Let f(x) = <math> <mrow> <mn>2</mn> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>x</mi> <mo>+</mo> <mn>4</mn> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>x</mi> <mo>−</mo> <mn>3</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mi>x</mi> <mo>∈</mo> <mrow> <mo>[</mo> <mrow> <mo>−</mo> <mn>1</mn> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mn>1</mn> </mrow> <mo>]</mo> </mrow> <mo>.</mo> <mtext> </mtext> <mtext> </mtext> <mi>I</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mrow> <mo>[</mo> <mrow> <mi>a</mi> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mi>b</mi> </mrow> <mo>]</mo> </mrow> </mrow> </math> is the range of the function f, then 4a – b is equal to:

f ( x ) = 2 c o s − 1 x + 4 c o t − 1 x − 3 x 2 − 2 x + 1 0 ∀ x ∈ [ − 1 , 1 ] ⇒ f ' ( x ) = − 2 1 − x 2 − 4 1 + x 2 − 6 x − 2 < 0 ∀ x ∈ [ − 1 , 1 ] So, f (x) is decreasing function and range of f (x) is [ f ( 1 ) , f ( − 1 ) ] , which is [ π + 5 , 5 π + 9 ] Now 4a – b = 4 (p + 5) - (5p + 9) = 11 - “π”

Let f(x) = <math> <mrow> <mn>2</mn> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>x</mi> <mo>+</mo> <mn>4</mn> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>x</mi> <mo>−</mo> <mn>3</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mi>x</mi> <mo>∈</mo> <mrow> <mo>[</mo> <mrow> <mo>−</mo> <mn>1</mn> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mn>1</mn> </mrow> <mo>]</mo> </mrow> <mo>.</mo> <mtext> </mtext> <mtext> </mtext> <mi>I</mi> <mi>f</mi> <mtext> </mtext> <mtext> </mtext> <mrow> <mo>[</mo> <mrow> <mi>a</mi> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mi>b</mi> </mrow> <mo>]</mo> </mrow> </mrow> </math> &nbsp;is the range of the function f, then 4a – b is equal to:

11 - “π”

11 + “π”

15 - “π”

Ask & Answer Question

Maths Class 12th Inverse Trigonometric Functions

Let f(x) = $2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1] . I f [a, b]$ is the range of the function f, then 4a – b is equal to:

Option 1 -

Option 2 -

11 - “π”

Option 3 -

11 + “π”

Option 4 -

15 - “π”

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 2

Detailed Solution:

$f (x) = 2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10 \forall x \in [- 1, 1]$

$\Rightarrow f' (x) = - \frac{2}{\sqrt[]{1 - x^{2}}} - \frac{4}{1 + x^{2}} - 6 x - 2 < 0 \forall x \in [- 1, 1]$

So, f (x) is decreasing function and range of f (x) is

$[f (1), f (- 1)],$ which is $[π + 5, 5 π + 9]$

Now 4a – b = 4 (p + 5) - (5p + 9) = 11 - “π”

<math> <mrow> <mi>f</mi> <mrow> <mo> (</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>2</mn> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>s</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>x</mi> <mo>+</mo> <mn>4</mn> <mi>c</mi> <mi>o</mi> <msup> <mrow> <mi>t</mi> </mrow> <mrow> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mi>x</mi> <mo>−</mo> <mn>3</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mo>∀</mo> <mi>x</mi> <mo>∈</mo> <mrow> <mo> [</mo> <mrow> <mo>−</mo> <mn>1</mn> <mo>, </mo> <mn>1</mn> </mrow> <mo>]</mo> </mrow> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>f</mi> <mo>'</mo> <mrow> <mo> (</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mo>−</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mroot> <mrow> <mn>1</mn> <mo>−</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mn>1</mn> <mo>+</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>−</mo> <mn>6</mn> <mi>x</mi> <mo>−</mo> <mn>2</mn> <mo><</mo> <mn>0</mn> <mo>∀</mo> <mi>x</mi> <mo>∈</mo> <mrow> <mo> [</mo> <mrow> <mo>−</mo> <mn>1</mn> <mo>, </mo> <mn>1</mn> </mrow> <mo>]</mo> </mrow> </mrow> </math> So, f (x) is decreasing function and range of f (x) is <math> <mrow> <mrow> <mo> [</mo> <mrow> <mi>f</mi> <mrow> <mo> (</mo> <mrow> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mo>, </mo> <mi>f</mi> <mrow> <mo> (</mo> <mrow> <mo>−</mo> <mn>1</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mo>]</mo> </mrow> <mo>, </mo> </mrow> </math> which is <math> <mrow> <mrow> <mo> [</mo> <mrow> <mi>π</mi> <mo>+</mo> <mn>5</mn> <mo>, </mo> <mn>5</mn> <mi>π</mi> <mo>+</mo> <mn>9</mn> </mrow> <mo>]</mo> </mrow> </mrow> </math> Now 4a – b = 4 (p + 5) - (5p + 9) = 11 - “π”

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Let f(x) = $2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1] . I f [a, b]$ is the range of the function f, then 4a – b is equal to:

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Let f(x) = 2 c o s − 1 x + 4 c o t − 1 x − 3 x 2 − 2 x + 1 0 , x ∈ [ − 1 , 1 ] . I f [ a , b ] is the range of the function f, then 4a – b is equal to:

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Let f(x) = $2 c o s^{- 1} x + 4 c o t^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1] . I f [a, b]$ is the range of the function f, then 4a – b is equal to: