Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The function \mathrm{f(x)=x^3-6 a x^2+5 x+2 } satisfies the condition of Lagrange's Mean Value Theorem over the interval [1,2] and the tangent to the curve  \mathrm{y=f(x) at x=\frac{7}{4} } is parallel to the chord that joins the points of intersection of the curve with ordinates x=1 and x=2. Then the value of ' a ' is

Option: 1

\frac{35}{16}


Option: 2

\frac{35}{48}


Option: 3

\frac{7}{16}


Option: 4

\frac{5}{16}


Answers (1)

best_answer

Given, \mathrm{f(x)=x^3-6 a x^2+5 x+2 }
Now, \mathrm{f(2)=8-24 a+10+2=20-24 a }
and \mathrm{f(1)=1-6 a+5+2=8-6 a }
Also,\mathrm{ f^{\prime}(x)=3 x^2-12 a x+5 }
From Lagrange's Mean Value Theorem, we have
\mathrm{\begin{aligned} & f^{\prime}(x)=\frac{f(2)-f(1)}{2-1} \\ & \Rightarrow f^{\prime}(x)=12-18 a \end{aligned} }
Also, \mathrm{f^{\prime}\left(\frac{7}{4}\right)=3 \times \frac{49}{16}-12 a \times \frac{7}{4}+5=12-18 a }
\mathrm{\Rightarrow \quad a=\frac{35}{48} }

Posted by

HARSH KANKARIA

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JoSAA 2025 reconstituted; NIT Rourkela to coordinate BTech counselling for IITs, NITs, IIITs
anam-khan

JEE Main 2025: BTech Information Technology cut-offs at NITs; know opening, closing ranks
anam-khan

JEE Main 2025: BTech Computer Science Engineering cut-offs at NITs went up this year

Latest Question