Category: Class 12 Mathematics Objective Questions

250+ TOP MCQs on Properties of Inverse Trigonometric Functions | Class 12 Maths

Mathematics Multiple Choice Questions on “Properties of Inverse Trigonometric Functions”.

1. sin-1⁡x in terms of cos-1⁡is ____________
a) cos-1⁡(sqrt{1+x^2})
b) cos-1⁡(sqrt{1-x^2})
c) cos-1⁡x
d) cos-1(frac{⁡1}{x})
Answer: b
Clarification: Let sin-1⁡x=y
⇒x=sin⁡y
⇒(x=sqrt{1-cos^2⁡y})
⇒(x^2=1-cos^2⁡y)
⇒(cos^2⁡y=1-x^2)
∴y=cos-1⁡(sqrt{1-x^2})=sin-1⁡x.

2. What is sec-1⁡x in terms of tan-1⁡?
a) tan-1⁡(sqrt{1+x^2})
b) tan-1⁡1+x2
c) tan-1⁡x
d) tan-1⁡(sqrt{x^2-1})
Answer: d
Clarification: Let sec-1⁡x=y
⇒x=sec⁡y
⇒x=(sqrt{1+tan^2⁡y})
⇒x2-1=tan2⁡y
∴y=tan-1⁡(sqrt{x^2-1})=sec-1⁡x.

3. What is the value of cos⁡(tan-1⁡((frac{4}{5})))?
a) (frac{5}{4})
b) (frac{5}{sqrt{41}})
c) (frac{sqrt{41}}{5})
d) (frac{4}{5})
Answer: b
Clarification: From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q3
tan-1⁡((frac{4}{5}))=cos-1⁡((frac{5}{sqrt{41}}))
cos⁡(tan-1⁡((frac{4}{5}))=cos⁡(cos-1⁡((frac{5}{sqrt{41}})))
=(frac{5}{sqrt{41}})

4. What is the solution of cot⁡(sin-1⁡x)?
a) (frac{sqrt{1-x^2}}{x})
b) x
c) (sqrt{1-x^2})
d) (sqrt{1+x^2})
Answer: a
Clarification: Let sin-1⁡x=y. From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q4
y=sin-1⁡x=cot-1⁡((frac{sqrt{1-x^2}}{x}))
∴cot⁡(sin-1⁡x)=(cot⁡(cot^{-1}(frac{sqrt{1-x^2}}{x}))=frac{sqrt{1-x^2}}{x}).

5. Which of the following formula is incorrect?
a) sin-1⁡x+sin-1⁡y=sin-1⁡{(xsqrt{1-y^2}+ysqrt{1-x^2})}
b) sin-1⁡x-sin-1⁡y=sin-1⁡{(xsqrt{1+y^2}+ysqrt{1+x^2})}
c) 2 tan-1⁡x=tan-1⁡((frac{2x}{1-x^2}))
d) 2 cos-1⁡x=cos-1⁡(3x-4x3)
Answer: b
Clarification: The formula sin-1⁡x-sin-1⁡y=sin-1⁡{(xsqrt{1+y^2}+ysqrt{1+x^2})} is incorrect. The correct formula is sin-1⁡x-sin-1⁡y=sin-1⁡{(xsqrt{1+y^2}-y sqrt{1-x^2})}.

6. Find the value of sin-1⁡((frac{5}{13}))+cos-1⁡((frac{3}{5})).
a) sin-1⁡((frac{63}{65}))
b) sin-1⁡1
c) 0
d) sin-1⁡((frac{64}{65}))
Answer: a
Clarification: From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q6
cos-1⁡((frac{3}{5}))=sin-1⁡((frac{4}{5}))
∴sin-1⁡((frac{5}{13}))+cos-1⁡((frac{3}{5}))=sin-1⁡((frac{5}{13}))+sin-1⁡((frac{4}{5}))
=sin-1⁡((frac{5}{13}sqrt{1-(frac{4}{5})^2}+frac{4}{5}sqrt{1-(frac{5}{13})^2}))
=(sin^{-1}(frac{5}{13}×frac{3}{5}+frac{4}{5}×frac{12}{13})=sin^{-1}(frac{15+48}{65})=sin^{-1}(frac{63}{65})).

7. Find the value of tan-1⁡((frac{1}{3}))+tan-1⁡((frac{1}{5}))+tan-1⁡(frac{1}{7})[/latex]
a) tan-1⁡((frac{4}{7}))
b) tan-1⁡((frac{9}{7}))
c) tan-1⁡((frac{7}{9}))
d) tan-1⁡1
Answer: c
Clarification: Using the formula tan-1⁡x+tan-1⁡y=tan-1⁡(frac{x+y}{1-xy}), we get
tan-1⁡((frac{1}{3}))+tan-1⁡((frac{1}{5}))=tan-1⁡(bigg(frac{frac{1}{3}+frac{1}{5}}{1-frac{1}{3}×frac{1}{5}}bigg))
= (tan^{-1}bigg(frac{frac{8}{15}}{frac{14}{15}}bigg)=tan^{-1}⁡(frac{8}{15}×frac{15}{14})=tan^{-1}⁡(frac{4}{7}))
=(tan^{-1}(frac{1}{3})+tan^{-1}(frac{1}{5})+tan^{-1}⁡(frac{1}{7})=tan^{-1}(frac{4}{7})+tan^{-1}⁡(frac{1}{7}))
=(tan^{-1}⁡bigg(frac{frac{4}{7} + frac{1}{7}}{1-frac{4}{7}×frac{1}{7}}bigg) = tan^{-1}⁡bigg(frac{frac{5}{7}}{frac{45}{49}}bigg)=tan^{-1}⁡(frac{5}{7}×frac{49}{45}))
=tan-1⁡((frac{7}{9})).

8. Find the value of sin-1⁡((frac{3}{5}))+sin-1⁡((frac{4}{5}))+cos-1⁡((frac{sqrt{3}}{2})).
a) (frac{π}{3})
b) (frac{2π}{3})
c) (frac{4π}{3})
d) (frac{π}{4})
Answer: b
Clarification: Using the formula sin-1⁡x+sin-1⁡y=sin-1⁡({x sqrt{1-y^2}+y sqrt{1-x^2}}), we get
sin-1⁡((frac{3}{5}))+sin-1⁡((frac{4}{5}))=sin-1(Big{ frac{3}{5} sqrt{1-(frac{4}{5})^2}+frac{4}{5} sqrt{1-(frac{3}{5})^2} Big})
=sin-1⁡((frac{3}{5})×(frac{3}{5})+(frac{4}{5})×(frac{4}{5}))=sin-1⁡((frac{25}{25})=frac{π}{2})
∴ sin-1⁡((frac{3}{5}))+sin-1⁡((frac{4}{5}))+cos-1⁡((frac{sqrt{3}}{2})=frac{π}{2}+frac{π}{6}=frac{3π+π}{6}=frac{2π}{3}).

9. What is the value of 2 tan-1⁡x in terms of sin-1⁡?
a) sec-1⁡x
b) 2 sec-1⁡x
c) 2 sec-1⁡((sqrt{1+x^2}))
d) sec-1⁡((sqrt{1+x^2}))
Answer: c
Clarification: Let 2 tan-1⁡x=y
⇒tan-1⁡x=(frac{y}{2})
From ∆ABC, we get
mathematics-questions-answers-properties-inverse-trigonometric-functions-q9
⇒tan-1⁡x=sec-1⁡(sqrt{1+x^2}=frac{y}{2})
⇒y=2 sec-1⁡((sqrt{1+x^2}))

10. sin-1⁡x+cos1⁡x= ___
a) (frac{π}{2})
b) π
c) (frac{π}{3})
d) 2π
Answer: a
Clarification: sin-1⁡x+cos-1⁡x=(frac{π}{2}); x∈[-1,1]

250+ TOP MCQs on Determinants – Adjoint and Inverse of a Matrix | Class 12 Maths

Mathematics Problems on “Determinants – Adjoint and Inverse of a Matrix”.

1. Which of the following is the adjoint of the matrix A=(begin{bmatrix}1&5\3&4end{bmatrix})?
a) (begin{bmatrix}4&-5\-3&-1end{bmatrix})
b) (begin{bmatrix}-4&5\-3&1end{bmatrix})
c) (begin{bmatrix}4&-5\-3&1end{bmatrix})
d) (begin{bmatrix}4&5\-3&1end{bmatrix})
Answer: c
Clarification: We have A11=(-1)1+1 4=4
A12=(-1)1+2 3=-3
A21=(1)2+1 5=-5
A22=(-1)2+2 1=1
∴adj A=(begin{bmatrix}A_{11}&A_{21}\A_{12}&A_{22}end{bmatrix})=(begin{bmatrix}4&-5\-3&1end{bmatrix}).

2. If A=(begin{bmatrix}5&-8\2&6end{bmatrix}), find A(adj A).
a) (begin{bmatrix}41&0\0&46end{bmatrix})
b) (begin{bmatrix}46&0\1&46end{bmatrix})
c) (begin{bmatrix}46&1\0&46end{bmatrix})
d) (begin{bmatrix}46&0\0&46end{bmatrix})
Answer: d
Clarification: Given that, A=(begin{bmatrix}5&-8\2&6end{bmatrix})
∴adj A=(begin{bmatrix}6&8\-2&5end{bmatrix})
A(adj A)=(begin{bmatrix}5&-8\2&6end{bmatrix}begin{bmatrix}6&8\-2&5end{bmatrix})
=(begin{bmatrix}5×6+(-8)×(-2)&5×8+5×(-8)\2×6+6×(-2)&2×8+6×5end{bmatrix})=(begin{bmatrix}46&0\0&46end{bmatrix}).

3. If A=(begin{bmatrix}1&0\9&4end{bmatrix}), then (adj A)A is ______________
a) (begin{bmatrix}-4&0\0&-4end{bmatrix})
b) (begin{bmatrix}4&0\1&4end{bmatrix})
c) (begin{bmatrix}4&0\0&4end{bmatrix})
d) (begin{bmatrix}4&0\0&-4end{bmatrix})
Answer: c
Clarification: Given that, A=(begin{bmatrix}1&0\9&4end{bmatrix})
We know that, A(adj A)=(adj A)A=|A|I
∴|A|=4-0=4
⇒A(adj A)=|A|I=(begin{bmatrix}4&0\0&4end{bmatrix}).

4. Which of the following is the formula for calculating the inverse of the matrix?
a) (frac{2}{|A|}) adj A
b) (frac{1}{|A|}) adj A
c) (frac{-1}{|A|}) adj A
d) (frac{1}{|2A|}) adj A
Answer: b
Clarification: The formula for calculating the inverse of the matrix is given by
A-1=(frac{1}{|A|}) adj A, where |A| is the determinant of the matrix and adj A is the adjoint of the matrix.

