Maximum or Minimum value MCQ Quiz in मराठी - Objective Question with Answer for Maximum or Minimum value - मोफत PDF डाउनलोड करा

We have,

Given function (F) = (x2 – 6x + 13)

It can be expressed as,

(x2 – 6x + 13) = (x2 – 6x + 9 + 4)

or, F = x² - 3x - 3x + 9 + 4

or, F = x(x - 3) - 3(x - 3) + 4

or, F = (x - 3)² + 4

Minimum value of function (F) will be occurs when,

x = 3

In that case,

F = Minimum Value = 0 + 4 = 4

Concept:

• 1 + 2 + 3 + ... + n = $\frac{n(n+1)}{2}$
• 12 + 22 + 32 + ...+ n² = $\frac{n(n+1)(2n+1)}{6}$
• 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($\frac{7(7+1)}{2}$)x + ($\frac{7(7+1)(2(7)+1)}{6}$)

⇒ f(x) = 7x2 - 2(7(4))x + ($\frac{7(8)(15)}{6}$)

⇒ 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).

From question, the expression given is:

$3cos\theta +5sin\left(\theta -\frac{\pi }{6}\right)$

$\Rightarrow 3cos\theta +5\left(sin\theta cos\frac{\pi }{6}-sin\frac{\pi }{6}cos\theta \right)$

$\Rightarrow 3cos\theta +5\left(\frac{\sqrt{3}}{2}sin\theta -\frac{1}{2}cos\theta \right)$

$\Rightarrow 3cos\theta -\frac{5}{2}cos\theta +\frac{5\sqrt{3}}{2}sin\theta$

$\therefore \frac{1}{2}cos\theta +\frac{5\sqrt{3}}{2}sin\theta$

The maximum value of a cos θ + b sin θ is $\sqrt{a^2+b^2}$.

The maximum value of the obtained equation $\frac{1}{2}cos\theta +\frac{5\sqrt{3}}{2}sin\theta$ is:

Concept:

The function f(x) has minimum value at x = a, if f '(a) = 0 and f "(a) > 0

Formula used:

• ${\log }_{b}a=\frac{\ln a}{\ln b}$
• $\frac{d}{dx}(\frac{p}{q})=\frac{p^{\prime }q-p{q}^{\prime }}{q^2}$

Calculation:

Given, f(x) = log10(x2 + 2x + 11) = $\frac{1}{\ln 10}$ ln (x2 + 2x + 11) 

⇒ f '(x) = $\frac{1}{\ln 10}$$\frac{1}{x^2+2x+11}$(x2 + 2x + 11)'

⇒ f '(x) = $\frac{1}{\ln 10}$$\frac{2x+2}{x^2+2x+11}$__(i)

⇒ f '(x) = 0 when 2x + 2 = 0 → x = -1

So, f '(-1) = 0

Now again differentiating (i),

⇒ f "(x) = $\frac{1}{\ln 10}$$(\frac{2(x^2+2x+11)-(2x+2)(2x+2)}{(x^2+2x+11{)}^2})$ 

⇒ f "(x) = $\frac{1}{\ln 10}$$(\frac{2x^2+4x+22-4x^2-8x-4}{(x^2+2x+11{)}^2})$  

⇒ f "(x) = $\frac{1}{\ln 10}$$(\frac{-2x^2-4x+18}{(x^2+2x+11{)}^2})$ 

⇒ f "(-1) = $\frac{1}{\ln 10}$$(\frac{-2-4(-1)+18}{(1+2(-1)+11{)}^2})$ 

⇒ f "(-1) = $\frac{1}{\ln 10}$$\times \frac{20}{100}$ 

⇒  f "(-1) = $\frac{1}{5\ln 10}$> 0

So, f '(-1) = 0 and f "(-1) > 0

⇒ f(x) has minimum value at x = -1

The minimum value of f(x) = f(-1) = log10((-1)2 + 2(-1) + 11)

⇒ The minimum value of f(x) = log10(10) = 1

∴ The correct option is (2).

Calculation:

Given:

f(θ) = 4(sin² θ + cos⁴ θ)

⇒ f(θ) = 4(1 - cos² θ + cos⁴ θ)

Which is a quadratic equation in cos²θ

$\Rightarrow \mathrm{f}\left(\theta \right)=4\left[1-{\cos }^2\theta (1-{\cos }^2\theta \right)]=4\left(1-{\sin }^2\theta {\cos }^2\theta \right)$

$\Rightarrow \mathrm{f}\left(\theta \right)=4\left(1-{\left(\sin \theta \cos \theta \right)}^2\right)=4-{\left(2\sin \theta \cos \theta \right)}^2=4-\mathrm{s}\mathrm{i}{\mathrm{n}}^{2}2\theta$

We know the range of sin 2θ:

-1 ≤ sin2θ ≤ 1

⇒ 0 ≤ sin² 2θ ≤ 1

⇒ -1 ≤ -sin²2θ ≤ 0

⇒ 3 ≤ 4 -sin²2θ ≤ 4

⇒ 3 ≤ f(θ) ≤ 4

Minimum value = 3

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

⇒ $\frac{\tan x{\sec }^{2}x}{\cot x{\text{cosec}}^{2}x}=\frac{q^2}{p^2}$

⇒ $\frac{{\tan }^{2}x{\sin }^{2}x}{{\text{cos}}^{2}x}=\frac{q^2}{p^2}$

⇒ ${\tan }^{4}x=\frac{q^2}{p^2}$

⇒ 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).

