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17 maths questions

Q2 Vectors 3D & Lines Inequality or Proof Involving Vectors View
2. $\vec { a } , \vec { b } , \vec { c } , \vec { d }$ are four distinct vectors satisfying the conditions $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$, then prove that $\vec { a } \cdot \vec { b } + \vec { c } \cdot \vec { d } \neq \vec { a } \cdot \vec { c } + \vec { b } \cdot \vec { d }$.
Sol. Given that $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$ $\Rightarrow \vec { a } \times ( \vec { b } - \vec { c } ) = ( \vec { c } - \vec { b } ) \times \vec { d } = \vec { d } \times ( \vec { b } - \vec { c } ) \Rightarrow \vec { a } - \vec { d } \| \vec { b } - \vec { c }$ $\Rightarrow ( \vec { a } - \vec { d } ) \cdot ( \vec { b } - \vec { c } ) \neq 0 \Rightarrow \vec { a } \cdot \vec { b } + \vec { d } \cdot \vec { c } \neq \vec { d } \cdot \vec { b } + \vec { a } \cdot \vec { c }$.
Q3 Permutations & Arrangements Combinatorial Proof or Identity Derivation View
3. Using permutation or otherwise prove that $\frac { n ^ { 2 } ! } { ( n ! ) ^ { n } }$ is an integer, where $n$ is a positive integer.
Sol. Let there be $\mathrm { n } ^ { 2 }$ objects distributed in n groups, each group containing n identical objects. So number of arrangement of these $n ^ { 2 }$ objects are $\frac { n ^ { 2 } ! } { ( n ! ) ^ { n } }$ and number of arrangements has to be an integer. Hence $\frac { \mathrm { n } ^ { 2 } } { ( \mathrm { n } ! ) ^ { \mathrm { n } } }$ is an integer.
Q4 3x3 Matrices Determinant and Rank Computation View
4. If M is a $3 \times 3$ matrix, where $\mathrm { M } ^ { \mathrm { T } } \mathrm { M } = \mathrm { I }$ and $\operatorname { det } ( \mathrm { M } ) = 1$, then prove that $\operatorname { det } ( \mathrm { M } - \mathrm { I } ) = 0$.
Sol. $\quad ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } = \mathrm { M } ^ { \mathrm { T } } - \mathrm { I } = \mathrm { M } ^ { \mathrm { T } } - \mathrm { M } ^ { \mathrm { T } } \mathrm { M } = \mathrm { M } ^ { \mathrm { T } } ( \mathrm { I } - \mathrm { M } )$
$$\Rightarrow \left| ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } \right| = | \mathrm { M } - \mathrm { I } | = \left| \mathrm { M } ^ { \mathrm { T } } \right| | \mathrm { I } - \mathrm { M } | = | \mathrm { I } - \mathrm { M } | \Rightarrow | \mathrm { M } - \mathrm { I } | = 0 .$$
Alternate: $\operatorname { det } ( \mathrm { M } - \mathrm { I } ) = \operatorname { det } ( \mathrm { M } - \mathrm { I } ) \operatorname { det } \left( \mathrm { M } ^ { \mathrm { T } } \right) = \operatorname { det } \left( \mathrm { MM } ^ { \mathrm { T } } - \mathrm { M } ^ { \mathrm { T } } \right)$
$$= \operatorname { det } \left( \mathrm { I } - \mathrm { M } ^ { \mathrm { T } } \right) = - \operatorname { det } \left( \mathrm { M } ^ { \mathrm { T } } - \mathrm { I } \right) = - \operatorname { det } ( \mathrm { M } - \mathrm { I } ) ^ { \mathrm { T } } = - \operatorname { det } ( \mathrm { M } - \mathrm { I } ) \Rightarrow \operatorname { det } ( \mathrm { M } - \mathrm { I } ) = 0 \text {. }$$
  1. If $\mathrm { y } ( \mathrm { x } ) = \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \mathrm { x } \cdot \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta$ then find $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $\mathrm { x } = \pi$.

Sol. $\mathrm { y } = \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \mathrm { x } \cdot \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta = \cos \mathrm { x } \int _ { \pi ^ { 2 } / 16 } ^ { \mathrm { x } ^ { 2 } } \frac { \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } \mathrm {~d} \theta$ so that $\frac { d y } { d x } = - \sin x \int _ { \pi ^ { 2 } / 16 } ^ { x ^ { 2 } } \frac { \cos \sqrt { \theta } } { 1 + \sin ^ { 2 } \sqrt { \theta } } d \theta + \frac { 2 x \cos x \cdot \cos x } { 1 + \sin ^ { 2 } x }$ Hence, at $\mathrm { x } = \pi , \frac { \mathrm { dy } } { \mathrm { dx } } = 0 + \frac { 2 \pi ( - 1 ) ( - 1 ) } { 1 + 0 } = 2 \pi$.
