Question

Let  f(x) = ax³ + bx² + cx + d  
∴ f(0) = 5  ⇒  d = 5
So, f(x) = ax³ + bx² + cx + 5
∴ f '(x) = 3ax² + 2bx + c

Now, f '(–2) = 0  ⇒  12a – 4b + c = 0            .....(1)
and y = f(x) passes through P (–2, 0), so 0 = – 8a + 4b – 2c + 5    ......(2)
Also, f '(0) = 3  ⇒  c = 3                .....(3)
∴  On solving, we get a    $\frac{-1}{2},\mathrm{b}=\frac{-3}{4}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{-1}{2}{\mathrm{x}}^3-\frac{3}{4}{\mathrm{x}}^2+3\mathrm{x}+5$

Required area $={\int }_{-1}^1 \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{-1}^1 \left(\frac{-1}{2}{\mathrm{x}}^3-\frac{3}{4}{\mathrm{x}}^2+3\mathrm{x}+5\right)\mathrm{dx}=2{\int }_0^1 \left(\frac{-3}{4}{\mathrm{x}}^2+5\right)\mathrm{dx}=\frac{19}{2}$

Let  f(x) = ax³ + bx² + cx + d  
∴ f(0) = 5  ⇒  d = 5
So, f(x) = ax³ + bx² + cx + 5
∴ f '(x) = 3ax² + 2bx + c

Now, f '(–2) = 0  ⇒  12a – 4b + c = 0            .....(1)
and y = f(x) passes through P (–2, 0), so 0 = – 8a + 4b – 2c + 5    ......(2)
Also, f '(0) = 3  ⇒  c = 3                .....(3)
∴  On solving, we get a    $\frac{-1}{2},\mathrm{b}=\frac{-3}{4}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{-1}{2}{\mathrm{x}}^3-\frac{3}{4}{\mathrm{x}}^2+3\mathrm{x}+5$

Equation of normal  at  Q(0, 5) is

$(\mathrm{y}-5)=\frac{-1}{3}(\mathrm{x}-0)\Rightarrow \mathrm{x}+3\mathrm{y}=15$

Let  f(x) = ax³ + bx² + cx + d  
∴ f(0) = 5  ⇒  d = 5
So, f(x) = ax³ + bx² + cx + 5
∴ f '(x) = 3ax² + 2bx + c

Now, f '(–2) = 0  ⇒  12a – 4b + c = 0            .....(1)
and y = f(x) passes through P (–2, 0), so 0 = – 8a + 4b – 2c + 5    ......(2)
Also, f '(0) = 3  ⇒  c = 3                .....(3)
∴  On solving, we get a    $\frac{-1}{2},\mathrm{b}=\frac{-3}{4}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{-1}{2}{\mathrm{x}}^3-\frac{3}{4}{\mathrm{x}}^2+3\mathrm{x}+5$

Clearly, from above graph, we get, number of solutions of equation |f(|x|)| = 3 are 4.