5. Find the inverse of the matrix A=(begin{bmatrix}8&5\4&1end{bmatrix}).
a) (begin{bmatrix}-frac{1}{12}&frac{5}{12}\frac{1}{3}&-frac{2}{3}end{bmatrix})
b) (begin{bmatrix}frac{1}{12}&frac{5}{12}\frac{1}{3}&-frac{2}{3}end{bmatrix})
c) (begin{bmatrix}-frac{1}{12}&frac{5}{12}\frac{1}{3}&frac{2}{3}end{bmatrix})
d) (begin{bmatrix}-frac{1}{12}&frac{5}{12}\-frac{1}{3}&-frac{2}{3}end{bmatrix})
Answer: a
Clarification: Give that, A=(begin{bmatrix}8&5\4&1end{bmatrix})
adj A=(begin{bmatrix}1&-5\-4&8end{bmatrix})
|A|=8×1-(-5)×(-4)=8-20=-12
A-1=(frac{1}{|A|}) adj A=(frac{1}{-12} begin{bmatrix}1&-5\-4&8end{bmatrix})=(begin{bmatrix}-frac{1}{12}&frac{5}{12}\frac{1}{3}&-frac{2}{3}end{bmatrix}).

6. Which of the below condition is incorrect for the inverse of a matrix A?
a) The matrix A must be a square matrix
b) A must be singular matrix
c) A must be a non-singular matrix
d) adj A≠0
Answer: b
Clarification: The matrix should not be a singular matrix. A square matrix is said to be singular |A|=0.
We know that, A-1=(frac{1}{|A|}) adj A,
Hence, if |A|=0 the inverse of the matrix does not exist.

7. Which of the below given matrices has the inverse (frac{1}{-6}begin{bmatrix}2&1\0&-3end{bmatrix})?
a) (begin{bmatrix}3&-1\0&2end{bmatrix})
b) (begin{bmatrix}-3&-1\0&2end{bmatrix})
c) (begin{bmatrix}-2&0\1&3end{bmatrix})
d) (begin{bmatrix}-3&-1\0&-2end{bmatrix})
Answer: b
Clarification: Consider the matrix (begin{bmatrix}-3&-1\0&2end{bmatrix})
adj A=(begin{bmatrix}2&1\0&-3end{bmatrix})
|A|=-6
∴A-1=(frac{1}{|A|}) adj A=(frac{1}{-6}begin{bmatrix}2&1\0&-3end{bmatrix}).

8. If A=(begin{bmatrix}-8&2\6&-3end{bmatrix}) and B=(begin{bmatrix}2&1\1&7end{bmatrix}). Find (AB)-1.
a) –(frac{1}{432}) (begin{bmatrix}-27&6\9&14end{bmatrix})
b) (frac{1}{432}) (begin{bmatrix}27&6\9&14end{bmatrix})
c) (frac{1}{432}) (begin{bmatrix}-27&6\9&14end{bmatrix})
d) (frac{-1}{432}) (begin{bmatrix}27&6\9&14end{bmatrix})
Answer: c
Clarification: Given that, A=(begin{bmatrix}-8&2\6&-3end{bmatrix}) and B=(begin{bmatrix}2&1\1&7end{bmatrix})
∴AB=(begin{bmatrix}-8×2+2×1&-8×1+2×7\6×2+(-3)×1&6×1+(-3)×7end{bmatrix})=(begin{bmatrix}-14&6\9&27end{bmatrix})
adj(AB)=(begin{bmatrix}27&-6\-9&-14end{bmatrix})
|AB|=27×(-14)-(-9)×(-6)=-378-54=-432
(AB)-1=(frac{1}{|AB|}) adj AB=(frac{1}{-432} begin{bmatrix}27&-6\-9&-14end{bmatrix})=(frac{1}{432} begin{bmatrix}-27&6\9&14end{bmatrix}).

9. Which of the following formula is incorrect?
a) A(adj A)=|A|I
b) |adj (A)|=|A|n-1, for an nth order matrix
c) A-1=(frac{1}{|A|}) adj A
d) A(adj A)=|A|n-1
Answer: d
Clarification: The formula A(adj A)=|A|n-1 is incorrect. The correct formula is A(adj A)=(adjA)A=|A|I.

10. A square matrix A is said to be non-singular if |A|≠0.
a) True
b) False
Answer: a
Clarification: The given statement is true. A square matrix A is said to be singular if |A|=0 and non-singular if A≠0.

Mathematics Problems,

250+ TOP MCQs on Integration as an Inverse Process of Differentiation | Class 12 Maths

Mathematics Quiz for Engineering Entrance Exams on “Integration as an Inverse Process of Differentiation”.

1. Find the integral of (8x^3+1).
a) 2x4+x+C
b) 2x6-5x+C
c) 2x4-x+C
d) 2x4+x2 C
Answer: a
Clarification: (int ,8x^{3+1} ,dx)
Using (int ,x^n ,dx=frac{x^{n+1}}{n+1}), we get
(int ,8x^{3+1} ,dx=int 8x^3 ,dx+int ,1 ,dx)
=(frac{8x^{3+1}}{3+1}+x)
=(frac{8x^4}{4}+x)
=2x4+x+C.

