For every pair of continuous function f, g : [0, 1] \to R such that max f(x) : x \in [0, 1] = max g(x) : x \in [0, 1]. The correct statement(s) is (are)

[f(c)]2 + 3f(c) = [g(c)]2 + 3g(c) for some c \in [0, 1], [f(c)]2 = [g(c)]2 for some c \in [0, 1]

[f(c)]2 + 3f(c) = [g(c)]2 + 3g(c) for some c \in [0, 1]

[f(c)]2 + f(c) = [g(c)]2 + 3g(c) for some c \in [0, 1]

[f(c)]2 + 3f(c) = [g(c)]2 + g(c) for some c \in [0, 1]

[f(c)]2 = [g(c)]2 for some c \in [0, 1]

Functions PYQ — JEE Advanced 2014 · Shift 1 · Q33 with Solution

View written solutionFree

Correct answer: A, D

Let $M_f=\max\{f(x):x\in[0,1]\},\qquad M_g=\max\{g(x):x\in[0,1]\}.$ Given $M_f=M_g=:M.$ Since $f,g$ are continuous on $[0,1]$ , by the Intermediate Value Theorem each function attains every value between its minimum and maximum.

We must check which statements are true for every such pair $f,g$ .

1. Option A

We need to show that for some $c\in[0,1]$ , $[f(c)]^2+3f(c)=[g(c)]^2+3g(c).$

Define $\phi(t)=t^2+3t.$ Then $\phi'(t)=2t+3.$ For all $t\ge -\tfrac32$ , $\phi$ is increasing, but we need a statement valid on all real numbers. So instead of monotonicity everywhere, use interval overlap.

Let $I_f=f([0,1]),\qquad I_g=g([0,1]).$ Since $f,g$ are continuous on $[0,1]$ , these are intervals: $I_f=[m_f,M],\qquad I_g=[m_g,M],$ for some minima $m_f,m_g$ . Hence both intervals contain the common value $M$ ; therefore $I_f\cap I_g\neq\varnothing.$ Take any $y\in I_f\cap I_g.$ Then there exist $u,v\in[0,1]$ such that

