Limits Continuity And Differentiability(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

LIMITS CONTINUITY AND DIFFERENTIABILITY MCQs

Total Questions : 30 | Page 2 of 3 pages

If $α$ is a repeated root of $a x^{2} + b x + c = 0$ then $lim x \to α \frac{sin (a x^{2} + b x + c)}{(x - α)^{2}}$ is

Discuss Question

Answer: Option B. ->

a

:
B

lim x \to α \frac{sin (a x^{2} + b x + c)}{(x - α)^{2}} = lim x \to α \frac{sin a (x - α) (x - α)}{(x - α)^{2}}

= lim x \to α \frac{sin a (x - α)^{2}}{a (x - α)^{2}} \times a = a

Question 12.

Let f be a function such that f(x+y)=f(x)+f(y) for all x and y and $f (x) = (2 x^{2} + 3 x) g (x)$ for all x where g(x) is continuous and g(0)=9 then f'(0) is equals to

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Answer: Option C. -> 27
:
C

f^{'} (x) = l i m h \to 0 \frac{f (x + h) - f (x)}{h}

= l i m h \to 0 \frac{f (x) + f (h) - f (x)}{h}

= l i m h \to 0 \frac{(2 h^{2} + 3 h) g (h)}{h}

= l i m h \to 0 (2 h + 3) g (h)

= 3 g (0)

= 27

Question 13.

$The value of l i m x \to \infty \frac{x + c o s x}{x + s i n x} i s$

-1
0
1
2

Discuss Question

Answer: Option C. -> 1
:
C

$l i m x \to \infty \frac{x + c o s x}{x + s i n x} [Putting x = \frac{1}{h}; a s x \to \infty, h \to 0] = l i m h \to 0 \frac{\frac{1}{h} + c o s (\frac{1}{h})}{\frac{1}{h} + s i n (\frac{1}{h})} = l i m h \to 0 \frac{1 + h c o s \frac{1}{h}}{1 + h c o s \frac{1}{h}} = \frac{1 + 0}{1 + 0} ⎡ ⎢⎢⎢ ⎣ \begin{matrix} ∵ - 1 \leq s i n \frac{1}{h} \leq 1 a n d - 1 \leq c o s \frac{1}{h} \leq 1, ∴ w h e r e h \to 0, h c o s \frac{1}{h} \to 0 a n d h s i n \frac{1}{h} \to 0 \end{matrix} ⎤ ⎥⎥⎥ ⎦ = 1$

Question 14.

$l i m x \to 0 \frac{(1 + x)^{\frac{1}{3}} - (1 - x)^{\frac{1}{3}}}{x} =$

Discuss Question

Answer: Option A. -> 2/3
:
A

$l i m x \to 0 \frac{(1 + x)^{\frac{1}{3}} - (1 - x)^{\frac{1}{3}}}{x} = l i m x \to 0 \frac{(1 + x)^{\frac{1}{3}} - (1 - x)^{\frac{1}{3}}}{(1 + x) - (1 - x)} .2 = 2. \frac{1}{3}$

Question 15.

If $l i m_{x \to 0}$ kx cosec x = $l i m_{x \to 0}$ x cosec kx , then k =

1
-1

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Answer: Option C. ->

:
C

$l i m_{x \to 0}$ kx cosec x = $l i m_{x \to 0}$ x cosec kx

= k $l i m_{x \to 0}$ $\frac{x}{s i n x}$ = $\frac{1}{k}$ $l i m_{x \to 0}$ $\frac{k x}{s i n k x}$ =k= $\frac{1}{k}$ =k=± 1

Question 16.

$The values of constants a and b so that l i m x \to \infty (\frac{x^{2} + 1}{x + 1} - a x - b) = \frac{1}{2}, a r e$

$a = 1, b = - \frac{3}{2}$
$a = - 1, b = \frac{3}{2}$
a=0, b=0
a =2, b= -1

Discuss Question

Answer: Option A. ->

a = 1, b = - \frac{3}{2}

:
A

$l i m x \to \infty (\frac{x^{2} + 1}{x + 1} - a x - b) = \frac{1}{2} \Rightarrow l i m x \to \infty \frac{(x^{2} + 1) - (a x + b) (x + 1)}{x + 1} = \frac{1}{2} \Rightarrow l i m x \to \infty \frac{x^{2} (1 - a) - (a + b) x - b + 1}{x + 1} = \frac{1}{2} \Rightarrow 1 - a = 0 a n d a + b = - \frac{1}{2} ∴ a = 1 b = - \frac{3}{2}$

Question 17.