Q7 Chain Rule Limit Involving Derivative Definition of Composed Functions View
7. If $f : [ - 1,1 ] \rightarrow R$ and $f ^ { \prime } ( 0 ) = \lim _ { n \rightarrow \infty } n f \left( \frac { 1 } { n } \right)$ and $f ( 0 ) = 0$. Find the value of $\lim _ { n \rightarrow \infty } \frac { 2 } { \pi } ( n + 1 ) \cos ^ { - 1 } \left( \frac { 1 } { n } \right) - n$.
Given that $0 < \left| \lim _ { \mathrm { n } \rightarrow \infty } \cos ^ { - 1 } \left( \frac { 1 } { \mathrm { n } } \right) \right| < \frac { \pi } { 2 }$. Sol. $\lim _ { n \rightarrow \infty } \frac { 2 } { \pi } ( n + 1 ) \cos ^ { - 1 } \frac { 1 } { n } - n = \lim _ { n \rightarrow \infty } n \left[ \frac { 2 } { \pi } \left( 1 + \frac { 1 } { n } \right) \cos ^ { - 1 } \frac { 1 } { n } - 1 \right]$ $= \lim _ { n \rightarrow \infty } n f \left( \frac { 1 } { n } \right) = f ^ { \prime } ( 0 )$ where $f ( x ) = \frac { 2 } { \pi } ( 1 + x ) \cos ^ { - 1 } x - 1$. Clearly, $\mathrm { f } ( 0 ) = 0$.
Now, $f ^ { \prime } ( x ) = \frac { 2 } { \pi } \left[ ( 1 + x ) \frac { - 1 } { \sqrt { 1 - x ^ { 2 } } } + \cos ^ { - 1 } x \right]$ $\Rightarrow \mathrm { f } ^ { \prime } ( 0 ) = \frac { 2 } { \pi } \left[ - 1 + \frac { \pi } { 2 } \right] = \frac { 2 } { \pi } \left[ \frac { \pi - 2 } { 2 } \right] = 1 - \frac { 2 } { \pi }$.
Q8 Proof Existence Proof View
8. If $\mathrm { p } ( \mathrm { x } ) = 51 \mathrm { x } ^ { 101 } - 2323 \mathrm { x } ^ { 100 } - 45 \mathrm { x } + 1035$, using Rolle's Theorem, prove that atleast one root lies between ( $45 ^ { 1 / 100 } , 46$ ). Sol. Let $\mathrm { g } ( \mathrm { x } ) = \int \mathrm { p } ( \mathrm { x } ) \mathrm { dx } = \frac { 51 \mathrm { x } ^ { 102 } } { 102 } - \frac { 2323 \mathrm { x } ^ { 101 } } { 101 } - \frac { 45 \mathrm { x } ^ { 2 } } { 2 } + 1035 \mathrm { x } + \mathrm { c }$ $= \frac { 1 } { 2 } \mathrm { x } ^ { 102 } - 23 \mathrm { x } ^ { 101 } - \frac { 45 } { 2 } \mathrm { x } ^ { 2 } + 1035 \mathrm { x } + \mathrm { c }$. Now $\mathrm { g } \left( 45 ^ { 1 / 100 } \right) = \frac { 1 } { 2 } ( 45 ) ^ { \frac { 102 } { 100 } } - 23 ( 45 ) ^ { \frac { 101 } { 100 } } - \frac { 45 } { 2 } ( 45 ) ^ { \frac { 2 } { 100 } } + 1035 ( 45 ) ^ { \frac { 1 } { 100 } } + \mathrm { c } = \mathrm { c }$ $\mathrm { g } ( 46 ) = \frac { ( 46 ) ^ { 102 } } { 2 } - 23 ( 46 ) ^ { 101 } - \frac { 45 } { 2 } ( 46 ) ^ { 2 } + 1035 ( 46 ) + \mathrm { c } = \mathrm { c }$. So $\mathrm { g } ^ { \prime } ( \mathrm { x } ) = \mathrm { p } ( \mathrm { x } )$ will have atleast one root in given interval.
Q9 Vectors 3D & Lines Volume of Pyramid/Tetrahedron Using Planes and Lines View
9. A plane is parallel to two lines whose direction ratios are $( 1,0 , - 1 )$ and $( - 1,1,0 )$ and it contains the point $( 1,1,1 )$. If it cuts coordinate axis at $\mathrm { A } , \mathrm { B } , \mathrm { C }$, then find the volume of the tetrahedron OABC .
