Let f(x) = <math> <mrow> <mrow> <mo>{</mo> <mrow> <mtable> <mtr columnalign="center"> <mtd columnalign="center"> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>−</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mi>x</mi> <mo>−</mo> <mn>7</mn> <mtext> </mtext> <mtext> </mtext> <mo>,</mo> <mi>x</mi> <mo>≤</mo> <mn>1</mn> </mtd> <mtd columnalign="center"> <mo>−</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mo>,</mo> <mi>x</mi> <mo>></mo> <mn>1</mn> </mtd> </mtr> </mtable> <mo>.</mo> </mrow> </mrow> </mrow> </math> Then the set of all values of b, for which f(x) has maximum value at x = 1, is:

f ( x ) = { x 3 − x 2 + 1 0 x − 7 , x ≤ 1 − 2 x + l o g 2 ( b 2 − 4 ) , x > 1 If f(x) has maximum value at x = 1 then f ( 1 ) ≤ f ( 1 ) ⇒ − 2 + l o g 2 ( b 2 − 4 ) ≤ 1 − 1 + 1 0 − 7 ⇒ l o g 2 ( b 2 − 4 ) ≤ 5 ⇒ 0 0 ⇒ b ∈ ( − ∞ , − 2 ) ∪ ( 2 , ∞ ) …….(i) A n d b 2 − 4 ≤ 3 2 ⇒ b ∈ [ − 6 , 6 ] …….(ii)From (i) and (ii) we get b ∈ [ − 6 , − 2 ) ∪ ( 2 , 6 ]

Let f(x) = <math> <mrow> <mrow> <mo>{</mo> <mrow> <mtable> <mtr columnalign="center"> <mtd columnalign="center"> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>−</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mi>x</mi> <mo>−</mo> <mn>7</mn> <mtext> </mtext> <mtext> </mtext> <mo>,</mo> <mi>x</mi> <mo>≤</mo> <mn>1</mn> </mtd> <mtd columnalign="center"> <mo>−</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mo>,</mo> <mi>x</mi> <mo>&gt;</mo> <mn>1</mn> </mtd> </mtr> </mtable> <mo>.</mo> </mrow> </mrow> </mrow> </math> &nbsp;Then the set of all values of b, for which f(x) has maximum value at x = 1, is:

Ask & Answer Question

Maxima and Minima Maths Applications of Derivatives Maths Class 12th

Let f(x) = ${\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 & - 2 x + l o g_{2} (b^{2} - 4), x > 1 \end{matrix} .$ Then the set of all values of b, for which f(x) has maximum value at x = 1, is:

Option 1 -

(-6, -2)

Option 2 -

(2, 6)

Option 3 -

$[- 6, - 2) \cup (2, 6]$

Option 4 -

$[- \sqrt[]{6}, - 2) \cup (2, \sqrt[]{6}]$

2 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 weeks ago

Correct Option - 3

Detailed Solution:

$f (x) = {\begin{matrix} x^{3} - x^{2} + 10 x - 7, & x \leq 1 \\ - 2 x + l o g_{2} (b^{2} - 4), & x > 1 \end{matrix}$

If f(x) has maximum value at x = 1 then

$f (1) \leq f (1) \Rightarrow - 2 + l o g_{2} (b^{2} - 4) \leq 1 - 1 + 10 - 7$

$\Rightarrow l o g_{2} (b^{2} - 4) \leq 5 \Rightarrow 0 < b^{2} - 4 \leq 32$

$\Rightarrow b^{2} - 4 > 0 \Rightarrow b \in (- \infty, - 2) \cup (2, \infty)$ …….(i)

$A n d b^{2} - 4 \leq 32 \Rightarrow b \in [- 6, 6]$ …….(ii)

From (i) and (ii) we get $b \in [- 6, - 2) \cup (2, 6]$

<math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mrow> <mo>{</mo> <mrow> <mtable> <mtr columnalign="center"> <mtd columnalign="center"> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>−</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mi>x</mi> <mo>−</mo> <mn>7</mn> <mo>,</mo> </mtd> <mtd columnalign="center"> <mi>x</mi> <mo>≤</mo> <mn>1</mn> </mtd> </mtr> <mtr columnalign="center"> <mtd columnalign="center"> <mo>−</mo> <mn>2</mn> <mi>x</mi> <mo>+</mo> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mo>,</mo> </mtd> <mtd columnalign="center"> <mi>x</mi> <mo>></mo> <mn>1</mn> </mtd> </mtr> </mtable> </mrow> </mrow> </mrow> </math>  If f(x) has maximum value at x = 1 then <math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mo>≤</mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mn>1</mn> </mrow> <mo>)</mo> </mrow> <mo>⇒</mo> <mo>−</mo> <mn>2</mn> <mo>+</mo> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mo>≤</mo> <mn>1</mn> <mo>−</mo> <mn>1</mn> <mo>+</mo> <mn>1</mn> <mn>0</mn> <mo>−</mo> <mn>7</mn> </mrow> </math> <math> <mrow> <mo>⇒</mo> <mi>l</mi> <mi>o</mi> <msub> <mrow> <mi>g</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mrow> <mo>(</mo> <mrow> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mo>≤</mo> <mn>5</mn> <mo>⇒</mo> <mn>0</mn> <mo><</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> <mo>≤</mo> <mn>3</mn> <mn>2</mn> </mrow> </math> <math> <mrow> <mo>⇒</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> <mo>></mo> <mn>0</mn> <mo>⇒</mo> <mi>b</mi> <mo>∈</mo> <mrow> <mo>(</mo> <mrow> <mo>−</mo> <mo>∞</mo> <mo>,</mo> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mo>∪</mo> <mrow> <mo>(</mo> <mrow> <mn>2</mn> <mo>,</mo> <mo>∞</mo> </mrow> <mo>)</mo> </mrow> </mrow> </math>         …….(i) <math> <mrow> <mi>A</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>−</mo> <mn>4</mn> <mo>≤</mo> <mn>3</mn> <mn>2</mn> <mo>⇒</mo> <mi>b</mi> <mo>∈</mo> <mrow> <mo>[</mo> <mrow> <mo>−</mo> <mn>6</mn> <mo>,</mo> <mn>6</mn> </mrow> <mo>]</mo> </mrow> </mrow> </math>                      …….(ii)From (i) and (ii) we get  <math> <mrow> <mi>b</mi> <mo>∈</mo> <mo stretchy="false">[</mo> <mo>−</mo> <mn>6</mn> <mo>,</mo> <mo>−</mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>∪</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>,</mo> <mn>6</mn> <mo stretchy="false">]</mo> </mrow> </math>

