Let f : R -&gt; R be a function defined by f(x) = <math> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>3</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </msup> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </msup> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>∈</mo> <mi>N</mi> <mo>.</mo> </mrow> </math> &nbsp;Then which of the following is NOT true?

For n1 = 3, n2 = 5, there exists <math> <mrow> <mi>α</mi> <mo>∈</mo> <mrow> <mo>(</mo> <mrow> <mn>3</mn> <mo>,</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> where f attains local maxima.

For n1 = 3, n2 = 4, there exists <math> <mrow> <mi>α</mi> <mo>∈</mo> <mrow> <mo>(</mo> <mrow> <mn>3</mn> <mo>,</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> where f attains local maxima.

For n1 = 4, n2 = 3, there exists <math> <mrow> <mi>α</mi> <mo>∈</mo> <mrow> <mo>(</mo> <mrow> <mn>3</mn> <mo>,</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> where f attains local minima.

For n1 = 4, n2 = 6, there exists <math> <mrow> <mi>α</mi> <mo>∈</mo> <mrow> <mo>(</mo> <mrow> <mn>3</mn> <mo>,</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> where f attains local maxima.

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Maths NCERT Exemplar Solutions Class 11th Chapter Three Maxima and Minima Maths Applications of Derivatives Maths Class 12th

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?

Option 1 -

For n₁ = 3, n₂ = 4, there exists $α \in (3, 5)$ where f attains local maxima.

Option 2 -

For n₁ = 4, n₂ = 3, there exists $α \in (3, 5)$ where f attains local minima.

Option 3 -

For n₁ = 3, n₂ = 5, there exists $α \in (3, 5)$ where f attains local maxima.

Option 4 -

For n₁ = 4, n₂ = 6, there exists $α \in (3, 5)$ where f attains local maxima.

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

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

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

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

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

<math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>⋅</mo> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mi>x</mi> <mo>−</mo> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>⋅</mo> <mfrac> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mi>x</mi> <mo>−</mo> <mn>5</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>⋅</mo> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>3</mn> </mrow> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mfrac> <mrow> <mrow> <mo>(</mo> <mrow> <mn>5</mn> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <mn>3</mn> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> <math> <mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>3</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mo>⋅</mo> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mn>5</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>−</mo> <mn>1</mn> </mrow> </msup> <mo>⋅</mo> <mrow> <mo>(</mo> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mrow> <mi>x</mi> <mo>−</mo> <mfrac> <mrow> <mrow> <mo>(</mo> <mrow> <mn>5</mn> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <mn>3</mn> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>n</mi> </mrow> <mrow> <mn>2</mn> <mi>l</mi> </mrow> </msub> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> </math> option (C) is incorrect, there will be minima.

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

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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:

Raj Pandey

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

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

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 $x = - \frac{2}{5}$

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

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

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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 : 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?

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Let f : R -> R be a function defined by f(x) = ( x − 3 ) n 1 ( x − 5 ) n 2 , n 1 , n 2 ∈ N . Then which of the following is NOT true?

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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?