Let be a function defined by . If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

Question

Let $f : R \to R$ be a function defined by $f (x) = | | x + 2 | - 2 | x | |$ . If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

Smart Achievers · Accepted Answer

(1)

We have, $f (x) = | | x + 2 | - 2 | x | |$

$\Rightarrow$ $f (x)$ $= {\begin{matrix} (- x + 2), & x < - 2 \\ (- 3 x - 2), & - 2 \leq x < \frac{- 2}{3} \\ (3 x + 2), & \frac{- 2}{3} \leq x < 0 \\ (- x + 2), & 0 \leq x < 2 \\ (x - 2), & x > 2 \end{matrix}$

Critical points are $x = - 2, \frac{- 2}{3}, 0, 2$

Number of local maxima = 1 = n

Number of local minima = 2 = m

$∴$ m + n = 2 + 1 = 3.

Solution

Q.
Let $f : R \to R$ be a function defined by $f (x) = | | x + 2 | - 2 | x | |$ . If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

1 3

2 4

3 5

4 2

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Solution

Q. Let f:R→R be a function defined by f(x)=||x+2|–2|x||. If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]

1 3

2 4

3 5

4 2

Ans. (1) We have, f(x)=||x+2|–2|x|| ⇒ f(x)={(–x+2),x<–2(–3x–2),–2≤x<–23(3x+2),–23≤x<0(–x+2),0≤x<2(x–2),x>2 Critical points are x=–2,–23,0,2 Number of local maxima = 1 = n Number of local minima = 2 = m ∴ m + n = 2 + 1 = 3.

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Q.
Let $f : R \to R$ be a function defined by $f (x) = | | x + 2 | - 2 | x | |$ . If m is the number of points of local minima and n is the number of points of local maxima of f, then m + n is [2025]