Home
Class 12
MATHS
find the equation of the tangent to the ...

find the equation of the tangent to the curve `y=-5x^2+6x+7` at the point `(1//2, 35//4)`

Text Solution

Verified by Experts

The correct Answer is:
A, D

`"For "alpha=1`
`y=|x-1|+|x-2|+x={{:(3-x,, xlt1),(1+x,,1le xlt2),(3x-3,,xge2):}`
`"For "alpha=0, y=3`

`A=(1)/(2)(2+3)xx1+(1)/(2)(2+3)xx1-overset(2)underset(0)int2sqrt(x)dx`
`rArr" "A=5-(8)/(3)sqrt(2)`
`therefore" "F(1)+(8)/(3)sqrt(2)=5`
`"For "alpha=0,y=|-1|+|-2|=3`

`A=6-overset(2)underset(0)int2sqrt(x)dx`
`rArr" "A=6-(8)/(3)sqrt(2)`
`therefore" "F(0)+(8)/(3)sqrt(2)=6`
Note : Solutions of the remaining parts are given in their respective chapters.
Promotional Banner

Topper's Solved these Questions

  • AREA

    CENGAGE PUBLICATION|Exercise Numerical Value Type|18 Videos
  • AREA

    CENGAGE PUBLICATION|Exercise Archives|10 Videos
  • AREA

    CENGAGE PUBLICATION|Exercise Linkded Comprehension Type|21 Videos
  • APPLICATIONS OF DERIVATIVES

    CENGAGE PUBLICATION|Exercise Subjective Type|2 Videos
  • BINOMIAL THEOREM

    CENGAGE PUBLICATION|Exercise Comprehension|11 Videos

Similar Questions

Explore conceptually related problems

Find the equation of the tangent to the curve y=(x-7)/((x-2)(x-3)) at the point where it cuts the x-axis.

Find the length of the tangent for the curve y=x^3+3x^2+4x-1 at point x=0.

Find the equations of the tangents drawn to the curve y^2-2x^3-4y+8=0 from the point (1,\ 2) .

Find the equations of the tangent and normal to the curve y=x^2-4x-5 at x= -2.

Find the equation of the tangent to the parabola y^(2)=8x at the point (2t^(2),4t) . Hence find the equation of the tangnet to this parabola, perpendicular to x+2y+7=0

If the equation of the tangent to the curve y^2=a x^3+b at point (2,3) is y=4x-5 , then find the values of a and b .

The equation of the tangent to the curve y^(2)=ax^(3)+b at the point (2,3) on it is y=4x-5 , find a and b.

The slope of the tangent to the curve (y-x^5)^2=x(1+x^2)^2 at the point (1,3) is

Find the length of tangent to the curve y=4x^3-2x^5 at (-1,1)

Find the equations of the tangent to the given curves at the indicated points: y=x^(4) -6x^(3)+13x^(2)-10x+5 1 at (0,5)