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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:

Option 1 -

1 7 + 3 2

Option 2 -

1 7 + 5 2

Option 3 -

5

Option 4 -

9 − 1 7 2

0 2 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • R

    Answered by

    Raj Pandey | Contributor-Level 9

    a month ago
    Correct Option - 1


    Detailed Solution:

    f ( x ) = | x 2 − 3 x − 2 | − x

    = | ( x − 3 − 1 7 2 ) ( x − 3 + 1 7 2 ) | − x

    ⇒ f ( x ) = [ x 2 − 4 x − 2 ; − 1 ≤ x ≤ 3 − 1 7 2 − x 2 + 2 x + 2 ; 3 − 1 7 2 < x ≤ 2 ]

    absolute minimum f ( 3 − 1 7 2 ) = − 3 + 1 7 2  

     absolute maximum = 3

    ∴ s u m     3 + − 3 + 1 7 2 = 3 + 1 7 2  

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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?
V
Vishal Baghel

f ' ( x ) = n 1 ⋅ f ( x ) x − 3 + n 2 ⋅ f ( x ) x − 5

= f ( x ) ⋅ ( n 1 + n 2 ) ( x − 3 ) ( x − 5 ) ( x − ( 5 n 1 + 3 n 2 ) n 1 + n 2 )

f ' ( x ) = ( x − 3 ) n 1 − 1 ⋅ ( x − 5 ) n 2 − 1 ⋅ ( n 1 + n 2 ) ( x − ( 5 n 1 + 3 n 2 ) n 1 + n 2 l )

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

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:
R
Raj Pandey

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 < b 2 − 4 ≤ 3 2

⇒ b 2 − 4 > 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 ]  

 

Assume e − 4 5 = 2 5 .  If x, y satisfy, y = ex and the minimum value of (x2 + y2) is expressed in the form of m n tehn ( 2 m − n ) 5  equals (where m & n are coprime natural numbers)
V
Vishal Baghel

OP2 = x2 = y2

y = ex, y’ = ex,

slope of normal =  − 1 e x

y x = − 1 e x    

− 1 e x = e x x ⇒ − x = e 2 x      

By hit and trial we get  x = − 2 5

P ≡ ( − 2 5 , e − 2 / 5 )

O P = 4 2 5 + e − 4 / 5 ⇒ O P 2 = 1 4 2 5 = 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
V
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:
V
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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