Let the function f:R→ℝ be defined by f(x)=x3−x2+(x−1)sin⁡x and let g:R→R be an arbitrary function. Let fg :ℝ→ℝ be the product function defined by (fg)(x)=f(x)g(x) . Then which of the following statements is/are TRUE ?

f:R→R f(x)=x2+sin⁡x(x−1) f1+=f1−=f(1)=0fg(x):f(x)⋅g(x) fg:R→R let fg⁡(x)=h(x)=f(x)⋅g(x) h:R→R option (c): h′(x)=f′(x)g(x)+f(x)g′(x) h′(1)=f′(1)g(1)+0 as f(1)=0,g′(x) exists ⇒ if g(x) is differentiable then h(x) is also differentiable (true) option (A) : If g(x) is continuous at x=1 then g1+=g1−=g1 h′1+=limh→0+ h(1+h)−h(1)hh′1+=limh→0+ f(1+h)g(1+h)−0h=f′(1)g(1)h′1−=limh→0+ f(1−h)g(1−h)−0−h=f′(1)g(1) So h(x)=f(x)⋅g(x) is differentiable at x=1 (True) option (B) (D): h′1+=limh→0+ h(1+h)−h(1)hh′1+=limh→0+ f(1+h)g(1+h)h=f′(1)g1+h′1−=limh→0+ f(1−h)g(1−h)−h=f′(1)⋅g1−⇒g1+=g1−so we cannot comment on the continuity and differentiability of the function gx

Let the function f:R→ℝ be defined by f(x)=x3−x2+(x−1)sin⁡x and let g:R→R be an arbitrary function. Let fg :ℝ→ℝ be the product function defined by (fg)(x)=f(x)g(x) . Then which of the following statements is/are TRUE ?

f:R→R f(x)=x2+sin⁡x(x−1) f1+=f1−=f(1)=0fg(x):f(x)⋅g(x) fg:R→R let fg⁡(x)=h(x)=f(x)⋅g(x) h:R→R option (c): h′(x)=f′(x)g(x)+f(x)g′(x) h′(1)=f′(1)g(1)+0 as f(1)=0,g′(x) exists ⇒ if g(x) is differentiable then h(x) is also differentiable (true) option (A) : If g(x) is continuous at x=1 then g1+=g1−=g1 h′1+=limh→0+ h(1+h)−h(1)hh′1+=limh→0+ f(1+h)g(1+h)−0h=f′(1)g(1)h′1−=limh→0+ f(1−h)g(1−h)−0−h=f′(1)g(1) So h(x)=f(x)⋅g(x) is differentiable at x=1 (True) option (B) (D): h′1+=limh→0+ h(1+h)−h(1)hh′1+=limh→0+ f(1+h)g(1+h)h=f′(1)g1+h′1−=limh→0+ f(1−h)g(1−h)−h=f′(1)⋅g1−⇒g1+=g1−so we cannot comment on the continuity and differentiability of the function gx

AI Mentor

Check Your IQ

Free Expert Demo

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

$Let the function f : R \to ℝ be defined by f (x) = x^{3} - x^{2} + (x - 1) \sin x and let g : R \to R be an$
$arbitrary function. Let fg : ℝ \to ℝ be the product function defined by (f g) (x) = f (x) g (x) . Then$
$which of the following statements is/are TRUE ?$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$If g is continuous at x = 1, then fg is differentiable at x = 1$

$If fg is differentiable at x = 1, then g is continuous at x = 1$

$If g is differentiable at x = 1, then f g is differentiable at x = 1$

$If fg is differentiable at x = 1, then g is differentiable at x = 1$

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{aligned} f : R \to R f (x) = (x^{2} + \sin x) (x - 1) f (1^{+}) = f (1^{-}) = f (1) = 0 \\ fg (x) : f (x) \cdot g (x) fg : R \to R \\ let fg (x) = h (x) = f (x) \cdot g (x) h : R \to R \\ option (c): h^{'} (x) = f^{'} (x) g (x) + f (x) g^{'} (x) \end{aligned}$

$\begin{matrix} h^{'} (1) = f^{'} (1) g (1) + 0 \\ (as f (1) = 0, g^{'} (x) exists\} \end{matrix}$

$\begin{array}{l} \Rightarrow if g (x) is differentiable then h (x) is also differentiable (true) \\ option (A) : If g (x) is continuous at x = 1 then g (1^{+}) = g (1^{-}) = g (1) \\ h^{'} (1^{+}) = lim_{h \to 0^{+}} \frac{h (1 + h) - h (1)}{h} \end{array}$

$\begin{array}{l} h^{'} (1^{+}) = lim_{h \to 0^{+}} \frac{f (1 + h) g (1 + h) - 0}{h} = f^{'} (1) g (1) \\ h^{'} (1^{-}) = lim_{h \to 0^{+}} \frac{f (1 - h) g (1 - h) - 0}{- h} = f^{'} (1) g (1) \\ So h (x) = f (x) \cdot g (x) is differentiable \\ at x = 1 (True) \end{array}$

$option (B) (D): h^{'} (1^{+}) = lim_{h \to 0^{+}} \frac{h (1 + h) - h (1)}{h}$

$\begin{array}{l} h^{'} (1^{+}) = lim_{h \to 0^{+}} \frac{f (1 + h) g (1 + h)}{h} = f^{'} (1) g (1^{+}) \\ h^{'} (1^{-}) = lim_{h \to 0^{+}} \frac{f (1 - h) g (1 - h)}{- h} = f^{'} (1) \cdot g (1^{-}) \\ \Rightarrow g (1^{+}) = g (1^{-}) \end{array}$