Category: Class 12 Mathematics Objective Questions

Mathematics Multiple Choice Questions on “Second Order Derivatives”.

1. Find the second order derivative of y=9 log⁡ t³.
a) (frac{27}{t^2})
b) –(frac{27}{t^2})
c) –(frac{1}{t^2})
d) –(frac{27}{2t^2})
Answer: b
Clarification: Given that, y=9 log⁡t³
(frac{dy}{dx}=9.frac{1}{t^3}.3t^2=frac{27}{t})
(frac{d^2 y}{dx^2}=27(-frac{1}{t^2})=-frac{27}{t^2}).

2. Find (frac{d^2y}{dx^2}), if y=tan²⁡x+3 tan⁡x.
a) sec²⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)
b) 2 sec²⁡⁡x tan⁡x (2 tan⁡x-sec⁡x+3)
c) 2 sec²⁡⁡x tan⁡x (2 tan⁡x+sec⁡x+3)
d) 2 sec²⁡⁡x tan⁡x (2 tan⁡x+sec⁡x-3)
Answer: c
Clarification: Given that, y=tan²⁡⁡x+3 tan⁡x
(frac{dy}{dx})=2 tan⁡x sec²⁡⁡x+3 sec²⁡x=sec²⁡⁡x (2 tan⁡x+3)
By using the u.v rule, we get
(frac{d^2 y}{dx^2}=frac{d}{dx}) (sec²⁡⁡x).(2 tan⁡x+3)+(frac{d}{dx}) (2 tan⁡x+3).sec²⁡⁡x
(frac{d^2 y}{dx^2})=2 sec²⁡⁡x tan⁡x (2 tan⁡x+3)+sec²⁡⁡x (2 sec⁡x tanx)
=2 sec²⁡x tan⁡x (2 tan⁡x+sec⁡x+3).

3. If y=6x²+3, then (left (frac{dy}{dx}right )^2=frac{d^2 y}{dx^2}).
a) True
b) False
Answer: b
Clarification: The given statement is false. Given that, y=6x²+3
(frac{dy}{dx})=12x
⇒(left (frac{dy}{dx}right )^2=(12x)^2=144x^2)
(frac{d^2 y}{dx^2}=frac{d}{dx}) (12x)=12
∴(left (frac{dy}{dx}right )^2≠frac{d^2 y}{dx^2})

4. Find the second order derivative of y=2e^2x-3 log⁡(2x-3).
a) 8e^2x+(frac{1}{(2x-3)^2})
b) 8e^2x–(frac{12}{(2x-3)^2})
c) e^2x+(frac{12}{(2x-3)^2})
d) 8e^2x+(frac{12}{(2x-3)^2})
Answer: d
Clarification: Given that, y=2e^2x-3 log⁡(2x-3)
(frac{dy}{dx})=4e^2x-3.(frac{1}{(2x-3)}).2=4e^2x–(frac{6}{(2x-3)})
(frac{d^2 y}{dx^2}=frac{d}{dx} (frac{dy}{dx}))=8e^2x+(frac{12}{(2x-3)^2})

5. Find (frac{d^2 y}{dx^2}), if y=2 sin^-1⁡(cos⁡x).
a) 0
b) sin^-1((frac{1}{cos⁡x}))
c) 1
d) -1
Answer: a
Clarification: Given that, y=2 sin^-1⁡(cos⁡x)
(frac{dy}{dx}=2.frac{1}{sqrt{1-cos^2⁡x}}).-sin⁡x=-2 (∵(sqrt{1-cos^2⁡x})=sin⁡x)
(frac{d^2 y}{dx^2})=(frac{d}{dx} (frac{dy}{dx})=frac{d}{dx}) (-2)=0

6. If y=log⁡(2x³), find (frac{d^2 y}{dx^2}).
a) –(frac{2}{x^2})
b) (frac{3}{x^2})
c) (frac{2}{x^2})
d) –(frac{3}{x^2})
Answer: d
Clarification: Given that, y=log⁡(2x³)
(frac{dy}{dx}=frac{1}{(2x^3)}.6x^2=frac{3}{x})
(frac{d^2 y}{dx^2}=-frac{3}{x^2})

7. Find (frac{d^2 y}{dx^2})-6 (frac{dy}{dx}) if y=4x⁴+2x.
a) ((4x^2+8x-1))
b) (12(4x^2+8x-1))
c) –(12(4x^2+8x-1))
d) (12(4x^2-8x-1))
Answer: d
Clarification: Given that, (y=4x^4+2x)
(frac{dy}{dx})=16x³+2
(frac{d^2 y}{dx^2})=48x²
(frac{d^2 y}{dx^2})-6 (frac{dy}{dx}=48x^2-96x^3-12)
=12(4x²-8x-1)

8. Find the second order derivative y=e^2x+sin^-1⁡e^x .
a) e^2x+(frac{e^x}{(1-e^2x)^{3/2}})
b) 4e^2x+(frac{1}{(1-e^2x)^{3/2}})
c) 4e^2x–(frac{e^x}{(1-e^2x)^{3/2}})
d) 4e^2x+(frac{e^x}{(1-e^2x)^{3/2}})
Answer: d
Clarification: Given that, y=e^2x+sin^-1⁡e^x
(frac{dy}{dx})=2e^2x+(frac{1}{sqrt{1-e^{2x}}} e^x)
(frac{d^2 y}{dx^2} = 4e^2x+bigg(frac{frac{d}{dx} (e^x) sqrt{1-e^{2x}} – frac{d}{dx} (sqrt{1-e^{2x}}).e^x}{(sqrt{1-e^{2x}})^2}bigg))
(=4e^{2x}+frac{(e^x sqrt{1-e^{2x}})-e^x left(frac{1}{2sqrt{1-e^{2x}}}.-2e^{2x}right)}{1-e^{2x}})
(=4e^{2x}+frac{(e^x (1-e^{2x})+e^{3x})}{(1-e^{2x})^{frac{3}{2}}})
(=4e^{2x}+frac{e^x (1-e^{2x}+e^{2x})}{(1-e^{2x})^{frac{3}{2}}})
4e^2x+(frac{e^x}{(1-e^2x)^{3/2}}).

9. Find the second order derivative of y=3x² 1 + log⁡(4x)
a) 3+(frac{1}{x^2})
b) 3-(frac{1}{x^2})
c) 6-(frac{1}{x^2})
d) 6+(frac{1}{x^2})
Answer: c
Clarification: Given that, y=3x²+log⁡(4x)
(frac{dy}{dx}=6x+frac{1}{4x}.4=6x+frac{1}{x}=frac{6x^2+1}{x})
(frac{d^2 y}{dx^2}=frac{frac{d}{dx} (6x^2+1).(x)-frac{d}{dx} (x).(6x^2+1)}{x^2} Big(using, frac{d}{dx} (frac{u}{v})=frac{(frac{d}{dx} (u).v-frac{d}{dx} (v).u)}{v^2}Big))
(frac{d^2 y}{dx^2}=frac{(12x.x-6x^2-1)}{x^2} )
(frac{d^2 y}{dx^2}=frac{6x^2-1}{x^2} = 6-frac{1}{x^2}).

10. Find the second order derivative if y=e^2x2.
a) 4e^2x2 (4x²+3)
b) 4e^2x2 (4x²-1)
c) 4e^2x2 (4x²+1)
d) e^2x2 (4x²+1)
Answer: c
Clarification: Given that, y=e^2x2
(frac{dy}{dx})=e^2x2.4x
By using u.v rule, we get
(frac{d^2 y}{dx^2}=frac{d}{dx} (e^{{2x}^2}).4x+frac{d}{dx} (4x).e^{{2x}^2})
16x² e^2x2+4e^2x2=4e^2x2 (4x²+1)

Mathematics Multiple Choice Questions on “Fundamental Theorem of Calculus-1”.

1. Find (int_0^8x ,dx).
a) 32
b) 34
c) 21
d) 24
Answer: a
Clarification: Let I=(int_0^8x ,dx)
F(x)=(int x ,dx=frac{x^2}{2})
Using the second fundamental theorem of calculus, we get
I=F(8)-f(0)
∴(int_0^8x ,dx=frac{8^2}{2}-0=32)

2. Find (int_0^{frac{π}{2}} ,5 ,sin⁡x ,dx).
a) -5
b) 9
c) 5
d) -9
Answer: c
Clarification: Let (I=int_0^{frac{π}{2}} ,5 ,sin⁡x ,dx)
F(x)=(int5 ,sin⁡x ,dx=-5 ,cos⁡x)
Applying the limits by using the fundamental theorem of calculus, we get
I=F((frac{π}{2}))-F(0)
∴(int_0^{frac{π}{2}} ,5 ,sin⁡x ,dx=-5[cos⁡frac{π}{2}-cos⁡0])
=-5[0-1]=5

3. Find the value of (int_4^5 ,log⁡x ,dx).
a) 5 log⁡5-log⁡4+1
b) 5 log⁡5-4 log⁡4-1
c) 4 log⁡5-4 log⁡4-1
d) 5-4 log⁡4-log⁡5
Answer: b
Clarification: Let I=(int_4^5 ,log⁡x ,dx).
F(x)=∫ log⁡x dx
By using the formula (int ,u.v dx=u int v ,dx-int u'(int ,v ,dx)), we get
(int log ⁡x ,dx=log⁡x int ,dx-int(log⁡x)’int ,dx)
F(x)=x log⁡x-∫ dx=x(log⁡x-1).
Applying the limits using the fundamental theorem of calculus, we get
I=F(5)-F(4)=(5 log⁡5-5)-(4 log⁡4-4)
=5 log⁡5-4 log⁡4-1.

4. Find (int_0^{frac{π}{4}} ,9 ,cos^2⁡x ,dx).
a) (frac{9}{2}left (frac{π}{6}-1right))
b) (frac{9}{4}left (frac{π}{2}+1right))
c) (frac{9}{4}left (frac{π}{2}-1right))
d) (left (frac{π}{2}-1right))
Answer: c
Clarification: Let I=(int_0^{frac{π}{4}} ,9 ,cos^2⁡x ,dx).
F(x)=(int ,9 ,cos^2⁡x ,dx)
=9(int(frac{1+cos⁡2x}{2})dx)
=(frac{9}{2} (x-frac{sin⁡2x}{2}))
Applying the limits, we get
I=(F(frac{π}{4})-F(0)=frac{9}{2} left (frac{π}{4}-frac{sin⁡2(frac{π}{4})}{2}right)-frac{9}{2} (0-frac{sin⁡0}{2}))
=(frac{9}{2}left (frac{π}{4}-frac{sin⁡π/2}{2}right )=frac{9}{4} (π/2-1))

5. Find (int_0^2 ,e^{2x} ,dx).
a) (frac{e^4-1}{6})
b) (frac{e^4+1}{2})
c) (frac{e-1}{2})
d) (frac{e^4-1}{2})
Answer: d
Clarification: Let (I=int_0^2 ,e^2x ,dx)
F(x)=(int e^{2x} dx)
=(frac{e^{2x}}{2})
Applying the limits, we get
I=F(2)-F(0)
=(frac{e^2(2)}{2}-frac{e^2(0)}{2}=frac{(e^4-1)}{2}).

6. Find (int_{π/4}^{π/2} ,2sinx ,sin⁡(cos⁡x) ,dx).
a) 2(1-cos⁡(frac{1}{sqrt{2}}))
b) (cos⁡(frac{1}{sqrt{2}})-cos⁡1)
c) 2(cos⁡(frac{1}{sqrt{2}})+1)
d) (cos⁡(frac{1}{sqrt{2}})+cos⁡1)
Answer: a
Clarification: Let (I=int_{π/4}^{π/2} ,2sinx ,sin⁡(cos⁡x) ,dx)
F(x)=(int 2 ,sin⁡x ,sin⁡(cos⁡x)dx)
Let cos⁡x=t
Differentiating w.r.t x, we get
sin⁡x dx=dt
∴(int 2 ,sin⁡x ,sin⁡(cos⁡x)dx=int 2 ,sin⁡t ,dt=-2 ,cos⁡t)
Replacing t with cos⁡x, we get
∴∫ 2 sin⁡x sin⁡(cos⁡x)dx=-2 cos⁡(cos⁡x)
By applying the limits, we get
(I=F(frac{π}{4})-F(frac{π}{2})=-2 cos⁡(frac{cos⁡π}{4})+2 cos⁡(frac{cos⁡π}{2}))
=2(1-cos⁡(frac{1}{sqrt{2}}))

7. Find (int_{-2}^1 ,5x^4 ,dx).
a) 54
b) 75
c) 33
d) 36
Answer: c
Clarification: (I=int_{-2}^1 ,5x^4 ,dx)
F(x)=(int5x^4 ,dx=5(frac{x^5}{5})=x^5)
Applying the limits by using the fundamental theorem of calculus, we get
I=F(1)-F(-2)
=(1)⁵-(-2)⁵=1+32=33.

8. Find (int_0^3 ,e^x ,dx).
a) e³+1
b) -e³-1
c) e³-1
d) 3e³-2
Answer: c
Clarification: Let I=(int_0^3 ,e^x ,dx)
F(x)=(int ,e^x ,dx=e^x)
Applying the limits, we get
I=F(3)-F(0)
=e³-e⁰=e³-1.

9. Find (int_0^{π/4} ,2 ,tan⁡x ,dx).
a) log⁡2
b) log⁡(sqrt{2})
c) 2 log⁡2
d) 0
Answer: a
Clarification: (I=int_0^{π/4} ,2 ,tan⁡x ,dx)
F(x)=∫ 2 tan⁡x dx
=2∫ tan⁡x dx
=2 log⁡|sec⁡x|
Therefore, by using the fundamental theorem of calculus, we get
I=F(π/4)-F(0)
(=2left(log⁡|sec frac{⁡π}{4}|-log⁡|sec⁡0|right)=2 log⁡sqrt{2}-log⁡1)
(=2 log⁡sqrt{2}=log⁡(sqrt{2})^2=log⁡2)
I=(frac{8}{3} log⁡2-frac{8}{3}-0+frac{1}{3}=frac{8}{3} log⁡2-frac{7}{3}).

10. Find (int_{-1}^1 ,2xe^x ,dx).
a) (frac{4}{e})
b) 4e
c) –(frac{4}{e})
d) -4e
Answer: a
Clarification: (I=int_{-1}^1 ,2xe^x ,dx)
F(x)=(int 2xe^x dx)
By using the formula, (int u.v ,dx=u int v ,dx-int u'(int v ,dx))
F(x)=2x(int e^x dx-int(2x)’int e^x ,dx)
=(2xe^x-int 2e^x dx)
=(2e^x (x-1))
Therefore, by using the fundamental theorem of calculus, we get
I=F(1)-F(-1)
I=2e¹ (1-1)-2e^-1 (-1-1)
I=(frac{4}{e}).

Mathematics Question Papers for Class 12 on “Product of Two Vectors-2”.

1. Evaluate the product ((2vec{a}+5vec{b}).(4vec{a}-5vec{b})).
a) (|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)
b) (8|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)
c) (8|vec{a}|^2-4vec{a}.vec{b}-15|vec{b}|^2)
d) (|vec{a}|^2+vec{a}.vec{b}-5|vec{b}|^2)
Answer: b
Clarification: To evaluate: ((2vec{a}+5vec{b}).(4vec{a}-5vec{b}))
=(2vec{a}.4vec{a}-2vec{a}.5vec{b}+3vec{b}.4vec{a}-3vec{b}.5vec{b})
=(8|vec{a}|^2+2vec{a}.vec{b}-15|vec{b}|^2)

2. Find the magnitude of (vec{a}) and (vec{b}) which are having the same magnitude and such that the angle between them is 60° and their scalar product is (frac{1}{4}).
a) (|vec{a}|=|vec{b}|=frac{1}{2√2})
b) (|vec{a}|=|vec{b}|=frac{1}{√2})
c) (|vec{a}|=|vec{b}|=frac{1}{2√3})
d) (|vec{a}|=|vec{b}|=frac{2}{√3})
Answer: a
Clarification: Given that: a) (|vec{a}|=|vec{b}|)
b) θ=60°
c) (vec{a}.vec{b}=frac{1}{4})
∴(|vec{a}||vec{b}| cos⁡θ=frac{1}{4})
=(|vec{a}|^2 cos⁡60°=frac{1}{4})
⇒(|vec{a}|^2=frac{1}{4}.frac{1}{2})
∴(|vec{a}|=|vec{b}|=frac{1}{2√2}).

3. If (vec{a}=hat{i}-hat{j}+3hat{k}, ,vec{b}=5hat{i}-2hat{j}+hat{k} ,and ,vec{c}=hat{i}-hat{j}) are such that (vec{a}+μvec{b}) is perpendicular to (vec{c}), then the value of μ.
a) (frac{7}{2})
b) –(frac{7}{2})
c) –(frac{3}{2})
d) (frac{7}{9})
Answer: b
Clarification: Given that: (vec{a}=hat{i}-hat{j}+3hat{k}, ,vec{b}=5hat{i}-2hat{j}+hat{k} ,and ,vec{c}=hat{i}-hat{j})
Also given, (vec{a}+μvec{b}) is perpendicular to (vec{c})
Therefore, ((vec{a}+μvec{b}).vec{c}=0)
i.e. ((hat{i}-hat{j}+3hat{k}+μ(5hat{i}-2hat{j}+hat{k})).(hat{i}-hat{j}))=0
(((1+5μ) ,hat{i}-(1+2μ) ,hat{j}+(μ+3) ,hat{k}).(hat{i}-hat{j}))=0
1+5μ+1+2μ=0
μ=-(frac{7}{2}).

4. Find the angle between (vec{a} ,and ,vec{b}) if (|vec{a}|=2,|vec{b}|=frac{1}{2√3}) and (vec{a}×vec{b}=frac{1}{2}).
a) (frac{2π}{3})
b) (frac{4π}{5})
c) (frac{π}{3})
d) (frac{π}{2})
Answer: c
Clarification: Given that, (|vec{a}|=2, ,|vec{b}|=frac{1}{2√3}) and (vec{a}×vec{b}=frac{1}{2})
We know that, (vec{a}×vec{b}=vec{a}.vec{b} ,sin⁡θ)
∴ (sin⁡θ=frac{(vec{a}×vec{b})}{|vec{a}|.|vec{b}|})
sin⁡θ=(frac{frac{1}{2}}{2×frac{1}{2√3}}=frac{sqrt{3}}{2})
θ=(sin^{-1}⁡frac{sqrt{3}}{2}=frac{π}{3})

5. Find the vector product of the vectors (vec{a}=2hat{i}+4hat{j}) and (vec{b}=3hat{i}-hat{j}+2hat{k}).
a) (hat{i}-19hat{j}-4hat{k})
b) (3hat{i}+19hat{j}-14hat{k})
c) (3hat{i}-19hat{j}-14hat{k})
d) (3hat{i}+5hat{j}+4hat{k})
Answer: c
Clarification: Given that, (vec{a}=2hat{i}+4hat{j}) and (vec{b}=3hat{i}-hat{j}+2hat{k})
Calculating the vector product, we get
(vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\2&4&-5\3&-1&2end{vmatrix})
=(hat{i}(8-5)-hat{j}(4-(-15))+hat{k}(-2-12))
=(3hat{i}-19hat{j}-14hat{k})

6. If (vec{a} ,and ,vec{b}) are two non-zero vectors then ((vec{a}-vec{b})×(vec{a}+vec{b}))=_________
a) (2(vec{a}×vec{b}))
b) ((vec{a}×vec{b}))
c) –(4(vec{a}×vec{b}))
d) (3(vec{a}×vec{b}))
Answer: a
Clarification: Consider ((vec{a}-vec{b})×(vec{a}+vec{b}))
=((vec{a}-vec{b})×vec{a}+(vec{a}+vec{b})×vec{b})
=(vec{a}×vec{a}-vec{b}×vec{a}+vec{a}×vec{b}-vec{b}×vec{b})
We know that, (vec{a}×vec{a}=0,vec{b}×vec{b}=0 ,and ,vec{a}×vec{b}=-vec{b}×vec{a})
∴ (vec{a}×vec{a}-vec{b}×vec{a}+vec{a}×vec{b}-vec{b}×vec{b}=0+2(vec{a}×vec{b})+0)
Hence, ((vec{a}-vec{b})×(vec{a}+vec{b})=2(vec{a}×vec{b}))

7. Find the product ((vec{a}+vec{b}).(7vec{a}-6vec{b})).
a) (2|vec{a}|^2+6vec{a}.vec{b}-3|vec{b}|^2)
b) (8|vec{a}|^2+5vec{a}.vec{b}-5|vec{b}|^2)
c) (2|vec{a}|^2+6vec{a}.vec{b}-8|vec{b}|^2)
d) (7|vec{a}|^2+vec{a}.vec{b}-6|vec{b}|^2)
Answer: d
Clarification: To evaluate: ((vec{a}+vec{b}).(7vec{a}-6vec{b}))
=(vec{a}.7vec{a}-vec{a}.6vec{b}+vec{b}.7vec{a}-6vec{b}.vec{b})
=(7|vec{a}|^2+vec{a}.vec{b}-6|vec{b}|^2)

8. Find the vector product of the vectors (vec{a}=-hat{j}+hat{k}) and (vec{b}=-hat{i}-hat{j}-hat{k}).
a) (2hat{i}-hat{j}+hat{k})
b) (2hat{i}-hat{j}-4hat{k})
c) (hat{i}+hat{j}-hat{k})
d) (2hat{i}-hat{j}-hat{k})
Answer: d
Clarification: Given that, (vec{a}=-hat{j}+hat{k}) and (vec{b}=-hat{i}-hat{j}-hat{k})
Calculating the vector product, we get
(vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\0&-1&1\-1&-1&-1end{vmatrix})
=(hat{i}(1-(-1))-hat{j}(0-(-1))+hat{k}(0-1))
=(2hat{i}-hat{j}-hat{k})

9. If (vec{a}=2hat{i}+3hat{j}+4hat{k}) and (vec{b}=4hat{i}-2hat{j}+3hat{k}). Find (|vec{a}×vec{b}|).
a) (sqrt{685})
b) (sqrt{645})
c) (sqrt{679})
d) (sqrt{689})
Answer: b
Clarification: Given that, (vec{a}=2hat{i}+3hat{j}+4hat{k}) and (vec{b}=4hat{i}-2hat{j}+3hat{k})
∴ (vec{a}×vec{b}=begin{vmatrix}hat{i}&hat{j}&hat{k}\2&3&4\4&-2&3end{vmatrix})
=(hat{i}(9—8)-hat{j}(6-16)+hat{k}(-4-12))
=(17hat{i}+10hat{j}-16hat{k})
∴(|vec{a}×vec{b}|=sqrt{17^2+10^2+(-16)^2})
=(sqrt{289+100+256})
=(sqrt{645})

