दो हजार बारह - (हिंदी अक्षरों को पढ़कर संख्याओं को विकल्पों में से पहचानकर लिखिए ।)

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दो हजार बारह - (हिंदी अक्षरों को पढ़कर संख्याओं को विकल्पों में से पहचानकर लिखिए ।)

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12a\dfrac{1}{2\sqrt{a}}

Question 21

Step 1: Write the given limit.

lim⁡x→ax−ax−a\lim_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a}

Step 2: Multiply numerator and denominator by the conjugate x+a\sqrt{x} + \sqrt{a}.

lim⁡x→a(x−a)(x+a)(x−a)(x+a)=lim⁡x→ax−a(x−a)(x+a)\lim_{x \to a} \frac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{(x - a)(\sqrt{x} + \sqrt{a})} = \lim_{x \to a} \frac{x - a}{(x - a)(\sqrt{x} + \sqrt{a})}

Step 3: Cancel the common factor x−ax - a (valid as x≠ax \neq a in the limit).

lim⁡x→a1x+a\lim_{x \to a} \frac{1}{\sqrt{x} + \sqrt{a}}

Step 4: Substitute x=ax = a.

1a+a=12a\frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}}

\dfrac{1{2\sqrt{a}}}

Question 22

Step 1: Rewrite tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}.

lim⁡x→0tan⁡x−xx3=lim⁡x→0sin⁡xcos⁡x−xx3=lim⁡x→0sin⁡x−xcos⁡xx3cos⁡x\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^3} = \lim_{x \to 0} \frac{\sin x - x \cos x}{x^3 \cos x}

Step 2: Since lim⁡x→0cos⁡x=1\lim_{x \to 0} \cos x = 1, the limit is lim⁡x→0sin⁡x−xcos⁡xx3\lim_{x \to 0} \frac{\sin x - x \cos x}{x^3}.

Step 3: Use Taylor expansions around x=0x = 0:

sin⁡x=x−x36+o(x3),cos⁡x=1−x22+o(x4)\sin x = x - \frac{x^3}{6} + o(x^3), \quad \cos x = 1 - \frac{x^2}{2} + o(x^4)

Step 4: Compute xcos⁡xx \cos x:

xcos⁡x=x(1−x22+o(x4))=x−x32+o(x5)x \cos x = x \left(1 - \frac{x^2}{2} + o(x^4)\right) = x - \frac{x^3}{2} + o(x^5)

Step 5: Numerator sin⁡x−xcos⁡x\sin x - x \cos x:

(x−x36+o(x3))−(x−x32+o(x5))=−x36+x32+o(x3)=(12−16)x3+o(x3)=13x3+o(x3)\left(x - \frac{x^3}{6} + o(x^3)\right) - \left(x - \frac{x^3}{2} + o(x^5)\right) = -\frac{x^3}{6} + \frac{x^3}{2} + o(x^3) = \left(\frac{1}{2} - \frac{1}{6}\right)x^3 + o(x^3) = \frac{1}{3}x^3 + o(x^3)

Step 6: Divide by x3x^3:

lim⁡x→013x3+o(x3)x3=13\lim_{x \to 0} \frac{\frac{1}{3}x^3 + o(x^3)}{x^3} = \frac{1}{3}

\dfrac{1{3}}

Question 23

Step 1: Substitute t=π2−xt = \frac{\pi}{2} - x. As x→π2x \to \frac{\pi}{2}, t→0t \to 0.

lim⁡x→π2(π2−x)tan⁡x=lim⁡t→0ttan⁡(π2−t)=lim⁡t→0tcot⁡t\lim_{x \to \frac{\pi}{2}} \left(\frac{\pi}{2} - x\right) \tan x = \lim_{t \to 0} t \tan\left(\frac{\pi}{2} - t\right) = \lim_{t \to 0} t \cot t

Step 2: Rewrite cot⁡t=1tan⁡t\cot t = \frac{1}{\tan t}.