\qquad g(v)=y.$$ Now define $$F(x)=\phi(f(x))-\phi(g(x)).$$ If we can find one point where $f=g$, then A is immediate. But from the above, we only know the two functions take a common value, perhaps at different points. So instead consider the function $$h(x)=f(x)-g(x).$$ This does not necessarily vanish. Thus we need another route. A better observation is that the statement asks for **some common input** $c$, so we examine $$F(x)=\big(f(x)^2+3f(x)\big)-\big(g(x)^2+3g(x)\big) =(f(x)-g(x))(f(x)+g(x)+3).$$ This is continuous. Let $a$ be a point where $f(a)=M$, and $b$ a point where $g(b)=M$. Then $$F(a)=M^2+3M-(g(a)^2+3g(a))=(M-g(a))(M+g(a)+3),$$ $$F(b)=f(b)^2+3f(b)-(M^2+3M)=(f(b)-M)(f(b)+M+3).$$ Now use the fact that for any real $t\le M$, $$M^2+3M-(t^2+3t)=(M-t)(M+t+3).$$ This is not always nonnegative because $M+t+3$ may be negative. So sign comparison is not guaranteed. Hence A is not obvious from endpoint sign arguments. Instead, note that option A is equivalent to existence of $c$ such that $$\phi(f(c))=\phi(g(c)).$$ Since $\phi(t)=t^2+3t$ has symmetry about $t=-\tfrac32$, equality means either $$f(c)=g(c)$$ or $$f(c)+g(c)=-3.$$ So A will hold if either of these happens for some $c$. Now define $$H(x)=f(x)-g(x),\qquad K(x)=f(x)+g(x)+3.$$ Then A asks whether $$H(c)K(c)=0$$ for some $c$. To test whether A must always hold, let us try to construct a counterexample. Take constants with same maximum, say $$f(x)\equiv 0,\qquad g(x)\equiv 0.$$ Then A holds. Try $$f(x)\equiv -10,\ g(x)\equiv -10,$$ again holds. If constants differ, maxima cannot be equal unless equal. Try nonconstant examples with same maximum but no point where $f=g$. For example, $$f(x)=x,\qquad g(x)=x^2$$ on $[0,1]$ have same maximum $1$, and they are equal at $x=0,1$ anyway. Similar examples tend to intersect due to same endpoint max. But not always. In fact, if two continuous functions on $[0,1]$ have the same maximum value $M$, then considering points where each attains $M$, one can connect values continuously and one gets either an intersection $f=g$ or a point where $f+g=-3$ if one function stays strictly above the other; thus A is plausible. More directly, option D (proved below) guarantees a point where $f(c)=g(c)$ or $f(c)=-g(c)$. If $f(c)=g(c)$, then A holds immediately. If $f(c)=-g(c)$, then $$f(c)^2+3f(c)=g(c)^2-3g(c),$$ not enough. So D alone does not imply A. Let us test A by examples. Take $$f(x)=x,\qquad g(x)=1-x.$$ Both have maximum $1$. Then $$x^2+3x=(1-x)^2+3(1-x)$$ $$x^2+3x=1-2x+x^2+3-3x=4-5x$$ $$8x=4\implies x=\tfrac12,$$ which lies in $[0,1]$. So A holds. Take $$f(x)=x-2,\qquad g(x)=-x-1.$$ Both have maximum $-1$? For $f$, max is $-1$ at $x=1$; for $g$, max is $-1$ at $x=0$. Then $$f(x)^2+3f(x)=g(x)^2+3g(x).$$ With $f=x-2$, $g=-x-1$: $$(x-2)^2+3(x-2)=(-x-1)^2+3(-x-1).$$ LHS $=x^2-x-2$, RHS $=x^2-x-2$, identically true. So A holds. These checks support A. We accept A as true. --- ## 2. Option B We need $$[f(c)]^2+f(c)=[g(c)]^2+3g(c)$$ for some $c$. Equivalently, $$f(c)^2+f(c)-g(c)^2-3g(c)=0.$$ We test by counterexample. Take $$f(x)=x,\qquad g(x)=1-x.$$ Both are continuous and have maximum $1$. Now solve $$x^2+x=(1-x)^2+3(1-x).$$ RHS: $$(1-x)^2+3(1-x)=1-2x+x^2+3-3x=x^2-5x+4.$$ So $$x^2+x=x^2-5x+4\implies 6x=4\implies x=\tfrac23.$$ This example satisfies B, so not a counterexample. Try a stronger example: $$f(x)=x-2,\qquad g(x)=-x-1.$$ Both have maximum $-1$. Then $$f^2+f=(x-2)^2+(x-2)=x^2-3x+2,$$ $$g^2+3g=(-x-1)^2+3(-x-1)=x^2-x-2.$$ Equality gives $$x^2-3x+2=x^2-x-2\implies -2x+4=0\implies x=2,$$ which is not in $[0,1]$. So no such $c$ exists. Hence **B is false**. --- ## 3. Option C We need $$[f(c)]^2+3f(c)=[g(c)]^2+g(c)$$ for some $c$. Again use the same counterexample: $$f(x)=x-2,\qquad g(x)=-x-1,$$ with common maximum $-1$. Compute: $$f^2+3f=(x-2)^2+3(x-2)=x^2-x-2,$$ $$g^2+g=(-x-1)^2+(-x-1)=x^2+x.$$ Equality gives $$x^2-x-2=x^2+x\implies -2x-2=0\implies x=-1,$$ not in $[0,1]$. So no such $c$ exists. Hence **C is false**. --- ## 4. Option D We need $$[f(c)]^2=[g(c)]^2$$ for some $c,$ i.e. $$f(c)=g(c)\quad\text{or}\quad f(c)=-g(c).$$ Consider the continuous function $$P(x)=f(x)^2-g(x)^2=(f(x)-g(x))(f(x)+g(x)).$$ We show $P$ must vanish somewhere. Let $a,b\in[0,1]$ be such that $$f(a)=M,\qquad g(b)=M.$$ Then $$P(a)=M^2-g(a)^2,$$ with $g(a)\le M$. Hence $$|g(a)|\le ?$$ This does not directly give sign of $P(a)$ if $M<0$. So we need a better argument. Use the same standard trick with maxima and minima? Let ranges be $$f([0,1])=[m_f,M],\qquad g([0,1])=[m_g,M].$$ If these intervals overlap in some value $y$, then there exist points where each takes value $y$, but not necessarily same point. So that alone is insufficient. Instead, test whether D can fail. We need continuous $f,g$ with same max and such that for all $x$, $$f(x)\ne g(x)\quad\text{and}\quad f(x)\ne -g(x).$$ That means both continuous functions $$u(x)=f(x)-g(x),\qquad v(x)=f(x)+g(x)$$ never vanish. So each has constant sign on $[0,1]$. Suppose $u$ has constant sign. - If $u>0$ everywhere, then $f(x)>g(x)$ for all $x$, so $$\max f > \max g,$$ contradiction. - If $u<0$ everywhere, similarly contradiction. Therefore $u$ must vanish somewhere. Hence there exists $c$ such that $$f(c)=g(c).$$ Then automatically $$f(c)^2=g(c)^2.$$ So D is true. This also gives a simpler proof for A: since there exists $c$ with $f(c)=g(c)$, substituting gives $$f(c)^2+3f(c)=g(c)^2+3g(c).$$ Thus A is true. --- ## 5. Final conclusion - **A is true** because same maximum implies $f-g$ cannot keep a strict sign on all of $[0,1]$, so for some $c$, $f(c)=g(c)$. - **B is false** by counterexample $f(x)=x-2$, $g(x)=-x-1$. - **C is false** by the same counterexample. - **D is true** because some $c$ satisfies $f(c)=g(c)$, hence $f(c)^2=g(c)^2$. Therefore the correct options are $$\boxed{A,\ D}.$$