2. Find ∫ 7x2-x3+2x dx.
a) (frac{7x^3}{3}+frac{x^4}{5}-frac{2x^2}{2}+C)
b) (frac{7x^3}{3}+frac{x^4}{4}+frac{2x^2}{2}+C)
c) (frac{7x^5}{9}-frac{x^4}{4}+frac{2x^2}{2}+C)
d) (frac{7x^3}{3}-frac{x^4}{4}+x^2+C)
Answer: d
Clarification: To find (int 7x^2-x^3+2x dx)
(int 7x^2-x^3+2x dx=int 7x^2 dx-int x^3 dx+2int x dx)
Using (int x^n dx=frac{x^{n+1}}{n+1}), we get
(int 7x^2-x^3+2x dx=frac{7x^{2+1}}{2+1}-frac{x^{3+1}}{3+1}+2(frac{x^{1+1}}{1+1}))
∴(int 7x^2-x^3+2x dx=frac{7x^3}{3}-frac{x^4}{4}+x^2+C)

3. Find the integral of 2 sin⁡2x+3.
a) sin⁡2x+3x+C
b) -cos⁡2x-3x3+C
c) -cos⁡2x+3x+C
d) cos⁡2x-3x+12+C
Answer: c
Clarification: To find ∫ 2 sin⁡2x+3 dx
(int ,2 ,sin⁡2x+3 ,dx=int ,2 ,sin⁡2x ,dx + int ,3 ,dx)
(int ,2 ,sin⁡2x+3 ,dx=2int ,sin⁡2x ,dx+3int ,dx)
(int ,2 ,sin⁡2x+3 ,dx=frac{-2 cos⁡2x}{2}+3x)
∴∫2 sin⁡2x+3 dx=-cos⁡2x+3x+C

4. Find the integral of (int 3e^x+frac{2}{x}+x^3 dx).
a) (3e^3x+frac{2}{x}-frac{x^4}{4}+c)
b) (3e^x+2 ,log⁡x+frac{x^4}{4}+c)
c) (e^x+2 ,log⁡x+frac{x^4}{4}+c)
d) (3e^x-frac{2}{x^2}+frac{x^4}{4}+c)
Answer: b
Clarification: To find (int ,3e^x+frac{2}{x}+x^3 ,dx)
(int ,3e^x+frac{2}{x}+x^3 dx=3int ,e^x ,dx+2int frac{1}{x} ,dx+int x^3 ,dx)
(int ,e^x ,dx=e^x)
(int frac{1}{x} dx=log⁡x)
∴(int 3e^x+frac{2}{x}+x^3 ,dx=3e^x+2 ,log⁡x+frac{x^4}{4}+c)

5. Find the integral of (frac{4x^4-3x^2}{x^3}).
a) 7x2-3 log⁡x3+C
b) 2x2-3 log⁡x+C
c) x2-log⁡x+C
d) 2x2+3 log⁡x+C
Answer: b
Clarification: To find (int frac{4x^4-3x^2}{x^3} dx)
(int frac{4x^4-3x^2}{x^3} ,dx=int frac{4x^4}{x^3} – frac{3x^2}{x^3} ,dx)
(int frac{4x^4-3x^2}{x^3} ,dx=int 4x dx-int frac{3}{x} dx)
(int frac{4x^4-3x^2}{x^3} ,dx=frac{4x^2}{2}-3 log⁡x)
∴ (int frac{4x^4-3x^2}{x^3} ,dx=2x^2-3 ,log⁡x+C).

6. Find (int ,3 ,cos⁡x+frac{1}{x} dx).
a) (3 ,sin⁡x-frac{1}{x}+C)
b) (2 ,sin⁡x+frac{1}{x^3}+C)
c) (3 ,sin⁡3x+frac{1}{x}+C)
d) (sin⁡x-frac{1}{x^2}+C)
Answer: a
Clarification: To find (int ,3 ,cos⁡x+frac{1}{x^2} dx)
(int ,3 ,cos⁡x+frac{1}{x^2} dx=3 int cos⁡x ,dx+int frac{1}{x^2} ,dx)
(int ,3 ,cos⁡x+frac{1}{x^2} dx=3 ,sin⁡x+int x^{-2} ,dx)
(int ,3 ,cos⁡x+frac{1}{x^2} dx=3 ,sin⁡x+frac{x^{-2+1}}{-2+1})
(int ,3 ,cos⁡x+frac{1}{x^2} dx=3 ,sin⁡x-frac{1}{x}+C)

7. Find (int (2+x)xsqrt{x} dx).
a) (frac{4x^{5/2}}{5}+frac{2x^{7/2}}{9}+C)
b) (frac{4x^{5/2}}{5}-frac{2x^{7/2}}{7}+C)
c) (frac{4x^{5/2}}{6}+frac{2x^{7/2}}{7}+C)
d) –(frac{4x^{5/2}}{5}+frac{2x^{7/2}}{7}+C)
Answer: c
Clarification: To find (int (2+x)xsqrt{x} dx)
(int ,(2+x)xsqrt{x} ,dx=int ,2xsqrt{x}+x^{5/2} ,dx)
(int ,(2+x)xsqrt{x} ,dx=int ,2x^{3/2} dx + int x^{5/2} dx)
(int ,(2+x)xsqrt{x} ,dx=frac{2x^{3/2+1}}{3/2+1}+frac{x^{5/2+1}}{5/2+1})
(int ,(2+x)xsqrt{x} ,dx=frac{4x^{5/2}}{5}+frac{2x^{7/2}}{7}+C)

8. Find (int ,7x^8-4e^{2x}-frac{2}{x^2} ,dx).
a) (frac{7x^4}{4}-2e^{2x}+frac{2}{x}+C)
b) (frac{7x^4}{4}+2e^{2x}+frac{2}{x}+C)
c) (frac{7x^4}{4}-2e^{2x} frac{2}{x^2}+C)
d) (frac{7x^4}{8}+2e^{2x}-frac{4}{x}+C)
Answer: a
Clarification: To find:(int 7x^8-4e^{2x}-frac{2}{x^2} dx)
(int ,7x^8-4e^{2x}-frac{2}{x^2} ,dx=int 7x^9 dx-4int e^{2x} dx-2int frac{1}{x}^2 dx)
(int ,7x^8-4e^{2x}-frac{2}{x^2} ,dx=frac{7x^{9+1}}{9+1}-frac{4e^{2x}}{2}-frac{2x^{-2+1}}{-2+1})
∴(int ,7x^8-4e^{2x}-frac{2}{x^2} dx=frac{7x^{10}}{10}-2e^{2x}+frac{2}{x}+C)

9. Find the integral (int sin⁡2x+e^3x-cos⁡3x dx).
a) –(frac{sin⁡2x}{2}+frac{e^{3x}}{3}-frac{sin⁡3x}{3}+C)
b) –(frac{cos⁡2x}{2}+frac{e^{3x}}{3}-frac{sin⁡3x}{3}+C)
c) (frac{cos⁡2x}{2}+frac{e^{3x}}{3}-frac{cos⁡3x}{3}+C)
d) –(frac{cos⁡2x}{2}-frac{e^{3x}}{3}+frac{cos⁡3x}{3}+C)
Answer: b
Clarification: To find (int ,sin⁡2x+e^{3x}-cos⁡3x ,dx)
(int sin⁡2x+e^{3x}-cos⁡3x ,dx=int ,sin⁡2x ,dx+int ,e^{3x} ,dx-int ,cos⁡3x ,dx)
(int sin⁡2x+e^{3x}-cos⁡3x ,dx=-frac{cos⁡2x}{2}+frac{e^{3x}}{3}-frac{sin⁡3x}{3}+C)

10. Find the integral of (ax2+b)2.
a) (frac{a^2 ,x^5}{5}+b^2 ,x+frac{2abx^3}{3}+C)
b) –(frac{a^2 ,x^5}{5}-b^2 ,x+frac{2abx^3}{3}+C)
c) (frac{b^2 ,x^5}{5}+b^2 x+frac{27x^3}{3}+C)
d) (frac{a^2 ,x^5}{5}+x+frac{2abx^3}{5}+C)
Answer: a
Clarification: To find (ax2+b)2
(int (ax^2+b)^2 dx=int (a^2 ,x^4+b^2+2ax^2 ,b) dx)
(int (ax^2+b)^2 dx=int ,a^2 ,x^4 ,dx+int ,b^2 ,dx+2int ,ax^2 ,b ,dx)
(int (ax^2+b)^2 dx=a^2 ,int ,x^4 ,dx+b^2 int ,dx+2abint ,x^2 ,dx)
(int (ax^2+b)^2 dx=a^2 (frac{x^5}{5})+b^2 x+2ab(frac{x^3}{3}))
(int (ax^2+b)^2 dx=frac{a^2 ,x^5}{5}+b^2 x+frac{2abx^3}{3}+C)

Mathematics Quiz for Engineering Entrance Exams,

250+ TOP MCQs on Linear First Order Differential Equations | Class 12 Maths

Mathematics Multiple Choice Questions for Engineering Entrance Exams on “Linear First Order Differential Equations – 2”.

1. A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?
a) dy/dx = [(xy + 2) ± √(1 + xy)]/ x2
b) dy/dx = [(xy – 2) ± √(1 + xy)]/ x2
c) dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
d) dy/dx = [(xy + 2) ± √(1 – xy)]/ x2
Answer: c
Clarification: The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2

2. A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?
a) xy = 2
b) xy = -1
c) x – y = 2
d) x + y = 2
Answer: d
Clarification: The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
Let, 1 – xy = t2
=> x(dy/dx) + y = -2t(dt/dx)
=> x2(dy/dx) = t2 – 1 – 2tx(dt/dx), so that (3) gives
t(x(dt/dx) – (t ± 1)) = 0
Hence, either t = 0
=>xy = 1 which is satisfied by (1, 1)
Or, x dt/dx = t ± 1
=> dx/x = dt/t ± 1
=> t ± 1 = cx
For x = 1, y = 1 and t = 0
=> c = ± 1, so the solution is
t = ± (x – 1) => t2 = (x – 1)2
Or, 1 – xy = x2 – 2x + 1
Or, x + y = 2
Thus, the two curves that satisfies are xy = 1 and x + y = 2

3. What will be the value of dy/dx – a/x * y = (x + 1)/x?
a) y = x/(1 – a) – 1/a + cxa
b) y = x/(1 + a) + 1/a + cxa
c) y = x/(1 – a) – 1/a – cxa
d) y = x/(1 + a) – 1/a + cxa
Answer: a
Clarification: dy/dx – a/x * y = (x + 1)/x …….(1)
Multiplying both sides of equation (1) by
e∫-a/xdx
= e-a log x
= elog x-a
= x-a
We get, x-ady/dx – x-a (a/x)y = x-a (x + 1)/x
Or, d/dx(y . x-a) = x-a + x-a – 1 …….(2)
Integrating both sides of (2) we get,
y. x-a = x-a + 1/(-a + 1) + x-a – 1 + 1/(-a -1 + 1) + c
= x-a.x/(1 – a) + x-a/-a + c
Or, y = x/(1 – a) – 1/a + cxa

4. What will be the differential equation form of √(a2 + x2)dy/dx + y = √(a2 + x2) – x?
a) a2 log (x + √(a2 – x2)) + c
b) a2 log (x + √( a2 + x2)) + c
c) a2 log (x – √( a2 + x2)) + c
d) a2 log (x – √( a2 – x2)) + c
Answer: b
Clarification: The given form of equation can be written as,
dy/dx + 1/√(a2 + x2) * y = (√(a2 + x2) – x)/√(a2 + x2) ……(1)
We have, ∫1/√(a2 + x2)dx = log(x + √(a2 + x2))
Therefore, integrating factor is,
e∫1/√(a2 + x2) = elog(x + √(a2 + x2))
= x + √(a2 + x2)
Therefore, multiplying both sides of (1) by x + √(a2 + x2) we get,
x + √(a2 + x2dy/dx + (x + √(a2 + x2))/ √(a2 + x2)*y = (x + √(a2 + x2))(√(a2 + x2) – x)/√(a2 + x2)
or, d/dx[x + √(a2 + x2)*y] = (a2 + x2) ………..(2)
Integrating both sides of (2) we get,
(x + √(a2 + x2) * y = a2∫dx/√(a2 + x2)
= a2 log (x + √(a2 + x2)) + c

5. What is the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?
a) 3x – y + log|2x + 3y – 3| = -c/3
b) 3x – y + log|2x + 3y – 3| = c/3
c) 3x + y + log|2x + 3y – 3| = -c/3
d) 3x – y – log|2x + 3y – 3| = c/3
Answer: a
Clarification: dy/dx = (6x + 9y – 7)/(2x + 3y – 6)
So, dy/dx = (3(2x + 3y) – 7)/(2x + 3x – 6) ……….(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx – 2)
Therefore, from (1) we get,
1/3(dz/dx – 2) = (3z – 7)/(z – 6)
Or, dz/dx = 2 + (3(3z – 7))/(z – 6)
= 11(z – 3)/(z – 6)
Or, (z – 6)/(z – 3) dz = 11 dx
Or, ∫(z – 6)/(z – 3) dz = ∫11 dx
Or, ∫(1 – 3/(z – 3)) dz = 11x + c
Or, z – log |z – 3| = 11x + c
Or, 2x + 3y – 11x – 3log|2x + 3y -3| = c
Or, 3y – 9x – 3log|2x + 3y – 3| = c
Or, 3x – y + log|2x + 3y – 3| = -c/3

6. A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the velocity of the particle?
a) 27cm/sec
b) 28 cm/sec
c) 29 cm/sec
d) 30 cm/sec
Answer: c
Clarification: Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t2 – 5t
Or, dv = 3t2 dt – 5tdt
Or, ∫dv = 3∫t2 dt – 5∫t dt
Or, v = t3 – (5/2)t2 + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t3 – (5/2)t2 + 5
Or, dx/dt = t3 – (5/2)t2 + 5 ………..(2)
Thus, the velocity of the particle at the end of 4 seconds,
= [v]t = 4 = (43 – (5/2)42 + 5 ) cm/sec [putting t = 4 in (2)]
= 29 cm/sec

7. A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the distance from the origin at the end of 4 seconds?
a) 30(4/3)
b) 30(2/3)
c) 30
d) Unpredictable
Answer: b
Clarification: Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t2 – 5t
Or, dv = 3t2 dt – 5t dt
Or, ∫dv = 3∫t2 dt – 5∫t dt
Or, v = t3 – (5/2)t2 + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t3 – (5/2)t2 + 5
Or, dx/dt = t3 – (5/2)t2 + 5 ………..(2)
Or, dx = t3 dt – (5/2)t2 dt + 5 dt
Integrating this we get,
x = (1/4)t4 – (5/2)t3/3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t4 – (5/6)t3 + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]t = 4 = (1/4)44 – (5/6)43 + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm

8. What is the solution of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?
a) cx±1/√2 = y/x + √(y2 – 2x2)/x2
b) cx±√2 = y/x + √(y2 + 2x2)/x2
c) cx±1/2√2 = y/x + √(y2 – 2x2)/x2
d) cx±1/√2 = y/x + √(y2 + 2x2)/x2
Answer: d
Clarification: Here, y2(dy/dx)2 + 4x2 + 4xy(dy/dx) = (y2 + 2x2)[1 + (dy/dx)2]
=>dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
=> v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±√dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2

9. What is the equation of the curve passing through (1, 0) of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?
a) √2x±1/√2 = y/x + √(y2 + 2x2)/x2
b) √2x±1/2√2 = y/x + √(y2 + 2x2)/x2
c) √2x√2 = y/x + √(y2 + 2x2)/x2
d) √2x = y/x + √(y2 + 2x2)/x2
Answer: a
Clarification: Here, y2(dy/dx)2 + 4x2 + 4xy(dy/dx) = (y2 + 2x2)[1 + (dy/dx)2]
=> dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
=> v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±∫dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2 (put v/√2 = tan t)
putting x = 1, y = 0, we get c = √2
So, the curve is given by,
√2x±1/√2 = y/x + √(y2 + 2x2)/x2

10. If, A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?
a) Parabola
b) Hyperbola
c) Ellipse
d) Circle
Answer: d
Clarification: Equation of the normal at a point P(x, y) is given by
Y – y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=>y(dy/dx) = ± √(k2 – y2) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k2 – y2) = ∫-dx
Or, x2 + y2 = k2 passes through (0, k)
Thus, it is a circle.