0 Upvotes 0 Downvotes

Similar Questions for you

Let f : R -> R be a function defined by f(x) = ${(x - 3)}^{n_{1}} {(x - 5)}^{n_{2}}, n_{1}, n_{2} \in N .$ Then which of the following is NOT true?

Vishal Baghel

$f' (x) = \frac{n_{1} \cdot f (x)}{x - 3} + n_{2} \cdot \frac{f (x)}{x - 5}$

$= \frac{f (x) \cdot (n_{1} + n_{2})}{(x - 3) (x - 5)} (x - \frac{(5 n_{1} + 3 n_{2})}{n_{1} + n_{2}})$

$f' (x) = {(x - 3)}^{n_{1} - 1} \cdot {(x - 5)}^{n_{2} - 1} \cdot (n_{1} + n_{2}) (x - \frac{(5 n_{1} + 3 n_{2})}{n_{1} + n_{2 l}})$

option (C) is incorrect, there will be minima.

The sum of the absolute minimum and the absolute maximum values of the friction f(x) = $| 3 x - x^{2} + 2 | - x$ in the interval [-1, 2] is:

Raj Pandey

$f (x) = | x^{2} - 3 x - 2 | - x$

$= | (x - \frac{3 - \sqrt[]{17}}{2}) (x - \frac{3 + \sqrt[]{17}}{2}) | - x$

$\Rightarrow f (x) = [\begin{matrix} x^{2} - 4 x - 2; - 1 \leq x \leq \frac{3 - \sqrt[]{17}}{2} & - x^{2} + 2 x + 2; \frac{3 - \sqrt[]{17}}{2} < x \leq 2 \end{matrix}]$

absolute minimum $f (\frac{3 - \sqrt[]{17}}{2}) = \frac{- 3 + \sqrt[]{17}}{2}$

absolute maximum = 3

$∴ s u m 3 + \frac{- 3 + \sqrt[]{17}}{2} = \frac{3 + \sqrt[]{17}}{2}$

Assume $e^{- \frac{4}{5}} = \frac{2}{5} .$ If x, y satisfy, y = e^x and the minimum value of (x² + y²) is expressed in the form of $\frac{m}{n}$ tehn $\frac{(2 m - n)}{5}$ equals (where m & n are coprime natural numbers)

Vishal Baghel

OP² = x² = y²

y = e^x, y’ = e^x,

slope of normal = $- \frac{1}{e^{x}}$

$\frac{y}{x} = - \frac{1}{e^{x}}$

$- \frac{1}{e^{x}} = \frac{e^{x}}{x} \Rightarrow - x = e^{2 x}$

By hit and trial we get $x = - \frac{2}{5}$

$P \equiv (- \frac{2}{5}, e^{- 2 / 5})$

$O P = \sqrt[]{\frac{4}{25} + e^{- 4 / 5}} \Rightarrow O P^{2} = \frac{14}{25} = \frac{m}{n}$

Let g(x) = ||x+2|-3|. If a denotes the number of relative minima, b denotes the number of relative maxima and c denotes the product of the zeros. Then the value of (a+2b-c) is

Vishal Baghel

c = -5, a = 2, b = 1

The lengths of the sides of a triangle are 10 + x², 10 + x² and 20 – 2x². If for x = k, the area of the triangle is maximum, then 3k² is equal to:

Vishal Baghel

CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 × CD × AB = 1/2 × 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Let f(x) = ${\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 & - 2 x + l o g_{2} (b^{2} - 4), x > 1 \end{matrix} .$ Then the set of all values of b, for which f(x) has maximum value at x = 1, is:

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Let f(x) = { x 3 − x 2 + 1 0 x − 7 , x ≤ 1 − 2 x + l o g 2 ( b 2 − 4 ) , x > 1 . Then the set of all values of b, for which f(x) has maximum value at x = 1, is:

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Let f(x) = ${\begin{matrix} x^{3} - x^{2} + 10 x - 7, x \leq 1 & - 2 x + l o g_{2} (b^{2} - 4), x > 1 \end{matrix} .$ Then the set of all values of b, for which f(x) has maximum value at x = 1, is: