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 -

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

4 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

a month 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>

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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

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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:

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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:

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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: