Let f:ℝ→ℝ be given by fx=x5+5x4+10x3+10x2+3x+1,x<0;x2−x+1,0≤x<1;23x3−4x2+7x−83,1≤x<3;x−2logex−2−x+103,x≥3. Then which of the following options is/are correct?

f is onto , f ' h a s a l o c a l m a x i m u m a t x = 1 , f ' is NOT differentiable at x = 1

Let f:ℝ→ℝ be given by fx=x5+5x4+10x3+10x2+3x+1,x

Let f : R → R f x = x 5 + 5 x 4 + 10 x 3 + 10 x 2 + 3 x + 1 , x < 0 x 2 − x + 1 , 0 ≤ x < 1 2 3 x 3 − 4 x 2 + 7 x − 8 3 , 1 ≤ x < 3 x − 2 ln x − 2 − x + 10 3 , x ≥ 3 = x + 1 5 − 2 x , x < 0 x 2 − x + 1 , 0 ≤ x ≤ 1 2 3 x 3 − 4 x 2 + 7 x − 8 3 , 1 ≤ x < 3 x − 2 ln x − 2 − x + 10 3 , x ≥ 3 f 1 x = 5 x + 1 4 − 2 , x < 0 2 x − 1 , 0 ≤ x < 1 2 x 2 − 8 x + 7 , 1 ≤ x < 3 1 + ln x − 2 − 1 , x ⩾ 3 f 11 x = 20 x + 1 3 , x < 0 2 0 ≤ x < 1 4 x − 8 1 ≤ x < 3 1 x − 2 x ⩾ 3 i f x → - ∞ t h e n f x → - ∞ a n d x → ∞ ⇒ f x → ∞ The range is = R and f is onto Clearly f 1 - 1 2 = 5 16 - 2 < 0 ∴ f i s n o t i n c r e a sin g i n - ∞ , 0 Also f 11 ( 1 - ) = 2 > 0 f 11 ( 1 + ) = - 4 < 0 ∣ ∴ f 1 ( x ) has local maximum at x = 1 f 1 1 ( 1 + ) = - 4 f 1 ( 1 - ) = 2 L . D ≠ R , D ∴ f 1 is not differentiable at x = 1

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let $f : ℝ \to ℝ$ be given by $f (x) = \{\begin{matrix} x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 3 x + 1, & x < 0; \\ x^{2} - x + 1, & 0 \leq x < 1; \\ \frac{2}{3} x^{3} - 4 x^{2} + 7 x - \frac{8}{3}, & 1 \leq x < 3; \\ (x - 2) \log_{e} (x - 2) - x + \frac{10}{3}, & x \geq 3. \end{matrix}$ Then which of the following options is/are correct?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

f is onto

f is increasing on $(- \infty, 0)$

$f'$ $h a s a l o c a l m a x i m u m a t x = 1$

$f'$ is NOT differentiable at x = 1

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let $f : R \to R$

$\begin{matrix} f (x) = \{\begin{cases} \begin{matrix} x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 3 x + 1, x < 0 \\ x^{2} - x + 1, 0 \leq x < 1 \end{matrix} \\ \begin{matrix} \frac{2}{3} x^{3} - 4 x^{2} + 7 x - \frac{8}{3}, 1 \leq x < 3 \\ (x - 2) \ln (x - 2) - x + \frac{10}{3}, x \geq 3 \end{matrix} \end{cases} \\ = \{\begin{matrix} {(x + 1)}^{5} - 2 x, & x < 0 \\ x^{2} - x + 1, & 0 \leq x \leq 1 \\ \frac{2}{3} x^{3} - 4 x^{2} + 7 x - \frac{8}{3}, & 1 \leq x < 3 \\ (x - 2) \ln (x - 2) - x + \frac{10}{3}, & x \geq 3 \end{matrix} \\ f^{1} (x) = \{\begin{matrix} 5 {(x + 1)}^{4} - 2, & x < 0 \\ 2 x - 1, & 0 \leq x < 1 \\ 2 x^{2} - 8 x + 7, & 1 \leq x < 3 \\ 1 + \ln (x - 2) - 1, & x ⩾ 3 \end{matrix} \\ f^{11} (x) = \{\begin{matrix} 20 {(x + 1)}^{3}, & x < 0 \\ 2 & 0 \leq x < 1 \\ 4 x - 8 & 1 \leq x < 3 \\ \frac{1}{x - 2} & x ⩾ 3 \end{matrix} \end{matrix}$

$i f x \to - \infty t h e n f (x) \to - \infty a n d x \to \infty \Rightarrow f (x) \to \infty$

The range is = R and f is onto

$\begin{array}{l} Clearly f^{1} (- \frac{1}{2}) = \frac{5}{16} - 2 < 0 ∴ f i s n o t i n c r e a \sin g i n (- \infty, 0) \\ Also f^{11} (1 -) = 2 > 0 \\ f^{11} (1 +) = - 4 < 0 ∣ \\ ∴ f^{1} (x) has local maximum at x = 1 \\ \begin{matrix} {(f^{1})}^{1} (1 +) = - 4 \\ (f^{1}) (1 -) = 2 \end{matrix}\} L . D \neq R, D ∴ f^{1} is not differentiable at x = 1 \end{array}$