Concept:

• cos2θ = 1 - 2sin2θ
• 0 ≤ sin2θ ≤ 1

Explanation:

We have  y = 4sin2θ − cos2θ

⇒ y = 4sin2θ − (1 - 2sin²θ) 

⇒ y = 4sin2θ − 1 + 2sin2θ

⇒ y = 6sin2θ − 1

⇒ sin2θ = $\frac{y+1}{6}$

Since, 0 ≤ sin2θ ≤ 1

Thus, 0 ≤ $\frac{y+1}{6}$ ≤ 1

⇒ 0 ≤ y + 1 ≤ 6

⇒ 0 - 1 ≤ y ≤ 6 -1

⇒ - 1 ≤ y ≤ 5 

that is, the minimum value of y is - 1 and the maximum value of y is 5

Now, we are given that l and m are the minima and maximum values of y respectively, then

l = - 1 and m = 5

Thus, let us check each option:

Option 1)

LHS = lm = -1 × 5 = - 5 = $\frac{5}{-1}=\frac{m}{l}$ = RHS

that option 1 is satisfied.

Option 2)

LHS = lm = -1 × 5 = - 5 ≠  $\frac{-1}{5}\ne \frac{l}{m}$  

that is option 2 is NOT satisfied.

Option 3)

LHS = l + m = -1 + 5 = 4 ≠ $\frac{-1}{5}\ne \frac{l}{m}$ = RHS

that is option 3 is NOT satisfied.

Option 4)

LHS = $\frac{lm}{l-m}=\frac{(-1)(5)}{-1+5}=\frac{-5}{4}$ ≠ 1 + 5 ≠ 1 + m

that is option 4 is NOT satisfied.

Concept:

Trigonometric Expressions of the form A sin x ± B cos x:

• We define sin y = $\frac{\mathrm{A}}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}$ and cos y = $\frac{\mathrm{A}}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}$, such that y = ${\tan }^{-1}{\displaystyle \frac{\mathrm{A}}{\mathrm{B}}}$.
• The expression simplifies to $\cos (\mathrm{x}\pm \mathrm{y})=\frac{\mathrm{C}}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}$.

Calculation:

Given that f(x) = 3 sin x + 4 cos x.

Dividing by $\frac{1}{\sqrt{3^2+4^2}}$, we get:

⇒ $\frac{\mathrm{f}(\mathrm{x})}{\sqrt{3^2+4^2}}$ = $\frac{3}{\sqrt{3^2+4^2}}$ sin x + $\frac{4}{\sqrt{3^2+4^2}}$ cos x

Let $\frac{3}{\sqrt{3^2+4^2}}$ = cos y and $\frac{4}{\sqrt{3^2+4^2}}$ = sin y.

⇒ $\frac{\mathrm{f}(\mathrm{x})}{5}$ = cos y sin x + sin y cos x

⇒ f(x) = 5[sin (x + y)]

Since -1 ≤ sin θ ≤ 1, the maximum value of f(x) is 5(1) = 5.

Concept:

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

Then the maximum value = $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

And the minimum value = - $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}$

Calculation:

15sin θ + 20cos θ

Here a = 15  and b = 20

Then the maximum value = $\sqrt{{15}^2+{20}^2}$

= $\sqrt{225+400}$

= √(625)

= 25 

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

Concept:

Trigonometric Expressions of the form A sin x ± B cos x:

• We define sin y = ${\displaystyle \frac{\mathrm{A}}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}}$ and cos y = ${\displaystyle \frac{\mathrm{A}}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}}$, such that y = ${\tan }^{-1}{\displaystyle \frac{\mathrm{A}}{\mathrm{B}}}$.
• The expression simplifies to $\cos (\mathrm{x}\pm \mathrm{y})={\displaystyle \frac{\mathrm{C}}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}}$.

Calculation:

Given that f(x) = 3 sin x + 4 cos x.

Dividing by ${\displaystyle \frac{1}{\sqrt{3^2+4^2}}}$, we get:

⇒ ${\displaystyle \frac{\mathrm{f}(\mathrm{x})}{\sqrt{3^2+4^2}}}$ = ${\displaystyle \frac{3}{\sqrt{3^2+4^2}}}$ sin x + ${\displaystyle \frac{4}{\sqrt{3^2+4^2}}}$ cos x

Let ${\displaystyle \frac{3}{\sqrt{3^2+4^2}}}$ = cos y and ${\displaystyle \frac{4}{\sqrt{3^2+4^2}}}$ = sin y.

⇒ ${\displaystyle \frac{\mathrm{f}(\mathrm{x})}{5}}$ = cos y sin x + sin y cos x

⇒ f(x) = 5[sin (x + y)]

Since -1 ≤ sin θ ≤ 1, the maximum value of f(x) is 5(1) = 5.