10. Find the angle between the vectors if (|vec{a}|=|vec{b}|=3sqrt{2}) and (vec{a}.vec{b}=9sqrt{3}).
a) (frac{π}{6})
b) (frac{π}{5})
c) (frac{π}{3})
d) (frac{π}{2})
Answer: a
Clarification: We know that, (vec{a}.vec{b}=|vec{a}|.|vec{b}| ,cos⁡θ)
Given that, (|vec{a}|=|vec{b}|=3sqrt{2} ,and ,vec{a}.vec{b}=9sqrt{3})
(cos⁡θ=frac{vec{a}.vec{b}}{|vec{a}|.|vec{b}|}=frac{9sqrt{3}}{(3sqrt{2})^2}=frac{sqrt{3}}{2})
(θ=cos^{-1}frac{sqrt{3}}{2}=frac{π}{6}).

Mathematics Question Papers for Class 12,

Mathematics Multiple Choice Questions on “Calculus Application – Tangents and Normals – 1”.

1. What is the equation of the tangent at a specific point of y² = 4ax at (0, 0)?
a) x = 0
b) x = 1
c) x = 2
d) x = 3
Answer: a
Clarification: Equation of the given parabola is y² = 4ax ……….(1)
Differentiating both side of (1) with respect to x we get,
2y(dy/dx) = 4a
Or dy/dx = 2a/y
Clearly dy/dx does not exist at (0, 0). Hence, the tangent to the parabola (1) at (0, 0) is parallel to y axis.
Again, the tangent passes through (0, 0). Therefore, the required tangent to the parabola (1) at (0, 0) is the y-axis and hence the required equation of the tangent is x = 0.

2. If X and Y are given as current co-ordinates, what is the equation of the tangent at a specific point of x³ – 3axy + y³ = 0 at (x, y)?
a) (x² – ay)X + (y² – ax)Y = -2axy
b) (x² – ay)X + (y² – ax)Y = 2axy
c) (x² – ay)X + (y² – ax)Y = axy
d) (x² – ay)X + (y² – ax)Y = -axy
Answer: c
Clarification: Equation of the given curve is, x³ – 3axy + y³ = 0 ……….(1)
Differentiating both sides with respect to x we get,
3x² – 3a(x(dy/dx) + y) + 3y²(dy/dx) = 0
Or dy/dx = (ay – x²)/(y² – ax)
So, it is clear that this can be written as,
Y – y = (dy/dx)(X – x)
Or Y – y = [(ay – x²)/(y² – ax)](X – x)
Simplifying the above equation by cross multiplication, we get,
(x² – ay)X + (y² – ax)Y = x³ – 3axy + y³ + axy
Using (1),
(x² – ay)X + (y² – ax)Y = axy

3. What will be the equation of the normal to the hyperbola xy = 4 at the point (2, 2)?
a) x + y = 0
b) x – y = 0
c) 2x – y = 0
d) x + 2y = 0
Answer: b
Clarification: Equation of the given hyperbola is, xy = 4 ……….(1)
Differentiating both side of (1) with respect to y, we get,
y*(dx/dy) + x(1) = 0
Or dx/dy = -(x/y)
Thus, the required equation of the normal to the hyperbola at (2, 2) is,
y – 2 = -[dx/dy]_{(2, 2)} (x – 2) = -(-2/2)(x – 2)
So, from here,
y – 2 = x – 2
Or x – y = 0

4. At which point does the normal to the hyperbola xy = 4 at (2, 2) intersects the hyperbola again?
a) (-2, -2)
b) (-2, 2)
c) (2, -2)
d) (0, 2)
Answer: a
Clarification: Equation of the given hyperbola is, xy = 4 ……….(1)
Differentiating both side of (1) with respect to y, we get,
y*(dx/dy) + x(1) = 0
Or dx/dy = -(x/y)
Thus, the required equation of the normal to the hyperbola at (2, 2) is,
y – 2 = -[dx/dy]_{(2, 2)} (x – 2) = -(-2/2)(x – 2)
So, from here,
y – 2 = x – 2
Or x – y = 0 ……….(2)
Solving the equation (1) and (2) we get,
x = 2 and y = 2 or x = -2 and y = -2
Thus, the line (2) intersects the hyperbola (1) at (2, 2) and (-2, -2).
Hence, the evident is that the normal at (2, 2) to the hyperbola (1) again intersects it at (-2, -2).

5. What will be the equation of the tangent to the circle x² + y² – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?
a) x + 2y – 9 = 0
b) x + 2y + 9 = 0
c) x + 2y – 10 = 0
d) x + 2y + 10 = 0
Answer: a
Clarification: The equation of any straight line perpendicular to the line 2x – y + 3 = 0 is,
x + 2y + k = 0 ……….(1)
Now, the co-ordinate of the center of the circle (3, -2) and its radius is,
√(9 + 4 – (-7) = 2√5
If straight line (1) be tangent to the given circle then, the perpendicular distance of the point (3, -2) from the line (1) = radius of the circle
Thus, ±(3 + 2(-2) + k)/√(1 + 4)
Or k – 1 = 2√5 * √5
So, k = 1 ± 10
= 11 or -9
Putting the value of k in (1) we get,
x + 2y + 11 = 0 and x + 2y – 9 = 0

