Trigonometric Function MCQ Quiz - Objective Question with Answer for Trigonometric Function - Download Free PDF

Answer (1)

Sol.

\(\lim _{x \rightarrow 0} \frac{\cos 2 x+a \cos ^{4} x-b}{x^{4}}=L\)

\(\lim _{x \rightarrow 0} \frac{2 \cos ^{2} x-1+a \cos ^{4} x-b}{x^{4}}=L \quad \ldots(1)\)

To get the finite value,

1 + a – b = 0 

⇒ \(a=b-1 \quad \quad ...(2)\)

Apply L Hospital 

\(\lim _{x \rightarrow 0} \frac{4 \cos x(-\sin x)+4 a \cos ^{3} x(-\sin x)}{4 x^{3}}\)

\(\lim _{x \rightarrow 0} \frac{4 \cos x+4 a \cos ^{3} x}{4 x^{3}}\left(\frac{-\sin x}{x}\right)\)

To get the finite value, a = –1

Also from (1)

b = 0 

∴ a = b = -1

Calculation

\(\lim_{x\rightarrow \frac{\pi}{2}} \left(\frac{1}{x-\frac{\pi}{2}}\int^{\left(\frac{\pi}{2}\right)^3}_{x^3} cos\left(t^{\frac{1}{3}}\right)dt\right)\)

Using L’hopital rule 

\(\Rightarrow\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\)

\(\Rightarrow\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \)

\(\Rightarrow\frac{3 \pi^2}{8} \)

Hence option 3 is correct

Calculation:

We have to find the value of \(\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}} = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}2\sin x - \sin x - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}2\sin x - \sin x + {\rm{\;}}1}}\)

\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{2\sin x(\sin x + 1) - 1\;(\sin x + 1)}}{{2\sin x(\sin x - 1) - 1\;(\sin x - 1)}}\)

\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{\left( {2\sin x - 1} \right)\;(\sin x + 1)}}{{\left( {2\sin x - 1} \right)\;(\sin x - 1)}}\)

\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{(\sin x + 1)}}{{(\sin x - 1)}} = \;\frac{{\left( {\frac{1}{2} + 1} \right)}}{{\left( {\frac{1}{2} - 1} \right)}} = \;\frac{{\left( {\frac{3}{2}} \right)}}{{\left( {\frac{{ - 1}}{2}} \right)}} = \; - 3\)

Explanation:

\(\rm \lim_{x\rightarrow \frac{\pi}{2}}(\sec \theta-\tan \theta)\)

= \(\rm \lim_{x\rightarrow \frac{\pi}{2}} \frac{1-sin\theta }{cos\theta} [\frac{0}{0}\) Form]

= \(\rm \lim_{x\rightarrow \frac{\pi}{2}}( \frac{-cos\theta}{-sin\theta}) =0\)  (using L' hospital rule)

∴ Option (b) is correct.

Calculation

\(\lim_{x\rightarrow \frac{\pi}{2}} \left(\frac{1}{x-\frac{\pi}{2}}\int^{\left(\frac{\pi}{2}\right)^3}_{x^3} cos\left(t^{\frac{1}{3}}\right)dt\right)\)

Using L’hopital rule 

\(\Rightarrow\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\)

\(\Rightarrow\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \)

\(\Rightarrow\frac{3 \pi^2}{8} \)

Hence option 3 is correct

Formula used:

\(\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{sin\ x}{x}}\right)=1\)

Calculation:

\(\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{\sqrt{1-cosx^2}}{(1-cosx)}}\right)\)

Since, 1 - cos 2θ = sin²θ

⇒ \(\rm \displaystyle\lim_ {x\rightarrow0}\left({\frac{√{2sin^2\frac{x^2}{2}}}{(2sin^2\frac{x}{2})}}\right)\)

⇒ \(\frac{1}{√ 2}\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{{\frac{x^2}{2}\times sin\frac{x^2}{2}}}{(sin^2\frac{x}{2})\times\frac{x^2}{2}}}\right)\)

∴  \(\frac{2}{√ 2}\rm \displaystyle\lim_ {x\rightarrow 0}\left(\frac{sin\frac{x^2}{2}}{\frac{x^2}{2}}\right)\times \left(\frac{\frac{x}{2}}{sin\frac{x}{2}}\right)^2\) = √2

Concept:

\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} =1\)

Trigonometry formulas:

1 - cos 2x = 2sin² x

1 + cos 2x = 2cos² x

Calculation:

Here, we have to find the value of the limit \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)

As we know, 1 - cos 2x = 2sin2 x

\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{1 - \cos 2x }{x^2} \)

= \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{2\sin^2 x }{x^2} \)

= 2 × \(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x }{x} × \lim_{x \rightarrow 0} \frac{\sin x }{x}\)

= 2 × 1 × 1

= 2

Concept:

sin(a) - sin(b) = 2 cos(\(\frac{a+b}{2}\)) sin(\(\frac{a-b}{2}\))......(1)

\(\displaystyle \lim_{ x\to 0}\frac{sinx}{x}=1\).........(2)

Explanation:

We are given \(\displaystyle \lim_{x \to 0}\) \(\frac{sin(2+x)-sin(2-x)}{x}\) 

\(⇒\displaystyle \lim_{ x\to 0} ~\frac{2cos(\frac{(2+x)+(2-x)}{2})sin(\frac{(2+x)-(2-x)}{2})}{x}\) (Using equation 1)

\(⇒\displaystyle \lim_{ x\to 0}~2cos(2)~\frac{sin(x)}{x}\)

\(⇒2 cos (2)\) \(\displaystyle \lim_{ x\to 0}\frac{sinx}{x}\) 

⇒ 2 cos (2) × 1 (Using equation 2) 

⇒ 2 cos 2

Calculation:

We have,

\(\mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\tan x\, -\, \sin x}{\sin^{3}x}\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\frac{\sin x}{\cos x}\, -\, \sin x}{\sin^{3}x}\)      (∵ tan x = sin x/cos x)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{\sin x(1-\cos x)}{\cos x\, \sin^{3}x}\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, \sin^{2}x}\)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, (1-\cos^{2}x)}\)      (∵ sin² x = 1 - cos² x)

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{(1-\cos x)}{\cos x\, (1+\cos x)(1-\cos x)}\)      

\(\Rightarrow \mathop {\lim }\limits_{{\rm{x\;}} \to 0} \frac{1}{\cos x\, (1+\cos x)}\)

\(\therefore \frac{1}{2}\) is the correct limit.

