range of sin x is [-1, 1], i.e.,

-1 ≤ sin x ≤ 1

Squaring the equalities, we get,

⇒ 0 ≤ sin² x ≤ 1

⇒ 0 ≤ 4sin² x ≤ 4

⇒ 1 ≤ 4sin² x + 1 ≤ 5

⇒ 1 ≤ f(x) ≤ 5

Hence, maximum value of f(x) is 5.

Concept:

-1 ≤ sin θ ≤ 1

sin 2 θ = 2 sin θ cos θ

Calculation

We have to find the minimum value of 1 − 4 sin x cos x

Let f(x) = 1 − 4 sin x cos x

= 1 – 2 × (2 sin x cos x)

= 1 – 2 sin 2x                   (∵sin 2θ = 2 sin θ cos θ)

For minimum value of f(x), sin 2x should be maximum

As we know that,

Maximum value of sin θ is 1, so maximum value of sin 2x is 1

Therefore, Minimum value of f(x) = 1 – 2 × 1 = -1

Concept:

-1 ≤ sin θ ≤ 1

0 ≤ sin² θ ≤ 1

sin 2 θ = 2 sin θ cos θ

Calculation

We have to find the maximum value of 1 − 4 sin² x cos² x

Let f(x) = 1 − 4 sin² x cos² x

= 1 – (2 sin x cos x)²

= 1 – sin² 2x                   (∵sin 2θ = 2 sin θ cos θ)

For maximum value of f(x), sin 2x should be minimum

As we know that,

Minimum value of sin² θ is 0

Therefore, Maximum value of f(x) = 1 – 0 = 1

Concept:

If the trigonometric ratio is in the form of 'asin θ + bcos θ'

Then the maximum value = \(\rm \sqrt {a^2+b^2}\)

And the minimum value = - \(\rm \sqrt {a^2+b^2}\)

Calculation:

15sin θ + 20cos θ

Here a = 15  and b = 20

Then the maximum value = \(\rm \sqrt {{15}^2+20^2}\)

= \(\sqrt{225+400}\)

= √(625)

= 25 

∴ The maximum value of 15sin θ + 20cos θ is 25.

Concept:

The minimum value of cos θ is –1 when θ = 180°. So, the range of values of cos θ is – 1 ≤ cos θ ≤ 1.

Calculation:

Since the minimum value of cosθ is -1.

So, the minimum value of 4 cosθ + 3 = 4(-1) + 3 

⇒ -4 + 3

⇒ -1

Hence, the minimum value of 4 cosθ + 3 is -1.

Concept:

a² + b² + 2ab = (a + b)²

\(\rm sin^2 x + \cos^2 x =1\)

2 sin x cos x = sin 2x

sin² x lies between [0, 1] ⇔  0 ≤ sin² x ≤ 1

Calculation:

To Find: maximum value of sin⁴ x + cos⁴ x

\(\text {Let f(x)}=\rm \sin^4 x + \cos^4 x\\= (sin^2 x)^2 + (\cos^2 x)^2\\= (sin^2 x)^2 + (\cos^2 x)^2+ 2sin^2 x \cos^2 x- 2sin^2 x \cos^2 x\)

\(\rm =(\sin^2 x + \cos^2 x)^2 - 2sin^2 x \cos^2 x\)            (∵ a² + b² + 2ab = (a + b)²)

\(\rm =1 - \frac{1}{2}\times 4sin^2 x \cos^2 x\)                              \(\rm (∵ sin^2 x + \cos^2 x =1)\)

\(\rm =1 - \frac{1}{2}\times (2\sin x \cos x)^2\)

\(\rm =1 - \frac{1}{2}\times \sin^2 2x\)                                        (∵ 2 sin x cos x = sin 2x)

As we know sin² x lies between [0, 1]

So 0 ≤ sin² 2x ≤ 1

We need to take the minimum value of sin x as there is a minus sign in front of the sine term.

So we need to take it as zero.

∴ Maximum value of f(x) = 1 - 0 = 1

Calculation:

Let f(θ) = 7cos2θ + 5sin2θ 

= 2cos2θ + 5cos2θ + 5sin2θ 

= 2cos2θ + 5(cos2θ + sin2θ)

= 2cos2θ + 5                        [∵ cos2θ + sin2θ = 1]

As we know, 0 ≤ cos2θ ≤ 1

⇒  0 ≤ 2cos2θ ≤ 2

⇒  0 + 5 ≤ 2cos2θ + 5 ≤ 2 + 5

⇒ 5 ≤ 2cos2θ + 5 ≤ 7

⇒ 5 ≤ f(θ) ≤ 7

So, Maximum value of function is 7.

Alternate Method

Concept:

If acos2θ + bsin2θ  where, a > b

a will be the maximum value and b will be the minimum value

Calculation:

7cos2θ + 5sin2θ 

According to concept 

acos2θ + bsin2θ  where, a > b

On comparing

⇒ a = 7, b = 5 where, 7 > 5

∴ maximum value of 7cos2θ + 5sin2θ is 7

Concept:

Maximum value of sin x is 1

Maximum value of cos x is 1

sin 2x = 2 sin x cos x

Calculation:

To Find: maximum value of sin x ⋅ cos x

Let f(x) = sin x ⋅ cos x

\(\rm = \dfrac{1}{2} \times (2\sin x \times \cos x) \\ = \dfrac{1}{2} \sin 2x\)

As we know, the maximum value of sin x is 1

Therefore the maximum value of sin 2x is 1

Hence maximum value of f(x) is \(\dfrac{1}{2} \times 1 = \dfrac{1}{2}\)

Concept:

• sin3x = 3sinx - 4sin³x
• cos3x = 4cos³x - 3cosx

Calculation:

f(x) = 3(sin x − cos x) + 4(cos3 x − sin3 x)

f(x) = 3sinx - 3cosx + 4cos³x - 4sin³x

f(x) = (3sinx - 4sin³x) + (4cos³x - 3cosx)

f(x) = sin3x + cos3x

f'(x) = 3cos3x - 3sin3x

Now, f'(x) = 0

⇒  3cos3x - 3sin3x = 0

⇒ cos3x - sin3x = 0

⇒ sin3x = cos3x

⇒ tan3x = 1

⇒ \(3x = \frac{\pi }{4}\)

Hence, the maximum value of given function is = \(sin\frac{\pi }{4}+cos\frac{\pi }{4}\)

= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)

∴ The correct option is (2)

Concept:

Value of sin x lies between [-1, 1]

Calculation:

Let f(x) = (5 + sin 2x)

As we know, value of sin x lies between [-1, 1]

⇒ -1 ≤ sin 2x ≤ 1

Adding 5 in above inequality, we get

⇒ 5 -1 ≤ 5 + sin 2x ≤ 5 + 1

⇒ 4 ≤ 5 + sin 2x ≤ 6

Therefore, Maximum value of 5 + sin 2x is 6 and Minimum value of 5 + sin 2x is 4.