If f(x)=(p2 - 1)[(tan)- 1 x]+ 4(q2 + 2 q - 3) (1/2 + x2) +(p + q)sgn(x2 - x + 2) is continuous in R and f(x1)=f(x2)∀ x1,x2∈ R , then largest value of |p + q| is [Note: sgn(y),[y] and y denote signum function, greatest integer function and fractional part function respectively.]

Question

If f(x)=(p2 - 1)[(tan)- 1 x]+ 4(q2 + 2 q - 3) (1/2 + x2) +(p + q)sgn(x2 - x + 2) is continuous in R and f(x1)=f(x2)∀ x1,x2∈ R , then largest value of |p + q| is [Note: sgn(y),[y] and y denote signum function, greatest integer function and fractional part function respectively.] (A) 0 (B) 2 (C) 4 (D) 5

Tardigrade Admin · Accepted Answer

Since f(x1)=f(x2)∀ x1,x2∈ R ∴ f(x) must be constant Hence p2-1=0⇒ p=±1 q2+2q-3=0⇒ q=-3,1 ∴ Largest possible value of |p + q|=4

$0$

$2$

$4$

$5$

Since $f (x_{1}) = f (x_{2}) \forall x_{1}, x_{2} \in R$
$∴ f (x)$ must be constant Hence
$p^{2} - 1 = 0 \Rightarrow p = \pm 1$
$q^{2} + 2 q - 3 = 0 \Rightarrow q = - 3, 1$
$∴$ Largest possible value of $∣ p + q ∣ = 4$

Q. If f(x)=(p2−1)[(tan)−1x]+ 4(q2+2q−3){2+x21​} +(p+q)sgn(x2−x+2) is continuous in R and f(x1​)=f(x2​)∀x1​,x2​∈R , then largest value of ∣p+q∣ is [Note: sgn(y),[y] and {y} denote signum function, greatest integer function and fractional part function respectively.]

0

2

4

5

Since f(x1​)=f(x2​)∀x1​,x2​∈R ∴f(x) must be constant Hence p2−1=0⇒p=±1 q2+2q−3=0⇒q=−3,1 ∴ Largest possible value of ∣p+q∣=4

$0$

$2$

$4$

$5$

Since $f (x_{1}) = f (x_{2}) \forall x_{1}, x_{2} \in R$
$∴ f (x)$ must be constant Hence
$p^{2} - 1 = 0 \Rightarrow p = \pm 1$
$q^{2} + 2 q - 3 = 0 \Rightarrow q = - 3, 1$
$∴$ Largest possible value of $∣ p + q ∣ = 4$