Sol. Let $( l , m , n )$ be the direction ratios of the normal to the required plane so that $l - n = 0$ and $- l + m = 0$ $\Rightarrow \mathrm { l } = \mathrm { m } = \mathrm { n }$ and hence the equation of the plane containing $( 1,1,1 )$ is $\frac { \mathrm { x } } { 3 } + \frac { \mathrm { y } } { 3 } + \frac { \mathrm { z } } { 3 } = 1$. Its intercepts with the coordinate axes are $\mathrm { A } ( 3,0,0 ) ; \mathrm { B } ( 0,3,0 ) ; \mathrm { C } ( 0,0,3 )$. Hence the volume of OABC $= \frac { 1 } { 6 } \left| \begin{array} { l l l } 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array} \right| = \frac { 27 } { 6 } = \frac { 9 } { 2 }$ cubic units.
Q10 Independent Events View
10. If A and B are two independent events, prove that $\mathrm { P } ( \mathrm { A } \cup \mathrm { B } ) . \mathrm { P } \left( \mathrm { A } ^ { \prime } \cap \mathrm { B } ^ { \prime } \right) \leq \mathrm { P } ( \mathrm { C } )$, where C is an event defined that exactly one of A and B occurs.
Sol. $P ( A \cup B ) . P \left( A ^ { \prime } \right) P \left( B ^ { \prime } \right) \leq ( P ( A ) + P ( B ) ) P \left( A ^ { \prime } \right) P \left( B ^ { \prime } \right)$ $= \mathrm { P } ( \mathrm { A } ) \cdot \mathrm { P } \left( \mathrm { A } ^ { \prime } \right) \mathrm { P } \left( \mathrm { B } ^ { \prime } \right) + \mathrm { P } ( \mathrm { B } ) \mathrm { P } \left( \mathrm { A } ^ { \prime } \right) \mathrm { P } \left( \mathrm { B } ^ { \prime } \right)$ $= \mathrm { P } ( \mathrm { A } ) \mathrm { P } \left( \mathrm { B } ^ { \prime } \right) ( 1 - \mathrm { P } ( \mathrm { A } ) ) + \mathrm { P } ( \mathrm { B } ) \mathrm { P } \left( \mathrm { A } ^ { \prime } \right) ( 1 - \mathrm { P } ( \mathrm { B } ) )$ $\leq \mathrm { P } ( \mathrm { A } ) \mathrm { P } \left( \mathrm { B } ^ { \prime } \right) + \mathrm { P } ( \mathrm { B } ) \mathrm { P } \left( \mathrm { A } ^ { \prime } \right) = \mathrm { P } ( \mathrm { C } )$.
Q11 Differential equations Solving Separable DEs with Initial Conditions View
11. A curve passes through $( 2,0 )$ and the slope of tangent at point $P ( x , y )$ equals $\frac { ( x + 1 ) ^ { 2 } + y - 3 } { ( x + 1 ) }$. Find the equation of the curve and area enclosed by the curve and the $x$-axis in the fourth quadrant.
Sol. $\frac { d y } { d x } = \frac { ( x + 1 ) ^ { 2 } + y - 3 } { x + 1 }$ or, $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { x } + 1 ) + \frac { \mathrm { y } - 3 } { \mathrm { x } + 1 }$ Putting $\mathrm { x } + 1 = \mathrm { X } , \mathrm { y } - 3 = \mathrm { Y }$ $\frac { d Y } { d X } = X + \frac { Y } { X }$ $\frac { d Y } { d X } - \frac { Y } { X } = X$ [Figure] I. $F = \frac { 1 } { X } \Rightarrow \frac { 1 } { X } \cdot Y = X + c$ $\frac { \mathrm { y } - 3 } { \mathrm { x } + 1 } = ( \mathrm { x } + 1 ) + \mathrm { c }$. It passes through $( 2,0 ) \Rightarrow \mathrm { c } = - 4$. So, $\mathrm { y } - 3 = ( \mathrm { x } + 1 ) ^ { 2 } - 4 ( \mathrm { x } + 1 )$ $\Rightarrow \mathrm { y } = \mathrm { x } ^ { 2 } - 2 \mathrm { x }$. ⇒ Required area $= \left| \int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 2 x \right) d x \right| = \left| \left[ \frac { x ^ { 3 } } { 3 } - x ^ { 2 } \right] _ { 0 } ^ { 2 } \right| = \frac { 4 } { 3 }$ sq. units.