6. What is the equation of the tangent to the parabola y² = 8x, which is inclined at an angle of 45° with the x axis?
a) x + y – 2 = 0
b) x + y + 2 = 0
c) x – y + 2 = 0
d) x – y – 2 = 0
Answer: c
Clarification: Equation of the given parabola is, y² = 8x ……….(1)
Differentiating both sides with respect to x,
2y(dy/dx) = 8
Or dy/dx = 4/y
Thus, equation of the tangent to the parabola (1) at (x₁, y₁) = (2t², 4t) is,
y – y₁ = [dy/dx]_{(x1, y1)} (x – 2t²)
y – 4t = [dy/dx]_{(2(t²), 4t)} (x – 2t²)
Putting the value of y = 4t in the equation dy/dx = 4/y, we get,
y – 4t = 4/4t(x – 2t²) ……….(2)
If the tangent to the parabola y² = 8x, which is inclined at an angle of 45° with the x axis,
Then, slope of tangent (2) = tan 45° = 1
Thus, 4/4t = 1
Or t = 1
Thus, required equation of the tangent is,
y– 4 = 1(x – 2)
Putting, t = 1 in (2),
So, x – y + 2 = 0

7. What will be the equation of the tangent to the circle x² + y² – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?
a) x – 2y + 11 = 0
b) x – 2y – 11 = 0
c) x + 2y + 11 = 0
d) x + 2y – 11 = 0
Answer: c
Clarification: The equation of any straight line perpendicular to the line 2x – y + 3 = 0 is,
x + 2y + k = 0 ……….(1)
Now, the co-ordinate of the center of the circle (3, -2) and its radius is,
√(9 + 4 – (-7) = 2√5
If straight line (1) be tangent to the given circle then, the perpendicular distance of the point (3, -2) from the line (1) = radius of the circle
Thus, ±(3 + 2(-2) + k)/√(1 + 4)
Or k – 1 = 2√5 * √5
So, k = 1 ± 10
= 11 or -9
Putting the value of k in (1) we get,
x + 2y + 11 = 0 and x + 2y – 9 = 0

8. What will be the equation of normal to the hyperbola 3x² – 4y² = 12 at the point (x₁, y₁)?
a) 3x₁y + 4y₁x + 7x₁y₁ = 0
b) 3x₁y + 4y₁x – 7x₁y₁ = 0
c) 3x₁y – 4y₁x – 7x₁y₁ = 0
d) 3x₁y – 4y₁x + 7x₁y₁ = 0
Answer: b
Clarification: Equation of the given hyperbola is, 3x² – 4y² = 12 ……….(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) – 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, the equation of the normal to the hyperbola (1) at the point (x₁, y₁) on it is,
y – y₁ = -[dx/dy]_{(x1, y1)} (x – x₁) = -4y₁/3x₁(x – x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0

9. What is the nature of the straight line x + y + 7 = 0 to the hyperbola 3x² – 4y² = 12 whose normal is at the point (x₁, y₁)?
a) Chord to hyperbola
b) Tangent to hyperbola
c) Normal to hyperbola
d) Segment to hyperbola
Answer: c
Clarification: Equation of the given hyperbola is, 3x² – 4y² = 12 ……….(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) – 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, the equation of the normal to the hyperbola (1) at the point (x₁, y₁) on it is,
y – y₁ = -[dx/dy]_{(x1, y1)} (x – x₁) = -4y₁/3x₁(x – x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0
Now, if possible, let us assume that the straight line
x + y + 7 = 0 ………..(2)
This line is normal to the hyperbola (1) at the point (x₁, y₁). Then, the equation (2) and (3) must be identical. Hence, we have,
3x₁/1 = 4y₁/1 = -7x₁y₁/7
So, x₁ = -4 and y₁ = -3
Now, 3x₁² – 4y₁² = 3(-4)² – 4(-3)² = 12
This shows the point (-4, -3) lies on the hyperbola (1).
Thus, it is evident that the straight line (3) is normal to the hyperbola (1).

10. What is the foot of the normal if the straight line x + y + 7 = 0 is normal to the hyperbola 3x² – 4y² = 12 whose normal is at the point (x₁, y₁)?
a) (4, 3)
b) (-4, 3)
c) (4, -3)
d) (-4, -3)
Answer: d
Clarification: Equation of the given hyperbola is, 3x² – 4y² = 12 ……….(1)
Differentiating both sides of (1) with respect to y we get,
3*2x(dy/dx) – 4*(2y) = 0
Or dx/dy = 4y/3x
Therefore, the equation of the normal to the hyperbola (1) at the point (x₁, y₁) on it is,
y – y₁ = -[dx/dy]_{(x1, y1)} (x – x₁) = -4y₁/3x₁(x – x₁)
Or 3x₁y + 4y₁x – 7x₁y₁ = 0
Now, if possible, let us assume that the straight line
x + y + 7 = 0 ………..(2)
This line is normal to the hyperbola (1) at the point (x₁, y₁). Then, the equation (2) and (3) must be identical. Hence, we have,
3x₁/1 = 4y₁/1 = -7x₁y₁/7
So, x₁ = -4 and y₁ = -3
Now, 3x₁² – 4y₁² = 3(-4)² – 4(-3)² = 12
This shows the point (-4, -3) lies on the hyperbola (1).
So, it’s the normal to the hyperbola.
Thus, it is evident that the straight line (3) is normal to the hyperbola (1); the co-ordinate foot is (-4, -3).

Mathematics Multiple Choice Questions on “Determinant – 1”.

1. Evaluate (begin{vmatrix}2&5\-1&-1end{vmatrix}).
a) 3
b) -7
c) 5
d) -2
Answer: a
Clarification: Expanding along R₁, we get
∆=2(-1)-5(-1)=-2+5
=3.

2. Evaluate (begin{vmatrix}5&-4\1&sqrt{3}end{vmatrix}).
a) 4(sqrt{3})+4
b) 4(sqrt{3})+5
c) 5(sqrt{3})+4
d) 5(sqrt{3})-4
Answer: c
Clarification: Evaluating along R₁, we get
∆=5((sqrt{3}))-(-4)1=5(sqrt{3})+4.

