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← JEE Advanced 2024 · Paper 1
MathematicsList match+3 / -1Question 17 of 51

JEE Advanced 2024 · Paper 1, Mathematics question 17

Let f:R→Rf : \mathbb{R} \to \mathbb{R} and g:R→Rg : \mathbb{R} \to \mathbb{R} be functions defined by

f(x)={x∣x∣sin⁡(1x),x≠0,0,x=0,andg(x)={1−2x,0≤x≤12,0,otherwise.f(x) = \begin{cases} x|x|\sin\left(\dfrac{1}{x}\right), & x \neq 0, \\ 0, & x = 0, \end{cases} \quad\text{and}\quad g(x) = \begin{cases} 1 - 2x, & 0 \le x \le \dfrac{1}{2}, \\ 0, & \text{otherwise}. \end{cases}

Let a,b,c,d∈Ra, b, c, d \in \mathbb{R}. Define the function h:R→Rh : \mathbb{R} \to \mathbb{R} by

h(x)=a f(x)+b(g(x)+g(12−x))+c (x−g(x))+d g(x),x∈R.h(x) = a\,f(x) + b\left(g(x) + g\left(\frac{1}{2} - x\right)\right) + c\,(x - g(x)) + d\,g(x),\quad x \in \mathbb{R}.

Match each entry in List-I to the correct entry in List-II.

The correct option is

List-I

  • P
    If a=0a = 0, b=1b = 1, c=0c = 0, and d=0d = 0, then
  • Q
    If a=1a = 1, b=0b = 0, c=0c = 0, and d=0d = 0, then
  • R
    If a=0a = 0, b=0b = 0, c=1c = 1, and d=0d = 0, then
  • S
    If a=0a = 0, b=0b = 0, c=0c = 0, and d=1d = 1, then

List-II

  • 1
    hh is one-one.
  • 2
    hh is onto.
  • 3
    hh is differentiable on R\mathbb{R}.
  • 4
    the range of hh is [0,1][0, 1].
  • 5
    the range of hh is {0,1}\{0, 1\}.
Choose one option
Answer and worked solution

Official answer: C

(P) h(x)=g(x)+g(12−x)h(x) = g(x) + g\left(\frac12 - x\right). On [0,12]\left[0, \frac12\right] both arguments lie in [0,12]\left[0,\frac12\right]: h=(1−2x)+2x=1h = (1-2x) + 2x = 1. Outside that interval both terms are 0. Range {0,1}\{0, 1\}. (P)→(5).

(Q) h=fh = f. For x≠0x \ne 0 it is differentiable, and f′(0)=lim⁡h→0∣h∣sin⁡1h=0f'(0) = \lim_{h\to0}|h|\sin\frac1h = 0. So ff is differentiable on R\mathbb{R}. (Q)→(3).

(R) h(x)=x−g(x)h(x) = x - g(x):

  • x<0x < 0: h=xh = x, values (−∞,0)(-\infty, 0);
  • 0≤x≤120 \le x \le \frac12: h=3x−1h = 3x - 1, values [−1,12]\left[-1, \frac12\right];
  • x>12x > \frac12: h=xh = x, values (12,∞)\left(\frac12, \infty\right).

The union is R\mathbb{R}, so hh is onto. It is not one-one, e.g. h(−12)=h(16)=−12h\left(-\frac12\right) = h\left(\frac16\right) = -\frac12. (R)→(2).

(S) h=gh = g, range [0,1][0, 1]. (S)→(4).

Option (C).