250+ TOP MCQs on Conditional Probability | Class 12 Maths

Mathematics Multiple Choice Questions on “Conditional Probability”.

1. If E and F are two events associated with the same sample space of a random experiment then P (E|F) is given by _________
a) P(E∩F) / P(F), provided P(F) ≠ 0
b) P(E∩F) / P(F), provided P(F) = 0
c) P(E∩F) / P(F)
d) P(E∩F) / P(E)

Answer: a
Clarification: E and F are two events associated with the same sample space of a random experiment.
The value of P (E|F) = (E∩F) / P(F), provided P(F) ≠ 0. We know that if P(F) = 0, then the value of P(E|F) will reach a value which is not defined hence it is wrong option. Also, P(E∩F) / P(F) and P(E∩F) / P(E) are wrong and do not equate to P(E|F).

2. Let E and F be events of a sample space S of an experiment, if P(S|F) = P(F|F) then value of P(S|F) is __________
a) 0
b) -1
c) 1
d) 2

Answer: c
Clarification: We know that P(S|F) = P(S∩F) / P(F). (By formula for conditional probability)
Which is equivalent to P(F) / P(F) = 1, hence the value of P(S|F) = 1.

3. Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E∩F) = 0.2, then P(E|F) ?
a) 2/3
b) 1/3
c) 3/4
d) 1/4

Answer: a
Clarification: We know that P(E|F) = P(E∩F) / P(F). (By formula for conditional probability)
Value of P(E∩F) is given to be 0.2 and value of P(F) is given to be 0.3.
P(E|F) = (0.2) / (0.3).
P(E|F) = 2 / 3.

4. Given that E and F are events such that P(E) = 0.5, P(F) = 0.4 and P(E∩F) = 0.3, then what will be the value of P(F|E)?
a) 2/5
b) 3/5
c) 3/4
d) 2/4

Answer: b
Clarification: We know that P(F|E) = P(E∩F) / P(E). (By formula for conditional probability)
Value of P(E∩F) is given to be 0.3 and value of P(E) is given to be 0.5.
P(F|E) = (0.3) / (0.5).
P(F|E) = 3 / 5.

5. Let E and F be events of a sample space S of an experiment, if P(S|F) = P(F|F), then value of P(F|F) is __________
a) 0
b) -1
c) 1
d) 2

Answer: c
Clarification: We know that P(S|F) = P(S∩F) / P(F). (By formula for conditional probability)
Which is equivalent to P(F|F) = P(F) / P(F) = 1, hence the value of P(F|F) = 1.

6. If P(A) = 7/11, P(B) = 6 / 11 and P(A∪B) = 8/11, then P(A|B) = ________
a) 3/5
b) 2/3
c) 1/2
d) 1

Answer: d
Clarification: We know that P(A|B) = P(A∩B) / P(B). (By formula for conditional probability)
Also P(A∪B) = P(A)+P(B) – P(A∩B). (By formula of probability)
(Rightarrow) 8/11 = 7/11 + 6/11 – P(A∩B)
(Rightarrow) P(A∩B) = 13/11 – 7/11
(Rightarrow) P(A∩B) = 6/11
P(A|B) = (6/11) / (6/11).
P(A|B) = 1.

7. If P(A) = 1/5, P(B) = 0, then what will be the value of P(A|B)?
a) 0
b) 1
c) Not defined
d) 1/5

Answer: c
Clarification: We know that P(A|B) = P(A∩B) / P(B). (By formula for conditional probability)
The value of P(B) = 0 in the given question. As the value of denominator becomes 0, the value of P(A|B) becomes un-defined.

8. If P(A) = 5/13, P(B) = 7/13 and P(A∩B) = 3/13, evaluate P(A|B).
a) 1/7
b) 3/7
c) 3/5
d) 2/7

Answer: b
Clarification: We know that P(A|B) = P(A∩B) / P(B). (By formula for conditional probability)
Which is equivalent to (3/13) / (7/13), hence the value of P(A|B) = 3/7.

250+ TOP MCQs on Calculus Application – Velocity | Class 12 Maths

Mathematics Problems for Engineering Entrance Exams on “Calculus Application – Velocity – 2”.

1. A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c = 0?
a) Uniform retardation
b) Uniform speed
c) Uniform acceleration
d) Uniform velocity
Answer: d
Clarification: We have, x = a + bt + ct2  ……….(1)
Let, v and f be the velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct   ……….(2)
And f = dv/dt = d(b + ct)/dt = c  ……….(3)
Clearly, when c = 0, then f = 0 that is, acceleration of the particle is zero.
Hence in this case the particle moves with an uniform velocity.

2. A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t2 – 4t + 5 cm/second. What will be the distance travelled by the particle during first 3 seconds after the start?
a) 21 cm
b) 22 cm
c) 23 cm
d) 24 cm
Answer: d
Clarification: Let, x be the distance travelled by the particle in time t seconds.
Then, v = dx/dt = 3t2 – 4t + 5
Or ∫dx = ∫ (3t2 – 4t + 5)dt
So, on integrating the above equation, we get,
x = t3 – 2t2 + 5t + c where, c is a constant.  ……….(1)
Therefore, the distance travelled by the particle at the end of 3 seconds,
= [x]t = 3 – [x]t = 0
= (33 – 2*32 + 5*3 + c) – c [using (1)]
= 24 cm.