Q12 Circles Circle Equation Derivation View
12. A circle touches the line $2 x + 3 y + 1 = 0$ at the point $( 1 , - 1 )$ and is orthogonal to the circle which has the line segment having end points $( 0 , - 1 )$ and $( - 2,3 )$ as the diameter.
Sol. Let the circle with tangent $2 \mathrm { x } + 3 \mathrm { y } + 1 = 0$ at $( 1 , - 1 )$ be $( x - 1 ) ^ { 2 } + ( y + 1 ) ^ { 2 } + \lambda ( 2 x + 3 y + 1 ) = 0$ or $x ^ { 2 } + y ^ { 2 } + x ( 2 \lambda - 2 ) + y ( 3 \lambda + 2 ) + 2 + \lambda = 0$. It is orthogonal to $\mathrm { x } ( \mathrm { x } + 2 ) + ( \mathrm { y } + 1 ) ( \mathrm { y } - 3 ) = 0$ Or $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + 2 \mathrm { x } - 2 \mathrm { y } - 3 = 0$ so that $\frac { 2 ( 2 \lambda - 2 ) } { 2 } \cdot \left( \frac { 2 } { 2 } \right) + \frac { 2 ( 3 \lambda + 2 ) } { 2 } \left( \frac { - 2 } { 2 } \right) = 2 + \lambda - 3 \Rightarrow \lambda = - \frac { 3 } { 2 }$. Hence the required circle is $2 x ^ { 2 } + 2 y ^ { 2 } - 10 x - 5 y + 1 = 0$.
Q13 Circles Circle-Related Locus Problems View
13. At any point $P$ on the parabola $y ^ { 2 } - 2 y - 4 x + 5 = 0$, a tangent is drawn which meets the directrix at $Q$. Find the locus of point R which divides QP externally in the ratio $\frac { 1 } { 2 } : 1$.
Sol. Any point on the parabola is $\mathrm { P } \left( 1 + \mathrm { t } ^ { 2 } , 1 + 2 \mathrm { t } \right)$. The equation of the tangent at P is $\mathrm { t } ( \mathrm { y } - 1 ) = \mathrm { x } - 1 + \mathrm { t } ^ { 2 }$ which meets the directrix $\mathrm { x } = 0$ at $\mathrm { Q } \left( 0,1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } \right)$. Let R be $( \mathrm { h } , \mathrm { k } )$. Since it divides QP externally in the ratio $\frac { 1 } { 2 } : 1 , \mathrm { Q }$ is the mid point of RP $\Rightarrow 0 = \frac { \mathrm { h } + 1 + \mathrm { t } ^ { 2 } } { 2 }$ or $\mathrm { t } ^ { 2 } = - ( \mathrm { h } + 1 )$ and $1 + \mathrm { t } - \frac { 1 } { \mathrm { t } } = \frac { \mathrm { k } + 1 + 2 \mathrm { t } } { 2 }$ or $\mathrm { t } = \frac { 2 } { 1 - \mathrm { k } }$ So that $\frac { 4 } { ( 1 - k ) ^ { 2 } } + ( h + 1 ) = 0$ Or $( k - 1 ) ^ { 2 } ( h + 1 ) + 4 = 0$. Hence locus is $( y - 1 ) ^ { 2 } ( x + 1 ) + 4 = 0$.
Q14 Indefinite & Definite Integrals Substitution Combined with Symmetry or Companion Integral View
14. Evaluate $\int _ { - \pi / 3 } ^ { \pi / 3 } \frac { \pi + 4 x ^ { 3 } } { 2 - \cos \left( | x | + \frac { \pi } { 3 } \right) } d x$.
Sol. $I = \int _ { - \pi / 3 } ^ { \pi / 3 } \frac { \left( \pi + 4 x ^ { 3 } \right) d x } { 2 - \cos \left( | x | + \frac { \pi } { 3 } \right) }$ $2 \mathrm { I } = \int _ { - \pi / 3 } ^ { \pi / 3 } \frac { 2 \pi \mathrm { dx } } { 2 - \cos \left( | \mathrm { x } | + \frac { \pi } { 3 } \right) } = \int _ { 0 } ^ { \pi / 3 } \frac { 2 \pi \mathrm { dx } } { 2 - \cos \left( \mathrm { x } + \frac { \pi } { 3 } \right) }$ $\mathrm { I } = \int _ { \pi / 3 } ^ { 2 \pi / 3 } \frac { 2 \pi \mathrm { dt } } { 2 - \cos \mathrm { t } } \Rightarrow \mathrm { I } = 2 \pi \int _ { \pi / 3 } ^ { 2 \pi / 3 } \frac { \sec ^ { 2 } \frac { \mathrm { t } } { 2 } \mathrm { dt } } { 1 + 3 \tan ^ { 2 } \frac { \mathrm { t } } { 2 } } = 2 \pi \int _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } \frac { 2 \mathrm { dt } } { 1 + 3 \mathrm { t } ^ { 2 } } = \frac { 4 \pi } { 3 } \int _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } \frac { \mathrm { dt } } { \left( \frac { 1 } { \sqrt { 3 } } \right) ^ { 2 } + \mathrm { t } ^ { 2 } }$ $\mathrm { I } = \frac { 4 \pi } { 3 } \sqrt { 3 } \left[ \tan ^ { - 1 } \sqrt { 3 } \mathrm { t } \right] _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } = \frac { 4 \pi } { \sqrt { 3 } } \left[ \tan ^ { - 1 } 3 - \frac { \pi } { 4 } \right] = \frac { 4 \pi } { \sqrt { 3 } } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$.