More from Functions

1. Option A

We need to show that for some

c\in[0,1]

[f(c)]^2+3f(c)=[g(c)]^2+3g(c).

Define

\phi(t)=t^2+3t.

Then

\phi'(t)=2t+3.

For all

t\ge -\tfrac32

\phi

is increasing, but we need a statement valid on all real numbers. So instead of monotonicity everywhere, use interval overlap.

Let

I_f=f([0,1]),\qquad I_g=g([0,1]).

Since

f,g

are continuous on

[0,1]

, these are intervals:

I_f=[m_f,M],\qquad I_g=[m_g,M],

for some minima

m_f,m_g

. Hence both intervals contain the common value

M

; therefore

I_f\cap I_g\neq\varnothing.

Take any

y\in I_f\cap I_g.

Then there exist

u,v\in[0,1]

such that

\qquad g(v)=y.$$ Now define $$F(x)=\phi(f(x))-\phi(g(x)).$$ If we can find one point where $f=g$, then A is immediate. But from the above, we only know the two functions take a common value, perhaps at different points. So instead consider the function $$h(x)=f(x)-g(x).$$ This does not necessarily vanish. Thus we need another route. A better observation is that the statement asks for **some common input** $c$, so we examine $$F(x)=\big(f(x)^2+3f(x)\big)-\big(g(x)^2+3g(x)\big) =(f(x)-g(x))(f(x)+g(x)+3).$$ This is continuous. Let $a$ be a point where $f(a)=M$, and $b$ a point where $g(b)=M$. Then $$F(a)=M^2+3M-(g(a)^2+3g(a))=(M-g(a))(M+g(a)+3),$$ $$F(b)=f(b)^2+3f(b)-(M^2+3M)=(f(b)-M)(f(b)+M+3).$$ Now use the fact that for any real $t\le M$, $$M^2+3M-(t^2+3t)=(M-t)(M+t+3).$$ This is not always nonnegative because $M+t+3$ may be negative. So sign comparison is not guaranteed. Hence A is not obvious from endpoint sign arguments. Instead, note that option A is equivalent to existence of $c$ such that $$\phi(f(c))=\phi(g(c)).$$ Since $\phi(t)=t^2+3t$ has symmetry about $t=-\tfrac32$, equality means either $$f(c)=g(c)$$ or $$f(c)+g(c)=-3.$$ So A will hold if either of these happens for some $c$. Now define $$H(x)=f(x)-g(x),\qquad K(x)=f(x)+g(x)+3.$$ Then A asks whether $$H(c)K(c)=0$$ for some $c$. To test whether A must always hold, let us try to construct a counterexample. Take constants with same maximum, say $$f(x)\equiv 0,\qquad g(x)\equiv 0.$$ Then A holds. Try $$f(x)\equiv -10,\ g(x)\equiv -10,$$ again holds. If constants differ, maxima cannot be equal unless equal. Try nonconstant examples with same maximum but no point where $f=g$. For example, $$f(x)=x,\qquad g(x)=x^2$$ on $[0,1]$ have same maximum $1$, and they are equal at $x=0,1$ anyway. Similar examples tend to intersect due to same endpoint max. But not always. In fact, if two continuous functions on $[0,1]$ have the same maximum value $M$, then considering points where each attains $M$, one can connect values continuously and one gets either an intersection $f=g$ or a point where $f+g=-3$ if one function stays strictly above the other; thus A is plausible. More directly, option D (proved below) guarantees a point where $f(c)=g(c)$ or $f(c)=-g(c)$. If $f(c)=g(c)$, then A holds immediately. If $f(c)=-g(c)$, then $$f(c)^2+3f(c)=g(c)^2-3g(c),$$ not enough. So D alone does not imply A. Let us test A by examples. Take $$f(x)=x,\qquad g(x)=1-x.$$ Both have maximum $1$. Then $$x^2+3x=(1-x)^2+3(1-x)$$ $$x^2+3x=1-2x+x^2+3-3x=4-5x$$ $$8x=4\implies x=\tfrac12,$$ which lies in $[0,1]$. So A holds. Take $$f(x)=x-2,\qquad g(x)=-x-1.