Concept:

\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\sin x}}{x}} = 1\)

Calculation:

Here, we have to find the value of the limit \(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\sin 4x}{\tan 2x}\)

\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\sin 4x}{\tan 2x}\)

= \(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\frac{\sin 4x}{4x} \times 4x}{\frac{\tan 2x}{2x} \times 2x}\)

= \(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{1 \times 4x}{1 \times 2x}\)

= \(\frac 4 2\)

= 2

Concept:

• \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)
• \(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = 1\)

Calculation:

Given: \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x{}^\circ }{\tan 3x{}^\circ }\)

We know that 180° = π Radian

∴ 1° = π /360 ⇒ x° = πx/180 

\(\\\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin {\rm{x}}^\circ }}{{\tan 3{\rm{x}}^\circ }}= \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\tan \frac{{3{\rm{\pi x}}}}{{180}}}}\ \)

= \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\frac{{\sin \frac{{{\rm{\pi x}}}}{{180}}}}{{\frac{{{\rm{\pi x}}}}{{180}}}}\; \times \frac{{{\rm{\pi x}}}}{{180}}}}{{\frac{{\tan \frac{{3{\rm{\pi x}}}}{{180}}}}{{\frac{{3{\rm{\pi x}}}}{{180}}}}\; \times \frac{{3{\rm{\pi x}}}}{{180}}}} = \;\)\(\mathop {\lim }\limits_{x \to 0} \frac{{\frac{{{\rm{\pi x}}}}{{180}}}}{{\frac{{3{\rm{\pi x}}}}{{180}}}} = \frac{1}{3}\)

∴ Option 2 is correct.

Concept:

Formula: \(\rm \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\)

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\) 

Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

Calculation:

We have to find the value of \(\rm \underset{x\to 0}{\mathop{\lim }}\,\frac{2\left( 1-\cos x \right)}{{{x}^{2}}}\)

\(\rm \underset{x\to 0}{\mathop{\lim }}\,\frac{2\left( 1-\cos x \right)}{{{x}^{2}}}\)                      Form of limit is (0/0)

Applying L-Hospital rule, we get

\(= \rm \lim_{x\rightarrow 0}\frac{2(0+\sin x)}{2x} \)

\(= \rm \lim_{x\rightarrow 0}\frac{\sin x}{x}\)

= 1

Concept:

Formula: \(\rm \lim_{x\rightarrow 0}\frac{\sin x}{x} = 1\)

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule:

\(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

Calculation:

We have to find the value of \(\rm \underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1-\cos x \right)}{\sqrt {x+1}-1}\)

\(\rm \underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1-\cos x \right)}{\sqrt {x+1}-1}\)                      Form of limit is (0/0)

Applying L-Hospital rule, we get

\(= \rm \lim_{x\rightarrow 0}\frac{(0+\sin x)}{\frac {1}{2\sqrt{x+1}}} \)

= 0

Concept:

L’ Hospital’s rule:

If \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{0}{0}\) then we have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = l \ne \frac{0}{0}\) where l is a finite value.

Calculation:

Given: \(\mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}} = \frac{0}{0}\)

By applying L hospital’s rule we get

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{1 - {{\cos }^3}4x}}{{{x^2}}} = \;\mathop {\lim }\limits_{x\; \to 0} \frac{{12{{\cos }^2}4x\sin 4x}}{{2x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x\; \to 0} \frac{{12{{\cos }^2}4x\sin 4x}}{{2x}} = \;\mathop {\lim }\limits_{x \to 0} \frac{{24\;\left( {{{\cos }^2}4x\sin 4x} \right)}}{{4x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{24\;\left( {{{\cos }^2}4x\sin 4x} \right)}}{{4x}} = 24 \times \;\mathop {\lim }\limits_{x \to 0} {\cos ^2}4x \times \;\mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x}}{{4x}}\)

Calculation:

We have to find the value of \(\mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}}\)

\( \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}\sin x\; - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}3\sin x\; + {\rm{\;}}1}} = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{2{{\sin }^2}x\; + {\rm{\;}}2\sin x - \sin x - {\rm{\;}}1}}{{2{{\sin }^2}x\; - {\rm{\;}}2\sin x - \sin x + {\rm{\;}}1}}\)

\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{2\sin x(\sin x + 1) - 1\;(\sin x + 1)}}{{2\sin x(\sin x - 1) - 1\;(\sin x - 1)}}\)

\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{\left( {2\sin x - 1} \right)\;(\sin x + 1)}}{{\left( {2\sin x - 1} \right)\;(\sin x - 1)}}\)

\( = \;\mathop {{\rm{lim}}}\limits_{x \to \frac{\pi }{6}} \;\frac{{(\sin x + 1)}}{{(\sin x - 1)}} = \;\frac{{\left( {\frac{1}{2} + 1} \right)}}{{\left( {\frac{1}{2} - 1} \right)}} = \;\frac{{\left( {\frac{3}{2}} \right)}}{{\left( {\frac{{ - 1}}{2}} \right)}} = \; - 3\)