Q15 Proof Direct Proof of an Inequality View
15. If $\mathrm { a } , \mathrm { b } , \mathrm { c }$ are positive real numbers, then prove that $[ ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) ] ^ { 7 } > 7 ^ { 7 } \mathrm { a } ^ { 4 } \mathrm {~b} ^ { 4 } \mathrm { c } ^ { 4 }$.
Sol. $( 1 + a ) ( 1 + b ) ( 1 + c ) = 1 + a b + a + b + c + a b c + a c + b c$ $\Rightarrow \frac { ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) - 1 } { 7 } \geq ( \mathrm { ab } . \mathrm { a } . \mathrm { b } . \mathrm { c } . \mathrm { abc } . \mathrm { ac } . \mathrm { bc } ) ^ { 1 / 7 } \quad ($ using $A M \geq G M )$ $\Rightarrow ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) - 1 > 7 \left( \mathrm { a } ^ { 4 } \cdot \mathrm {~b} ^ { 4 } \cdot \mathrm { c } ^ { 4 } \right) ^ { 1 / 7 }$ $\Rightarrow ( 1 + \mathrm { a } ) ( 1 + \mathrm { b } ) ( 1 + \mathrm { c } ) > 7 \left( \mathrm { a } ^ { 4 } \cdot \mathrm {~b} ^ { 4 } \cdot \mathrm { c } ^ { 4 } \right) ^ { 1 / 7 }$ $\Rightarrow ( 1 + \mathrm { a } ) ^ { 7 } ( 1 + \mathrm { b } ) ^ { 7 } ( 1 + \mathrm { c } ) ^ { 7 } > 7 ^ { 7 } \left( \mathrm { a } ^ { 4 } \cdot \mathrm {~b} ^ { 4 } \cdot \mathrm { c } ^ { 4 } \right)$.
Q16 Chain Rule Continuity Conditions via Composition View
16. $\mathrm { f } ( \mathrm { x } ) = \begin{cases} \mathrm { b } \sin ^ { - 1 } \left( \frac { \mathrm { x } + \mathrm { c } } { 2 } \right) , & - \frac { 1 } { 2 } < \mathrm { x } < 0 \\ \frac { 1 } { 2 } , & \mathrm { x } = 0 \\ \frac { \mathrm { e } ^ { \frac { \mathrm { a } } { 2 } \mathrm { x } } - 1 } { \mathrm { x } } , & 0 < \mathrm { x } < \frac { 1 } { 2 } \end{cases}$
If $\mathrm { f } ( \mathrm { x } )$ is differentiable at $\mathrm { x } = 0$ and $| \mathrm { c } | < \frac { 1 } { 2 }$ then find the value of ' a ' and prove that $64 \mathrm {~b} ^ { 2 } = \left( 4 - \mathrm { c } ^ { 2 } \right)$. Sol. $f \left( 0 ^ { + } \right) = f \left( 0 ^ { - } \right) = f ( 0 )$ Here $\mathrm { f } \left( 0 ^ { + } \right) = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \mathrm { x } } = \lim _ { \mathrm { x } \rightarrow \infty } \frac { \mathrm { e } ^ { \frac { \mathrm { ax } } { 2 } } - 1 } { \frac { \mathrm { ax } } { 2 } } \cdot \frac { \mathrm { a } } { 2 } = \frac { \mathrm { a } } { 2 }$. $\Rightarrow \mathrm { b } \sin ^ { - 1 } \frac { \mathrm { c } } { 2 } = \frac { \mathrm { a } } { 2 } = \frac { 1 } { 2 } \Rightarrow \mathrm { a } = 1$. $\mathrm { L } \mathrm { f } ^ { \prime } \left( 0 _ { - } \right) = \lim _ { \mathrm { h } \rightarrow 0 ^ { - } } \frac { \mathrm { b } \sin ^ { - 1 } \frac { ( \mathrm {~h} + \mathrm { c } ) } { 2 } - \frac { 1 } { 2 } } { \mathrm {~h} } = \frac { \mathrm { b } / 2 } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } }$ $R f ^ { \prime } \left( 0 _ { + } \right) = \lim _ { h \rightarrow 0 ^ { + } } \frac { \frac { e ^ { h / 2 } - 1 } { h } - \frac { 1 } { 2 } } { h } = \frac { 1 } { 8 }$ Now $\mathrm { L } ^ { \prime } \left( 0 _ { - } \right) = \mathrm { Rf } ^ { \prime } \left( 0 _ { + } \right) \Rightarrow \frac { \frac { \mathrm { b } } { 2 } } { \sqrt { 1 - \frac { \mathrm { c } ^ { 2 } } { 4 } } } = \frac { 1 } { 8 }$ $4 b = \sqrt { 1 - \frac { c ^ { 2 } } { 4 } } \Rightarrow 16 b ^ { 2 } = \frac { 4 - c ^ { 2 } } { 4 } \Rightarrow 64 b ^ { 2 } = 4 - c ^ { 2 }$.