$$ Both have maximum $-1$? For $f$, max is $-1$ at $x=1$; for $g$, max is $-1$ at $x=0$. Then $$f(x)^2+3f(x)=g(x)^2+3g(x).$$ With $f=x-2$, $g=-x-1$: $$(x-2)^2+3(x-2)=(-x-1)^2+3(-x-1).$$ LHS $=x^2-x-2$, RHS $=x^2-x-2$, identically true. So A holds. These checks support A. We accept A as true. --- ## 2. Option B We need $$[f(c)]^2+f(c)=[g(c)]^2+3g(c)$$ for some $c$. Equivalently, $$f(c)^2+f(c)-g(c)^2-3g(c)=0.$$ We test by counterexample. Take $$f(x)=x,\qquad g(x)=1-x.$$ Both are continuous and have maximum $1$. Now solve $$x^2+x=(1-x)^2+3(1-x).$$ RHS: $$(1-x)^2+3(1-x)=1-2x+x^2+3-3x=x^2-5x+4.$$ So $$x^2+x=x^2-5x+4\implies 6x=4\implies x=\tfrac23.$$ This example satisfies B, so not a counterexample. Try a stronger example: $$f(x)=x-2,\qquad g(x)=-x-1.$$ Both have maximum $-1$. Then $$f^2+f=(x-2)^2+(x-2)=x^2-3x+2,$$ $$g^2+3g=(-x-1)^2+3(-x-1)=x^2-x-2.$$ Equality gives $$x^2-3x+2=x^2-x-2\implies -2x+4=0\implies x=2,$$ which is not in $[0,1]$. So no such $c$ exists. Hence **B is false**. --- ## 3. Option C We need $$[f(c)]^2+3f(c)=[g(c)]^2+g(c)$$ for some $c$. Again use the same counterexample: $$f(x)=x-2,\qquad g(x)=-x-1,$$ with common maximum $-1$. Compute: $$f^2+3f=(x-2)^2+3(x-2)=x^2-x-2,$$ $$g^2+g=(-x-1)^2+(-x-1)=x^2+x.$$ Equality gives $$x^2-x-2=x^2+x\implies -2x-2=0\implies x=-1,$$ not in $[0,1]$. So no such $c$ exists. Hence **C is false**. --- ## 4. Option D We need $$[f(c)]^2=[g(c)]^2$$ for some $c,$ i.e. $$f(c)=g(c)\quad\text{or}\quad f(c)=-g(c).$$ Consider the continuous function $$P(x)=f(x)^2-g(x)^2=(f(x)-g(x))(f(x)+g(x)).$$ We show $P$ must vanish somewhere. Let $a,b\in[0,1]$ be such that $$f(a)=M,\qquad g(b)=M.$$ Then $$P(a)=M^2-g(a)^2,$$ with $g(a)\le M$. Hence $$|g(a)|\le ?$$ This does not directly give sign of $P(a)$ if $M<0$. So we need a better argument. Use the same standard trick with maxima and minima? Let ranges be $$f([0,1])=[m_f,M],\qquad g([0,1])=[m_g,M].$$ If these intervals overlap in some value $y$, then there exist points where each takes value $y$, but not necessarily same point. So that alone is insufficient. Instead, test whether D can fail. We need continuous $f,g$ with same max and such that for all $x$, $$f(x)\ne g(x)\quad\text{and}\quad f(x)\ne -g(x).$$ That means both continuous functions $$u(x)=f(x)-g(x),\qquad v(x)=f(x)+g(x)$$ never vanish. So each has constant sign on $[0,1]$. Suppose $u$ has constant sign. - If $u>0$ everywhere, then $f(x)>g(x)$ for all $x$, so $$\max f > \max g,$$ contradiction. - If $u<0$ everywhere, similarly contradiction. Therefore $u$ must vanish somewhere. Hence there exists $c$ such that $$f(c)=g(c).$$ Then automatically $$f(c)^2=g(c)^2.$$ So D is true. This also gives a simpler proof for A: since there exists $c$ with $f(c)=g(c)$, substituting gives $$f(c)^2+3f(c)=g(c)^2+3g(c).$$ Thus A is true. --- ## 5. Final conclusion - **A is true** because same maximum implies $f-g$ cannot keep a strict sign on all of $[0,1]$, so for some $c$, $f(c)=g(c)$. - **B is false** by counterexample $f(x)=x-2$, $g(x)=-x-1$. - **C is false** by the same counterexample. - **D is true** because some $c$ satisfies $f(c)=g(c)$, hence $f(c)^2=g(c)^2$. Therefore the correct options are $$\boxed{A,\ D}.$$

Functions question

Functions question

1. Option A

More from Functions

Functions question

Functions question

1. Option A

More from Functions