tcot⁡t=ttan⁡t=tcos⁡tsin⁡tt \cot t = \frac{t}{\tan t} = \frac{t \cos t}{\sin t}

Step 3: lim⁡t→0tcos⁡tsin⁡t=lim⁡t→0(tsin⁡t⋅cos⁡t)=(lim⁡t→0tsin⁡t)(lim⁡t→0cos⁡t)=1⋅1=1\lim_{t \to 0} \frac{t \cos t}{\sin t} = \lim_{t \to 0} \left( \frac{t}{\sin t} \cdot \cos t \right) = \left( \lim_{t \to 0} \frac{t}{\sin t} \right) \left( \lim_{t \to 0} \cos t \right) = 1 \cdot 1 = 1

1

Question 24

Step 1: Factor the numerator using 1−u3=(1−u)(1+u+u2)1 - u^3 = (1 - u)(1 + u + u^2) where u=cos⁡xu = \cos x.

1−cos⁡3x=(1−cos⁡x)(1+cos⁡x+cos⁡2x)1 - \cos^3 x = (1 - \cos x)(1 + \cos x + \cos^2 x) lim⁡x→01−cos⁡3xx2=lim⁡x→0(1−cos⁡x)(1+cos⁡x+cos⁡2x)x2=lim⁡x→0[1−cos⁡xx2⋅(1+cos⁡x+cos⁡2x)]\lim_{x \to 0} \frac{1 - \cos^3 x}{x^2} = \lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x + \cos^2 x)}{x^2} = \lim_{x \to 0} \left[ \frac{1 - \cos x}{x^2} \cdot (1 + \cos x + \cos^2 x) \right]

Step 2: Evaluate each limit separately.

lim⁡x→01−cos⁡xx2=12,lim⁡x→0(1+cos⁡x+cos⁡2x)=1+1+12=3\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \quad \lim_{x \to 0} (1 + \cos x + \cos^2 x) = 1 + 1 + 1^2 = 3

Step 3: Multiply the limits.

12×3=32\frac{1}{2} \times 3 = \frac{3}{2}

\dfrac{3{2}}

Question 25

Step 1: Use Taylor expansion for sin⁡x\sin x around x=0x = 0.

sin⁡x=x−x36+o(x3)\sin x = x - \frac{x^3}{6} + o(x^3)

Step 2: Numerator x−sin⁡xx - \sin x:

x−sin⁡x=x−(x−x36+o(x3))=x36+o(x3)x - \sin x = x - \left(x - \frac{x^3}{6} + o(x^3)\right) = \frac{x^3}{6} + o(x^3)

Step 3: Divide by x3x^3:

lim⁡x→0x−sin⁡xx3=lim⁡x→0x36+o(x3)x3=16\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{6} + o(x^3)}{x^3} = \frac{1}{6}

\dfrac{1{6}}

Question 26

Step 1: Rewrite tan⁡x−sin⁡x\tan x - \sin x.

tan⁡x−sin⁡x=sin⁡xcos⁡x−sin⁡x=sin⁡x(1cos⁡x−1)=sin⁡x⋅1−cos⁡xcos⁡x\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x \left( \frac{1}{\cos x} - 1 \right) = \sin x \cdot \frac{1 - \cos x}{\cos x} lim⁡x→0tan⁡x−sin⁡xx3=lim⁡x→0sin⁡x(1−cos⁡x)x3cos⁡x=lim⁡x→0[sin⁡xx⋅1−cos⁡xx2⋅1cos⁡x]\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3 \cos x} = \lim_{x \to 0} \left[ \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{\cos x} \right]

Step 2: Evaluate each limit separately.

lim⁡x→0sin⁡xx=1,lim⁡x→01−cos⁡xx2=12,lim⁡x→01cos⁡x=1\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \quad \lim_{x \to 0} \frac{1}{\cos x} = 1

Step 3: Multiply the limits.

1⋅12⋅1=121 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2}

\dfrac{1{2}}

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Question 21 Step 1: Write the given limit. _x a sqrt(x) - sqrt(a)x - a Step 2: Multiply numerator and denominator by the conjugate sqrt(x) + sqrt(a).

