arthham

JEE Advanced 2014 · Continuity and Differentiability

Let f1:R→Rf_1 : R \to R, f2:[0,∞)→Rf_2 : [0, \infty) \to R, f3:R→Rf_3 : R \to R and f4:R→[0,∞)f_4 : R \to [0, \infty) be defined by f1(x)={∣x∣if x<0,exif x≥0;f_1(x) = \begin{cases} |x| & \text{if } x < 0, \\ e^{x} & \text{if } x \ge 0; \end{cases} f2(x)=x2f_2(x) = x^{2}; f3(x)={sin⁡xif x<0,xif x≥0;f_3(x) = \begin{cases} \sin x & \text{if } x < 0, \\ x & \text{if } x \ge 0; \end{cases} and f4(x)={f2(f1(x))if x<0,f2(f1(x))−1if x≥0.f_4(x) = \begin{cases} f_2(f_1(x)) & \text{if } x < 0, \\ f_2(f_1(x)) - 1 & \text{if } x \ge 0. \end{cases} | List-I | List-II | | --- | --- | | (P) f4f_4 is | (1) Onto but not one-one | | (Q) f3f_3 is | (2) Neither continuous nor one-one | | (R) f2∘f1f_2 \circ f_1 is | (3) Differentiable but not one-one | | (S) f2f_2 is | (4) Continuous and one-one |

Answer: D. P→1P \to 1; Q→3Q \to 3; R→2R \to 2; S→4S \to 4

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

Let f₁ : R → R, f₂ : [0, ∞) → R, f₃ : R → R and f₄ : R → [0, ∞) be defined by f₁(x) = |x| if x < 0, eˣ if x ≥ 0; f₂(x) = x²; f₃(x) = sin x if x < 0, x if x ≥ 0; and f₄(x) = f₂(f₁(x)) if x < 0, f₂(f₁(x)) - 1 if x ≥ 0. | List-I | List-II | | --- | --- | | (P) f₄ is | (1) Onto but not one-one | | (Q) f₃ is | (2) Neither continuous nor one-one f1(x)={∣x∣x<0exx≥0f2(x)=x2f_1(x) = \begin{cases} |x| & x<0 \\ e^x & x\ge 0 \end{cases} \quad f_2(x) = x^2 f3(x)={sin⁡xx<0xx≥0f4(x)={f2(f1(x))x<0f2(f1(x))−1x≥0f_3(x) = \begin{cases} \sin x & x<0 \\ x & x\ge 0 \end{cases} \quad f_4(x) = \begin{cases} f_2(f_1(x)) & x<0 \\ f_2(f_1(x))-1 & x\ge 0 \end{cases} For each: continuity, differentiability, one-one / onto\text{For each: continuity, differentiability, one-one / onto} check the break point x=0\text{check the break point } x=0 (f2∘f1)(x)={(−x)2=x2x<0(ex)2=e2xx≥0(f_2\circ f_1)(x) = \begin{cases} (-x)^2 = x^2 & x<0 \\ (e^x)^2 = e^{2x} & x\ge 0 \end{cases} lim⁡x→0−x2=0≠1=lim⁡x→0+e2x⇒discontinuous\lim_{x\to 0^-}x^2 = 0 \neq 1 = \lim_{x\to 0^+}e^{2x} \Rightarrow \text{discontinuous} (R)→(2) Neither continuous nor one-one(R) \to (2)\ \text{Neither continuous nor one-one} f3(0)=0,lim⁡x→0sin⁡x=0=lim⁡x→0x⇒continuousf_3(0)=0,\quad \lim_{x\to0}\sin x = 0 = \lim_{x\to0}x \Rightarrow \text{continuous} left derivative cos⁡0=1=right derivative\text{left derivative } \cos 0 = 1 = \text{right derivative} f3 not one-one (sin repeats on x<0)f_3 \text{ not one-one (sin repeats on } x<0) (Q)→(3) Differentiable but not one-one(Q) \to (3)\ \text{Differentiable but not one-one} f2(x)=x2 on [0,∞): continuous, strictly increasingf_2(x)=x^2 \text{ on } [0,\infty):\ \text{continuous, strictly increasing} (S)→(4) Continuous and one-one(S) \to (4)\ \text{Continuous and one-one} f4(x)={x2x<0e2x−1x≥0f_4(x) = \begin{cases} x^2 & x<0 \\ e^{2x}-1 & x\ge 0 \end{cases} lim⁡x→0−x2=0=lim⁡x→0+(e2x−1)⇒continuous\lim_{x\to0^-}x^2 = 0 = \lim_{x\to0^+}(e^{2x}-1) \Rightarrow \text{continuous} range=[0,∞)=codomain⇒onto; not one-one\text{range} = [0,\infty) = \text{codomain} \Rightarrow \text{onto; not one-one} (P)→(1) Onto but not one-one(P) \to (1)\ \text{Onto but not one-one} P→1; Q→3; R→2; S→4P\to1;\ Q\to3;\ R\to2;\ S\to4 Answer: option (4)\boxed{\text{Answer: option (4)}}

More from Continuity and Differentiability

All 35 Continuity and Differentiability questions →

Continuity and Differentiability: formulas and what the chapter covers →