Verified Solution • 2025–26 Session moderate

QJEEM22Ma3662: JEE Main 2022 Mathematics - Class 12 JEEM22Ma3662 Solution: The curve touches the -axis at the point and cuts the -axis at…

The curve y(x)=ax3+bx2+cx+5y(x)=a x^{3}+b x^{2}+c x+5 touches the xx -axis at the point P(−2,0)\mathrm{P}(-2,0) and cuts the yy -axis at the point QQ , where y′y^{\prime} is equal to 3 . Then the local maximum value of y(x)y(x) is:

Multiple Choice Options

  • A 274\frac{27}{4}
  • B 294\frac{29}{4}
  • C 374\frac{37}{4}
  • D 92\frac{9}{2}

Step-by-Step Verified Solution

f(x)=y=ax3+bx2+cx+5…f(x)=y=a x^{3}+b x^{2}+c x+5 \quad \ldots (i)

dydx=3ax2+2bx+c…(ii)\frac{d y}{d x}=3 a x^{2}+2 b x+c \quad \ldots (ii)

Touches xx -axis at P(−2,0)P(-2,0)

⇒y∣x=−2=0⇒−8a+4b−2c+5=0…...(iii)\left.\Rightarrow y\right|_{x=-2}=0 \Rightarrow-8 a+4 b-2 c+5=0 \quad \ldots ...(iii)

Touches xx -axis at P(−2,0)P(-2,0) also implies

dydx∣x=−2=0⇒12a−4b+c=0…...(iv)\left.\frac{d y}{d x}\right|_{x=-2}=0 \Rightarrow 12 a-4 b+c=0 \quad \ldots...(iv)

y=f(x)y=f(x) cuts yy -axis at (0,5)(0,5)

Given, dydx∣x=0=c=3\left.\frac{d y}{d x}\right|_{x=0}=c=3

From (iii), (iv) and (v)

amp;a=−12,b=−34,c=3amp;⇒f(x)=−x22−34x2+3x+5amp;f′(x)=−32x2−32x+3amp;=−32(x+2)(x−1)amp;f′(x)=0 at x=−2 and x=1\begin{aligned}&a=-\frac{1}{2}, b=-\frac{3}{4}, c=3 \\\\&\Rightarrow f(x)=\frac{-x^{2}}{2}-\frac{3}{4} x^{2}+3 x+5 \\\\&f^{\prime}(x)=\frac{-3}{2} x^{2}-\frac{3}{2} x+3 \\\\&=\frac{-3}{2}(x+2)(x-1) \\\\&f^{\prime}(x)=0 \text { at } x=-2 \text { and } x=1\end{aligned}

By first derivative test x=1x=1 in point of local maximum Hence local maximum value of f(x)f(x) is f(1)f(1)

i.e., 274\frac{27}{4}

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