Let f(x)=2^x-x^2, x \in \mathbbR . If m and n are respectively the number of points at which the curves y=f(x) and y=f^\prime(x) intersect the x -axis, then the value of m+n is \underline\hspace2cm .

Application of Derivatives PYQ — JEE Main 2024 · 29 Jan · Shift 1 · Q59 with Solution

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Correct answer: 5

We need:
- $m$ = number of points where $y=f(x)$ cuts/intersects the $x$ -axis, i.e. number of real solutions of $f(x)=2^x-x^2=0.$
- $n$ = number of points where $y=f'(x)$ cuts/intersects the $x$ -axis, i.e. number of real solutions of $f'(x)=0.$

Part A: Find $m$

We solve $2^x-x^2=0 \iff 2^x=x^2.$

Let $g(x)=2^x-x^2.$ We look for real zeros.

1. Check some obvious values

At $x=2$ : $g(2)=2^2-2^2=4-4=0.$
At $x=4$ : $g(4)=2^4-4^2=16-16=0.$ So $x=2,4$ are two roots.

Also,

At $x=-2$ : $g(-2)=2^{-2}-(-2)^2=\frac14-4<0.$
At $x=0$ : $g(0)=1>0.$ Hence there is at least one root in $(-2,0)$ .

So far, we have at least 3 roots.

2. Show there cannot be more than 3 roots

Use derivatives.

$g'(x)=2^x\ln 2-2x,$ $g''(x)=2^x(\ln 2)^2-2.$

Now consider $g''(x)=0$ :

\iff 2^x=\frac{2}{(\ln 2)^2}.$$ Since $2^x$ is strictly increasing, this equation has exactly one real solution. Therefore $g''$ changes sign only once, so $g'$ has exactly one local minimum and hence $g'$ can have at most two zeros. Therefore $g$ can have at most three real zeros. Since we already found three real zeros (one in $(-2,0)$, and $x=2,4$), it follows that $g(x)=0$ has exactly 3 real roots. Hence, $$m=3.$$ --- ## Part B: Find $n$ Now $$f'(x)=2^x\ln 2-2x.$$ We need the number of real solutions of $$2^x\ln 2-2x=0.$$ Let $$h(x)=2^x\ln 2-2x.$$ ### 1. Check obvious value At $x=1$, $$h(1)=2\ln 2-2<0$$ because $\ln 2<1$. At $x=0$, $$h(0)=\ln 2>0.$$ So there is one root in $(0,1)$. At $x=2$, $$h(2)=4\ln 2-4<0.$$ At $x=4$, $$h(4)=16\ln 2-8>0.$$ So there is another root in $(2,4)$. Thus there are at least 2 roots. ### 2. Show there cannot be more than 2 roots Differentiate: $$h'(x)=2^x(\ln 2)^2-2.$$ This is exactly the same expression as $g''(x)$ above. Since $2^x$ is strictly increasing, the equation $$h'(x)=0$$ has exactly one real solution. Thus $h$ has exactly one critical point, which is a minimum. Therefore $h$ can cross the $x$-axis at most twice. Since we already found two roots, it follows that $h(x)=0$ has exactly 2 real roots. Hence, $$n=2.$$ --- ## Final calculation $$m+n=3+2=5.$$ So the required integer is $$\boxed{5}.$$

More from Application of Derivatives

Part A: Find

m

We solve

2^x-x^2=0 \iff 2^x=x^2.

Let

g(x)=2^x-x^2.

We look for real zeros.

1. Check some obvious values

x=2

g(2)=2^2-2^2=4-4=0.

x=4

g(4)=2^4-4^2=16-16=0.

x=2,4

are two roots.

Also,

x=-2

g(-2)=2^{-2}-(-2)^2=\frac14-4<0.

x=0

g(0)=1>0.

Hence there is at least one root in

(-2,0)

So far, we have at least 3 roots.

2. Show there cannot be more than 3 roots

Use derivatives.

g'(x)=2^x\ln 2-2x,

g''(x)=2^x(\ln 2)^2-2.

Now consider

g''(x)=0

\iff 2^x=\frac{2}{(\ln 2)^2}.$$ Since $2^x$ is strictly increasing, this equation has exactly one real solution. Therefore $g''$ changes sign only once, so $g'$ has exactly one local minimum and hence $g'$ can have at most two zeros. Therefore $g$ can have at most three real zeros. Since we already found three real zeros (one in $(-2,0)$, and $x=2,4$), it follows that $g(x)=0$ has exactly 3 real roots. Hence, $$m=3.$$ --- ## Part B: Find $n$ Now $$f'(x)=2^x\ln 2-2x.$$ We need the number of real solutions of $$2^x\ln 2-2x=0.$$ Let $$h(x)=2^x\ln 2-2x.$$ ### 1. Check obvious value At $x=1$, $$h(1)=2\ln 2-2<0$$ because $\ln 2<1$. At $x=0$, $$h(0)=\ln 2>0.$$ So there is one root in $(0,1)$. At $x=2$, $$h(2)=4\ln 2-4<0.$$ At $x=4$, $$h(4)=16\ln 2-8>0.$$ So there is another root in $(2,4)$. Thus there are at least 2 roots. ### 2. Show there cannot be more than 2 roots Differentiate: $$h'(x)=2^x(\ln 2)^2-2.$$ This is exactly the same expression as $g''(x)$ above. Since $2^x$ is strictly increasing, the equation $$h'(x)=0$$ has exactly one real solution. Thus $h$ has exactly one critical point, which is a minimum. Therefore $h$ can cross the $x$-axis at most twice. Since we already found two roots, it follows that $h(x)=0$ has exactly 2 real roots. Hence, $$n=2.$$ --- ## Final calculation $$m+n=3+2=5.$$ So the required integer is $$\boxed{5}.$$

Application of Derivatives question

Application of Derivatives question

Part A: Find $m$

1. Check some obvious values

2. Show there cannot be more than 3 roots

More from Application of Derivatives

Application of Derivatives question

Application of Derivatives question

Part A: Find $m$

1. Check some obvious values

2. Show there cannot be more than 3 roots

More from Application of Derivatives

Part A: Find mmm

1. Check some obvious values

2. Show there cannot be more than 3 roots

More from Application of Derivatives

Part A: Find mmm

1. Check some obvious values

2. Show there cannot be more than 3 roots

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Part A: Find $m$

Part A: Find $m$