3. Evaluate (begin{vmatrix}-sinθ&-1\1&sin⁡θend{vmatrix}).
a) cos²⁡θ
b) -cos²⁡θ
c) cos⁡2θ
d) cos⁡θ
Answer: a
Clarification: Expanding along R₁, we get
∆=-sinθ(sinθ)-(-1)1=-sin²⁡θ+1=cos²⁡θ.

4. Evaluate (begin{vmatrix}i&-1\-1&-iend{vmatrix}).
a) 4
b) 3
c) 2
d) 0
Answer: d
Clarification: Expanding along R₁, we get
∆=-i(i)-(-1)(-1)=-i²-1=-(-1)-1=0.

5. Evaluate (begin{vmatrix}1&1&-2\3&4&5\-1&2&1end{vmatrix}).
a) -6
b) -34
c) 34
d) 22
Answer: b
Clarification: ∆=(begin{vmatrix}1&1&-2\3&4&5\-1&2&1end{vmatrix})
Expanding along the first row, we get
∆=1(begin{vmatrix}4&5\2&1end{vmatrix})-1(begin{vmatrix}3&5\-1&1end{vmatrix})-2(begin{vmatrix}3&4\-1&2end{vmatrix})
=1(4-5(2))-1(3-5(-1))-2(6-4(-1))
=(4-10)-(3+5)-2(6+4)
=-6-8-20=-34.

6. Evaluate (begin{vmatrix}5&4&3\3&4&1\5&6&1end{vmatrix}).
a) 4
b) -24
c) -8
d) 8
Answer: c
Clarification: Expanding along the first row, we get
∆=5(begin{vmatrix}4&1\6&1end{vmatrix})-4(begin{vmatrix}3&1\5&1end{vmatrix})+3(begin{vmatrix}3&4\5&6end{vmatrix})
=5(4-6)-4(3-5)+3(18-20)
=5(-2)-4(-2)+3(-2)=-10+8-6=-8.

7. Evaluate (begin{vmatrix}8x+1&2x-2\x^2-1&3x+5end{vmatrix}).
a) -2x³-26x²+45x+3
b) -2x³+26x²+45x+3
c) -2x³+26x²+45x-3
d) -2x³-26x²-45x+3
Answer: b
Clarification: Expanding along the first row, we get
∆=8x+1(3x+5)-(2x-2)(x²-1)
=(24x²+43x+5)-(2x³-2x²-2x+2)
=-2x³+26x²+45x+3.

8. If A=(begin{bmatrix}2&5&9\6&1&3\4&8&2end{bmatrix}), find |A|.
a) 352
b) 356
c) 325
d) 532
Answer: a
Clarification: Given that, A=(begin{bmatrix}2&5&9\6&1&3\4&8&2end{bmatrix})
⇒|A|=(begin{vmatrix}2&5&9\6&1&3\4&8&2end{vmatrix})
Evaluating along the first row, we get
∆=2(begin{vmatrix}1&3\8&2end{vmatrix})-5(begin{vmatrix}6&3\4&2end{vmatrix})+9(begin{vmatrix}6&1\4&8end{vmatrix})
∆=2(2-24)-5(12-12)+9(48-4)
∆=2(-22)-0+9(44)
∆=-44+9(44)=44(-1+9)=352

9. Evaluate (begin{vmatrix}sqrt{3}&sqrt{2}\-1&2sqrt{3}end{vmatrix}).
a) 6-3(sqrt{2})
b) 6-(sqrt{2})
c) 6+3(sqrt{2})
d) 6+(sqrt{2})
Answer: d
Clarification: ∆=(begin{vmatrix}sqrt{3}&sqrt{2}\-1&2sqrt{3}end{vmatrix})
∆=((sqrt{3})×2(sqrt{3}))+(sqrt{2})
∆=6+(sqrt{2}).

10. Find the value of x if (begin{vmatrix}3&x\2&x^2 end{vmatrix})=(begin{vmatrix}5&3\3&2end{vmatrix}).
a) x=1, –(frac{1}{3})
b) x=-1, –(frac{1}{3})
c) x=1, (frac{1}{3})
d) x=-1, (frac{1}{3})
Answer: a
Clarification: Given that (begin{vmatrix}3&x\2&x^2 end{vmatrix})=(begin{vmatrix}5&3\3&2end{vmatrix})
⇒3x²-2x=5(2)-3(3)
⇒3x²-2x=1
Solving for x, we get
x=1, –(frac{1}{3}).

Mathematics Multiple Choice Questions on “Mean Value Theorem”.

1. Function f should be _____ on [a,b] according to Rolle’s theorem.
a) continuous
b) non-continuous
c) integral
d) non-existent
Answer: a
Clarification: According to Rolle’s theorem, if f : [a,b] → R is a function such that
i) f is continuous on [a,b]
ii) f is differentiable on (a,b)
iii) f(a) = f(b) then there exists at least one point c ∈ (a,b) such that f’(c) = 0

2. Function f is differential on (a,b) according to Rolle’s theorem.
a) True
b) False
Answer: a
Clarification: According to Rolle’s theorem, if f : [a,b] → R is a function such that
i) f is continuous on [a,b]
ii) f is differentiable on (a,b)
iii) f(a) = f(b) then there exists at least one point c ∈ (a,b) such that f’(c) = 0

3. What is the relation between f(a) and f(b) according to Rolle’s theorem?
a) Equals to
b) Greater than
c) Less than
d) Unequal
Answer: a
Clarification: According to Rolle’s theorem, if f : [a,b] → R is a function such that
i) f is continuous on [a,b]
ii) f is differentiable on (a,b)
iii) f(a) = f(b) then there exists at least one point c ∈ (a,b) such that f’(c) = 0