If the derivative of the function $f (x) = b x^{2} + a x + 4; x \geq - 1 a x^{2} + b; x < - 1^{^{'}}$ , is everywhere continuous, then

a = 2, b = 3
a = 3, b = 2
a = - 2, b = - 3
a = - 3, b = - 2

Discuss Question

Answer: Option A. -> a = 2, b = 3
:
A

W e h a v e, f (x) = {\begin{matrix} a x^{2} + b, & x < - 1 b x^{2} + a x + 4, & x \geq - 1 \end{matrix} ∴ f^{^{'}} (x) = {\begin{matrix} 2 a x, & < - 1 2 b x + a, & x \geq - 1 \end{matrix} S i n c e, f (x) i s d i f f e r e n t i a b l e a t x = - 1, t h e r e f o r e i t i s c o n t i n u o u s a t x = - 1 a n d h e n c e, l i m x \to - 1 - f (x) = l i m x \to - 1 + f (x) \Rightarrow a + b = b - a + 4 \Rightarrow a = 2 a n d a l s o, l i m x \to - 1 - f (x) = l i m x \to - 1 + f (x) \Rightarrow - 2 a = - 2 b + a \Rightarrow 3 a = 2 b \Rightarrow b = 3 (∵ a = 2) H e n c e, a = 2, b = 3

Question 18.

The value of $lim x \to \frac{π}{2} [s i n^{- 1} s i n x]$ ,[x] is the greatest integer function of x, is

1
$\frac{π}{2}$
0
$\frac{1}{2}$

Discuss Question

Answer: Option A. -> 1
:
A

$lim x \to \frac{π}{2} [s i n^{- 1} s i n x]$
= $lim x \to \frac{π}{2} [x]$
= 1

Question 19.

$The value of {lim}_{x \to 0} \frac{1 - c o s^{3} x}{x s i n x c o s x}$

Discuss Question

Answer: Option C. -> 3/2
:
C

${lim}_{x \to 0} \frac{1 - c o s^{3} x}{x s i n x c o s x} = {lim}_{x \to 0} \frac{(1 - c o s x) (1 + c o s x + c o s^{2} x)}{x s i n x c o s x} = {lim}_{x \to 0} \frac{2 s i n^{2} (\frac{x}{2})}{2 s i n (\frac{x}{2}) c o s (\frac{x}{2}) . x} \times \frac{(1 + c o s x + c o s^{2} x)}{c o s x} = {lim}_{x \to 0} \frac{s i n (\frac{x}{2})}{2 (\frac{x}{2})} \times \frac{1 + c o s x + c o s^{2} x}{c o s (\frac{x}{2}) c o s x} = \frac{1}{2} \times 3 = \frac{3}{2}$

Question 20.

$I f f (x) = {\begin{matrix} x^{2} - 3, & 2 < x < 3 2 x + 5, & 3 < x < 4 \end{matrix}$ , the equation whose roots are $lim x \to 3 - f (x)$ and $lim x \to 3 + f (x) i s$

$x^{2} - 12 x + 36 = 0$
$x^{2} - 26 x + 66 = 0$
$x^{2} - 17 x + 66 = 0$
$x^{2} - 22 x + 121 = 0$

Discuss Question

Answer: Option C. ->

x^{2} - 17 x + 66 = 0

:
C

$f (x) = {\begin{matrix} x^{2} - 3, & 2 < x < 3 2 x + 5, & 3 < x < 4 \end{matrix}$
$∴ lim x \to 3 - f (x) = lim x \to 3 - (x^{2} - 3) = 6$
and $lim x \to 3 + f (x) = lim x \to 3 + (2 x + 5) = 11$
Hence, the required equation will be
$x^{2}$ - (sum of roots)x + (Products of roots) = 0