Q17 Stationary points and optimisation Prove an inequality using calculus-based optimisation View
17. Prove that $\sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } \cdot ( \mathrm { x } + 1 ) } { \pi } \forall \mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$. (Justify the inequality, if any used).
Sol. Let $\mathrm { f } ( \mathrm { x } ) = 3 \mathrm { x } ^ { 2 } + ( 3 - 2 \pi ) \mathrm { x } - \pi \sin \mathrm { x }$ $\mathrm { f } ( 0 ) = 0 , \mathrm { f } \left( \frac { \pi } { 2 } \right) = - \mathrm { ve }$ $\mathrm { f } ^ { \prime } ( \mathrm { x } ) = 6 \mathrm { x } + 3 - 2 \pi - \pi \cos \mathrm { x }$ $\mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) = 6 + \pi \sin \mathrm { x } > 0$ $\Rightarrow \mathrm { f } ^ { \prime } ( \mathrm { x } )$ is increasing function in $\left[ 0 , \frac { \pi } { 2 } \right]$ ⇒ there is no local maxima of $\mathrm { f } ( \mathrm { x } )$ in $\left[ 0 , \frac { \pi } { 2 } \right]$ ⇒ graph of $\mathrm { f } ( \mathrm { x } )$ always lies below the x -axis [Figure] in $\left[ 0 , \frac { \pi } { 2 } \right]$. $\Rightarrow \mathrm { f } ( \mathrm { x } ) \leq 0$ in $\mathrm { x } \in \left[ 0 , \frac { \pi } { 2 } \right]$. $3 \mathrm { x } ^ { 2 } + 3 \mathrm { x } \leq 2 \pi \mathrm { x } + \pi \sin \mathrm { x } \Rightarrow \sin \mathrm { x } + 2 \mathrm { x } \geq \frac { 3 \mathrm { x } ( \mathrm { x } + 1 ) } { \pi }$.
Q18 Matrices Linear System and Inverse Existence View
18. $\mathrm { A } = \left[ \begin{array} { l l l } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right] , \mathrm { B } = \left[ \begin{array} { c c c } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right] , \mathrm { U } = \left[ \begin{array} { c } \mathrm { f } \\ \mathrm { g } \\ \mathrm { h } \end{array} \right] , \mathrm { V } = \left[ \begin{array} { c } \mathrm { a } ^ { 2 } \\ 0 \\ 0 \end{array} \right]$. If there is vector matrix X , such that $\mathrm { AX } = \mathrm { U }$ has infinitely many solutions, then prove that $\mathrm { BX } = \mathrm { V }$ cannot have a unique solution. If afd $\neq 0$ then prove that $\mathrm { BX } = \mathrm { V }$ has no solution.
