The function f(x) = x³ – 6x² + ax + b is such that f(2) = f(4) = 0.

Consider two statements.

(S1) there exists $x_{1}, x_{2} \in (2, 4), x_{1} < x_{2}, s u c h t h a t f' (x_{1}) = - 1 a n d f' (x_{2}) = 0 .$

(S2) there exists $x_{3}, x_{4} \in (2, 4), x_{3}, x_{4},$ such that f is decreasing in (2, x₄), increasing in $(x_{4}, 4) a n d 2 f' (x_{3}) = \sqrt[]{3} f (x_{4}) .$

Option 1 -

Both (S1) and (S2) are true

Option 2 -

(S1) is true and (S2) is false

Option 3 -

(S1) is false and (S2) is true

Option 4 -

Both (S1) and (S2) are false

10 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 1

Detailed Solution:

$f (2) = f (4) = 0 a n d f (x) = x^{3} - 6 x^{2} + a x + b$

$\Rightarrow f (x) = (x - 2) (x - 4) x = x^{3} - 6 x^{2} + 8 x$

->a = 8, b = 0

$f' (x) = 0 \Rightarrow x = 2 \pm \frac{2}{\sqrt[]{3}}, x_{4} = 2 + \frac{2}{\sqrt[]{3}}, f (x_{4}) = \frac{- 16}{3 \sqrt[]{3}}$

$f' (x_{3}) = \frac{\sqrt[]{3}}{2} f (x_{4}) = \frac{- 8}{3}$

<math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>0</mn> <mtext> </mtext> <mtext> </mtext> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mtext> </mtext> <mtext> </mtext> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>−</mo> <mn>6</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mi>a</mi> <mi>x</mi> <mo>+</mo> <mi>b</mi> </mrow> </math>   <math> <mrow> <mo>⇒</mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>2</mn> </mrow> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>4</mn> </mrow> <mo>)</mo> </mrow> <mi>x</mi> <mo>=</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>−</mo> <mn>6</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>8</mn> <mi>x</mi> </mrow> </math>               ->a = 8, b = 0 <math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mn>0</mn> <mo>⇒</mo> <mi>x</mi> <mo>=</mo> <mn>2</mn> <mo>±</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> <mo>,</mo> <msub> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> <mo>=</mo> <mn>2</mn> <mo>+</mo> <mfrac> <mrow> <mn>2</mn> </mrow> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> <mo>,</mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mn>1</mn> <mn>6</mn> </mrow> <mrow> <mn>3</mn> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math>            <math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mroot> <mrow> <mn>3</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <mo>−</mo> <mn>8</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

0 Upvotes 0 Downvotes

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

1 Answer

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