दो हजार बारह - (हिंदी अक्षरों को पढ़कर संख्याओं को विकल्पों में से पहचानकर लिखिए ।)
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Question 21 Step 1: Write the given limit. _x a sqrt(x) - sqrt(a)x - a Step 2: Multiply numerator and denominator by the conjugate sqrt(x) + sqrt(a). _x a (sqrt(x) - sqrt(a))(sqrt(x) + sqrt(a))(x - a)(sqrt(x) + sqrt(a)) = _x a (x - a)/((x - a)(sqrt(x) + a)) Step 3: Cancel the common factor x - a (valid as x ≠ a in the limit). _x a (1)/(sqrt(x) + a) Step 4: Substitute x = a. (1)/(sqrt(a) + a) = (1)/(2sqrt(a)) (1)/(2sqrt(a)) Question 22 Step 1: Rewrite x = ( x)/( x). _x 0 ( x - x)/(x^3) = _x 0 ( x)/( x) - xx^3 = _x 0 ( x - x x)/(x^3 x) Step 2: Since _x 0 x = 1, the limit is _x 0 ( x - x x)/(x^3). Step 3: Use Taylor expansions around x = 0: x = x - (x^3)/(6) + o(x^3), x = 1 - (x^2)/(2) + o(x^4) Step 4: Compute x x: x x = x (1 - (x^2)/(2) + o(x^4)) = x - (x^3)/(2) + o(x^5) Step 5: Numerator x - x x: (x - (x^3)/(6) + o(x^3)) - (x - (x^3)/(2) + o(x^5)) = -(x^3)/(6) + (x^3)/(2) + o(x^3) = ((1)/(2) - (1)/(6))x^3 + o(x^3) = (1)/(3)x^3 + o(x^3) Step 6: Divide by x^3: _x 0 (1)/(3)x^3 + o(x^3)x^3 = (1)/(3) (1)/(3) Question 23 Step 1: Substitute t = ()/(2) - x. As x ()/(2), t 0. _x ()/(2) (()/(2) - x) x = _t 0 t (()/(2) - t) = _t 0 t t Step 2: Rewrite t = (1)/( t). t t = (t)/( t) = (t t)/( t) Step 3: _t 0 (t t)/( t) = _t 0 ( (t)/( t) · t ) = ( _t 0 (t)/( t) ) ( _t 0 t ) = 1 · 1 = 1 1 Question 24 Step 1: Factor the numerator using 1 - u^3 = (1 - u)(1 + u + u^2) where u = x. 1 - ^3 x = (1 - x)(1 + x + ^2 x) _x 0 (1 - ^3 x)/(x^2) = _x 0 ((1 - x)(1 + x + ^2 x))/(x^2) = _x 0 [ (1 - x)/(x^2) · (1 + x + ^2 x) ] Step 2: Evaluate each limit separately. _x 0 (1 - x)/(x^2) = (1)/(2), _x 0 (1 + x + ^2 x) = 1 + 1 + 1^2 = 3 Step 3: Multiply the limits. (1)/(2) × 3 = (3)/(2) (3)/(2) Question 25 Step 1: Use Taylor expansion for x around x = 0. x = x - (x^3)/(6) + o(x^3) Step 2: Numerator x - x: x - x = x - (x - (x^3)/(6) + o(x^3)) = (x^3)/(6) + o(x^3) Step 3: Divide by x^3: _x 0 (x - x)/(x^3) = _x 0 (x^3)/(6) + o(x^3)x^3 = (1)/(6) (1)/(6) Question 26 Step 1: Rewrite x - x. x - x = ( x)/( x) - x = x ( (1)/( x) - 1 ) = x · (1 - x)/( x) _x 0 ( x - x)/(x^3) = _x 0 ( x (1 - x))/(x^3 x) = _x 0 [ ( x)/(x) · (1 - x)/(x^2) · (1)/( x) ] Step 2: Evaluate each limit separately. _x 0 ( x)/(x) = 1, _x 0 (1 - x)/(x^2) = (1)/(2), _x 0 (1)/( x) = 1 Step 3: Multiply the limits. 1 · (1)/(2) · 1 = (1)/(2) (1)/(2)