Question

Consider f(x)=x21+x2 and g(x)=px where p∈R. Then which of the following statements is (are) CORRECT?

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Solution

The correct options are

**A** If f(x)=g(|x|) has exactly three distinct solutions, then number of all possible real values of p is 1.

**B** If f(x)=g(|x|) has exactly five distinct solutions, then range of all possible real values of p is (0,12).

**C** Number of distinct tangents that can be drawn to the curve y=f(x) from the origin is 3.

f(x)≥0

limx→∞f(x)=1

So, range of f is [0,1)

Clearly, f(x)=g(|x|) has x=0 as one solution.

f(x)=g(|x|) has three solutions iff each branch of g(|x|)=p|x|, p>0 is tangent to y=f(x)

⇒f′(x)=p

⇒2x(1+x2)2=p

Intersection point of f(x)=g(|x|):

x21+x2=2x2(1+x2)2

⇒1+x2=2

⇒x=±1

⇒p=12 as p>0

If p>12, y=f(x) will not intersect y=g(|x|) and hence, there will be no solution.

For five distinct solutions, p∈(0,12)

f(x)≥0

limx→∞f(x)=1

So, range of f is [0,1)

Clearly, f(x)=g(|x|) has x=0 as one solution.

f(x)=g(|x|) has three solutions iff each branch of g(|x|)=p|x|, p>0 is tangent to y=f(x)

⇒f′(x)=p

⇒2x(1+x2)2=p

Intersection point of f(x)=g(|x|):

x21+x2=2x2(1+x2)2

⇒1+x2=2

⇒x=±1

⇒p=12 as p>0

If p>12, y=f(x) will not intersect y=g(|x|) and hence, there will be no solution.

For five distinct solutions, p∈(0,12)

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