Sol. $\mathrm { AX } = \mathrm { U }$ has infinite solutions $\Rightarrow | \mathrm { A } | = 0$ $\left| \begin{array} { c c c } \mathrm { a } & 0 & 1 \\ 1 & \mathrm { c } & \mathrm { b } \\ 1 & \mathrm {~d} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm { ab } = 1$ or $\mathrm { c } = \mathrm { d }$ and $\left| \mathrm { A } _ { 1 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 0 & \mathrm { f } \\ 1 & \mathrm { c } & \mathrm { g } \\ 1 & \mathrm {~d} & \mathrm {~h} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } ; \left| \mathrm { A } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { f } & 1 \\ 1 & \mathrm {~g} & \mathrm {~b} \\ 1 & \mathrm {~h} & \mathrm {~b} \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h }$ $\left| \mathrm { A } _ { 3 } \right| = \left| \begin{array} { l l l } \mathrm { f } & 0 & 1 \\ \mathrm {~g} & \mathrm { c } & \mathrm { b } \\ \mathrm { h } & \mathrm { d } & \mathrm { b } \end{array} \right| = 0 \Rightarrow \mathrm {~g} = \mathrm { h } , \mathrm { c } = \mathrm { d } \Rightarrow \mathrm { c } = \mathrm { d }$ and $\mathrm { g } = \mathrm { h }$ $\mathrm { BX } = \mathrm { V }$ $| \mathrm { B } | = \left| \begin{array} { l l l } \mathrm { a } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ \mathrm { f } & \mathrm { g } & \mathrm { h } \end{array} \right| = 0 \quad$ (since $\mathrm { C } _ { 2 }$ and $\mathrm { C } _ { 3 }$ are equal) $\quad \Rightarrow \mathrm { BX } = \mathrm { V }$ has no unique solution. and $\left| \mathrm { B } _ { 1 } \right| = \left| \begin{array} { l l l } \mathrm { a } ^ { 2 } & 1 & 1 \\ 0 & \mathrm {~d} & \mathrm { c } \\ 0 & \mathrm {~g} & \mathrm {~h} \end{array} \right| = 0 ($ since $\mathrm { c } = \mathrm { d } , \mathrm { g } = \mathrm { h } )$ $\left| \mathrm { B } _ { 2 } \right| = \left| \begin{array} { c c c } \mathrm { a } & \mathrm { a } ^ { 2 } & 1 \\ 0 & 0 & \mathrm { c } \\ \mathrm { f } & 0 & \mathrm {~h} \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { cf } = \mathrm { a } ^ { 2 } \mathrm { df } \quad ($ since $\mathrm { c } = \mathrm { d } )$
$$\left| \mathrm { B } _ { 3 } \right| = \left| \begin{array} { c c c } \mathrm { a } & 1 & \mathrm { a } ^ { 2 } \\ 0 & \mathrm {~d} & 0 \\ \mathrm { f } & \mathrm {~g} & 0 \end{array} \right| = \mathrm { a } ^ { 2 } \mathrm { df }$$
since if $\operatorname { adf } \neq 0$ then $\left| \mathrm { B } _ { 2 } \right| = \left| \mathrm { B } _ { 3 } \right| \neq 0$. Hence no solution exist.
Q19 Probability Definitions Combinatorial Probability View
19. A bag contains 12 red balls and 6 white balls. Six balls are drawn one by one without replacement of which atleast 4 balls are white. Find the probability that in the next two draws exactly one white ball is drawn. (leave the answer in terms of ${ } ^ { \mathrm { n } } \mathrm { C } _ { \mathrm { r } }$ ).
Sol. Let $\mathrm { P } ( \mathrm { A } )$ be the probability that atleast 4 white balls have been drawn. $P \left( A _ { 1 } \right)$ be the probability that exactly 4 white balls have been drawn. $\mathrm { P } \left( \mathrm { A } _ { 2 } \right)$ be the probability that exactly 5 white balls have been drawn. $P \left( A _ { 3 } \right)$ be the probability that exactly 6 white balls have been drawn. $\mathrm { P } ( \mathrm { B } )$ be the probability that exactly 1 white ball is drawn from two draws. $\mathrm { P } ( \mathrm { B } / \mathrm { A } ) = \frac { \sum _ { \mathrm { i } = 1 } ^ { 3 } \mathrm { P } \left( \mathrm { A } _ { \mathrm { i } } \right) \mathrm { P } \left( \mathrm { B } / \mathrm { A } _ { \mathrm { i } } \right) } { \sum _ { \mathrm { i } = 1 } ^ { 3 } \mathrm { P } \left( \mathrm { A } _ { \mathrm { i } } \right) } = \frac { \frac { { } ^ { 12 } \mathrm { C } _ { 2 } { } ^ { 6 } \mathrm { C } _ { 4 } } { { } ^ { 18 } \mathrm { C } _ { 6 } } \cdot \frac { { } ^ { 10 } \mathrm { C } _ { 1 } { } ^ { 2 } \mathrm { C } _ { 1 } } { { } ^ { 12 } \mathrm { C } _ { 2 } } + \frac { { } ^ { 12 } \mathrm { C } _ { 1 } { } ^ { 6 } \mathrm { C } _ { 5 } } { { } ^ { 18 } \mathrm { C } _ { 6 } } \cdot \frac { { } ^ { 11 } \mathrm { C } _ { 1 } { } ^ { 1 } \mathrm { C } _ { 1 } } { { } ^ { 12 } \mathrm { C } _ { 2 } } } { \frac { { } ^ { 12 } \mathrm { C } _ { 2 } { } ^ { 6 } \mathrm { C } _ { 4 } } { { } ^ { 18 } \mathrm { C } _ { 6 } } + \frac { { } ^ { 12 } \mathrm { C } _ { 1 } { } ^ { 6 } \mathrm { C } _ { 5 } } { { } ^ { 18 } \mathrm { C } _ { 6 } } + \frac { { } ^ { 12 } \mathrm { C } _ { 0 } { } ^ { 6 } \mathrm { C } _ { 6 } } { { } ^ { 18 } \mathrm { C } _ { 6 } } }$ $= \frac { { } ^ { 12 } \mathrm { C } _ { 2 } { } ^ { 6 } \mathrm { C } _ { 4 } { } ^ { 10 } \mathrm { C } _ { 1 } { } ^ { 2 } \mathrm { C } _ { 1 } + { } ^ { 12 } \mathrm { C } _ { 1 } { } ^ { 6 } \mathrm { C } _ { 5 } { } ^ { 11 } \mathrm { C } _ { 1 } { } ^ { 1 } \mathrm { C } _ { 1 } } { { } ^ { 12 } \mathrm { C } _ { 2 } \left( { } ^ { 12 } \mathrm { C } _ { 2 } { } ^ { 6 } \mathrm { C } _ { 4 } + { } ^ { 12 } \mathrm { C } _ { 1 } { } ^ { 6 } \mathrm { C } _ { 5 } + { } ^ { 12 } \mathrm { C } _ { 0 } { } ^ { 6 } \mathrm { C } _ { 6 } \right) }$
Q20 Proof Coplanarity and Relative Position of Planes View
20. Two planes $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ pass through origin. Two lines $\mathrm { L } _ { 1 }$ and $\mathrm { L } _ { 2 }$ also passing through origin are such that $\mathrm { L } _ { 1 }$ lies on $\mathrm { P } _ { 1 }$ but not on $\mathrm { P } _ { 2 } , \mathrm {~L} _ { 2 }$ lies on $\mathrm { P } _ { 2 }$ but not on $\mathrm { P } _ { 1 }$. A, B, C are three points other than origin, then prove that the permutation $\left[ \mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } \right]$ of $[ \mathrm { A } , \mathrm { B } , \mathrm { C } ]$ exists such that
(i). $\quad \mathrm { A }$ lies on $\mathrm { L } _ { 1 } , \mathrm {~B}$ lies on $\mathrm { P } _ { 1 }$ not on $\mathrm { L } _ { 1 } , \mathrm { C }$ does not lie on $\mathrm { P } _ { 1 }$.
(ii). $\quad \mathrm { A } ^ { \prime }$ lies on $\mathrm { L } _ { 2 } , \mathrm {~B} ^ { \prime }$ lies on $\mathrm { P } _ { 2 }$ not on $\mathrm { L } _ { 2 } , \mathrm { C } ^ { \prime }$ does not lie on $\mathrm { P } _ { 2 }$.
Sol. A corresponds to one of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ and
B corresponds to one of the remaining of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ and
C corresponds to third of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$. Hence six such permutations are possible eg One of the permutations may $\mathrm { A } \equiv \mathrm { A } ^ { \prime } ; \mathrm { B } \equiv \mathrm { B } ^ { \prime } , \mathrm { C } \equiv \mathrm { C } ^ { \prime }$ From the given conditions:
A lies on $\mathrm { L } _ { 1 }$.
B lies on the line of intersection of $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ and ' C ' lies on the line $\mathrm { L } _ { 2 }$ on the plane $\mathrm { P } _ { 2 }$. Now, $\mathrm { A } ^ { \prime }$ lies on $\mathrm { L } _ { 2 } \equiv \mathrm { C }$. $\mathrm { B } ^ { \prime }$ lies on the line of intersection of $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 } \equiv \mathrm {~B}$ $\mathrm { C } ^ { \prime }$ lie on $\mathrm { L } _ { 1 }$ on plane $\mathrm { P } _ { 1 } \equiv \mathrm {~A}$. Hence there exist a particular set [ $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ ] which is the permutation of $[ \mathrm { A } , \mathrm { B } , \mathrm { C } ]$ such that both (i) and
(ii) is satisfied. Here $\left[ \mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } \right] \equiv [ \mathrm { CBA } ]$.