Question 16 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Sept. 24, 2021 by Teachoo

Find the equation(s) of the tangent(s) to the curve y = (x
^{
3
}
−1)(x − 2) at the points where the curve intersects the x – axis.
OR
Find the intervals in which the function f(x) = −3 log (1 + x) + 4 log (2 + x) − 4/(2 + x) is strictly increasing or strictly decreasing.
This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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Transcript

Question 16
Find the equation(s) of the tangent(s) to the curve y = (x3 1)(x 2) at the points where the curve intersects the x axis. We need to find equation of tangent to curve at which y cuts x-axis
First, let us find the points at which y cuts x-axis
Since y = 0 on points on x-axis
Putting y = 0 in equation of y
0 = (x3 1)(x 2)
(x3 1)(x 2) = 0
x3 1 = 0
x3 = 1
x3 = 13
x = 1
x 2 = 0
x = 2
So, x = 1, 2
Thus, points are (1, 0) and (2, 0)
Now, to find equation of tangent, we find slope
We know that
Slope of tangent = /
= (( ^3 1)( 2))^
= ( ^3 ( 2) 1( 2))^
= ( ^4 2 ^3 +2)^
= 4 ^3 2 3 ^2 1
= 4 ^3 6 ^2 1
For point (1, 0)
Slope of tangent
= 4 ^3 6 ^2 1
Putting x = 1
= 4(1) ^3 6(1) ^2 1
= 4 6 1
= 3
Equation of tangent passing through (1, 0) with slope 3 is
_1= ( 1)
0= 3 ( 1)
= +
For point (2, 0)
Slope of tangent
= 4 ^3 6 ^2 1
Putting x = 2
= 4(2) ^3 6(2) ^2 1
= 4 8 6 4 1
= 32 24 1 = 7
Equation of tangent passing through (2, 0) with slope 7 is
_1= ( 1)
0=7 ( 2)
=
Question 16Theory
We follow the following steps
Step 1: Calculating f (x)
Step 2: Putting f (x) = 0
Step 3: Drawing graphs and finding interval
Step 4: Finding in which interval f (x) > 0 or f (x) < 0
Find the intervals in which the function f(x) = 3 log (1 + x) + 4 log (2 + x) 4/(2 + ) is strictly increasing or strictly decreasing.
Given f(x) = 3 log (1 + x) + 4 log (2 + x) 4/(2 + )
Step 1:
Calculating f ( )
^ ( ) = (" 3 log (1 + x) + 4 log (2 + x) " 4/(2 + ))^
^ ( ) = (" 3 log (1 + x)" )^ + ("4 log (2 + x) " )^ (4/(2 + ))^
^ ( ) = ( 3)/((1 + )) + 4/(2 + ) 4 1 (2+ )^( 2)
^ ( ) = ( 3)/((1 + )) + 4/(2 + ) + 4/(2 + )^2
^ ( ) = ( 3(2 + )^2 + 4(2 + )(1 + ) + 4(1 + ))/((1 + ) (2 + )^2 )
^ ( ) = ( 3(4 + ^2 + 4 )+ 4(2(1 + ) + (1 + )) + 4 + 4 )/((1 + ) (2 + )^2 )
^ ( ) = ( 12 3 ^2 12 + 4 ^2 + 12 + 8 + 4 + 4 )/((1 + ) (2 + )^2 )
Step 2:
Putting ^ ( ) = 0
4 3 = 0
4 = 3
= 3/4
^ ( ) = ( ^2 + 4 )/((1 + ) (2 + )^2 )
^ ( ) = ( ( + 4))/((1 + ) (2 + )^2 )
Step 2:
Putting ^ ( ) = 0
( ( + 4))/((1 + ) (2 + )^2 )= 0
( + 4)= 0
So, x = 0, x = 4
Step 3:
Drawing graph
Before drawing graph, first we find domain of function
Given f(x) = 3 log (1 + x) + 4 log (2 + x) 4/(2 + )
Now,
log (1 + x) is possible if 1 + x > 0 i.e x > 1
log (2 + x) is possible if 2 + x > 0 i.e. x > 2
4/(2 + ) is possible if x + 2 0, i.e. x 2
So, domain will be which satisfies all 3 conditions
Domain is x > 1 i.e. ( 1, )
Since domain is ( 1, )
x = 4 is not possible
x = 0 is the only solution
Now, making graph
We make graph from 1 to infinity
and plot point 0
Point = 0 divides the line into 2 disjoint intervals
i.e. ( 1, 0] & [0 , )

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