3. A particle moving in a straight line traverses a distance x in time t. If t = x2/2 + x, then which one is correct?
a) The retardation of the particle is the cube of its velocity
b) The acceleration of the particle is the cube of its velocity
c) The retardation of the particle is the square of its velocity
d) The acceleration of the particle is the square of its velocity
Answer: a
Clarification: We have, t = x2/2 + x
Therefore, dt/dx = 2x/2 + 1 = x + 1
Thus, if v be the velocity of the particle at time t, then
v = dx/dt = 1/(dt/dx)
= 1/(x + 1) = (x + 1)-1
Thus dv/dt = d((x + 1)-1)/dt
= (-1)(x + 1)-2 d(x + 1)/dt
= -1/(x + 1)2 * dx/dt
As, 1/(x + 1) = dx/dt,
So, -(dx/dt)2(dx/dt)
Or dv/dt = -v2*v  [as, dx/dt = v]
= -v3
We know, dv/dt = acceleration of a particle.
As, dv/dt is negative, so there is a retardation of the particle.
Thus, the retardation of the particle = -dv/dt = v3 = cube of the particle.

4. A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?
a) 30 cm
b) 31 cm
c) 32 cm
d) 33 cm
Answer: b
Clarification: We assume that the particle moves with uniform acceleration 2f m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b  ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft2 + bt + a   ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21  ……….(3)
16f + 4b + a = 43  ……….(4)
49f + 7b + a = 91   ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t2 + 5t + 7
Therefore, the distance described by the particle in 3 seconds,
= [x]t = 3 = (32 + 5*3 + 7)m = 31m

5. A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds, respectively.What is the velocity of the particle in 3 seconds?
a) 11m/sec
b) 31 cm/sec
c) 21m/sec
d) 41m/sec
Answer: a
Clarification: We assume that the particle moves with uniform acceleration 2f m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b   ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft2 + bt + a   ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21   ……….(3)
16f + 4b + a = 43   ……….(4)
49f + 7b + a = 91  ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t2 + 5t + 7
Putting t = 3, f = 1 and b = 5 in (1),
We get, the velocity of the particle in 3 seconds,
= [v]t = 3 = (2*1*3 + 5)m/sec = 11m/sec.

6. A particle moves in a horizontal straight line under retardation kv3, where v is the velocity at time t and k is a positive constant. If initial velocity be u and x be the displacement at time,then which one is correct?
a) 1/v2 = 1/2u2 + 2kt
b) 1/v2 = 1/2u2 – 2kt
c) 1/v2 = 1/u2 – 2kt
d) 1/v2 = 1/u2 + 2kt
Answer: d
Clarification: Since the particle is moving in a straight line under a retardation kv3, hence, we have,
dv/dt = -kv3   ……….(1)
Or dv/v3 = -k dt
Or ∫v-3 dv = -k∫dt
Or v-3+1/(-3 + 1) = -kt – c [c = constant of integration]
Or 1/2v2 = kt + c   ……….(2)
Given, u = v when, t = 0; hence, from (2) we get,
1/2u2 = c
Thus, putting c = 1/2u2 in (2) we get,
1/2v2 = kt + 1/2u2
Or 1/v2 = 1/u2 + 2kt

7. The distance x of a particle moving along a straight line from a fixed point on the line at time t after start is given by t = ax2 + bx + c (a, b, c are positive constants). If v be the velocity of the particle and u(≠0) be the initial velocity of the particle then which one is correct?
a) 1/v2 + 1/u2 = 4at
b) 1/v2 + 1/u2 = -4at
c) 1/v2 – 1/u2 = 4at
d) 1/v2 – 1/u2 = -4at
Answer: c
Clarification: We have, t = ax2 + bx + c  ……….(1)
Differentiating both sides of (1) with respect to x we get,
dt/dx = d(ax2 + bx + c)/dx = 2ax + b
Thus, v = velocity of the particle at time t
= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)-1  ……….(2)
Initially, when t = 0 and v = u, let x = x0; hence, from (1) we get,
ax02 + bx0 + c = 0
Or ax02 + bx0 = -c   ……….(3)
And from (2) we get, u = 1/(2ax0 + b)
Thus, 1/v2 – 1/u2 = (2ax + b)2 – (2ax0 + b)2
= 4a2x2 + 4abx – 4a2x02 – 4abx0
= 4a2x2 + 4abx – 4a(ax02 – bx0)
= 4a2x2 + 4abx – 4a(-c)   [using (3)]
= 4a(ax2 + bx + c)
Or 1/v2 – 1/u2 = 4at

8. The distance s of a particle moving along a straight line from a fixed-pointO on the line at time t seconds after start is given by x = (t – 1)2(t – 2)2. What will be the distance of the particle from O when its velocity is zero?
a) 4/27 units
b) 4/23 units
c) 4/25 units
d) 4/35 units
Answer: a
Clarification: Let v be the velocity of the particle at time t seconds after start (that is at a distance s from O). Then,
v = ds/dt = d[(t – 1)2(t – 2)2]/dt
Or v = (t – 2)(3t – 4)
Clearly, v = 0, when (t – 2)(3t – 4) = 0
That is, when t = 2
Or 3t – 4 = 0 i.e., t = 4/3
Now, s = (t – 1)(t – 2)2
Therefore, when t = 4/3, then s = (4/3 – 1)(4/3 – 2)2 = 4/27
And when t = 2, then s =(2 – 1)(2 – 2)2 = 0
Therefore, the velocity of the particle is zero, when its distance from O is 4/27 units and when it is at O.