JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q7

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MathematicsSingle correct+4 / -1Question 7 of 90

The curve $y(x)=ax^3+bx^2+cx+5$ touches the $x$ -axis at the point $P(-2,0)$ and cuts the $y$ -axis at the point Q, where $y'$ is equal to 3. Then the local maximum value of $y(x)$ is :

\dfrac{27}4

\dfrac{29}4

\dfrac{37}4

\dfrac92

Answer and worked solution

Official answer: A

Conditions. $y'(0)=c=3$ , $y(-2)=0$ and $y'(-2)=0$ give $a=-\frac12$ , $b=-\frac34$ .

Maximum. $y'=-\frac32x^2-\frac32x+3=-\frac32(x+2)(x-1)$ ; at $x=1$ , $y=-\frac12-\frac34+3+5=\frac{27}4$ .

(A)

Topics:Maxima and minima Tangents and normals

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JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 7