<p><strong>Let <math><mrow><mi>λ</mi></mrow></math> be the largest value of <math><mrow><mi>λ</mi></mrow></math> for which the function <math><mrow><msub><mrow><mi>f</mi></mrow><mrow><mi>λ</mi></mrow></msub><mrow><mo>(</mo><mrow><mi>x</mi></mrow><mo>)</mo></mrow><mo>=</mo><mn>4</mn><mi>λ</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>−</mo><mn>3</mn><mn>6</mn><mi>λ</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mn>3</mn><mn>6</mn><mi>x</mi><mo>+</mo><mn>4</mn><mn>8</mn></mrow></math> is increasing for all <math><mrow><mi>x</mi><mo>∈</mo><mi>R</mi><mo>.</mo></mrow></math> Then <math><mrow><msub><mrow><mi>f</mi></mrow><mrow><mi>λ</mi></mrow></msub><mo>.</mo><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo><mo>+</mo><msub><mrow><mi>f</mi></mrow><mrow><mi>λ</mi></mrow></msub><mo>.</mo><mo stretchy="false">(</mo><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></math> is equal to</strong></p>

? fλ (x)=4λx3−36λx2+36x+48fλ (x)=12 (λx2−6λx+3)For increasing fλ (x)≥0fλ* (x)=43x3−12x2+36x+48∴fλ* (1)+fλ* (−1)=7312−112=72

Ask & Answer Question

Maths NCERT Exemplar Solutions Class 12th Chapter Three Application of Derivatives Ncert Solutions Maths class 12th Maths Class 12th

Let $λ$ be the largest value of $λ$ for which the function $f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$ is increasing for all $x \in R .$ Then $f_{λ} . (1) + f_{λ} . (- 1)$ is equal to

Option 1 -

Option 2 -

Option 3 -

Option 4 -

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$? f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$

$f_{λ} (x) = 12 (λ x^{2} - 6 λ x + 3)$

For increasing $f_{λ} (x) \geq 0$

$f_{λ}^{*} (x) = \frac{4}{3} x^{3} - 12 x^{2} + 36 x + 48 ∴ f_{λ}^{*} (1) + f_{λ}^{*} (- 1) = 73 \frac{1}{2} - 1 \frac{1}{2} = 72$

0 Upvotes 0 Downvotes

Similar Questions for you

If the normal to the curve y(x) = ∫(from 2 to x) (2t² - 15t + 10)dt at a point (a, b) is parallel to the line x + 3y = -5, a > 1, then the value of |a + 6b| is equal to..........

Raj Pandey

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406|

...more

If the tangent to the curve, y = f(x) = xlog? x, (x > 0) at a point (c, f(c)) is parallel to the line - segment joining the points (1,0) and (e, e) then c is equal to:

Raj Pandey

f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
c = e^ (1/ (e-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:

Raj Pandey

$C D = \sqrt[]{{(10 + x^{2})}^{2} - {(10 - x^{2})}^{2}} = 2 \sqrt[]{10} | x |$

Area

$= \frac{1}{2} \times C D \times A B = \frac{1}{2} \times 2 \sqrt[]{10} | x | (20 - 2 x^{2})$

$\Rightarrow 10 - x^{2} = 2 x$

3x² = 10

x = k

3k² = 10

Let F₁(A,B,C) = $(A \land \sim B) \lor$ $[\sim C \land (A \lor B)] \lor \sim A a n d F_{2} (A, B) = (A \lor B) \lor (B \to \sim A)$ be two logical expressions. Then:

Vishal Baghel

By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So F₂ (A, B) be a tautology.

The triangle of maximum area that can be inscribed in a given circle of radius ‘r’ is:

Vishal Baghel

From option let it be isosceles where AB = AC then

$x = \sqrt[]{r^{2} - {(h - r)}^{2}}$

= $\sqrt[]{r^{2} - h^{2} - r^{2} + 2 r h}$

$x = \sqrt[]{2 h r - h^{2}} . . . . . . . . (i)$

Now ar $(Δ A B C) = Δ = \frac{1}{2} B C \times A L$

$Δ = \frac{1}{2} \times 2 \sqrt[]{2 h r - a^{2}} \times h$

then $x = \sqrt[]{2 \times \frac{3 r}{2} \times r - \frac{9 r^{2}}{4}} = \frac{\sqrt[]{3}}{2} r f r o m (i)$

$\Rightarrow B C = \sqrt[]{3} r$

So $A B = \sqrt[]{h^{2} + x^{2}} = \sqrt[]{\frac{9 r^{2}}{4} + \frac{3 r^{2}}{4}} = \sqrt[]{3} r$ .

Hence $Δ$ be equilateral having each side of length $\sqrt[]{3} r .$

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 $λ$ be the largest value of $λ$ for which the function $f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$ is increasing for all $x \in R .$ Then $f_{λ} . (1) + f_{λ} . (- 1)$ is equal to

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

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 λ be the largest value of λ for which the function fλ(x)=4λx3−36λx2+36x+48 is increasing for all x∈R. Then fλ.(1)+fλ.(−1) is equal to

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

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 $λ$ be the largest value of $λ$ for which the function $f_{λ} (x) = 4 λ x^{3} - 36 λ x^{2} + 36 x + 48$ is increasing for all $x \in R .$ Then $f_{λ} . (1) + f_{λ} . (- 1)$ is equal to