4. Does Rolle’s theorem applicable if f(a) is not equal to f(b)?
a) Yes
b) No
c) Under particular conditions
d) May be
Answer: b
Clarification: According to Rolle’s theorem, if f : [a,b] → R is a function such that
i) f is continuous on [a,b]
ii) f is differentiable on (a,b)
iii) f(a) = f(b) then there exists at least one point c ∈ (a,b) such that f’(c) = 0

5. Another form of Rolle’s theorem for the differential condition is _____
a) f is differentiable on (a,ah)
b) f is differentiable on (a,a-h)
c) f is differentiable on (a,a/h)
d) f is differentiable on (a,a+h)
Answer: d
Clarification: According to Rolle’s theorem, if f : [a,a+h] → R is a function such that
i) f is continuous on [a,a+h]
ii) f is differentiable on (a,a+h)
iii) f(a) = f(a+h) then there exists at least one θ c ∈ (0,1) such that f’(a+θh) = 0

6. Another form of Rolle’s theorem for the continuous condition is _____
a) f is continuous on [a,a-h]
b) f is continuous on [a,h]
c) f is continuous on [a,a+h]
d) f is continuous on [a,ah]
Answer: c
Clarification: According to Rolle’s theorem, if f : [a,a+h] → R is a function such that
i) f is continuous on [a,a+h]
ii) f is differentiable on (a,a+h)
iii) f(a) = f(a+h) then there exists at least one θ c ∈ (0,1) such that f’(a+θh) = 0

7. What is the relation between f(a) and f(h) according to another form of Rolle’s theorem?
a) f(a) < f(a+h)
b) f(a) = f(a+h)
c) f(a) = f(a-h)
d) f(a) > f(a+h)
Answer: b
Clarification: According to Rolle’s theorem, if f : [a,a+h] → R is a function such that
i) f is continuous on [a,a+h]
ii) f is differentiable on (a,a+h)
iii) f(a) = f(a+h) then there exists at least one θ c ∈ (0,1) such that f’(a+θh) = 0

8. Function f is not continuous on [a,b] to satisfy Lagrange’s mean value theorem.
a) False
b) True
Answer: a
Clarification: According to Lagrange’s mean value theorem, if f : [a,b] → R is a function such that
i) f is continuous on [a,b]
ii) f is differentiable on (a,b) then there exists a least point c ∈ (a,b) such that f’(c) = (frac {f(b)-f(a)}{b-a}).

9. What are/is the conditions to satify Lagrange’s mean value theorem?
a) f is continuous on [a,b]
b) f is differentiable on (a,b)
c) f is differentiable and continuous on (a,b)
d) f is differentiable and non-continuous on (a,b)
Answer: c
Clarification: According to Lagrange’s mean value theorem, if f : [a,b] → R is a function such that
i) f is continuous on [a,b]
ii) f is differentiable on (a,b) then there exists a least point c ∈ (a,b) such that f’(c) = (frac {f(b)-f(a)}{b-a}).

10. Function f is differentiable on [a,b] to satisfy Lagrange’s mean value theorem.
a) True
b) False
Answer: a
Clarification: According to Lagrange’s mean value theorem, if f : [a,b] → R is a function such that f is differentiable on (a,b) then there exists a least point c ∈ (a,b) such that f’(c) = (frac {f(b)-f(a)}{b-a}). This shows Function f is differentiable on [a,b].

11. Lagrange’s mean value theorem is also called as _____
a) Euclid’s theorem
b) Rolle’s theorem
c) a special case of Rolle’s theorem
d) the mean value theorem
Answer: d
Clarification: Lagrange’s mean value theorem is also called the mean value theorem and Rolle’s theorem is just a special case of Lagrange’s mean value theorem when f(a) = f(b).

12. Rolle’s theorem is a special case of _____
a) Euclid’s theorem
b) another form of Rolle’s theorem
c) Lagrange’s mean value theorem
d) Joule’s theorem
Answer: c
Clarification: Rolle’s theorem is just a special case of Lagrange’s mean value theorem when f(a) = f(b) and Lagrange’s mean value theorem is also called the mean value theorem.

13. Is Rolle’s theorem applicable to f(x) = tan x on [ (frac {pi }{4}, frac {5pi }{4}) ]?
a) Yes
b) No
Answer: b
Clarification: Given function is f(x) = tan x on [ (frac {pi }{4}, frac {5pi }{4}) ]
F(x) = tan x is not defined at x on [ (frac {pi }{4}, frac {5pi }{4}) ]
So, f(x) is not continuous on [ (frac {pi }{4}, frac {5pi }{4}) ].
Hence, Rolle’s theorem is not applicable.

14. What is the formula for Lagrange’s theorem?
a) f’(c) = (frac {f(a)+f(b)}{b-a})
b) f’(c) = (frac {f(b)-f(a)}{b-a})
c) f’(c) = (frac {f(a)+f(b)}{b+a})
d) f’(c) = (frac {f(a)-f(b)}{b+a})
Answer: b
Clarification: According to Lagrange’s mean value theorem, if f : [a,b] → R is a function such that f is differentiable on (a,b) then the formula for Lagrange’s theorem is f’(c) = (frac {f(b)-f(a)}{b-a}).

15. Find ’C’ using Lagrange’s mean value theorem, if f(x) = e^x, a = 0, b = 1.
a) e^e-1
b) e-1
c) log(_e^{e+1})
d) log(_e^{e-1})
Answer: d
Clarification: Given f(x) = e^x, a = 0, b = 1
f’(c) = (frac {f(b)-f(a)}{b-a})
e^c = (frac {e-1}{1-0})
e^c = e – 1
C = log(_e^{e-1})